COLLEGE ALGEBRA VERSION (0.9 BY CARL STITZ, PH.D. L~AKELAND COMMUTNITY COLLEGE JEFF ZEAGER, PH.D . L~ORAIN COUTNTY COMMUTNITY COLLEGE AUGUST 10, 2009   TABLE OF CONTENTS I elations and1Fuctions 1.1 The Carttsian Coordinate P1lane.. . .. 1.1.1 Distance in the Plane.. .. .. .. 1.1.2 Exercises..... .. .. .. .. .. .. 1.1.3 Answers... .. .. .. .. .. .... 1.2 Relations..... .. .. .. .. .. .. .. .. 1.2.1 Exercise:..... .. .. .. .. .. .. 1.2.2 Answers.. .. .. .. .. .. .. .. 1.3 Graphs of E(qnati(ons.... .. .. .. .. .. 1.3.1 Exercise:..... .. .. .. .. .. .. 1.3.2 Answers... .. .. .. .. .. .... 1.4 Jntro~dnction to nunction.. .. .. .... 1.4.1 Exercises.... .. .. .. .. .. .. 1.4.2 Answers... .. .. .. .. .. .... 1-5 Fu nction Notation.... .. .. .. .. .. .. 1.J.1 Exercises..... .. .. .. .. .. .. 1.5.2 Answers.. .. .. .. .. .. .. .. 1.6 Function Anithnnetic.... .. .. .. .. .. 1.6.1 Exercises..... .. .. .. .. .. .. 1.6.2 Answers... .. .. .. .. .. .... 1.7 Graph: )f Fnnctions.... .. .. .. .. .. 1...1 Gene~ral Functioni Behavior .... 1.7.2 Exercise:..... .. .. .. .. .. .. 1.7.3 Answrs... .. .. .. .. .. .... 1.8 rfr ansftriiations..... .. .. .. .. .. .. 1.8.1 Exercises.... .. .. .. .. .. .. 1.8.2 Answers... .. .. .. .. .. .... 2 Linear and Qadratic unctions 2.1 Linear Fu nction:..... .. .. .. .. .. .. 2. 1.1 Exercise:..... .. .. .. .. .. .. 2.1.2 rAnsw ers... .. .. .. .. .. .... 1 1 6 10 12 14 18 20 22 27 28 31 37 40 41 47 49 50 55 57 58 64 71 74 76 96 98 101 .101 .113 .116  IV Iv TABLE OF CONTENTS 2.2 AbsolutcValue Functions . 117 2.2.1 Exercies: . 125 2.2.2 Answers. .. .. .. .... .... .... .... .... .... .... .....126 2.3 Quadiratic Functions. .. .. ... .... .... .... .... .... .... .....129 2.3.1 Exercies:.. .. ... .... .... .... .... .... ... .... .....138 2.3.2 Answers. .. .. .. .... .... .... .... .... .... .... .....140 2.4 incqualities:s. .. .. ... .... .... .... .... .... .... .... .... ..143 2.4.1 Exercies:.. .. ... .... .... .... .... .... ... .... .....156 2.4.2 Answvers. .. .. .. .... .... .... .... .... .... .... .....157 2.5 Regression... .. .. .. ... .... ... .... .... .... .... .... .....159 2.5.1 Exercies .. .. ... .... .... .... .... .... ... .... .....164 2..2 Answers. .. .. .. .... .... .... .... .... .... .... .....167 3 olynomial Functions 169 3.1 Craphs of Polynomials. .. .. .. .... .... .... .... .... .... .....169 3.1.1 Exercises .. .. ... .... .... .... .... .... ... .... .....180 3.1.2 Answers. .. .. .. .... .... .... .... .... .... .... .....183 3.2 The Factor Theorem andI Thc Rcemaindcer Thcorcm. .. .. .. .... .... .....187 3.2.1 Exercie: .. .. ... .... .... .... .... .... ... .... .....195 3.2.2 Answers. .. .. .. .... .... .... .... .... .... .... .....196 3.3 Real Zcros of Polynomial........ ....... .. .. .. .. .. .. .. .... .... ....197 3.3.1 For Thosc Wishing to usc Gaphinig Calculator. .. .. ... ... .... ..198 3.3.2 For Those 'Wishing NT to us: Craphing Calculato~r. .. .. ..........201 3.3.3 Excrcises .. .. ... .... .... .... .... ... .... .... .....207 3.3.4rAnswecrs. .. .. .. .... .... .... .... .... .... .... .....208 3.4 Complex Zeros and the Fundamental 1horem)f Algebra.. .. .... .... ....209 3.4.1 Exercises .. .. ... .... .... .... .... ... .... .... .....217 3.4.2 Answe ~rs. .. .. .. .... .... .... .... .... .... .... .....218 4 Rational notions 219 4.1 introduction to)Rational Functions. .. .. ... ... .... .... .... .... ..219 4.1..1Exercises .. .. ... .... .... .... .... ... .... .... .....230 4.1.2 Answers. .. .. .. .... .... .... .... .... .... .... .....232 4.2 Craphs of Rational Functions. .. .. ... .... .... ... .... .... .....234 4.2.1 Exercises .. .. ... .... .... .... .... ... .... .... .....247 4.2.2 Answe ~rs. .. .. .. .... .... .... .... .... .... .... .....249  TABLE OF CONTENTS 5 Frther Toics in Functions 5.1 Function Composition.. .. ... .... .... ... .. 5.1.1 Exercises...... .... .. .. .. .. .. .. .. .. 5.1-2 Answers....... .. .. .. .. .. .. .. .. .. ... 5.2 lnverse Functions........... .. .. .. .. .. .. .. .. 5.2.1 Exercises...... .... .. .. .. .. .. .. .. .. 5.2.2 Answer............ .. .. .. .. .. .. .. ... 5.3 Othe~r AIge~braic Functions.... .. .. .. .. .. .. .. .. 5.3.1 Exercises...... .... .. .. .. .. .. .. .. .. 5.3.2 Answers....... .... .. .. .. .. .. .. .. . 6 xponential and Loarithmic Functions 6.1 introdunction to Exponential. and Logarith:amic Functions 6.1.1 Exercises...... .... .. .. .. .. .. .. .. .. 6.1.2 Answers....... .... .. .. .. .. .. .. .. . 6.2 Properties of Logarit.ims........ .. .. .. .. .. .. .. . 6.2.1 Exercises. .. ....... .. .. .. .. .. .. .. .. 6.2.2 Answers....... .... .. .. .. .. .. .. .. . 6.3 ExponentiaI Equatio)nsand Inequalities.... .. .. .. .. 6.3.1 Exercises. .. ....... .. .. .. .. .. .. .. .. 6.3.2 Answers....... .... .. .. .. .. .. .. .. . 6.4 Logarithmic Equa tions and Inequalities... .. .. .. .. . 6.4. 1 Exercises...... .... .. .. .. .. .. .. .. .. 6.4.2 iAnswcrs....... .... .. .. .. .. .. .. .. . 6.5 Ap)plica tions of Exponentil a~ndI Logarithmic Functions 6.5.1 Applications of ExponentialI Enctions. .. .. .. 6.5.2 Applications of Logaritlh ns..... .. .. .. .. .. . 6.5.3 Exercises...... .... .. .. .. .. .. .. .. .. 6 ..4 Answers....... .... .. .. .. .. .. .. .. . 7.1 introdunction to Conies........ .. .. .. .. .. .. .. . 7.2 Circles.......... . . .... .. .. .. .. .. .. .. .. 7.2.1 Exercises...... .... .. .. .. .. .. .. .. .. 7.2.2 Answers....... .... .. .. .. .. .. .. .. . 7.3 Paraolas......... . .... .. .. .. .. .. .. .. .. 7 .4.1 Exercise:...... .... .. .. .. .. .. .. .. .. 7.4.2 -An~es................. V 265 265 275 277 279 295 296 297 307 310 315 315 328 330 332 340 342 343 351 352 353 360 362 363 363 371 376 380 383 383 386 390 391 392 400 401 402 411  VI VI TABLE OF CONTENTS 7.5.2 Answers . 427 8 ystems of Euations and atrices 429 8.1 Sys~tems of Linear Eqnuations: Causiian Elimination .. .. ... .... .... ....429 8.1.1 Exercises .. .. ... .... .... .... .... .... ... .... .....442 8.1.2 Answers. .. .. .. .... .... .... .... .... .... .... .....443 8.2 Systemns of Linear Equatiorns: Au gmentedI Matrices. .. .. ... .... ... .....444 8.2.1 Exercises.. .. ... .... .... .... .... .... ... .... .....451 8.2.2 Answers. .. .. .. .... .... .... .... .... .... .... .....453 8.3 Matrix Arithmetic .. .. ... ... .... .... .... .... .... .... ....454 8.3.1 Exercises .. .. ... .... .... .... .... .... .... .... ....467 8.3.2 rAnswers. .. .. .. .... .... .... .... .... .... .... .....470 8.4 Systemsn of Linear Eqnuations: Mvatrix Inverses........ .. .. .. .. .. .. .. ... ..471 8.4.1 Exercises .. .. ... .... .... .... .... .... ... .... .....483 8.4.2 Answers. .. .. .. .... .... .... .... .... .... .... .....485 8.5Determ--inants andICramners Rnie .. .. .. .... .... .... .... ... .....486 8.5.1 Definition and Properties ofI lhe Deternminant .. .. .. .... .... .....486 8.5.2 Cramer's Rnie and Maitrix djoints .. .. ... .... .... .... .....490 8.5.3 Exercises:.. .. ... .... .... .... .... .... ... .... .....495 8.5 .4 Answers. .. .. .. .... .... .... .... .... .... .... .....498 8.6 Partial Eraction ecomposition .. .. .. .... .... ... .... .... .... ..499 8.6.1 Exercises .. .. ... .... .... .... .... .... ... .... .....507 8.6.2 Answers. .. .. .. .... .... .... .... .... .... .... .....508 8.7 Systemsof Non-Linear Eqnations and Inequailities:.. .. ... .... .... .....509 8.7.1 Exercises .. .. ... .... .... .... .... ... .... .... .....518 847.2 Answers. .. .. ... ... .... .... .... .... .... .... .....520 9 equences and the inomnial lieorem 523 9.1 Sequnces:.. .. .. .... .... .... .... .... ... .... .... .... ..523 9.1.1 Exercises .. .. ... .... .... .... .... ... .... .... .....531 9.1.2 Anwers. .. .. ... .... .... .... ... .... .... .... .....533 9.2 Sunnnation Notationi............ .. .. .. .. .. .............. . .. .. .. .. .......534 9.2.1 Exercises .. .. ... .... .... .... .... .... ... .... .....542 9.2.2 An:svers. .. .. .. .... .... .... .... .... .... .... .....543 9.3 MathematicalIindnction. .. .. .. ... ... ... ... .... ... ... ... ...544 9.3.1 Exercises .. .. ... .... .... .... .... .... .... .... ....549 9.3.2 Se'lected Aiiswers. .. .. .. ... ... ... ... ... ... ... .... ...550 9.4 The Bino)m ial Theorem ......... .. .. .. .. .. .. .. .. .. .. ... ... ....552 9.4.1 Exercises .. .. ... .... .... .... .... ... .... .... .....561 9.4.2Anwers. .. .. ... .... .... .... ... .... .... .... .....562 Index 563  CHAPTER 1 RELATIONS AND FUNCTIONS 1.1 THE CARTESIAN COORDINATE PLANE In order to visualize the pure excitement that is Algebra, we need to unite Algebra and Geometry. Simply put, we must find a way to draw algebraic things. Let's start with possibly the greatest mathematical achievement of all time: the Cartesian Coordinate Plane.1 Imagine two real number lines crossing at a right angle at 0 as below. 4- 3- 2- 1- |I |I |I |I 'I |I |I |I |I -4 -3 -2 -1 1 2 3 4 -1- -2- -3- -4- The horizontal number line is usually called the x-axis while the vertical number line is usually called the y-axis.2 As with the usual number line, we imagine these axes extending off indefinitely in both directions. Having two number lines allows us to locate the position of points off of the number lines as well as points on the lines themselves. 'So named in honor of Ren6 Descartes. 2The labels can vary depending on the context of application.  2 RELATIONS AND FUNCTIONS For example, consider the point P below on the left. To use the numbers on the axes to label this point, we imagine dropping a vertical line from the x-axis to P and extending a horizontal line from the y-axis to P. We then describe the point P using the ordered pair (2, -4). The first number in the ordered pair is called the abscissa or x-coordinate and the second is called the ordinate or y-coordinate.3 Taken together, the ordered pair (2, -4) comprise the Cartesian coordinates of the point P. In practice, the distinction between a point and its coordinates is blurred; for example, we often speak of 'the point (2, -4).' We can think of (2, -4) as instructions on how to reach P from the origin by moving 2 units to the right and 4 units downwards. Notice that the order in the ordered pair is important - if we wish to plot the point (-4, 2), we would move to the left 4 units from the origin and then move upwards 2 units, as below on the right. y y 4 4 3 3 (-4, 2) 2 --- - - - - - - - -2 - 1 1 I I I I I I I I )I I I I I I I I ) -4 -3 -2 -1 1 3 4 x -4 -3 -2 -1 1 3 4 x -1 -1 -2-- -2-- -3 ---3 - -4 ----. P -4 ---- P(2,-4) EXAMPLE 1.1.1. Plot the following points: A(5, 8), B (- 2, 3), C(-5.8, -3), D(4.5,-1), E(5,0), F(0, 5), G(-7, 0), H(0, -9), 0(0, 0).4 SOLUTION. To plot these points, we start at the origin and move to the right if the x-coordinate is positive; to the left if it is negative. Next, we move up if the y-coordinate is positive or down if it is negative. If the x-coordinate is 0, we start at the origin and move along the y-axis only. If the y-coordinate is 0 we move along the x-axis only. 3Again, the names of the coordinates can vary depending on the context of the application. If, for example, the horizontal axis represented time we might choose to call it the t-axis. The first number in the ordered pair would then be the t-coordinate. 4The letter 0 is almost always reserved for the origin.  1.1 THE CARTESIAN COORDINATE PLANE 3 9 8 7 6 5 F(0, 5) 4 3 2 A A(5, 8) B (-2,3) 1- G(-7, 0) O(0, 0) E(5, 0) I I I O i I I I I I " I II " I 'I 'I 'I -9 -8 -7 -6 -5 -4 -3 -2 -1 C(-5.8, -3) -1 - 2 - -3- -4- -5- -6- -7- -8- -94 1 2 3 4 5 6 - D(4.5,-1) H(0, -9) D- When we speak of the Cartesian Coordinate Plane, we mean the set of all possible ordered pairs (x, y) as x and y take values from the real numbers. Below is a summary of important facts about Cartesian coordinates. Important Facts about the Cartesian Coordinate Plane " (a, b) and (c, d) represent the same point in the plane if and only if a = c and b = d. " (x, y) lies on the x-axis if and only if y = 0. " (x, y) lies on the y-axis if and only if x= 0. " The origin is the point (0, 0). It is the only point common to both axes.  4 RELATIONS AND FUNCTIONS The axes divide the plane into four regions called quadrants. They are labeled with Roman numerals and proceed counterclockwise around the plane: Quadrant II < 0, y>0 4- 3- 2- 1- Quadrant I x>0, y>0 ' -4 -3 -2 - Quadrant III x<0, y<0 1 -1- -2- -3- -4- 1 2 3 4 x Quadrant IV x>0, y<0 For example, (1, 2) lies in Quadrant I, (-1, 2) in Quadrant II, (-1, -2) in Quadrant III, and (1, -2) in Quadrant IV. If a point other than the origin happens to lie on the axes, we typically refer to the point as lying on the positive or negative x-axis (if y = 0) or on the positive or negative y-axis (if x = 0). For example, (0, 4) lies on the positive y-axis whereas (-117, 0) lies on the negative x-axis. Such points do not belong to any of the four quadrants. One of the most important concepts in all of mathematics is symmetry.5 There are many types of symmetry in mathematics, but three of them can be discussed easily using Cartesian Coordinates. DEFINITION 1.1. Two points (a, b) and (c, d) in the plane are said to be " symmetric about the x-axis if a = c and b = -d " symmetric about the y-axis if a = -c and b = d " symmetric about the origin if a = -c and b = -d 5According to Carl. Jeff thinks symmetry is overrated.  1.1 THE CARTESIAN COORDINATE PLANE 5 Schematically, y Q(-x,y) P(x, y) 00 0 x R(-x, -y) S(x, -y) In the above figure, P and S are symmetric about the x-axis, as are Q and R; P and Q are symmetric about the y-axis, as are R and S; and P and R are symmetric about the origin, as are Q and S. EXAMPLE 1.1.2. Let P be the point (-2, 3). Find the points which are symmetric to P about the: 1. x-axis 2. y-axis 3. origin Check your answer by graphing. SOLUTION. The figure after Definition 1.1 gives us a good way to think about finding symmetric points in terms of taking the opposites of the x- and/or y-coordinates of P(-2, 3). 1. To find the point symmetric about the x-axis, we replace the y-coordinate with its opposite to get (-2, -3). 2. To find the point symmetric about the y-axis, we replace the x-coordinate with its opposite to get (2, 3). 3. To find the point symmetric about the origin, we replace the x- and y-coordinates with their opposites to get (2, -3). y . 3. P(-2,3) 2 (2,3) 1- -3 -2 -1 1 2 3 X -1. -2. . -3. . (-2, -3) (2, -3)D  6 RELATIONS AND FUNCTIONS One way to visualize the processes in the previous example is with the concept of reflections. If we start with our point (-2, 3) and pretend the x-axis is a mirror, then the reflection of (-2, 3) across the x-axis would lie at (-2, -3). If we pretend the y-axis is a mirror, the reflection of (-2, 3) across that axis would be (2, 3). If we reflect across the x-axis and then the y-axis, we would go from (-2, 3) to (-2, -3) then to (2, -3), and so we would end up at the point symmetric to (-2, 3) about the origin. We summarize and generalize this process below. Reflections To reflect a point (x, y) about the: " x-axis, replace y with -y. " y-axis, replace x with -x. " origin, replace x with -x and y with -y. 1.1.1 DISTANCE IN THE PLANE Another important concept in geometry is the notion of length. If we are going to unite Algebra and Geometry using the Cartesian Plane, then we need to develop an algebraic understanding of what distance in the plane means. Suppose we have two points, P (x1, y1) and Q (x2, Y2), in the plane. By the distance d between P and Q, we mean the length of the line segment joining P with Q. (Remember, given any two distinct points in the plane, there is a unique line containing both points.) Our goal now is to create an algebraic formula to compute the distance between these two points. Consider the generic situation below on the left. Q (x2,y2) Q (x2,y2) d di --------------------------l P (x1, y1) P (x1, y1) (x2, Y1) With a little more imagination, we can envision a right triangle whose hypotenuse has length d as drawn above on the right. From the latter figure, we see that the lengths of the legs of the triangle are x2 - il and lY2 - Yil so the Pythagorean Theorem gives us x2 x1 2 _+ Y2 - 12d2 (x2 - x1)2 + (y2 - Y1)2 d2 (Do you remember why we can replace the absolute value notation with parentheses?) By extracting the square root of both sides of the second equation and using the fact that distance is never negative, we get  1.1 THE CARTESIAN COORDINATE PLANE 7 EQUATION 1.1. The Distance Formula: The distance d between the points P (x, y1) and Q (2,y2) is: d = (z2- i)2+ (y2 - y1)2 It is not always the case that the points P and Q lend themselves to constructing such a triangle. If the points P and Q are arranged vertically or horizontally, or describe the exact same point, we cannot use the above geometric argument to derive the distance formula. It is left to the reader to verify Equation 1.1 for these cases. EXAMPLE 1.1.3. Find and simplify the distance between P(-2, 3) and Q(1, -3). SOLUTION. d(X = 2- i) + (y2 - #1)2 = v /(1 -(2))2 + (-3 -3)2 = v9+36 = 3v/5 So, the distance is 3v5. D EXAMPLE 1.1.4. Find all of the points with x-coordinate 1 which are 4 units from the point (3, 2). SOLUTION. We shall soon see that the points we wish to find are on the line x we'll just view them as points of the form (1, y). Visually, 1, but for now y A 3- 2- 1- * (3, 2) distance is 4 units / 3 (1,y) -1 - -2 - -3- We require that the distance from (3, 2) to (1, y) be 4. The Distance Formula, Equation 1.1, yields  8 RELATIONS AND FUNCTIONS d = (z2 - i) + (y2 - 91)2 4 = f(l - 3)2 +(y - 2)2 4 =f4 + -(y - 2)2 42 ( 4+(y-2)2) 16 4 + (y - 2)2 12 = (y -2)2 (y - 2)2 = 12 y-2 = +i12v y - 2 = +2v/3 y = 2+2v/3 squaring both sides extracting the square root We obtain two answers: (1, 2 + 2v/3) and (1, 2 - 2v/). The reader is encouraged to think about why there are two answers. D Related to finding the distance between two points is the problem of finding the midpoint of the line segment connecting two points. Given two points, P (x1Yi1) and Q (x2, Y2), the midpoint, M, of P and Q is defined to be the point on the line segment connecting P and Q whose distance from P is equal to its distance from Q. Q (2,y2) M P (x1, y1) If we think of reaching M by going 'halfway over' and 'halfway up' we get the following formula. EQUATION 1.2. The Midpoint Formula: The midpoint M of the line segment connecting P(x1,y1) and Q (2,y2) is: M (x+x2-Y1+Y2 2 '2 If we let d denote the distance between P and Q, we leave it as an exercise to show that the distance between P and M is d/2 which is the same as the distance between M and Q. This suffices to show that Equation 1.2 gives the coordinates of the midpoint.  1.1 THE CARTESIAN COORDINATE PLANE 9 EXAMPLE 1.1.5. Find the midpoint of the line segment connecting P(-2, 3) and Q(1, -3). SOLUTION. M (x1+x2 Y1+Y2) 2 2 (-2) +1 3 +(-3)) 2 2 ) 1 0 - ( 0 The midpoint is (- , 0 . An interesting application6 of the midpoint formula follows. EXAMPLE 1.1.6. Prove that the points (a, b) and (b, a) are symmetric about the line y = x. SOLUTION. By 'symmetric about the line y = z', we mean that if a mirror were placed along the line y = x, the points (a, b) and (b, a) would be mirror images of one another. (You should compare and contrast this with the other types of symmetry presented back in Definition 1.1.) Schematically, (a, b) (b,a) From the figure, we see that this problem amounts to showing that the midpoint of the line segment connecting (a, b) and (b, a) lies on the line y = x. Applying Equation 1.2 yields M (a+b b+a 2 '2 (a+b a+b 2 '2 Since the x and y coordinates of this point are the same, we find that the midpoint lies on the line y =x, as required. D 6This is a key concept in the development of inverse functions. See Section 5.2  10 RELATIONS AND FUNCTIONS 1.1.2 EXERCISES 1. Plot and label the points A(-3, -7), B(1.3, -2), C(7, 10), D(0, 8), E(-5.5, 0), F(-8, 4), G(9.2, -7.8) and H(7, 5) in the Cartesian Coordinate Plane given below. 9- 88- 7- 6- 5- 4- 3- 2-|- 1- I I I I I I I I I I I I I I I I I I I -9 -8 -7 -6 -5 -4 -3 -2 -1 1 2314 5 3 ;7 8 9 -1- -2- -3- -4- -5- -6- -7- -8- -9- 2. For each point given in Exercise I above " Identify the quadrant or axis in/on which the point lies. " Find the point symmetric to the given point about the x-axis. " Find the point symmetric to the given point about the y-axis. " Find the point symmetric to the given point about the origin.  1.1 THE CARTESIAN COORDINATE PLANE 11 3. For each of the following pairs of points, find the distance d between them and find the midpoint M of the line segment connecting them. (a) (1, 2), (-3, 5) (d) (- 4) 2) (b) (3, -10), (-1 2)(e)(2 3)(-8- (c) ,4 , 2)-1(f) (0, 0), (X, y) 4. Find all of the points of the form (x, -1) which are 4 units from the point (3, 2). 5. Find all of the points on the y-axis which are 5 units from the point (-5, 3). 6. Find all of the points on the x-axis which are 2 units from the point (-1, 1). 7. Find all of the points of the form (x, -x) which are 1 unit from the origin. 8. Let's assume for a moment that we are standing at the origin and the positive y-axis points due North while the positive x-axis points due East. Our Sasquatch-o-meter tells us that Sasquatch is 3 miles West and 4 miles South of our current position. What are the coordinates of his position? How far away is he from us? If he runs 7 miles due East what would his new position be? 9. Verify the Distance Formula 1.1 for the cases when: (a) The points are arranged vertically. (Hint: Use P(a, y1) and Q(a, y2).) (b) The points are arranged horizontally. (Hint: Use P(xi, b) and Q(x2, b).) (c) The points are actually the same point. (You shouldn't need a hint for this one.) 10. Verify the Midpoint Formula by showing the distance between P(zi, Yi) and M and the distance between M and Q(z2, Y2) are both half of the distance between P and Q. 11. Show that the points A(-3, 1), B(4, 0) and C(0, -3) are the vertices of a right triangle. 12. Find a point D(x, y) such that the points A(-3, 1), B(4, 0), C(0, -3) and D are the corners of a square. Justify your answer. 13. The world is not flat.? Thus the Cartesian Plane cannot possibly be the end of the story. Discuss with your classmates how you would extend Cartesian Coordinates to represent the three dimensional world. What would the Distance and Midpoint formulas look like, assuming those concepts make sense at all? 7There are those who disagree with this statement. Look them up on the Internet some time when you're bored.  12 RELATIONS AND FUNCTIONS 1.1.3 ANSWERS 1. The required points A(-3, -7), B(1.3, -2), C(7, 10), D(0, 8), E(-5.5, 0), F(-8, 4), G(9.2, -7.8), and H(7, 5) are plotted in the Cartesian Coordinate Plane below. y 9+I 8i 7- D(o,8g) 611 H(7, 5) 0 F(-8, 4) 4. 3. 0(, o 2+- I E(-5.5, J) I I I I I I I I -9 -8 -7 -6 -5 -4 -3 -2 -1 i 2 3 4 5 3 7 8 9~ -1- -2- -3. -4. B(1.3, -2) -5+I " A(-3) -7) -6- -7. -8- 0 G (9.2) - 7.8) -9+  1.1 THE CARTESIAN COORDINATE PLANE 2. (a) The point A(-3, -7) is " in Quadrant III " symmetric about x-axis with (-3, 7) " symmetric about y-axis with (3, -7) " symmetric about origin with (3, 7) (b) The point B(1.3, -2) is " in Quadrant IV " symmetric about x-axis with (1.3, 2) " symmetric about y-axis with (-1.3, -2) " symmetric about origin with (-1.3, 2) (c) The point C(7r, 10) is " in Quadrant I " symmetric about x-axis with (7r, - 10) " symmetric about y-axis with (-7r, 10) " symmetric about origin with (-r, -io) (d) The point D(0, 8) is 13 (e) The point E(-5.5, 0) is " on the negative x-axis " symmetric about x-axis with (-5.5, 0) " symmetric about y-axis with (5.5, 0) " symmetric about origin with (5.5, 0) (f) The point F(-8,4) is " in Quadrant II " symmetric about x-axis with (-8, -4) " symmetric about y-axis with (8,4) " symmetric about origin with (8, -4) (g) The point G(9.2, -7.8) is " in Quadrant IV " symmetric about x-axis with (9.2, 7.8) " symmetric about y-axis with (-9.2, -7.8) " symmetric about origin with (-9.2, 7.8) (h) The point H(7, 5) is " in Quadrant I " symmetric about x-axis with (7, -5) " symmetric about y-axis with (-7, 5) " symmetric about origin with (-7, -5) " " 0 " on the positive y-axis symmetric about x-axis with (0, -8) symmetric about y-axis with (0, 8) symmetric about origin with (0, -8) 3. (a) d 5, M (-1, (b) d =4v/-0O, M = (1, -4) (c) d = v26, M = 1, 3 4. (3 + v, -2), (3 - v'7, -2) 5. (0,3) (d) d3 = , M (2'4) (e) d =3v/5, M = -2- ) (f) d = /2 + y2, M = x 6. (-1+ 3,0), (- ' 2 2 3 -1- V3,0) 2 ' 2 8. (-3, -4), 5 miles, (4, -4)  14 RELATIONS AND FUNCTIONS 1.2 RELATIONS We now turn our attention to sets of points in the plane. DEFINITION 1.2. A relation is a set of points in the plane. Throughout this text we will see many different ways to describe relations. In this section we will focus our attention on describing relations graphically, by means of the list (or roster) method and algebraically. Depending on the situation, one method may be easier or more convenient to use than another. Consider the set of points below 4- 3- 2- 1- ( (4, 3) (-2, 1) e0 I I I I ' I I I I I -4 -3 -2 -1 -1 - -2 - 1 2 3 4 X ), -3) -3 (( -4- These three points constitute a relation. Let us call this relation R. Above, we have a graphical description of R. Although it is quite pleasing to the eye, it isn't the most portable way to describe R. The list (or roster) method of describing R simply lists all of the points which belong to R. Hence, we write: R = {(-2, 1), (4, 3), (0, -3)}.1 The roster method can be extended to describe infinitely many points, as the next example illustrates. EXAMPLE 1.2.1. Graph the following relations. 1. A = {(0, 0), (-3,1), (4,2), (-3, 2)} 2. HLS1={(x,3) :-2 -4} {(2, y) : y < 5} {(-2, y) : -3 < y < 4} {(X, y) : x < 3} (e) {(x, y) (f) {(xy) (g) {(xy) (h) {(x, y) : y < 4} : x < 3, y < 2} : 3> 0, y <-4} : _- 21F i r< y < } 3. Describe the following relations using the roster method. 4- 34 . 2- " 1- -4 -3-2-1 0 -1- y 431.:::; y 11L (c) Thegaph 2 3 i (c) The graph of relation C (a) The graph of relation A A 1- A' 2 1 -3- o (b) The graph of relation B -4 --1- The -3- The graph of relation D (d)  1.2 RELATIONS 19 y 5- 4- 3- 2- 1- -1- -1 y 1- 1 "A 10 . .. .. . . . . . . . . . . . . . 1 2 3 4 53....7....... (f) The graph of relation F (e) The graph of relation E 4. Graph the following lines. (a) x -2 (b) y = 3 5. What is another name for the line x 0? For y= 0? 6. Some relations are fairly easy to describe in words or with the roster method but are rather difficult, if not impossible, to graph. Discuss with your classmates how you might graph the following relations. Please note that in the notation below we are using the ellipsis, ..., to denote that the list does not end, but rather, continues to follow the established pattern indefinitely. For the first two relations, give two examples of points which belong to the relation and two points which do not belong to the relation. (a) {(x, y) : x is an odd integer, and y is an even integer.} (b) {(x, 1) : x is an irrational number } (c) {(1,0), (2, 1), (4,2), (8,3), (16,4), (32,5),.. .} (d) {. .. ,(-3,9),(-2,4),(-1,1), (, 0 ,(1,1),(2,4),(3,9),. ..}  LZ) 9 9 I I I Q LN IQ IQ +J P C) [ I I ta IQ - eQ + 0L LD )- 4.U IQ IQ " - 0+ PC1C - 0 f -" ~ IQ ~ ~ -~ " z 0 LL t C)1 -N IQ k " LN IQ Q P " I 0 IQ QJ P C1 i i i i i i i 1 1 1 1 1i i i> 4- IQ. Q1J ..S . IQ t 0 I 0i 0 z H 0 -0  1.2 RELATIONS 2 21 (e) (g) 3 2: i ._..: y 2-- -1Ii 1 2 3 -3-2- 1 2 3 (f) (h) y 3 1 1 2 3 .0-- 1 2 3: 7- 6- 5- 2- 1- 171 -3 -2 -1 1 3: 3. (a) (b) (c) (d) (e) (f) 4. (a) A B C D £ F {Q) Y) :3x:> -2} {Q(v)Y) :y > O} {(x,y) -3 <3x < 2} {Q(v)Y) 3x > O~y > o} 2, 1), (- 1,2), (0, 3), (1,4)} -3 -1 I: Si and the line x= 1 D- Suppose a relation F describes y as a function of x. The sets of x- and y-coordinates are given special names. DEFINITION 1.5. Suppose F is a relation which describes y as a function of x. " The set of the x-coordinates of the points in F is called the domain of F. " The set of the y-coordinates of the points in F is called the range of F. We demonstrate finding the domain and range of functions given to us either graphically or via the roster method in the following example. EXAMPLE 1.4.4. Find the domain and range of the following functions 1. F = {(-3, 2), (0, 1), (4, 2), (5, 2)} 2. G is the function graphed below:  34 RELATIONS AND FUNCTIONS y I: The graph of G SOLUTION. The domain of F is the set of the x-coordinates of the points in F: {-3, 0, 4, 5} and the range of F is the set of the y-coordinates: {1, 2}.2 To determine the domain and range of G, we need to determine which x and y values occur as coordinates of points on the given graph. To find the domain, it may be helpful to imagine collapsing the curve to the x-axis and determining the portion of the x-axis that gets covered. This is called projecting the curve to the x-axis. Before we start projecting, we need to pay attention to two subtle notations on the graph: the arrowhead on the lower left corner of the graph indicates that the graph continues to curve downwards to the left forever more; and the open circle at (1, 3) indicates that the point (1, 3) isn't on the graph, but all points on the curve leading up to that point are on the curve. y y down I: I 1/ 1 project up The graph of G The graph of G 2When listing numbers in a set, we list each number only once, in increasing order.  1.4 INTRODUCTION TO FUNCTIONS 35 We see from the figure that if we project the graph of G to the x-axis, we get all real numbers less than 1. Using interval notation, we write the domain of G is (-oc, 1). To determine the range of G, we project the curve to the y-axis as follows: y y pro >3 ject right 2 1- -1 project left i 1 :01 -1 1 I: -1- The graph of G The graph of G Note that even though there is an open circle at (1, 3), we still include the y value of 3 in our range, since the point (-1, 3) is on the graph of G. We see that the range of G is all real numbers less than or equal to 4, or, in interval notation: (-oc, 4].Q All functions are relations, but not all relations are functions. Thus the equations which described the relations in Section1.2 may or may not describe y as a function of x. The algebraic representation of functions is possibly the most important way to view them so we need a process for determining whether or not an equation of a relation represents a function. (We delay the discussion of finding the domain of a function given algebraically until Section 1.5.) EXAMPLE 1.4.5. Determine which equations represent y as a function of x: 1. x3 _+y2 1 2. 2 +y3 i 3. x2y = 1 -3y SOLUTION. For each of these equations, we solve for y and determine whether each choice of x will determine only one corresponding value of y. 1. 3 _ 2 y2 y2 1 1-x3 1-x3 i-x3 extract square roots  36 RELATIONS AND FUNCTIONS If we substitute x = 0 into our equation for y, we get: y = + 1 -03 = +1, so that (0, 1) and (0, -1) are on the graph of this equation. Hence, this equation does not represent y as a function of x. 2. x2 + y 3 1 3 - 1 - x2 1-x2 y= 3 1 - z2 For every choice of x, the equation y = 1 - x2 returns equation describes y as a function of x. only one value of y. Hence, this 3. z2y 12y + 3y y (x2 + 3) 1 - 3y 1 1 factor 1 x2 + 3 For each choice of x, there is only one value for y, so this equation describes y as a function of x. Q Of course, we could always use our graphing calculator to verify our responses to the previous example. For example, if we wanted to verify that the first equation does not represent y as a function of x, we could enter the equation for y into the calculator as indicated below and graph. Note that we need to enter both solutions - the positive and the negative square root - for y. The resulting graph clearly fails the Vertical Line Test, so does not represent y as a function of x. Plot1 Plot2 PlOt3 \Y1 = (1-X^3) \Y2E= -' (1-X^3) %Y 4= x=o Y=o  1.4 INTRODUCTION TO FUNCTIONS 37 1.4.1 EXERCISES 1. Determine which of the following relations represent y as a function of x. Find the domain and range of those relations which are functions. (a) {(-3, 9), (-2, 4), (-1, 1), (0, 0), (1,1), (2,4), (3,9)} (b) {(-2, 2), (-2, -1), (3, 5), (3, -4)} (c) {(x, y) : x is an odd integer, and y is an even integer} (d) {(x, 1) : x is an irrational number} (e) {(1,0), (2,1), (4,2), (8,3), (16,4), (32,5), ... } (f) {..., (-3,9), (-2,4), (-1, 1), (0,0), (1,1), (2,4), (3,9), ... } (g) {(-2, y) : -3 < y < 4} (h) {(x, 3) : -2 -4 -3 -2 " -i i x -1- (b) (d) -- >>. 3. 2. 1. 2 ' i.s 4 - 3 - 2 - 1 _ : 1 1 - - - - - - - - - - J i i i 1 i i i -3 -2 - -1. -2. -3. 2 3  38 RELATIONS AND FUNCTIONS (e) (g) 3- 2-- - I 4. 3. " 2. " " 1. . 1- 1 2 3 4 5 6 7 8 9 i i i i i 30 (f) -4 -3 -2 -1 e x -1. 4 3- 2- 1- I I (h) 4 In 3- 2- 1- -4 -3 -2 -1 1 2 3 4 x 4p lp :01 -5 -4 -3 -2 -1 -1- - 2 03 x 3. Determine which of the following relations represent y as a function of x. (a) y x3 -x (b) y x-2 (c) x3y =-4 (d) 9 _2- 22 1 (e) y =2 2 (f) y =--6 4. Explain why the height h of a Sasquatch is a function of its age N in years. Given that a Sasquatch is 2 feet tall at birth, experiences growth spurts at ages 3, 23 and 57, and lives to be about 150 years old with a maximum height of 9 feet, sketch a rough graph of the height function. 5. Explain why the population P of Sasquatch in a given area is a function of time t. What would be the range of this function? 6. Explain why the relation between your classmates and their email addresses may not be a function. What about phone numbers and Social Security Numbers? 7. The process given in Example 1.4.5 for determining whether an equation of a relation rep- resents y as a function of x breaks down if we cannot solve the equation for y in terms of x.  1.4 INTRODUCTION TO FUNCTIONS 39 However, that does not prevent us from proving that an equation which fails to represent y as a function of x actually fails to do so. What we really need is two points with the same x-coordinate and different y-coordinates which both satisfy the equation so that the graph of the relation would fail the Vertical Line Test 1.1. Discuss with your classmates how you might find such points for the relations given below. (a) x3 +Y 3- 3xy = 0 (c) y2 = x3 + 3x2 (b) x4 = X2 + y2 (d) (x2 +y2)2 = X3 +y3  40 RELATIONS AND FUNCTIONS 1.4.2 ANSWERS 1. (a) Function domain = {-3, -2, -1, 0, 1, 2 ,3} range = {0, 1, 4, 9} (b) Not a function (c) Not a function (d) Function domain = { x is irrational} range = {1} (e) Function domain = {x: x = 2n for some whol number n} range = {y: y is any whole number} (f) Function domain {x: x is any integer} range {y: y - n2 for some integer n (g) Not a function 2. (a) Function domain = {-4, -3, -2, -1, 0, 1} range = {-1, 0, 1, 2, 3, 4} (b) Not a function (c) Function domain = (-oc, oc) range = [1,oo) (d) Not a function e (e) Function domain = [2, oc) range = [0, oo) (f) Function domain (-oc, oc) } range (0,4] (g) Not a function (h) Function domain = [-5, 3) range = (-2, -1) U [0,4) (h) Function domain =[-2, 4) range = {3} 3. (a) Function (b) Function (c) Function (d) Not a fur (e) Function (f) Not a fu nction nction  1.5 FUNCTION NOTATION 41 1.5 FUNCTION NOTATION In Definition 1.4, we described a function as a special kind of relation - one in which each x- coordinate is matched with only one y-coordinate. In this section, we focus more on the process by which the x is matched with the y. If we think of the domain of a function as a set of inputs and the range as a set of outputs, we can think of a function f as a process by which each input x is matched with only one output y. Since the output is completely determined by the input x and the process f, we symbolize the output with function notation: 'f(x)', read 'f of x.' In this case, the parentheses here do not indicate multiplication, as they do elsewhere in algebra. This could cause confusion if the context is not clear. In other words, f(x) is the output which results by applying the process f to the input x. This relationship is typically visualized using a diagram similar to the one below. f x y = f (z) Domain Range (Inputs) (Outputs) The value of y is completely dependent on the choice of x. For this reason, x is often called the independent variable, or argument of f, whereas y is often called the dependent variable. As we shall see, the process of a function f is usually described using an algebraic formula. For example, suppose a function f takes a real number and performs the following two steps, in sequence 1. multiply by 3 2. add 4 If we choose 5 as our input, in step 1 we multiply by 3 to get (5)(3) = 15. In step 2, we add 4 to our result from step 1 which yields 15+ 4 = 19. Using function notation, we would write f (5) = 19 to indicate that the result of applying the process f to the input 5 gives the output 19. In general, if we use c for the input, applying step 1 produces 3x. Following with step 2 produces 3x + 4 as our final output. Hence for an input cv, we get the output f(cv) =3cv + 4. Notice that to check our formula for the case cv= 5, we replace the occurrence of cv in the formula for f(cv) with 5 to get f (5) =3(5) + 4 =15 + 4 =19, as required.  42 RELATIONS AND FUNCTIONS EXAMPLE 1.5.1. Suppose a function g is described by applying the following steps, in sequence 1. add 4 2. multiply by 3 Determine g(5) and find an expression for g(x). SOLUTION. Starting with 5, step 1 gives 5 + 4 = 9. Continuing with step 2, we get (3)(9) = 27. To find a formula for g(x), we start with our input x. Step 1 produces x + 4. We now wish to multiply this entire quantity by 3, so we use a parentheses: 3(x + 4) = 3x + 12. Hence, g(x) = 3x + 12. We can check our formula by replacing x with 5 to get g(5) = 3(5) + 12 = 15 + 12 = 27 /. D Most of the functions we will encounter in College Algebra will be described using formulas like the ones we developed for f(x) and g(x) above. Evaluating formulas using this function notation is a key skill for success in this and many other math courses. EXAMPLE 1.5.2. For f (x) = -2 + 3x + 4, find and simplify 1. f (-1), f(0), f (2) 2. f(2x), 2f(x) 3. f(x + 2), f (x) + 2, f (x) + f (2) SOLUTION. 1. To find f(-1), we replace every occurrence of x in the expression f(x) with -1 f(-1) = -(-1)2 + 3(-1) + 4 =-(1) +(-3)+ 4 = 0 Similarly, f (0) = -(0)2 + 3(0) + 4 = 4, and f (2) =_-(2)2 + 3(2) + 4 = -4 + 6 + 4 = 6. 2. To find f(2x), we replace every occurrence of x with the quantity 2x f(2x) = -(2x)2 + 3(2x) + 4 = -(4x2) + (6x) + 4 =-4x2 +6x+4 The expression 2f(x) means we multiply the expression f(x) by 2 2f (x) =2 (-2+ 3x +4) = -2x2+6x+8  1.5 FUNCTION NOTATION 43 Note the difference between the answers for f(2x) and 2f(x). For f(2x), we are multiplying the input by 2; for 2f(x), we are multiplying the output by 2. As we see, we get entirely different results. Also note the practice of using parentheses when substituting one algebraic expression into another; we highly recommend this practice as it will reduce careless errors. 3. To find f(x + 2), we replace every occurrence of x with the quantity x + 2 f (x+ 2) -(x + 2)2 + 3(x + 2) + 4 - (z2 + 4x + 4) + (3x + 6) + 4 -2-4x-4+3x+6+4 = - 2-x+6 To find f(x) + 2, we add 2 to the expression for f(x) f(x)+2 = (-z2 +3x+4) +2 = -2 +3x+6 Once again, we see there is a dramatic difference between modifying the input and modifying the output. Finally, in f(x) + f(2) we are adding the value f(2) to the expression f(x). From our work above, we see f(2) = 6 so that f(x) + f(2) (-z2 +3x+4) +6 -2 +3x + 10 Notice that f (x+ 2), f(x) + 2 and f(x) + f(2) are three different expressions. Even though function notation uses parentheses, as does multiplication, there is no general 'distributive property' of function notation. D 2x Suppose we wish to find r(3) for r(x) = 2 Substitution gives 2(3) 6 (3)2 -9 0' which is undefined. The number 3 is not an allowable input to the function r; in other words, 3 is not in the domain of r. Which other real numbers are forbidden in this formula? We think back to arithmetic. The reason r(3) is undefined is because substitution results in a division by 0. To determine which other numbers result in such a transgression, we set the denominator equal to 0 and solve 2 -9 = 0 92 9 =9v extract square roots z = 3  44 RELATIONS AND FUNCTIONS As long as we substitute numbers other than 3 and -3, the expression r(x) is a real number. Hence, we write our domain in interval notation as (-oc, -3) U (-3,3) U (3, ox). When a formula for a function is given, we assume the function is valid for all real numbers which make arithmetic sense when substituted into the formula. This set of numbers is often called the implied domain' of the function. At this stage, there are only two mathematical sins we need to avoid: division by 0 and extracting even roots of negative numbers. The following example illustrates these concepts. EXAMPLE 1.5.3. Find the domain2 of the following functions. 2 1. f (X) 4x 1-4x x:-3 2. g(x)= V4 -3x 3. h(x) = 4 - 3x 4. r(x) =_ 4 6- x+3 5. I(x) 3x2 3: SOLUTION. 1. In the expression for f, there are two denominators. We need to make sure neither of them is 0. To that end, we set each denominator equal to 0 and solve. For the 'small' denominator, we get 3 - 3 = 0 or x= 3. For the 'large' denominator 4x 1- x:-3 0 4x 1 = - (-3 (1)(x-3) x:-3 - fclear denominators 4x -3 = x So we get two real numbers which make denominators 0, namely x= -1 and x= 3. Our domain is all real numbers except -1 and 3: (-oc, -1) U (-1, 3) U (3, oc). 1or, 'implicit domain' 2The word 'implied' is, well, implied.  1.5 FUNCTION NOTATION 45 2. The potential disaster for g is if the radicand3 is negative. To avoid this, we set 4 - 3x > 0 4-3x > 0 4 > 3x 4 3 - Hence, as long as 3x < , the expression 4 - 3x >0, and the formula g(x) returns a real number. Our domain is (-oc, 4]. 3. The formula for h(x) is hauntingly close to that of g(x) with one key difference - whereas the expression for g(x) includes an even indexed root (namely a square root), the formula for h(x) involves an odd indexed root (the fifth root.) Since odd roots of real numbers (even negative real numbers) are real numbers, there is no restriction on the inputs to h. Hence, the domain is (-oc, oc). 4. To find the domain of r, we notice that we have two potentially hazardous issues: not only do we have a denominator, we have a square root in that denominator. To satisfy the square root, we set the radicand x + 3 > 0 so 3:> -3. Setting the denominator equal to zero gives 6- x+3 = 0 6 = x+3 62 = (x+3) 36 = :x+3 33 = x Since we squared both sides in the course of solving this equation, we need to check our answer. Sure enough, when 3 = 33, 6 - xz + 3 = 6 - V3 -= 0, and so 3 = 33 will cause problems in the denominator. At last we can find the domain of r: we need 3 > -3, but x # 33. Our final answer is [-3, 33) U (33, oc). 5. It's tempting to simplify I(x) = 3x2 = 3x, and, since there are no longer any denominators, claim that there are no longer any restrictions. However, in simplifying I(x), we are assuming x # 0, since °ois undefined.4 Proceeding as before, we find the domain of I to be all real numbers except 0: (-oc, 0) U (0, oc). D It is worth reiterating the importance of finding the domain of a function before simplifying, as evidenced by the function I in the previous example. Even though the formula -I(x) simplifies to 3x, it would be inaccurate to write I(x) = 3x without adding the stipulation that : # 0. It would be analogous to not reporting taxable income or some other sin of omission. 3The 'radicand' is the expression 'inside' the radical. 4More precisely, the fraction j is an 'indeterminant form'. Much time will be spent in Calculus wrestling with such creatures.  46 RELATIONS AND FUNCTIONS Our next example shows how a function can be used to model real-world phenomena. EXAMPLE 1.5.4. The height h in feet of a model rocket above the ground t seconds after lift off is given by h(t) -5t2 +100t, if 0 < t < 20 0, if t > 20 Find and interpret h(10) and h(60). SOLUTION. There are a few qualities of h which may be off-putting. The first is that, unlike previous examples, the independent variable is t, not x. In this context, t is chosen because it represents time. The second is that the function is broken up into two rules: one formula for values of t between 0 and 20 inclusive, and another for values of t greater than 20. To find h(10), we first notice that 10 is between 0 and 20 so we use the first formula listed: h(t) = -5t2 + loot. Hence, h(10) = -5(10)2 + 100(100) = 500. In terms of the model rocket, this means that 10 seconds after lift off, the model rocket is 500 feet above the ground. To find h(60), we note that 60 is greater than 20, so we use the rule h(t) = 0. This function returns a value of 0 regardless of what value is substituted in for t, so h(60) = 0. This means that 60 seconds after lift off, the rocket is 0 feet above the ground; in other words, a minute after lift off, the rocket has already returned to earth. D The type of function in the previous example is called a piecewise-defined function, or 'piecewise' function for short. Many real-world phenomena (e.g. postal rates,5 income tax formulas6) are modeled by such functions. Also note that the domain of h in the above example is restricted to t > 0. For example, h(-3) is not defined because t = -3 doesn't satisfy any of the conditions in any of the function's pieces. There is no inherent arithmetic reason which prevents us from calculating, say, -5(-3)2 + 100(-3), it's just that in this applied setting, t = -3 is meaningless. In this case, we say h has an applied domain7 of [0, oc) 5See the United States Postal Service website http://www.usps.com/prices/first-class-mail-prices.htm 6See the Internal Revenue Service's website http://www.irs.gov/pub/irs-pdf/i1040tt.pdf 7or, 'explicit domain'  1.5 FUNCTION NOTATION 47 1.5.1 EXERCISES 1. Suppose f is a function that takes a real number x and performs the following three steps in the order given: (1) square root; (2) subtract 13; (3) make the quantity the denominator of a fraction with numerator 4. Find an expression for f(x) and find its domain. 2. Suppose g is a function that takes a real number x and performs the following three steps in the order given: (1) subtract 13; (2) square root; (3) make the quantity the denominator of a fraction with numerator 4. Find an expression for g(x) and find its domain. 3. Suppose h is a function that takes a real number x and performs the following three steps in the order given: (1) square root; (2) make the quantity the denominator of a fraction with numerator 4; (3) subtract 13. Find an expression for h(x) and find its domain. 4. Suppose k is a function that takes a real number x and performs the following three steps in the order given: (1) make the quantity the denominator of a fraction with numerator 4; (2) square root; (3) subtract 13. Find an expression for k(z) and find its domain. 5. For f (x) 9= 2 - 3x + 2, find and simplify the following: (a) f(3) (b) f(-1) (c) f(j) (d) f(4x) (e) 4f (x) (f) f(-x) (g) f(-4) (h) f (x)-4 (i) f (2) 6. Repeat Exercise 5 above for f(x) =2 7. Find the (implied) domain of the function. (a) f (x) = 4 - 13x3 + 56x2 - 19 t (b) s(t) t-8 t/i8 (c) Q(r) r-8 (d) b(0) = - v/6-s8 (e) a (y) 3 __ y- (f) A (x) =v/z - -7 + v/9 - x 1 (g) g(v)~ 1 4-2 (h) u(w) =5 8 5 -w91  48 RELATIONS AND FUNCTIONS 2 if x< -1 8. Let f (x) = /1 -x2 if -1 1 (a) f(4) (d) f(0) (b) f (-3) (e) f (-1) (c) f(1) (f) f(-0.999) 9. The population of Sasquatch in Portage County can be modeled by the function P(t) 150t t + 15 where t = 0 represents the year 1803. What is the applied domain of P? What range "makes sense" for this function? What does P(0) represent? What does P(205) represent? 10. Recall that the integers is the set of numbers Z = {... , -3, -2, -1, 0, 1, 2, 3,. ..}.8 The greatest integer of x, [x], is defined to be the largest integer k with k 1 SOLUTION. We proceed as before: finding intercepts, testing for symmetry and then plotting additional points as needed. To find the x-intercepts, as before, we set f(x) = 0. The twist is that we have two formulas for f(x). For x < 1, we use the formula f(x) = 4 - 2. Setting f(x) = 0 gives 0 = 4 - 2, so that x +=i2. However, of these two answers, only x= -2 fits in the domain x < 1 for this piece. This means the only x-intercept for the x < 1 region of the x-axis is (-2, 0). For x >1, f(x) = x - 3. Settingf (x) = 0 gives 0 = x - 3, or x= 3. Since x= 3 satisfies the inequality x > 1, we get (3, 0) as another x-intercept. Next, we seek the y-intercept. Notice that x = 0 falls in the domain x < 1. Thus f(0) = 4 - 02 = 4 yields the y-intercept (0, 4). As far as symmetry is concerned, you can check that the equation y = 4 - 2 is symmetric about the y-axis; unfortunately, this equation (and its symmetry) is valid only for x < 1. You can also verify y = x - 3 possesses none of the symmetries discussed in the Section 1.3. When plotting additional points, it is important to keep in mind the restrictions on x for each piece of the function. The sticking point for this function is x= 1, since this is where the equations change. When x= 1, we use the formula f(x) = x - 3, so the point on the graph (1, f(1)) is (1, -2). However, for all values less than 1, we use the formula f(x) = 4 -9x2. As we have discussed earlier in Section 1.2, there is no real number which immediately precedes x= 1 on the number line. Thus for the values x= 0.9, x = 0.99, x= 0.999, and so on, we find the corresponding y values using the formula f (x) = 4 - 2. Making a table as before, we see that as the x values sneak up to x= 1 in this fashion, the f(x) values inch closer and closer1 to 4 - 12 = 3. To indicate this graphically, we use an open circle at the point (1, 3). Putting all of this information together and plotting additional points, we get y z: f () (x, f )) 0.9 3.19 (0.9, 3.19) 0.99 L 3.02 (0.99, 3.02) 0.999 L 3.002 (0.999, 3.002) El 'We've just stepped into Calculus here!  60 RELATIONS AND FUNCTIONS In the previous two examples, the x-coordinates of the x-intercepts of the graph of y = f(x) were found by solving f (x) = 0. For this reason, they are called the zeros of f. DEFINITION 1.7. The zeros of a function f are the solutions to the equation f(x) = 0. In other words, x is a zero of f if and only if (x, 0) is an x-intercept of the graph of y = f(x). Of the three symmetries discussed in Section 1.3, only two are of significance to functions: symmetry about the y-axis and symmetry about the origin.2 Recall that we can test whether the graph of an equation is symmetric about the y-axis by replacing x with -x and checking to see if an equivalent equation results. If we are graphing the equation y = f(x), substituting -x for x results in the equation y = f(-x). In order for this equation to be equivalent to the original equation y = f(x) we need f (-x) = f (x). In a similar fashion, we recall that to test an equation's graph for symmetry about the origin, we replace x and y with -x and -y, respectively. Doing this substitution in the equation y = f(x) results in -y = f(-x). Solving the latter equation for y gives y = -f(-x). In order for this equation to be equivalent to the original equation y = f(x) we need -f(-x) = f(x), or, equivalently, f(-x) = -f(x). These results are summarized below. Steps for testing if the graph of a function possesses symmetry The graph of a function f is symmetric: " About the y-axis if and only if f(-x) = f(x) for all x in the domain of f. " About the origin if and only if f(-x) = -f(x) for all x in the domain of f. For reasons which won't become clear until we study polynomials, we call a function even if its graph is symmetric about the y-axis or odd if its graph is symmetric about the origin. Apart from a very specialized family of functions which are both even and odd,3 functions fall into one of three distinct categories: even, odd, or neither even nor odd. EXAMPLE 1.7.3. Analytically determine if the following functions are even, odd, or neither even nor odd. Verify your result with a graphing calculator. 5 5z 1. f (X) - 2 4. i(x) - 35 2-x 2x-z 5x2 x 2. g(x) - 5. j(x) = - _ -1 2 - 2 100 5x 3. h(x) = SOLUTION. The first step in all of these problems is to replace z with -z and simplify. 2Why are we so dismissive about symmetry about the x-axis for graphs of functions? 3Any ideas?  1.7 GRAPHS OF FUNCTIONS 1. 61 f(x) f(-x) f(-x) f(-x) 5 2 5 2 - (-x)2 5 2 2-x2 f (x) Hence, f is even. The graphing calculator furnishes the following: ploki P102 P103 %.Y105/(2-X2)1 %-Y2= %Y3= This suggests4 the graph of f is symmetric about the y-axis, as expected. 2. g(x) g(-x) g(-x) 5x 2 2 - x2 5(-x) 2 - (-x)2 -5x 2 2 It doesn't appear that g(-x) is equivalent to g(x). To prove this, we check with an x value. After some trial and error, we see that g(1) = 5 whereas g(-1) = -5. This proves that g is not even, but it doesn't rule out the possibility that g is odd. (Why not?) To check if g is odd, we compare g( x) with -g(x) -g(x) 5x 2-x -5x 2-x2 g(-x) g(x) Hence, g is odd. Graphically, 4'Suggests' is about the extent of what it can do.  62 RELATIONS AND FUNCTIONS Ploit Pot2 Plot3 %Y1=5X/(2-X2) %-Y2= The calculator indicates the graph of g is symmetric about the origin, as expected. 3. h(x) h(-x) h(-x) 5x 2 3 5(-x) 2 - (-x)3 -5x 2 +x3 Once again, h( -x) doesn't appear to be equivalent to h(x). We check with an x value, for example, h(1) = 5 but h(-1) odd. (Why?) Graphically, . This proves that h is not even and it also shows h is not Ploit Pot2 Plot3 %Y1 5X/(2-X^3)E -Y 2= I C I The graph of h appears to be neither symmetric about the y-axis nor the origin. 4. i(x) 4-x) z(-x) 5x 2x-x3 5(-x) 2(-x) - (-x)3 -5x -2x + x3 The expression i( -x) doesn't appear to be equivalent to i(x). However, after checking some x values, for example x = 1 yields i(1) = 5 and i(-1) = 5, it appears that i(-x) does, in fact, equal i(x). However, while this suggests i is even, it doesn't prove it. (It does, however,  1.7 GRAPHS OF FUNCTIONS 63 prove i is not odd.) To prove i(-x) = i(x), we need to manipulate our expressions for i(x) and i(-x) and show they are equivalent. A clue as to how to proceed is in the numerators: in the formula for i(x), the numerator is 5x and in i(-x) the numerator is -5x. To re-write i(x) with a numerator of -5x, we need to multiply its numerator by -1. To keep the value of the fraction the same, we need to multiply the denominator by -1 as well. Thus i(x) 5x 2x x (-1)5x (-1) (2x - x3) -5x -2x + x3 Hence, i(x) = i(-x), so i is even. The calculator supports our conclusion. Plot1 Plot2 Plot3 %Y1=5X/(2X-X^3)E N-Y2= N-Y3= 5. fix) j(-x) j(-x) 2 _ x X100 1 (-x)2 1-0 2 __ 100 1 The expression for j(-x) doesn't seem to be equivalent to j(x), so we check using x = 1 to get j(1) = -l and j(-1) =l . This rules out j being even. However, it doesn't rule out j being odd. Examining -j(x) gives fi(x) j(x) j(x) 2 __ i - -1 100 (X2 __ -12+ +-1 100 1) 100 1)149 The expression -j(x) doesn't seem to match j(-x) either. Testing x = 2 gives j(2) = and j(-2) = , so j is not odd, either. The calculator gives:  64 RELATIONS AND FUNCTIONS Ploki P102 P103 %.Y 1 0X'2-X/ 100-1 %-Y2_ %-Y3_ %-Y 4 %-Y5_ %-Y6_ %-Y7_ The calculator suggests that the graph of j is symmetric about the y-axis which would imply that j is even. However, we have proven that is not the case. Q There are two lessons to be learned from the last example. The first is that sampling function values at particular x values is not enough to prove that a function is even or odd - despite the fact that j(-1) = -j(1), j turned out not to be odd. Secondly, while the calculator may suggest mathematical truths, it is the algebra which proves mathematical truths.5 1.7.1 GENERAL FUNCTION BEHAVIOR The last topic we wish to address in this section is general function behavior. As you shall see in the next several chapters, each family of functions has its own unique attributes and we will study them all in great detail. The purpose of this section's discussion, then, is to lay the foundation for that further study by investigating aspects of function behavior which apply to all functions. To start, we will examine the concepts of increasing, decreasing, and constant. Before defining the concepts algebraically, it is instructive to first look at them graphically. Consider the graph of the function f given on the next page. Reading from left to right, the graph 'starts' at the point (-4, -3) and 'ends' at the point (6, 5.5). If we imagine walking from left to right on the graph, between (-4, -3) and (-2, 4.5), we are walking 'uphill'; then between (-2, 4.5) and (3, -8), we are walking 'downhill'; and between (3, -8) and (4, -6), we are walking 'uphill' once more. From (4, -6) to (5, -6), we 'level off', and then resume walking 'uphill' from (5, -6) to (6, 5.5). In other words, for the x values between -4 and -2 (inclusive), the y-coordinates on the graph are getting larger, or increasing, as we move from left to right. Since y = f(x), the y values on the graph are the function values, and we say that the function f is increasing on the interval [-4, -2]. Analogously, we say that f is decreasing on the interval [-2, 3] increasing once more on the interval [3, 4], constant on [4, 5], and finally increasing once again on [5, 6]. It is extremely important to notice that the behavior (increasing, decreasing or constant) occurs on an interval on the x-axis. When we say that the function f is increasing on [-4, -2] we do not mention the actual y values that f attains along the way. Thus, we report where the behavior occurs, not to what extent the behavior occurs.6 Also notice that we do not say that a function is increasing, decreasing or constant at a single x value. In fact, we would run 50r, in other words, don't rely too heavily on the machine! 6The notions of how quickly or how slowly a function increases or decreases are explored in Calculus.  1.7 GRAPHS OF FUNCTIONS 65 into serious trouble in our previous example if we tried to do so because x= -2 is contained in an interval on which f was increasing and one on which it is decreasing. (There's more on this issue and many others in the exercises.) y 7 - (6, 5.5) (-2, 4.5) 5 4 3 1 -4 -3 -2 -1 1 2 3 4 5 6 7 x -1 - -2 - -3- (-4, -3) -4- -5- (4, -6) -6 -7 (5, -6) -8 -9 (3, -8) The graph of y = f(x) We're now ready for the more formal algebraic definitions of what it means for a function to be increasing, decreasing or constant. DEFINITION 1.8. Suppose f is a function defined on an interval I. We say f is: " increasing on I if and only if f(a) < f(b) for all real numbers a, b in I with a < b. " decreasing on I if and only if f(a) > f(b) for all real numbers a, b in I with a < b. " constant on I if and only if f(a) = f(b) for all real numbers a, b in I. It is worth taking some time to see that the algebraic descriptions of increasing, decreasing, and constant as stated in Definition 1.8 agree with our graphical descriptions given earlier. You should look back through the examples and exercise sets in previous sections where graphs were given to see if you can determine the intervals on which the functions are increasing, decreasing or constant. Can you find an example of a function for which none of the concepts in Definition 1.8 apply? Now let's turn our attention to a few of the points on the graph. Clearly the point (-2, 4.5) does not have the largest y value of all of the points on the graph of f - indeed that honor goes to  66 RELATIONS AND FUNCTIONS (6, 5.5) - but (-2, 4.5) should get some sort of consolation prize for being 'the top of the hill' between xc= -4 and xc= 3. We say that the function f has a local maximuml at the point (-2, 4.5), because the y-coordinate 4.5 is the largest y-value (hence, function value) on the curve 'near'8 x = -2. Similarly, we say that the function f has a local minimum9 at the point (3, -8), since the y-coordinate -8 is the smallest function value near xc= 3. Although it is tempting to say that local extrema10 occur when the function changes from increasing to decreasing or vice versa, it is not a precise enough way to define the concepts for the needs of Calculus. At the risk of being pedantic, we will present the traditional definitions and thoroughly vet the pathologies they induce in the exercises. We have one last observation to make before we proceed to the algebraic definitions and look at a fairly tame, yet helpful, example. If we look at the entire graph, we see the largest y value (hence the largest function value) is 5.5 at x = 6. In this case, we say the maximum- of f is 5.5; similarly, the minimum12 of f is -8. We formalize these concepts in the following definitions. DEFINITION 1.9. Suppose f is a function with f(a) = b. " We say f has a local maximum at the point (a, b) if and only if there is an open interval I containing a for which f(a) > f(x) for all x in I different than a. The value f(a) = b is called 'a local maximum value of f' in this case. " We say f has a local minimum at the point (a, b) if and only if there is an open interval I containing a for which f(a) < f(x) for all x in I different than a. The value f(a) = b is called 'a local minimum value of f' in this case. " The value b is called the maximum of f if b > f(cx) for all x in the domain of f. " The value b is called the minimum of f if b < f(cx) for all x in the domain of f. It's important to note that not every function will have all of these features. Indeed, it is possible to have a function with no local or absolute extrema at all! (Any ideas of what such a function's graph would have to look like?) We shall see in the exercises examples of functions which have one or two, but not all, of these features, some that have instances of each type of extremum and some functions that seem to defy common sense. In all cases, though, we shall adhere to the algebraic definitions above as we explore the wonderful diversity of graphs that functions provide to us. Here is the 'tame' example which was promised earlier. It summarizes all of the concepts presented in this section as well as some from previous sections so you should spend some time thinking deeply about it before proceeding to the exercises. 7Also called 'relative maximum'. 8We will make this more precise in a moment. 9Also called a 'relative minimum'. io'Maxima' is the plural of 'maximum' and 'mimima' is the plural of 'minimum'. 'Extrema' is the plural of 'extremum' which combines maximum and minimum. "1Sometimes called the 'absolute' or 'global' maximum. "2Again, 'absolute' or 'global' minimum can be used.  1.7 GRAPHS OF FUNCTIONS 67 EXAMPLE 1.7.4. Given the graph of y = f(x) below, answer all of the following questions. 4- 2- 1 - (-2, 0) (0, 3) (2,0) (4, -3) (-4, -3) 1. Find the domain of f. 2. Find the range of f. 3. Determine f(2). 4. List the x-intercepts, if any exist. 5. List the y-intercepts, if any exist. 6. Find the zeros of f. 7. Solve f(x) < 0. 8. Determine the number of solutions to the equation f(x) = 1. 9. List the intervals on which f is increasing. 10. List the intervals on which f is decreasing. 11. List the local maximums, if any exist. 12. List the local minimums, if any exist. 13. Find the maximum, if it exists. 14. Find the minimum, if it exists. 15. Does f appear to be even, odd, or neither? SOLUTION. 1. To find the domain of f, we proceed as in Section 1.4. By projecting the graph to the x-axis, we see the portion of the x-axis which corresponds to a point on the graph is everything from -4 to 4, inclusive. Hence, the domain is [-4, 4]. 2. To find the range, we project the graph to the y-axis. We see that the y values from -3 to 3, inclusive, constitute the range of f. Hence, our answer is [-3, 3]. 3. Since the graph of f is the graph of the equation y = f(x), f(2) is the y-coordinate of the point which corresponds to x= 2. Since the point (2, 0) is on the graph, we have f(2) = 0.  68 RELATIONS AND FUNCTIONS 4. The x-intercepts are the points on the graph with y-coordinate 0, namely (-2, 0) and (2, 0). 5. The y-intercept is the point on the graph with x-coordinate 0, namely (0, 3). 6. The zeros of f are the x-coordinates of the x-intercepts of the graph of y = f(x) which are x = -2,2. 7. To solve f(x) < 0, we look for the x values of the points on the graph where the y-coordinate is less than 0. Graphically, we are looking where the graph is below the x-axis. This happens for the x values from -4 to -2 and again from 2 to 4. So our answer is [-4, -2) U (2, 4]. 8. To find where f(x) = 1, we look for points on the graph where the y-coordinate is 1. Even though these points aren't specified, we see that the curve has two points with a y value of 1, as seen in the graph below. That means there are two solutions to f(x) = 1. y 4- ----> I: -4- 9. As we move from left to right, the graph rises from (-4, -3) to (0, 3). This means f is increasing on the interval [-4, 0]. (Remember, the answer here is an interval on the x-axis.) 10. As we move from left to right, the graph falls from (0, 3) to (4, -3). This means f is decreasing on the interval [0, 4]. (Remember, the answer here is an interval on the x-axis.) 11. The function has its only local maximum at (0, 3). 12. There are no local minimums. Why don't (-4, -3) and (4, -3) count? Let's consider the point (-4, -3) for a moment. Recall that, in the definition of local minimum, there needs to be an open interval I which contains x= -4 such that f(-4) < f(x) for all x in I different from -4. But if we put an open interval around x= -4 a portion of that interval will lie outside of the domain of f. Because we are unable to fulfill the requirements of the definition  1.7 GRAPHS OF FUNCTIONS 69 for a local minimum, we cannot claim that f has one at (-4, 3). The point (4, 3) fails for the same reason - no open interval around x = 4 stays within the domain of f. 13. The maximum value of f is the largest y-coordinate which is 3. 14. The minimum value of f is the smallest y-coordinate which is -3. 15. The graph appears to be symmetric about the y-axis. This suggests13 that f is even. El. With few exceptions, we will not develop techniques in College Algebra which allow us to determine the intervals on which a function is increasing, decreasing or constant or to find the local maximums and local minimums analytically; this is the business of Calculus.14 When we have need to find such beasts, we will resort to the calculator. Most graphing calculators have 'Minimum' and 'Maximum' features which can be used to approximate these values, as demonstrated below. EXAMPLE 1.7.5. Let f(x) 15x ~2 . Use a graphing calculator to approximate the intervals on x2 + 3 which f is increasing and those on which it is decreasing. Approximate all extrema. SOLUTION. Entering this function into the calculator gives Ploit Pot2 Plot3 Y1=15X/(X^2+3) V lYi V ?3= Using the Minimum and Maximum features, we get Maximum X=1.732051 Y=4.330127 Minimum X= -1.73205 Y= -4.330127? 13but does not prove 14Although, truth be told, there is only one step of Calculus involved, followed by several pages of algebra.  70 RELATIONS AND FUNCTIONS To two decimal places, f appears to have its only local minimum at (-1.73, -4.33) and its only local maximum at (1, 73, 4.33). Given the symmetry about the origin suggested by the graph, the relation between these points shouldn't be too surprising. The function appears to be increasing on [-1.73, 1.73] and decreasing on (-oc, -1.73] U [1.73, c). This makes -4.33 the (absolute) minimum and 4.33 the (absolute) maximum. Q EXAMPLE 1.7.6. Find the points on the graph of y = (x -3)2 which are closest to the origin. Round your answers to two decimal places. SOLUTION. Suppose a point (x, y) is on the graph of y = (x - 3)2. Its distance to the origin, (0, 0), is given by d = (x - 0)2 + (y - 0)2 = VIzX2 +y2 r 2 + [(x - 3)2]2 Since y = (x - 3)2 = x2 + (x - 3)4 Given a value for x, the formula d = fz2 + (x - 3)4 is the distance from (0, 0) to the point (x, y) on the curve y = (x - 3)2. What we have defined, then, is a function d(x) which we wish to minimize over all values of x. To accomplish this task analytically would require Calculus so as we've mentioned before, we can use a graphing calculator to find an approximate solution. Using the calculator, we enter the function d(x) as shown below and graph. Ploit Pot2 Plot3 \Y1 ^T(X2+(X-3)^ 4) %-Y2= \ %-Y3= _ %NY s = Minimum NY6= x=2.OOOOOO .Y=2.23606 . Using the Minimum feature, we see above on the right that the (absolute) minimum occurs near x = 2. Rounding to two decimal places, we get that the minimum distance occurs when x = 2.00. To find the y value on the parabola associated with x = 2.00, we substitute 2.00 into the equation to get y = (x - 3)2 = (2.00 - 3)2 = 1.00. So, our final answer is (2.00, 1.00).15 (What does the y value listed on the calculator screen mean in this problem?) Q 15It seems silly to list a final answer as (2.00, 1.00). Indeed, Calculus confirms that the exact answer to this problem is, in fact, (2, 1). As you are well aware by now, the author is a pedant, and as such, uses the decimal places to remind the reader that any result garnered from a calculator in this fashion is an approximation, and should be treated as such.  1.7 GRAPHS OF FUNCTIONS 71 1.7.2 EXERCISES 1. Sketch the graphs of the following functions. State the domain of the function, identify any intercepts and test for symmetry. (a) f (x) x - 2 3 (b) f (x) =N/5 -x (C) f () = +/1 (d ) f (x ) = z2 1 2. Analytically determine if the following functions are even, odd or neither. (a) f (x) = 7x (b) f (x) = 7x + 2 1 (c) f 1)= 3 (d) f (x)=4 (e) f (x)=0 (f) f (W = z -x 4 +x:2 +9 (gz)fW =z -X5 -x 3 +x (h) f(z) (i) f(z) 3:4 + 32+x:+1 5-3x (j) f W = x 2 _ 3. Use your graphing calculator to approximate the local and absolute extrema of the following functions. Approximate the intervals on which the function is increasing and those on which it is decreasing. Round your answers to two decimal places. (a) f (x) = z - 3x3 - 24x2 + 283+ 48 (b) f (x) = 32/3(: - 4) (c) f () (d) f(z) 9-x2 3: 9-x2 4. Sketch the graphs of the following piecewise-defined functions. (a) f (x) =( -2x - 4 if x < 0 3x if 3x>0 32 if (c) f () = 3 - x if 4 if 1 - if - (d) f(:)~ = f - :< -2 -2<3:<2 3:>2 -65 5. Let f(x) = [3:], the greatest integer function defined in Exercise 10 in Section 1.5. (a) Graph y = f(x). Be careful to correctly describe the behavior of the graph near the integers. (b) Is f even, odd, or neither? Explain. (c) Discuss with your classmates which points on the graph are local minimums, local max- imums or both. Is f ever increasing? Decreasing? Constant?  72 RELATIONS AND FUNCTIONS 6. Use your graphing calculator to show that the following functions do not have any extrema, neither local nor absolute. (a) f(x) = x3 + x- 12 (b) f ( V) = -5x + 2 7. In Exercise 9 in Section 1.5, we saw that the population of Sasquatch in Portage County 1 50t could be modeled by the function P(t) = ,+ ' where t = 0 represents the year 1803. Use your graphing calculator to analyze the general function behavior of P. Will there ever be a time when 200 Sasquatch roam Portage County? 8. One of the most important aspects of the Cartesian Coordinate Plane is its ability to put Algebra into geometric terms and Geometry into algebraic terms. We've spent most of this chapter looking at this very phenomenon and now you should spend some time with your classmates reviewing what we've done. What major results do we have that tie Algebra and Geometry together? What concepts from Geometry have we not yet described algebraically? What topics from Intermediate Algebra have we not yet discussed geometrically? 9. It's now time to "thoroughly vet the pathologies induced" by the precise definitions of local maximum and local minimum. We'll do this by providing you and your classmates a series of exercises to discuss. You will need to refer back to Definition 1.8 (Increasing, Decreasing and Constant) and Definition 1.9 (Maximum and Minimum) during the discussion. (a) Consider the graph of the function f given below. 3 2 -2 -1 1 2 x -1 -2 -3.. i. Show that f has a local maximum but not a local minimum at the point (-1, 1). ii. Show that f has a local minimum but not a local maximum at the point (1, 1). iii. Show that f has a local maximum AND a local minimum at the point (0, 1). iv. Show that f is constant on the interval [-1, 1] and thus has both a local maximum AND a local minimum at every point (x, f(x)) where -1 i i i I i i i i )I -2 -1 1 2 3 4 5 yf (x) -2 -1 1 2 3 4 5 shift left 2 units subtract 2 from each x-coordinate 9y=g(x) = f(x +2) Note that while the ranges of f and g are the same, the domain of g is [-2, 3] whereas the domain of f is [0, 5]. In general, when we shift the graph horizontally, the range will remain the same, but the domain could change. If we set out to graph j (x) = f(x - 2), we would find ourselves adding  1.8 TRANSFORMATIONS 79 2 to all of the x values of the points on the graph of y = f(x) to effect a shift to the right 2 units. Generalizing, we have the following result. THEOREM 1.3. Horizontal Shifts. Suppose f is a function and h is a positive number. " To graph y = f(x + h), shift the graph of y = f(x) left h units by subtracting h from the x-coordinates of the points on the graph of f. " To graph y = f(x - h), shift the graph of y = f(x) right h units by adding h to the x-coordinates of the points on the graph of f. In other words, Theorem 1.3 says adding to or subtracting from the input to a function amounts to shifting the graph left or right, respectively. Theorems 1.2 and 1.3 present a theme which will run common throughout the section: changes to the outputs from a function affect the y-coordinates of the graph, resulting in some kind of vertical change; changes to the inputs to a function affect the x-coordinates of the graph, resulting in some kind of horizontal change. EXAMPLE 1.8.1. 1. Graph f(x) =Vx. Plot at least three points. 2. Use your graph in 1 to graph g(x) = - 1. 3. Use your graph in 1to graph j(x) = x - 1. 4. Use your graph in 1 to graph m(x) x= x + 3 - 2. SOLUTION. 1. Owing to the square root, the domain of f is x > 0, or [0, oc). We choose perfect squares to build our table and graph below. From the graph we verify the domain of f is [0, oo) and the range of f is also [0, oc). x fx) (x, fx)) 2-- (1, 1 0 0 (0,0)1- 1 1 (1,1) ( o) ________ ____ ___12 3 4 x 4 2 (4,2) is tht X 0. I we ompar theformua fo g=x)withf =) 2. The domain of g is the same as the domain of f, since the only condition on both functions is that zx> 0. If we compare the formula for g(x) with f (x), we see that g(x) = f(x) - 1. In other words, we have subtracted 1 from the output of the function f. By Theorem 1.2, we know that in order to graph g, we shift the graph of f down one unit by subtracting 1 from each of the y-coordinates of the points on the graph of f. Applying this to the three points we have specified on the graph, we move (0, 0) to (0, -1), (1, 1) to (1, 0), and (4, 2) to (4, 1). The rest of the points follow suit, and we connect them with the same basic shape as before. We confirm the domain of g is [0, oo) and find the range of g to be [-1, oo).  80 RELATIONS AND FUNCTIONS y (4, 2) 2-- (1,21) 4 (0, 0) 1 2 3 4 'x y 2 (4, 1) __(1, 0) 1 2 3 4 ' (0, -1) y = 9g(x) = fi - 1 shift down 1 unit subtract 1 from each y-coordinate 3. Solving x - 1 > 0 gives x > 1, so the domain of j is [1, oc). To graph j, we note that j(x) = f(x - 1). In other words, we are subtracting 1 from the input of f. According to Theorem 1.3, this induces a shift to the right of the graph of f. We add 1 to the x-coordinates of the points on the graph of f and get the result below. The graph reaffirms the domain of j is [1, o) and tells us that the range is [0, oo). y (4, 2) 2-- (1, 1) 1-- (0, 0) 2 1 2 3 4 5 x y =f(x)~ =xi y (5, 2) 2- (2,1) 1- (1,0) 2 3 4 5 x y = j(x)= =x-1 shift right 1 unit add 1 to each x-coordinate 4. To find the domain of m, we solve x + 3 > 0 and get [-3, oc). Comparing the formulas of f (x) and m(x), we have m(x) = f (xz +3) - 2. We have 3 being added to an input, indicating a horizontal shift, and 2 being subtracted from an output, indicating a vertical shift. We leave it to the reader to verify that, in this particular case, the order in which we perform these transformations is immaterial; we will arrive at the same graph regardless as to which transformation we apply first.2 We follow the convention 'inputs first', and to that end we first tackle the horizontal shift. Letting m1(x) = f(x + 3) denote this intermediate step, Theorem 1.3 tells us that the graph of y = m1(x) is the graph of f shifted to the left 3 units. Hence, we subtract 3 from each of the x-coordinates of the points on the graph of f. 2- 1- (0, 0) (4,2) (1, 1 2- (-2, 1 1- (-3,0 -3 -2 -1 -1- -2. y -(1,2 -3 - f-1 1) 2 4x 1 2 3 4 x shift left 3 units subtract 3 from each x-coordinate ymi(x) =f(x+3) = x+3 2We shall see in the next example that order is generally important when applying more than one transformation to a graph. 3We could equally have chosen the convention 'outputs first'.  1.8 TRANSFORMATIONS 81 Since m(x) = f (x + 3) - 2 and f (x + 3) = m1(x), we have m(x) = m1(x) - 2. We can apply Theorem 1.2 and obtain the graph of m by subtracting 2 from the y-coordinates of each of the points on the graph of m1(x). The graph verifies that the domain of m is [-3, o) and we find the range of m is [-2, oc). y y (1, 2) 2 2 (-2, 1 1 1 (-3, 0 I I I I(1, 0) -3 -2-1 1 2 3 4 x -3 -2 -1 1 2 3 4 x -1. (-2, -1) - -2 -2 shift down 2 units (-3, -2) y = mi(x)=f (x + 3) = x + 3 subtract 2 from each y-coordinate y = m(x) = mi(x) - 2 = x+3 - 2 Keep in mind that we can check our answer to any of these kinds of problems by showing that any of the points we've moved lie on the graph of our final answer. For example, we can check that (-3, -2) is on the graph of m, by computing m(-3) = V(-3) + 3 - 2 = vO - 2 = -2 / D We now turn our attention to reflections. We know from Section 1.1 that to reflect a point (x, y) across the x-axis, we replace y with -y. If (x, y) is on the graph of f, then y = f(x), so replacing y with -y is the same as replacing f(x) with -f(x). Hence, the graph of y = -f(x) is the graph of f reflected across the x-axis. Similarly, the graph of y = f(-x) is the graph of f reflected across the y-axis. Returning to inputs and outputs, multiplying the output from a function by -1 reflects its graph across the x-axis, while multiplying the input to a function by -1 reflects the graph across the y-axis. THEOREM 1.4. Reflections. Suppose f is a function. " To graph y = -f(x), reflect the graph of y = f(x) across the x-axis by multiplying the y-coordinates of the points on the graph of f by -1. " To graph y = f(-x), reflect the graph of y = f(x) across the y-axis by multiplying the x-coordinates of the points on the graph of f by -1. Applying Theroem 1.4 to the graph of y = f(x) given at the beginning of the section, we can graph y =-f(x) by reflecting the graph of f about the x-axis 4The expressions -f(x) and f(-x) should look familiar - they are the quantities we used in Section 1.7 to test if a function was even, odd, or neither. The interested reader is invited to explore the role of reflections and symmetry of functions. What happens if you reflect an even function across the y-axis? What happens if you reflect an odd function across the y-axis? What about the x-axis?  82 RELATIONS AND FUNCTIONS 5- 4. 3- 2- (0,1). -1. -2. -3. -4. -5. (5, 5) (2, 3) (4, 3) 1 2 3 4 5 x 5- 4- 3- 2- 1- (0, -1) -2- -3- -4- -5- y 1 2 3 4 5 x (4,-3) (2(,-3) (5,-5) y = -f(x) reflect across x-axis multiply each y-coordinate by -1 y=f(x) By reflecting the graph of f across the y-axis, we obtain the graph of y = f(-x). 5- 4- 3- 2- (0, 1)4 (5, 5) (2, 3) (4, 3) (-5, 5) 5- 4 - (-2, 3) 3 - (-4,3) 2- y (0, 1) -5 -4 -3 -2 -1I 1 2 3 4 5 a i i i i i i i i i i i : O- reflect across y-axis multiply each x-coordinate by -1 -5 -4 -3 -2 -i 1 2 3 4 5 y=f(x) y=f(-x) With the addition of reflections, it is now more important than ever to transformations, as the next example illustrates. EXAMPLE 1.8.2. Let f(x) =xV1. Use the graph of f from Example 1.8.1 functions below. Also, state their domains and ranges. consider the order of to graph the following 1. g(x) =,/-X 2. j(x) v/3-x 3. m(x) 3 - V/7 SOLUTION. 1. The mere sight of Vx usually causes alarm, if not panic. When we discussed domains in Section 1.5, we clearly banished negatives from the radicals of even roots. However, we must remember that x is a variable, and as such, the quantity -x isn't always negative. For example, if x= -4, -x= 4, thus =-x /-(-4) = 2 is perfectly well-defined. To find the  1.8 TRANSFORMATIONS 83 domain analytically, we set -x > 0 which gives x < 0, so that the domain of g is (-oc, 01. Since g(x) = f(-x), Theorem 1.4 tells us the graph of g is the reflection of the graph of f across the y-axis. We can accomplish this by multiplying each x-coordinate on the graph of f by -1, so that the points (0, 0), (1, 1), and (4, 2) move to (0, 0), (-1, 1), and (-4,2), respectively. Graphically, we see that the domain of g is (-oc, 0] and the range of g is the same as the range of f, namely [0, oc). (4,2) 2 2- (-4,2) 1 1 (0 0) (1, 1) (0,0) -4 -3 - 2 -1 1 2 3 4 x reflect across y-axis-4321 1 2 3 4 x y = f(x) x multiply each x-coordinate by -1 y= g(x) = f(-x) =/ 2. To determine the domain of j(x) = 3 - x, we solve 3 - x > 0 and get x < 3, or (-oc, 3]. To determine which transformations we need to apply to the graph of f to obtain the graph of j, we rewrite j(x) =_v-x + 3 = f(-x + 3). Comparing this formula with f(x) = V/7, we see that not only are we multiplying the input x by -1, which results in a reflection across the y-axis, but also we are adding 3, which indicates a horizontal shift to the left. Does it matter in which order we do the transformations? If so, which order is the correct order? Let's consider the point (4, 2) on the graph of f. We refer to the discussion leading up to Theorem 1.3. We know f(4) = 2 and wish to find the point on y = j(x) = f(-x + 3) which corresponds to (4, 2). We set -x + 3 = 4 and solve. Our first step is to subtract 3 from both sides to get -x= 1. Subtracting 3 from the x-coordinate 4 is shifting the point (4, 2) to the left. From -x= 1, we then multiply" both sides by -1to get x= -1. Multiplying the x-coordinate by -1 corresponds to reflecting the point about the y-axis. Hence, we perform the horizontal shift first, then follow it with the reflection about the y-axis. Starting with f(x) = V/7, we let j1(x) be the intermediate function which shifts the graph of f 3 units to the left, j1(x) = f (x + 3). y y (4,2) (1,2) 2 - - 2 (1, 1) (-2, 1 1- - 1(- ,0 -4 -3 -2-1 1 2 3 4 xshift left 3 units -4 -3 -2-1 1 2 3 4 x y = f (x) =x subtract 3 from each x-coordinate y = j1 (x) = f (x + 3) x=/ + 3 To obtain the function j, we reflect the graph of j1 about y-axis. Theorem 1.4 tells us we have j(x) =ji(-z). Putting it all together, we have j(x) =ji(-z) =f (-z + 3) =v-z + 3, 5Or divide - it amounts to the same thing.  84 RELATIONS AND FUNCTIONS which is what we want.' From the graph, we confirm the domain of j is (-oc, 3] and we get the range is [0, oo). y (1, 2) 2 (-2,1 (-3,0) -4 -3 -2 -1 1 2 3 4 x y = j1(x) = x + 3 y 1) (3, 0) 4 x reflect across y-axis multiply each x-coordinate by -1 j (x) = jl (-x) _ -x + 3 3. The domain of m works out to be the domain of f, [0, oc). Rewriting m(x) =-v x+ 3, we see m(x) = -f(x) + 3. Since we are multiplying the output of f by -1 and then adding 3, we once again have two transformations to deal with: a reflection across the x-axis and a vertical shift. To determine the correct order in which to apply the transformations, we imagine trying to determine the point on the graph of m which corresponds to (4,2) on the graph of f. Since in the formula for m(x), the input to f is just x, we substitute to find m(4) = -f(4) + 3 = -2 + 3 =1. Hence, (4, 1) is the corresponding point on the graph of m. If we closely examine the arithmetic, we see that we first multiply f(4) by -1, which corresponds to the reflection across the x-axis, and then we add 3, which corresponds to the vertical shift. If we define an intermediate function m1(x) =-f(x) to take care of the reflection, we get 3 2 1 (0, 0) -1- -2- y (4, 2) (1,1) 1 2 3 4 ' y 3-- 2-- 1 (0, 0) 1 2 3 4 X -2. .1-i (4, -2) y= m1(x = -f (W = -vi reflect across x-axis multiply each y-coordinate by -1 To shift the graph of mi up 3 units, we set m(x) we put it all together, we get m(z) = m1(z) + 3: graph that the range of m is (-oo, 3]. m1(x) + 3. Since m1(x) =_-f(x), when -f (x) + 3 = - x+ 3. We see from the 6If we had done the reflection first, then ji(x) = f(-x). Following this by a shift left would give us j(x) ji(x + 3) = f(-(x + 3)) = f(-x - 3) =_ -x - 3 which isn't what we want. However, if we did the reflection first and followed it by a shift to the right 3 units, we would have arrived at the function j (x). We leave it to the reader to verify the details.  1.8 TRANSFORMATIONS 85 y A 31 2 -- (0, 3) 2- 1- y (1, 2) (4, 1) 1- (0, 0) -1- -2- 1 2 3 4 x (1, -1) (4, -2) 1 2 3 4 x -1- shift up 3 units add 3 to each y-coordinate -2. =TnIW = -v,'x y~m(x) =mi(x)+3= -V~x+3 El We now turn our attention to our last class of transformations, scalings. Suppose we wish to graph the function g(x) = 2f(x) where f(x) is the function whose graph is given at the beginning of the section. From its graph, we can build a table of values for g as before. y T (5, 5) 54 4. 3. 2- (0,1).A (2, 3) (4, 3) 1 2 3 4 5 x y = f(x) 0 (0,1) 1 2 (0, 2) 2 (2, 3) 3 6 (2, 6) 4 (4, 3) 3 6 (4, 6) 5 (5, 5) 5 10 (5,10) In general, if (a, b) is on the graph of f, then f(a) = b so that g(a) = 2f(a) = 2b puts (a, 2b) on the graph of g. In other words, to obtain the graph of g, we multiply all of the y-coordinates of the points on the graph of f by 2. Multiplying all of the y-coordinates of all of the points on the graph of f by 2 causes what is known as a 'vertical scaling7 by a factor of 2', and the results are 7Also called a 'vertical stretch', 'vertical expansion' or 'vertical dilation' by a factor of 2.  86 RELATIONS AND FUNCTIONS given below. y lo y (5, 10) 9- 8- 7- 6- 5- 4- 3- 2- (0, 1) (5, 5) (2, 3) (4, 3) (4, 6) (0, I 1 1 1 1i a 1 2 3 4 5 X vertical scaling by a factor of 2 x y=f(x) multiply each y-coordinate by 2 y=2f(x) If we wish to graph y = jf(x), we multiply the all of the y-coordinates of the points on the graph of f by 2. This creates a 'vertical scaling8 by a factor of 1' as seen below. 5-- 4-- 3-- 2-- (01) , y (5, 5) (2, 3) (4, 3) 5- 4- 3. 2. y - () (5,() - (4,4) I I I I I 1 2 3 4 5 ' 1- (0, j) 1 2 3 4 5 x vertical scaling by a factor of 1 y = f(x) multiply each y-coordinate by j These results are generalized in the following theorem. y = jf(x) THEOREM 1.5. Vertical Scalings. Suppose f is a function and a > 0. To graph y = aff(x), multiply all of the y-coordinates of the points on the graph of f by a. We say the graph of f has been vertically scaled by a factor of a. " If a > 1, we say the graph of f has undergone a vertical stretch (expansion, dilation) by a factor of a. " If 0 < a < 1, we say the graph of f has undergone a vertical shrink (compression, contraction) by a factor of}. A few remarks about Theorem 1.5 are in order. First, a note about the verbiage. To the authors, the words 'stretch', 'expansion', and 'dilation' all indicate something getting bigger. Hence, 'stretched 8Also called 'vertical shrink,"vertical compression' or 'vertical contraction' by a factor of 2.  1.8 TRANSFORMATIONS 87 by a factor of 2' makes sense if we are scaling something by multiplying it by 2. Similarly, we believe words like 'shrink', 'compression' and 'contraction' all indicate something getting smaller, so if we scale something by a factor of 2, we would say it 'shrinks by a factor of 2' - not 'shrinks by a factor of 1.' This is why we have written the descriptions 'stretch by a factor of a' and 'shrink by a factor of }' in the statement of the theorem. Second, in terms of inputs and outputs, Theorem 1.5 says multiplying the outputs from a function by positive number a causes the graph to be vertically scaled by a factor of a. It is natural to ask what would happen if we multiply the inputs of a function by a positive number. This leads us to our last transformation of the section. Referring to the graph of f given at the beginning of this section, suppose we want to graph g(x) = f(2x). In other words, we are looking to see what effect multiplying the inputs to f by 2 has on its graph. If we attempt to build a table directly, we quickly run into the same problem we had in our discussion leading up to Theorem 1.3, as seen in the table on the left below. We solve this problem in the same way we solved this problem before. For example, if we want to determine the point on g which corresponds to the point (2, 3) on the graph of f, we set 2x = 2 so that x= 1. Substituting x = 1 into g(x), we obtain g(1) = f(2. 1) = f(2) = 3, so that (1, 3) is on the graph of g. Continuing in this fashion, we obtain the table on the lower right. x (x, f (x)) f (x) g(x)= f (2x) (X, g x)) 0 (0,1) 1 f(2-O)=f(O)=1 (0,1) 2 (2,3) 3 f(2.2)=f(4)=3 (2,3) 4 (4,3) 3 f(2-4)=f(8)=? 5 (5,5) 5 f (2.5)=f (10)=? x 2x g(x) = f (2x) (x,g(x)) 0 0 g(0) = f (0) = 1 (0,0) 1 2 g(1) = f (2) = 3 (1,3) 2 4 g(2) = f (4) = 3 (2,3) 2 5 g (2)=f (5) = 5 (j,5) In general, if (a, b) is on the graph of f, then f(a) = b. Hence g (1)2= f (2.-2) = f(a) = b so that (2, b) is on the graph of g. In other words, to graph g we divide the x-coordinates of the points on the graph of f by 2. This results in a horizontal scaling9 by a factor of 1. y y (5,5) ^ (5, 5) 5 -- 5-- 4 -- 4-- (2,3) (1,3) 3 3-- (4,3) (2,3) 2 2-- (0,1) (0,1) 1 2 3 4 5 x horizontal scaling by a factor of 21 2 3 4 5 x y = f(x) multiply each x-coordinate by } y = g(x) = f(2x) If, on the other hand, we wish to graph y = f (ix), we end up multiplying the x-coordinates of the points on the graph of f by 2 which results in a horizontal scaling10 by a factor of 2, as demonstrated below. 9Also called 'horizontal shrink,"horizontal compression' or 'horizontal contraction' by a factor of 2. 1oAlso called 'horizontal stretch,"horizontal expansion' or 'horizontal dilation' by a factor of 2.  88 RELATIONS AND FUNCTIONS 5- 4- 3- 2- (0, 1)4 (5, 5) (2, 3) (4, 3) 5- 4- 3- 2- (0, 1) (10, 5) (4, 3) (8,3) I I I I I I I I I I I 1 2 3 4 5 6 7 8 9 10 ox horizontal scaling by a factor of 2 1 2345678 9 10x y=f(x) multiply each x-coordinate by 2 9=9(x) f ( x) We have the following theorem. THEOREM 1.6. Horizontal Scalings. Suppose f is a function and b > 0. To graph y = f(bx), divide all of the x-coordinates of the points on the graph of f by b. We say the graph of f has been horizontally scaled by a factor of . " If 0 < b < 1, we say the graph of f has undergone a horizontal stretch (expansion, dilation) by a factor of . " If b > 1, we say the graph of f has undergone a vertical shrink (compression, contraction) by a factor of b. Theorem 1.6 tells us that if we multiply the input to a function by b, the resulting graph is scaled horizontally by a factor of 1 since the x-values are divided by b to produce corresponding points on the graph of f(bx). The next example explores how vertical and horizontal scalings sometimes interact with each other and with the other transformations introduced in this section. EXAMPLE 1.8.3. Let f(x) =x1. Use the graph of f from Example 1.8.1 to graph the following functions below. Also, state their domains and ranges. 1. g(x) = 3V 2. j(x) = v9x 3. m(x) 1 - +3 SOLUTION. 1. First we note that the domain of g is [0, oc) for the usual reason. Next, we have g(x) = 3f(x) so by Theorem 1.5, we obtain the graph of g by multiplying all of the y-coordinates of the points on the graph of f by 3. The result is a vertical scaling of the graph of f by a factor of 3. We find the range of g is also [0, oc).  1.8 TRANSFORMATIONS 89 y y (4, 6) 6--6 - - 5 - 5-- 4 -- 4-- 3 - 3 (4, 2) (1, 3) 2 2 (1,1) 1 1 (0, 0) (0,0) 1 2 3 4 x vertical scale by a factor of 3 1 2 3 4 x y = f(x) = multiply each y-coordinate by 3 y= g(x) = 3f(x) = 3/ 2. To determine the domain of j, we solve 9x > 0 to find x > 0. Our domain is once again [0, oc). We recognize j(x) = f(9x) and by Theorem 1.6, we obtain the graph of j by dividing the x-coordinates of the points on the graph of f by 9. From the graph, we see the range of j is also [0,oo). Y y (4, 2) 2 2 - 2 (1,1) ( 9, 2) 1 1- (0,0) (0,0) 1 2 3 4 1 2 3 4 x horizontal scale by a factor of x y = f (x) multiply each x-coordinate by } y = j(x) = f (9x) =/9x 3. Solving x23 0 gives x > -3, so the domain of m is [-3, oc). To take advantage of what we know of transformations, we rewrite m(x) =- x + 2 + 1, or m(x) =_-f (2x + 2) + 1. Focusing on the inputs first, we note that the input to f in the formula for m(x) is 2x + 2. Multiplying the x by 2 corresponds to a horizontal stretch by a factor of 2, and adding the corresponds to+a shift to the left by . As before, we resolveAwhich to+prform firstIby thinking about how we would find the point on m corresponding to a point on f, in this case, (4, 2). To use f(4) =2, we solve jz+ j 4. Our first step is to subtract the j (the horizontal shift) to obtain jz = . Next, we multiply by 2 (the horizontal stretch) and obtain x 5. We define two intermediate functions to handle first the shift, then the stretch. In accordance with Theorem 1.3, m1(x)= f (z =x +2! + will shift the graph of f to the left j units.  90 RELATIONS AND FUNCTIONS 2- 1- (0, 0) (4, 2) (1 1) 2 2' ,i) y . 0 i i. . i. i . . . -3 -2 -lit 2 3 4 5 -2f -20) - 11 shift left 2 units subtract 2 from each x-coordinate MI(X) f(x+2) V +2 Next, m2(x) = mi (ix) 2x + will, according to Theorem 1.6, horizontally stretch the graph of mi by a factor of 2. 2- (-2,1) y - ~ 2 2- (-l, 1)- y(5,2) I I i I I I I I I I A I I I I I I I I ij, -3 -2 -1 (-,0) 1 2 -2 -1 (-3, 0) 1 2 l1- -2t ml W X + 2 -2-|- horizontal scale by a factor of 2 multiply each x-coordinate by 2 y m2(x)= ml ( x) We now examine what's happening to the outputs. From m(x) =_-f (x + 2) + 1, we see the output from f is being multiplied by -1 (a reflection about the x-axis) and then a 1 is added (a vertical shift up 1). As before, we can determine the correct order by looking at how the point (4,2) is moved. We have already determined that to make use of the equation f (4) = 2, we need to substitute x= 5. We get m(5) =-f ((5) + ) + 1 =-f(4) + 1 =-2 + 1 =-1. We see that f (4) (the output from f) is first multiplied by -1 then the 1 is added meaning we first reflect the graph about the x-axis then shift up 1. Theorem 1.4 tells us m3(x) = -m2(z) will handle the reflection. y y (5, 2) 2- (-l, 1) 2- (-3, 0) 1- 6 I I I I I I I I 4 .k i i i i i i i i jo -2 -1 (-3, 0) 1 3 4 -1- -2- -2 -1 (-1, -1 - -2. 1 2 3 4 5 (5-2) reflect across x-axis multiply each y-coordinate by -1 ym2 (x) V2 +2 2 - M3(X) -m2(x) -1/ 2X + 2  1.8 TRANSFORMATIONS 91 Finally, to handle the vertical shift, Theorem 1.2 gives m(x) = m3(x) + 1, and we see that the range of m is (-oc, 1]. y (5, 2) y 2- (-l, 1) (-3, 1) 1,10){1 1 9 i i i i i i i i : p -2 -1 (-3, 0) 1 2 3 4 5 -2 -1 -1- 2 3 4 5 ' (5,-i) -2. .+ y=m2 (X) = 2 +A -2- shift up 1 unit add 1 to each y-coordinate y = m(x) = m3(x) + 1 = - 2x + 2 + 1 D- Some comments about Example 1.8.3 are in order. First, recalling the properties of radicals from Intermediate Algebra, we know that the functions g and j are the same, since j and g have the same domains and j(x) = v- x= v- 9 = 3 = g(x). (We invite the reader to verify that the all of the points we plotted on the graph of g lie on the graph of j and vice-versa.) Hence, for f (x) = x4, a vertical stretch by a factor of 3 and a horizontal shrink by a factor of 9 result in the same transformation. While this kind of phenomenon is not universal, it happens commonly enough with some of the families of functions studied in College Algebra that it is worthy of note. Secondly, to graph the function m, we applied a series of four transformations. While it would have been easier on the authors to simply inform the reader of which steps to take, we have strived to explain why the order in which the transformations were applied made sense. We generalize the procedure in the theorem below. THEOREM 1.7. Transformations. Suppose f is a function. To graph g(x) = Af(Bx+ H) + K 1. Subtract H from each of the x-coordinates of the points on the graph of f. This results in a horizontal shift to the left if H > 0 or right if H < 0. 2. Divide the x-coordinates of the points on the graph obtained in Step 1 by B. This results in a horizontal scaling, but may also include a reflection about the y-axis if B < 0. 3. Multiply the y-coordinates of the points on the graph obtained in Step 2 by A. This results in a vertical scaling, but may also include a reflection about the x-axis if A < 0. 4. Add K to each of the y-coordinates of the points on the graph obtained in Step 3. This results in a vertical shift up if K > 0 or down if K < 0. Theorem 1.7 can be established by generalizing the techniques developed in this section. Suppose (a, b) is on the graph of f. Then f(a) = b, and to make good use of this fact, we set Bx + H = a and solve. We first subtract the H (causing the horizontal shift) and then divide by B. If B  92 RELATIONS AND FUNCTIONS is a positive number, this induces only a horizontal scaling by a factor of . If B < 0, then we have a factor of -1 in play, and dividing by it induces a reflection about the y-axis. So we have x = a-H as the input to g which corresponds to the input x= a to f. We now evaluate g (-H) = Af (B. a--H + H) + K = A f (a) + K = Ab + K. We notice that the output from f is first multiplied by A. As with the constant B, if A > 0, this induces only a vertical scaling. If A < 0, then the -1 induces a reflection across the x-axis. Finally, we add K to the result, which is our vertical shift. A less precise, but more intuitive way to paraphrase Theorem 1.7 is to think of the quantity Bx+ H is the 'inside' of the function f. What's happening inside f affects the inputs or x-coordinates of the points on the graph of f. To find the x-coordinates of the corresponding points on g, we undo what has been done to x in the same way we would solve an equation. What's happening to the output can be thought of as things happening 'outside' the function, f. Things happening outside affect the outputs or y-coordinates of the points on the graph of f. Here, we follow the usual order of operations agreement: we first multiply by A then add K to find the corresponding y-coordinates on the graph of g. EXAMPLE 1.8.4. Below is the complete graph of y f(x). Use it to graph g(x) 4-3f(1-2x) y (0,3) 2-- 1 -- (-2,0) (2,0) -4 -3 2 -1 1 3 4 x -1-- -2-- -3-- (-4, -3) (4, -3) SOLUTION. We use Theorem 1.7 to track the five 'key points' (-4, -3), (-2,0), (0,3), (2, 0) and (4, -3) indicated on the graph of f to their new locations. We first rewrite g(x) in the form presented in Theorem 1.7, g(x)= -2f(-2x +1) +2. We set -2x + 1 equal to the x-coordinates of the key points and solve. For example, solving -2x + 1 = -4, we first subtract 1 to get -2x = -5 then divide by -2, to get -r, = 2 . tracting the 1 is a horizontal shift to the left 1 unit. Dividing by -2 can be thought of as a two step process: dividing by 2 which compresses the graph horizontally by a factor of 2 followed by dividing (multiplying) by -1 which causes a reflection across the y-axis. We summarize the results in the table below.  1.8 TRANSFORMATIONS 93 (a, f (a)) a -2x + 1 = a X (-4,-3) -4 -2x + 1= -4 x = j (-2,0) -2 -2x + 1= -2 x = j (0,3) 0 -2x + 1= 0 x = 1 (2,0) 2 -2x+ 1= 2 x = -1 (4,-3) 4 -2x+ 1= 4 x = -3 Next, we take each of the x values and substitute them into g(x) =_-2f(-2x + 1) + 2 to get the corresponding y-values. Substituting x = 2, and using the fact that f(-4) -3, we get 5 3 (5' 3 3 9 13 g - -f (-2 - +1) +2= f(-4)+2= -(-3) +2= +2= We see the output from f is first multiplied by - 2. Thinking of this as a two step process, multiplying by 2 then by -1, we see we have a vertical stretch by a factor of 2 followed by a reflection across the x-axis. Adding 2 results in a vertical shift up 2 units. Continuing in this manner, we get the table below. X g(x) (x,g(x)) 5 13 (5 13l 2_ 2 (22) 2 2 (--,2) 3 13 -)13\ To graph g, we plot each of the points in the table above and connect them in the same order and fashion as the points to which they correspond. Plotting f and g side-by-side gives y ' y (a2 6 6 5 5 4 4 (0,3) 3 2 (- ,)2) - (,2) 1 (-2, 0) (2, 0) -4 -3 2 -1 1 3 4 x -4 -3-2 -1 2 3 4 x -1- -1- -2- -2- (-4, -3) (4, -3) - -4- -4-  94 RELATIONS AND FUNCTIONS The reader is strongly encouraged1 to graph the series of functions which shows the gradual trans- formation of the graph of f into the graph of g. We have outlined the sequence of transformations in the above exposition; all that remains is to plot all five intermediate stages. Q Our last example turns the tables and asks for the formula of a function given a desired sequence of transformations. If nothing else, it is a good review of function notation. EXAMPLE 1.8.5. Let f(x) = x2. Find and simplify the formula of the function g(x) whose graph is the result of f undergoing the following sequence of transformations. Check your answer using a graphing calculator. 1. Vertical shift up 2 units 2. Reflection across the y-axis 3. Horizontal shift right 1 unit 4. Horizontal stretch by a factor of 2 SOLUTION. We build up to a formula for g(x) using intermediate functions as we've seen in previous examples. We let g1 take care of our first step. Theorem 1.2 tells us gi(x) = f(x)+2 = x2+2. Next, we reflect the graph of g1 about the x-axis using Theorem 1.4: g2(x) = -g1(x) = - (x2 + 2) -z2 - 2. We shift the graph to the right 1 unit, according to Theorem 1.3, by setting g3(x) g2(x - 1) = -(x - 1)2 - 2 = -x2 + 2x - 3. Finally, we induce a horizontal stretch by a factor of 2 using Theorem 1.6 to get g(x) = g3 ('X) (,X)2+2 (ix) - 3 which yields g(x) = - x2+ x - 3. We use the calculator to graph the stages below to confirm our result. y = f(x) =X 2 y = g1(x) = x2 + 2 shift up 2 units add 2 to each y-coordinate y reflect across x-axis multiply each y-coordinate by -1 = g1(x) = f(x) + 2 = x2 + 2 y = 2(X) =-g1(x) z2 -2 "You really should do this once in your life.  1.8 TRANSFORMATIONS 95 y = g2(X) - -t 2 shift right 1 unit add 1 to each n-coordinate J = g3(x) = g2(x - 1) x2 +2x-3 horizontal stretch by a factor of 2 multiply each x-coordinate by 2 y = g3(x) x2 + 2x- 3 y = g(x) = g3 (jz) x2 + x -3 H We have kept the viewing window the same in all of the graphs above. This had the undesirable consequence of making the last graph look 'incomplete' in that we cannot see the original shape of f(x) = x2. Altering the viewing window results in a more complete graph of the transformed function as seen below. y = g(x) This example brings our first chapter to a close. In the chapters which lie ahead, be on the lookout for the concepts developed here to resurface as we study different families of functions.  96 RELATIONS AND FUNCTIONS 1.8.1 EXERCISES 1. Use the graph of y previous graph.) S(x) below to graph the following. (Note: Each graph builds upon the (1, 3) 3- 2- 1 (-2, 0) 1 (2, 0) (-1, -3) The graph of y = S(x) (a) (b) (c) (d) y y y y S(z + 1) S(-x+1) jS(-x +1) 2S(-x+1)+1 2. The complete graph of y f(x) is given below. Use it to graph the following functions. 2- 1- (0,3) i i i i i 30 -3 -2 (-3, 0) -1 2 3 (3,0) -1- (a) (b) (c) (d) (e) (f) g(x) h(x) j(x) b(x) c(x) f(x)+3 f (X) -2) f ( + 4) f (z+ 1) -i1 jf(xz) (g) (h) (i) (j) (k) (1) m(z) n(x) : q(x) = -2f (x) f (3x) = -f (3x) :4f (-3)-6 4+f(1 -2x) - f (x4) -3  1.8 TRANSFORMATIONS 97 3. The graph of y = (x is given below on the left and the graph of y = g(x) is given on the right. Find a formula for g based on transformations of the graph of f. Check your answer by confirming that the points shown on the graph of g satisfy the equation y = g(x). 5-5 4 4 -11-10-9-8-7-6-5-4-3-2-1 1 2 3 4 5 6 7 8 *X -11-10-9-8-7-6-5-4- -2-11 1 2 3 4 5 6 7 8 X g= g~g(x) 4. For many common functions, the properties of algebra make a horizontal scaling the same as a vertical scaling by (possibly) a different factor. For example, we stated earlier that 9x = 39/x. With the help of your classmates, find the equivalent vertical scaling produced by the horizontal scalings y = (2x)3, y =|5xl, y = V/27x and y =(j)2. What about y=(-2x)3,y=| -5,y =/-27) and y2? 5. We mentioned earlier in the section that, in general, the order in which transformations are applied matters, yet in our first example with two transformations the order did not matter. (You could perform the shift to the left followed by the shift down or you could shift down and then left to achieve the same result.) With the help of your classmates, determine the situations in which order does matter and those in which it does not. 6. What happens if you reflect an even function across the y-axis? 7. What happens if you reflect an odd function across the y-axis? 8. What happens if you reflect an even function across the x-axis? 9. What happens if you reflect an odd function across the x-axis? 10. How would you describe symmetry about the origin in terms of reflections? 11. As we saw in Example 18.5, the viewing window on the graphing calculator affects how we see the transformations done to a graph. Using two different calculators, find viewing windows so that f~x) =9z on the one calculator looks like g(x) =392 on the other.  98 98 RELATIONS AND FUNCTIONS 1.8.2 ANSWERS 1. (a) (c) 1 3) 2- 1- (1,0) (3,0) .9 1 Ik I i '41 (-3, 0)( -2- 1 2 3 (2-2) l S(-x + 1 (-2-3) (d) 1 y S~x + 1) (b) (1, 1) (3, 1) (- -1 3 -1. (1, 0) (3, 0) (-1,I y 'S(-x + 1)+1 (2-3) y S(-x + 1) 2.(a) g (x) f (x)+ 3 (0, 6) 4 3T (b) (x) f ( y (-3,3) 2- 1- x (3, 3) ( I I I I I I -3 -2 -1 1 2 3 -1-  1.8 TRANSFORMATIONS 99 (x - 3) y (,3) 3 2 1 -3 -2 -1 1 2 3 x C-3, 0) -1 C 3 ,0) (h) k(x) f(23x y x(0,3) 46 I I I I I- I I I I 1 -4 -3 -2 -1 (-2,0) 1 2 3 4 'x (2,o) -11 ()a(x) f (x +4) (i m W --~(3x) 3- 2- 1- y (-1,0o) (1,0) a i ' -1. -(0)-41 x + i i i i i + i -7 -6 -5 -4 -3 -2 -1 (-70 1,-12) 1 x yJ ~ )=4 x-3 (3,6) (-4, -1) -2 -1 -l-I (2, -i) Ay 2 x -3 -2 -1 (-3,0) 1 2 3 ' (3, 0) -1- (g) d(x)_ -2f (x) y (-3,0) (0-6) (6, -6) (3, 0) x I(~-6)  100 (k) p(x) 4 + f(1 -2x) - f (-2x +1)+ 4 (1) q RELATIONS AND FUNCTIONS (x)-f(5-) -3-2f (2x+2) -3 -10-9-8-7-6-5-4-3-2-1 1 2 X (- x 3. g(x) =-2 3x +3 - 1or g~x)=23-x -3- 1  CHAPTER 2 LINEAR AND QUADRATIC FUNCTIONS 2.1 LINEAR FUNCTIONS We now begin the study of families of functions. Our first family, linear functions, are old friends as we shall soon see. Recall from Geometry that two distinct points in the plane determine a unique line containing those points, as indicated below. P (x0, yo) Q (xi, yi) To give a sense of the 'steepness' of the line, we recall we can compute the slope of the line using the formula below. EQUATION 2.1. The slope m of the line containing the points P (xo, yo) and Q (xi, yi) is: Y1 - Yo m =, 1 - zxo provided x # zo. A couple of notes about Equation 2.1 are in order. First, don't ask why we use the letter 'm' to represent slope. There are many explanations out there, but apparently no one really knows for sure.1 Secondly, the stipulation x1 # x ensures that we aren't trying to divide by zero. The reader is invited to pause to think about what is happening geometrically; the anxious reader can skip along to the next example. EXAMPLE 2.1.1. Find the slope of the line containing the following pairs of points, if it exists. Plot each pair of points and the line containing them. 'See www.nathforum.org or www.mathworld.wvolfram.corn for discussions on this topic.  102 LINEAR AND QUADRATIC FUNCTIONS 1. P(0, 0), Q(2, 4) 4. P(-3, 2), Q(4, 2) 2. P(-1, 2), Q(3, 4) 5. P(2, 3), Q(2,-1) 3. P(-2, 3), Q(2,-3) 6. P(2, 3), Q(2.1,-1) SOLUTION. In each of these examples, we apply the slope formula, Equation 2.1. 4- Q 3- 4-0 _4 22- _ PFl 2 3 4 X y 4 Q 4-2 2 _13- 2.m 3- (-1) 4 2 P -1 1 2 3 X 4 P 3 2 1 -3-3 -6 3 3. - ______ ___ - 3-- - - - tt + - 2- (-2) 4 2 -3 -2 -1 1 2 3x -3 Q -4  2.1 LINEAR FUNCTIONS 103 y 3- 2- P 1- -1 - 3 5.m = 2-2 -1 - 3 6. m = 2.1 - 2 -4 which is undefined 0 ' -4 - 40 0.1 1 + Q -1- -2- -3 - y x D- A few comments Example 2.1.1 are in order. First, for reasons which will be made clear soon, if the slope is positive then the resulting line is said to be increasing. If it is negative, we say the line is decreasing. A slope of 0 results in a horizontal line which we say is constant, and an undefined slope results in a vertical line.2 Second, the larger the slope is in absolute value, the steeper the line. You may recall from Intermediate Algebra that slope can be described as the ratio 'rise For run example, in the second part of Example 2.1.1, we found the slope to be 1. We can interpret this as a rise of 1 unit upward for every 2 units to the right we travel along the line, as shown below. 4- 3- 1. -'up 1' 'over 2' i ' I I I 3 -1 1 2 3 x Using more formal notation, given points (x0, yo) and (x1, yi), we use the Greek letter delta 'A' to write Ay =_Yi - yo and Ax =z1 - x. In most scientific circles, the symbol A means 'change in'. 2Some authors use the unfortunate moniker 'no slope' when a slope is undefined. It's easy to confuse the notions of 'no slope' with 'slope of 0'. For this reason, we will describe slopes of vertical lines as 'undefined'.  104 LINEAR AND QUADRATIC FUNCTIONS Hence, we may write Ay M Ax' which describes the slope as the rate of change of y with respect to x. Rates of change abound in the 'real world,' as the next example illustrates. EXAMPLE 2.1.2. At 6 AM, it is 24°F; at 10 AM, it is 32°F. 1. Find the slope of the line containing the points (6, 24) and (10, 32). 2. Interpret your answer to the first part in terms of temperature and time. 3. Predict the temperature at noon. SOLUTION. 1. For the slope, we have m = 3-2-_24 8 2. 2. Since the values in the numerator correspond to the temperatures in °F, and the values in 2 20F the denominator correspond to time in hours, we can interpret the slope as 2 - or slope 1 1 hour' or 2°F per hour. Since the slope is positive, we know this corresponds to an increasing line. Hence, the temperature is increasing at a rate of 2°F per hour. 3. Noon is two hours after 10 AM. Assuming a temperature increase of 2°F per hour, in two hours the temperature should rise 4°F. Since the temperature at 10 AM is 32°F, we would expect the temperature at noon to be 32 + 4 = 36°F. Now it may well happen that in the previous scenario, at noon the temperature is only 33°F. This doesn't mean our calculations are incorrect. Rather, it means that the temperature change throughout the day isn't a constant 2°F per hour. Mathematics is often used to describe, or model, real world phenomena. Mathematical models are just that: models. The predictions we get out of the models may be mathematically accurate, but may not resemble what happens in the real world. We will discuss this more thoroughly in Section 2.5. In Section 1.2, we discussed the equations of vertical and horizontal lines. Using the concept of slope, we can develop equations for the other varieties of lines. Suppose a line has a slope of m and contains the point (x0, yo). Suppose (x, y) is another point on the line, as indicated below. (x,y) (x, yo)  2.1 LINEAR FUNCTIONS 105 We have y - yo m = x- o m(x-x0) y - yo y- yo = m(x-x0) y = m(x-xo)+yo. We have just derived the point-slope form of a line.3 EQUATION 2.2. The point-slope form of the line with slope m containing the point (x0, yo) is the equation y = m (x - x0) + yo. EXAMPLE 2.1.3. Write the equation of the line containing the points (-1, 3) and (2, 1). SOLUTION. In order to use Equation 2.2 we need to find the slope of the line in question. So we use Equation 2.1 to get m = -3 - . We are spoiled for choice for a point (x0, yo). useEqutio 2. togetm =Ax =2-(-1) We'll use (-1, 3) and leave it to the reader to check that using (2, 1) results in the same equation. Substituting into the point-slope form of the line, we get y = m(x-xo)+yo 2 y = (x- (-1))+3 3 2 2 y =-3- 3 +3 3 3 2 7 y = 3-- + -. 3 3 We can check our answer by showing that both (-1, 3) and (2, 1) are on the graph of y =-x + 3 algebraically, as we did in Section 1.3. In simplifying the equation of the line in the previous example, we produced another form of a line, the slope-intercept form. This is the familiar y = mx + b form you have probably seen in Intermediate Algebra. The 'intercept' in 'slope-intercept' comes from the fact that if we set x= 0, we get y = b. In other words, the y-intercept of the line y = mx + b is (0, b). EQUATION 2.3. The slope-intercept form of the line with slope m and y-intercept (0, b) is the equation y = mx + b. Note that if we have slope m = 0, we get the equation y = b which matches our formula for a horizontal line given in Section 1.2. The formula given in Equation 2.3 can be used to describe all lines except vertical lines. All lines except vertical lines are functions (why?) and so we have finally reached a good point to introduce linear functions. 3We can also understand this equation in terms of applying transformations to the function I(x) =x. See the exercises.  106 LINEAR AND QUADRATIC FUNCTIONS DEFINITION 2.1. A linear function is a function of the form f(x) = mx + b, where m and b are real numbers with m -f 0. The domain of a linear function is (-oc, oo). For the case m = 0, we get f(x) = b. These are given their own classification. DEFINITION 2.2. A constant function is a function of the form f(x)=b, where b is real number. The domain of a constant function is (-oc, oo). Recall that to graph a function, f, we graph the equation y = f(x). Hence, the graph of a linear function is a line with slope m and y-intercept (0, b); the graph of a constant function is a horizontal line (with slope m = 0) and a y-intercept of (0, b). Now think back to Section 1.7.1, specifically Definition 1.8 concerning increasing, decreasing and constant functions. A line with positive slope was called an increasing line because a linear function with m > 0 is an increasing function. Similarly, a line with a negative slope was called a decreasing line because a linear function with m < 0 is a decreasing function. And horizontal lines were called constant because, well, we hope you've already made the connection. EXAMPLE 2.1.4. Graph the following functions. Identify the slope and y-intercept. 1. f(z)=3 3. f (x) 3 42x z2_4 4. f (X) x2-2 2. f(x)=3x-I SOLUTION. 1. To graph f(x) = 3, we graph y = 3. This is a horizontal line (m = 0) through (0, 3). 2. The graph of f(x) = 3x - 1 is the graph of the line y = 3x - 1. Comparison of this equation with Equation 2.3 yields m = 3 and b = -1. Hence, our slope is 3 and our y-intercept is (0, -1). To get another point on the line, we can plot (1, f(1)) = (1, 2). y 4 1 -3 -2 -1 1 2 3 x f(x)=3 4. 3. 2. -2-1 4- 1 2x f(x) = 3x - 1  2.1 LINEAR FUNCTIONS 3. At first glance, the function f(x)= 3-42x does not fit the form in Definition 2.1 but rearranging we get f(x) =3-2x = - = -1x+ 1. Weidentify m = -1 and b = our graph is a line with a slope of -j and a y-intercept of (0, j). Plotting an point, we can choose (1, f(1)) to get (1, 4). 4. If we simplify the expression for f, we get 107 after some j. Hence, additional f() = x- zx- 2 Qy--2(z +2) x+2. If we were to state f(x) = x + 2, we would be committing a sin of omission. Remember, to find the domain of a function, we do so before we simplify! In this case, f has big problems when x= 2, and as such, the domain of f is (-oc, 2) U (2, oc). To indicate this, we write f(x) = x + 2, x -f 2. So, except at x = 2, we graph the line y = x + 2. The slope m = 1 and the y-intercept is (0, 2). A second point on the graph is (1, f(1)) = (1, 3). Since our function f is not defined at x= 2, we put an open circle at the point that would be on the line y = x + 2 when x= 2, namely (2, 4). 4- y 2 -3 -2 -1 1 3 x 3-2 f () - 2x 4 y 3, 2 1- -1 1 2 3 x 2_4 f (x)=x2 x+2 D- The last two functions in the previous example showcase some of the difficulty in defining a linear function using the phrase 'of the form' as in Definition 2.1, since some algebraic manipulations may be needed to rewrite a given function to match 'the form.' Keep in mind that the domains of linear and constant functions are all real numbers, (-oc, oc), and so while f(x) =§- simplified to a formula f(x) = x + 2, f is not considered a linear function since its domain excludes x= 2. However, we would consider 2x2 + 2 f(x)=a z2 + 1 to be a constant function since its domain is all real numbers (why?) and 2x2 + 2 fxx)= 2 = 2 The following example uses linear functions to model some basic economic relationships.  108 LINEAR AND QUADRATIC FUNCTIONS EXAMPLE 2.1.5. The cost, C, in dollars, to produce x PortaBoy4 game systems for a local retailer is given by C(x) = 80x + 150 for x > 0. 1. Find and interpret C(10). 2. How many PortaBoys can be produced for $15,000? 3. Explain the significance of the restriction on the domain, x > 0. 4. Find and interpret C(0). 5. Find and interpret the slope of the graph of y = C(x). SOLUTION. 1. To find C(10), we replace every occurrence of x with 10 in the formula for C(z) to get C(10) = 80(10) + 150 = 950. Since x represents the number of PortaBoys produced, and C(x) represents the cost, in dollars, C(10) = 950 means it costs $950 to produce 10 PortaBoys for the local retailer. 2. To find how many PortaBoys can be produced for $15,000, we set the cost, C(x), equal to 15000, and solve for x C(x) = 15000 80x + 150 = 15000 80x = 14850 14850 x - go 185.625 80 Since we can only produce a whole number amount of PortaBoys, we can produce 185 PortaBoys for $15,000. 3. The restriction x > 0 is the applied domain, as discussed in Section 1.5. In this context, x represents the number of PortaBoys produced. It makes no sense to produce a negative quantity of game systems.' 4. To find C(0), we replace every occurrence of x with 0 in the formula for C(x) to get C(0) 80(0) + 150 = 150. This means it costs $150 to produce 0 PortaBoys. The $150 is often called the fixed or start-up cost of this venture. (What might contribute to this cost?) 4The similarity of this name to PortaJohn is deliberate. 5Actually, it makes no sense to produce a fractional part of a game system, either, as we saw in the previous part of this example. This absurdity, however, seems quite forgivable in some textbooks but not to us.  2.1 LINEAR FUNCTIONS 109 5. If we were to graph y = C(x), we would be graphing the portion of the line y = 80x + 150 for x > 0. We recognize the slope, m = 80. Like any slope, we can interpret this as a rate of change. In this case, C(x) is the cost in dollars, while x measures the number of PortaBoys so Ay AC 80 $80 m xAx AX 1 1PortaBoy In other words, the cost is increasing at a rate of $80 per PortaBoy produced. This is often called the variable cost for this venture. D The next example asks us to find a linear function to model a related economic problem. EXAMPLE 2.1.6. The local retailer in Example 2.1.5 has determined that the number of PortaBoy game systems sold in a week, x, is related to the price of each system, p, in dollars. When the price was $220, 20 game systems were sold in a week. When the systems went on sale the following week, 40 systems were sold at $190 a piece. 1. Find a linear function which fits this data. Use the weekly sales, x, as the independent variable and the price p, as the dependent variable. 2. Find a suitable applied domain. 3. Interpret the slope. 4. If the retailer wants to sell 150 PortaBoys next week, what should the price be? 5. What would the weekly sales be if the price were set at $150 per system? SOLUTION. 1. We recall from Section 1.5 the meaning of 'independent' and 'dependent' variable. Since x is to be the independent variable, and p the dependent variable, we treat x as the input variable and p as the output variable. Hence, we are looking for a function of the form p(x) = mx + b. To determine m and b, we use the fact that 20 PortaBoys were sold during the week the price was 220 dollars and 40 units were sold when the price was 190 dollars. Using function notation, these two facts can be translated as p(20) = 220 and p(40) = 190. Since m represents the rate of change of p with respect to x, we have Ap 190 - 220 -30 Ac 40-20 20 -1.5. We now have determined p(x) =-1.5xv+ b. To determine b, we can use our given data again. Using p(20) = 220, we substitute xc= 20 into p(x) = 1.5x + b and set the result equal to 220: -1.5(20) + b = 220. Solving, we get b = 250. Hence, we get p(x) = -1.5x + 250. We can check our formula by computing p(20) and p(40) to see if we get 220 and 190, respectively. Incidentally, this equation is sometimes called the price-demand6 equation for this venture. 60r simply the demand equation  110 LINEAR AND QUADRATIC FUNCTIONS 2. To determine the applied domain, we look at the physical constraints of the problem. Cer- tainly, we can't sell a negative number of PortaBoys, so x > 0. However, we also note that the slope of this linear function is negative, and as such, the price is decreasing as more units are sold. Another constraint, then, is that the price, p(x) > 0. Solving -1.5x + 250 > 0 results 500 in -1.5x> -250 or x <0 = 166.6. Since x represents the number of PortaBoys sold in a -3 week, we round down to 166. As a result, a reasonable applied domain for p is [0, 166]. 3. The slope m = -1.5, once again, represents the rate of change of the price of a system with respect to weekly sales of PortaBoys. Since the slope is negative, we have that the price is decreasing at a rate of $1.50 per PortaBoy sold. (Said differently, you can sell one more PortaBoy for every $1.50 drop in price.) 4. To determine the price which will move 150 PortaBoys, we find p(150) That is, the price would have to be $25. -1.5(150)+250 = 25. 5. If the price of a PortaBoy were set at $150, we have p(x) = 150, or, -1.5x+250 = 150. Solving, we get -1.5x= -100 or x= 66.6. This means you would be able to sell 66 PortaBoys a week if the price were $150 per system. Q Not all real-world phenomena can be modeled using linear functions. Nevertheless, it is possible to use the concept of slope to help analyze non-linear functions using the following: DEFINITION 2.3. Let f be a function defined on the interval [a, b]. The average rate of change of f over [a, b] is defined as: Af _f (b) - f (a) Ax b-a Geometrically, if we have the graph of y = f(x), the average rate of change over [a, b] is the slope of the line which connects (a, f(a)) and (b, f(b)). This is called the secant line through these points. For that reason, some textbooks use the notation msec for the average rate of change of a function. Note that for a linear function m = msec, or in other words, its rate of change over an interval is the same as its average rate of change.  2.1 LINEAR FUNCTIONS 111 y = f(x) f(b)) The graph of y = f(x) and its secant line through (a, f(a)) and (b, f(b)) The interested reader may question the adjective 'average' in the phrase 'average rate of change.' In the figure above, we can see that the function changes wildly on [a, b], yet the slope of the secant line only captures a snapshot of the action at a and b. This situation is entirely analogous to the average speed on a trip. Suppose it takes you 2 hours to travel 100 miles. Your average speed is 100 miles= 2hours 50 miles per hour. However, it is entirely possible that at the start of your journey, you traveled 25 miles per hour, then sped up to 65 miles per hour, and so forth. The average rate of change is akin to your average speed on the trip. Your speedometer measures your speed at any one instant along the trip, your instantaneous rates of change, and this is one of the central themes of Calculus.7 When interpreting rates of change, we interpret them the same way we did slopes. In the context of functions, it may be helpful to think of the average rate of change as: change in outputs change in inputs EXAMPLE 2.1.7. The revenue of selling x units at a price p per unit is given by the formula R = xp. Suppose we are in the scenario of Examples 2.1.5 and 2.1.6. 1. Find and simplify an expression for the weekly revenue R as a function of weekly sales, x. 2. Find and interpret the average rate of change of R over the interval [0, 50]. 3. Find and interpret the average rate of change of R as x changes from 50 to 100 and compare that to your result in part 2. 4. Find and interpret the average rate of change of weekly revenue as weekly sales increase from 100 PortaBoys to 150 PortaBoys. SOLUTION. 7Here we go again...  112 LINEAR AND QUADRATIC FUNCTIONS 1. Since R = p, we substitute p(x) 1.5x + 250 from Example 2.1.6 to get R(x) x(-1.5x + 250) 1.5x2 + 250x 2. Using Definition 2.3, we get the average rate of change is AR R(50) - R(0) Ax 50-0 8750 - 0 = 175. 50-0 Interpreting this slope as we have in similar situations, we conclude that for every additional PortaBoy sold during a given week, the weekly revenue increases $175. 3. The wording of this part is slightly different than that in Definition 2.3, but its meaning is to find the average rate of change of R over the interval [50, 100]. To find this rate of change, we compute AR _ R(100) - R(50) Ax 100 - 50 10000 - 8750 _ 25. 50 In other words, for each additional PortaBoy sold, the revenue increases by $25. Note while the revenue is still increasing by selling more game systems, we aren't getting as much of an increase as we did in part 2 of this example. (Can you think of why this would happen?) 4. Translating the English to the mathematics, we are being asked to find the average rate of change of R over the interval [100, 150]. We find AR _ R(150) - R(100) Ax 150 - 100 3750 - 10000 50 125. This means that we are losing $125 dollars of weekly revenue for each additional PortaBoy sold. (Can you think why this is possible?) We close this section with a new look at difference quotients, first introduced in Section 1.5. If we wish to compute the average rate of change of a function f over the interval [x, x + h], then we would have Af -f (x+h) - f (x) -f (x+h) - f (x) Ax (x+h)-x h As we have indicated, the rate of change of a function (average or otherwise) is of great importance in Calculus.8 8So, we are not torturing you with these for nothing.  2.1 LINEAR FUNCTIONS 113 2.1.1 EXERCISES 1. Find both the point-slope form and the slope-intercept form of the line with the given slope which passes through the given point. (a) m = j, P(-1,4) (c) m = -5, P(v/3, 2V3) (b) m = 2v/2, P(0, -3) (d) m = 678, P(-1, -12) 2. Find the slope-intercept form of the line which passes through the given points. (a) P(0,0), Q(-3,5) (c) P(5,0), Q(0,-8) (b) P(-1,-2), Q(3,-2) (d) P(3,-5), Q(7,4) 3. Water freezes at 00 Celsius and 320 Fahrenheit and it boils at 100°C and 212°F. (a) Find a linear function F that expresses temperature in the Fahrenheit scale in terms of degrees Celsius. Use this function to convert 20°C into Fahrenheit. (b) Find a linear function C that expresses temperature in the Celsius scale in terms of degrees Fahrenheit. Use this function to convert 110°F into Celsius. (c) Is there a temperature n such that F(n) = C(n)? 4. A salesperson is paid $200 per week plus 5% commission on her weekly sales of x dollars. Find a linear function that represents her total weekly pay in terms of x. What must her weekly sales be in order for her to earn $475.00 for the week? 5. Find all of the points on the line y = 2x + 1 which are 4 units from the point (-1, 3). 6. Economic forces beyond anyone's control have changed the cost function for PortaBoys to C(x) = 105x + 175. Rework Example 2.1.5 with this new cost function. 7. In response to the economic forces in the exercise above, the local retailer sets the selling price of a PortaBoy at $250. Remarkably, 30 units were sold each week. When the systems went on sale for $220, 40 units per week were sold. Rework Examples 2.1.6 and 2.1.7 with this new data. What difficulties do you encounter? 8. Legend has it that a bull Sasquatch in rut will howl approximately 9 times per hour when it is 40°F outside and only 5 times per hour if it's 70°F. Assuming that the number of howls per hour, N, can be represented by a linear function of temperature Fahrenheit, find the number of howls per hour he'll make when it's only 20°F outside. What is the applied domain of this function? Why? 9. (Parallel Lines) Recall from Intermediate Algebra that parallel lines have the same slope. (Please note that two vertical lines are also parallel to one another even though they have an undefined slope.) In the exercises below, you are given a line and a point which is not on that line. Find the line parallel to the given line which passes through the given point.  114 LINEAR AND QUADRATIC FUNCTIONS (a) y = 3x + 2, P(0, 0) (b) y =-6x + 5, P(3,2) 10. (Perpendicular Lines) Recall from Intermediate Algebra that two non-vertical lines are per- pendicular if and only if they have negative reciprocal slopes. That is to say, if one line has slope mi and the other has slope m2 then mi1- m2 = -1. (You will be guided through a proof of this result in the next exercise.) Please note that a horizontal line is perpendicular to a vertical line and vice versa, so we assume mi # 0 and m2 $ 0. In the exercises below, you are given a line and a point which is not on that line. Find the line perpendicular to the given line which passes through the given point. (a) y = 3x+ 2, P(0,0) (b) y=-6x + 5, P(3,2) 11. We shall now prove that y = mix + bi is perpendicular to y = m2x + b2 if and only if mi " -= -1. To make our lives easier we shall assume that mi > 0 and m2 < 0. We can also "move" the lines so that their point of intersection is the origin without messing things up, so we'll assume bi = b2 = 0. (Take a moment with your classmates to discuss why this is okay.) Graphing the lines and plotting the points 0(0, 0) , P(1, i) and Q(1, m2) gives us the following set up. y 0 Q x The line y = mix will be perpendicular to the line y = m2x if and only if AOPQ is a right triangle. Let di be the distance from 0 to P, let d2 be the distance from 0 to Q and let d3 be the distance from P to Q. Use the Pythagorean Theorem to show that AOPQ is a right triangle if and only if mi1- m2 = -1 by showing d2 + d= d3 if and only if mi - m2 = -1.  2.1 LINEAR FUNCTIONS 115 12. The function defined by I(x) = x is called the Identity Function. (a) Discuss with your classmates why this name makes sense. (b) Show that the point-slope form of a line (Equation 2.2) can be obtained from I using a sequence of the transformations defined in Section 1.8. 13. Compute the average rate of change of the given function over the specified interval. (a) f (x) x3, [-1,2] (d) f (x) =9z2, [-3, 3] 1 z+4 (b) f (x) -, [1,5] (e) f (x) =, [5, 7] (c) f(x) =V/, [0,16] (f) f (x) = 3x2 + 2x - 7, [-4, 2] 14. Compute the average rate of change of the given function over the interval [x, x + h]. Here we assume [x, x + h] is in the domain of each function. (a) f(x) = zx3 (c) f(4) = (b) f(x) =1- (d) f (x) = 3x2 + 2x - 7 15. With the help of your classmates find several "real-world" examples of rates of change that are used to describe non-linear phenomena.  116 LINEAR AND QUADRATIC FUNCTIONS 2.1.2 ANSWERS 1. (a) y-4=}(x+ 1) y=7 + 7 (b) y + 3 = v/25(x-0) y=-v x-3 (c) y - 2v =3-5(x- 3/) y =-5x+73v/ (d) y + 12 = 678(x+ 1) y = 678x + 666 2. (a) y (b) y 3 -2 (c) y (d) y=x 8 47 3. (a) F(C) = (b) C(F) = (c) F(-40) 5C+32 (F- 160 =9 490 --40 - C(-40). 4. W(x) = 200 + .05x, She must make $5500 in weekly sales. 5. (-1, -1) and (1,2) 8. N (T ) =2-T + 4 Having a negative number of howls makes no sense and since N(107.5) = 0 we can put an upper bound of 107.50 on the domain. The lower bound is trickier because there's nothing other than common sense to go on. As it gets colder, he howls more often. At some point it will either be so cold that he freezes to death or he's howling non-stop. So we're going to say that he can withstand temperatures no lower than -60° so that the applied domain is [-60,107.5]. 9. (a) y = 3x (b) y - -6x + 20 10. (a) y --3x 22 _- _13 12. (a) 2- (-1) (b) y =6x + 2 32 - (-3)2 (d) 3-(-3) 3 0 (b) ___ 1 5-1 5 1- 0o (c) -16 16-0 4 13. (a) 3x2 + 3xh + h2 -1 (b) x h z(z + h) (e) 7 - 5-3- 8 (f) (3(2)2 + 2(2) - 7) - (3(-4)2 + 2(-4) - 7) 2 - (-4) (c) 6x + 3h + 2 -7 -4 d (x - 3)(x + h - 3)  2.2 ABSOLUTE VALUE FUNCTIONS 117 2.2 ABSOLUTE VALUE FUNCTIONS There are a few ways to describe what is meant by the absolute value Iz| of a real number x. You may have been taught that Iz| is the distance from the real number x to the 0 on the number. So, for example, |51 = 5 and | - 5| = 5, since each is 5 units from 0 on the number line. distance is 5 units distance is 5 units -5 -4 -3 -2 -1 0 1 2 3 4 5 Another way to define absolute value is by the equation Iz| =v2. Using this definition, we have |5| = V(5)2= 25 = 5 and | - 5| = (-5)2 = 25 = 5. The long and short of both of these procedures is that Iz| takes negative real numbers and assigns them to their positive counterparts while it leaves positive numbers alone. This last description is the one we shall adopt, and is summarized in the following definition. DEFINITION 2.4. The absolute value of a real number x, denoted Izis given by -{ , if zx< 0 1XI x, if x > 0 In Definition 2.4, we define Iz| using a piecewise-defined function. (See page 46 in Section 1.5.) To check that this definition agrees with what we previously understood as absolute value, note that since 5 > 0, to find |51 we use the rule Iz| = x, so 15 = 5. Similarly, since -5 < 0, we use the rule Iz|v= -, so that | - 5| = -(-5) = 5. This is one of the times when it's best to interpret the expression '-x' as 'the opposite of x' as opposed to 'negative x.' Before we embark on studying absolute value functions, we remind ourselves of the properties of absolute value. THEOREM 2.1. Properties of Absolute Value: Let a, b, and x be real numbers and let n be an integer.a Then " Product Rule: lab= la||b| " Power Rule: la"| = |al" whenever an is defined a a * Quotient Rule: - - , provided b - 0 b |b' " c= 0 if and only if x = 0. " For c>0, cv =cif and only if =cor x=-c. " For c < 0, Iz|v= c has no solution. "Recall that this means nr= 0, +1, +2,... The proof of the Product and Quotient Rules in Theorem 2.1 boils down to checking four cases: when both a and b are positive; when they are both negative; when one is positive and the other  118 LINEAR AND QUADRATIC FUNCTIONS is negative; when one or both are zero. For example, suppose we wish to show lab| =la|bl. We need to show this equation is true for all real numbers a and b. If a and b are both positive, then so is ab. Hence, |al = a, |b| = b, and lab| = ab. Hence, the equation lab| = la||b| is the same as ab = ab which is true. If both a and b are negative, then ab is positive. Hence, |al = -a, |b| = -b, and lab| = ab. The equation lab| = la||b| becomes ab = (-a)(-b), which is true. Suppose a is positive and b is negative. Then ab is negative, and we have lab| = -ab, |al = a and |b| = -b. The equation lab = la||b| reduces to -ab = a(-b) which is true. A symmetric argument shows the equation lab = la||b| holds when a is negative and b is positive. Finally, if either a or b (or both) are zero, then both sides of lab| = la||b| are zero, and so the equation holds in this case, too. All of this rhetoric has shown that the equation lab| = la||b| holds true in all cases. The proof of the Quotient Rule is very similar, with the exception that b -f 0. The Power Rule can be shown by repeated application of the Product Rule. The last three properties can be proved using Definition 2.4 and by looking at the cases when x > 0, in which case Ix = x, or when x < 0, in which case Ix = -x. For example, if c > 0, and Ix =c, then if x> 0, we havex zx Iz| =c. If, on the other hand, x < 0, then -x x= z|= c, so x= -c. The remaining properties are proved similarly and are left for the exercises. To graph functions involving absolute value we make liberal use of Definition 2.4, as the next example illustrates. EXAMPLE 2.2.1. Graph each of the following functions. Find the zeros of each function and the x- and y-intercepts of each graph, if any exist. From the graph, determine the domain and range of each function, list the intervals on which the function is increasing, decreasing, or constant, and find the relative and absolute extrema, if they exist. 1. f(x) x= I 3. h(x) =x -3 2. g(x) =I - 3| 4. i(x) = 4 - 2|3x + 1| SOLUTION. 1. To find the zeros of f, we set f(x) = 0. We get Ix|= 0, which, by Theorem 2.1 gives us x= 0. Since the zeros of f are the x-coordinates of the x-intercepts of the graph of y = f(x), we get (0, 0) is our only x-intercept. To find the y-intercept, we set x= 0, and find y = f(0) = 0, so that (0, 0) is our y-intercept as well.' With Section 2.1 under our belts, we can use Definition 2.4 to get l -if x < 0 f(z) -Iz x, if x> 0 Hence, for x < 0, we are graphing the line y = -x; for x > 0, we have the line y = x. Proceeding as we did in Section 1.7, we get 1Actually, since functions can have at most one y-intercept (why?), as soon as we found (0, 0) as the x-intercept, we knew this was also the y-intercept.  2.2 ABSOLUTE VALUE FUNCTIONS 119 y 4 3 2 1 -3 -2 -1 1 2 3 x f(x) =ll, x < 0 4- 3- 2- 1- i i i IF i i i 30 -3 -2 -1 1 2 f(x) =x, x > 0 3 x Notice we have an 'open circle' at (0, 0) in the graph when x < 0. As we have seen before, this is due to the fact the points on y = -x approach (0, 0) as the x-values approach 0. Since x is required to be strictly less than zero on this stretch, however, the open circle is drawn. However, notice that when x > 0, we get to fill in the point at (0, 0), which effectively 'plugs' the hole indicated by the open circle. Hence, we get 4- 3- 2- 1- -3 -2 -1 1 2 3 x f(x) = lol By projecting the graph to the x-axis, we see that the domain is (-oc, oc). Projecting to the y-axis gives us the range [0, oc). The function is increasing on [0, oc) and decreasing on (-oc, 0]. The relative minimum value of f is the same as the absolute minimum, namely 0 which occurs at (0,0). There is no relative maximum value of f. There is also no absolute maximum value of f, since the y values on the graph extend infinitely upwards. 2. To find the zeros of g, we set g(x) =Ix - that x= 3. Hence, the x-intercept is (3,' y = g(0) = |0 - 3| 1=3, and so (0, 3) isc Definition 2.4 to rewrite as g(x) =_Ix - 3|= - 3| = 0. By Theorem 2.1, we get x - 3 = 0 so 0). To find our y-intercept, we set x = 0 so that our y-intercept. To graph g(x) =Ix - 3|, we use x 3), - 3, if if I 3: 3<0 3>0 Simplifying, we get g(x) = l -x + 3, x - 3, if if 3:<3 3:>3  120 LINEAR AND QUADRATIC FUNCTIONS As before, the open circle we introduce at (3, 0) from the graph of y= -x+ 3 is filled by the point (3, 0) from the line y = x - 3. We determine the domain as (-oc, oc) and the range as [0, oc). The function g is increasing on [3, oc) and decreasing on (-oc, 3]. The relative and absolute minimum value of g is 0 which occurs at (3, 0). As before, there is no relative or absolute maximum value of g. 3. Setting h(x) = 0 to look for zeros gives Ix - 3 = 0. Before we can use any of the properties in Theorem 2.1, we need to isolate the absolute value. Doing so gives |x|= 3 so that x= 3 or x= -3. As a result, we have a pair of x-intercepts: (-3,0) and (3,0). Setting x = 0 gives y = h(0) = |0| - 3 = -3, so our y-intercept is (0, -3). As before, we rewrite the absolute value in h to get h (x) ={--3, if x<0 X- 3,if x > 0 Once again, the open circle at (0, -3) from one piece of the graph of h is filled by the point (0, -3) from the other piece of h. From the graph, we determine the domain of h is (-oc, oo) and the range is [-3, oc). On [0, oc), h is increasing; on (-oc, 0] it is decreasing. The relative minimum occurs at the point (0, -3) on the graph, and we see -3 is both the relative and absolute minimum value of h. Also, h has no relative or absolute maximum value. yY 4 2 - 1 I I I I ) 1 2 3 4 5 g(x) = lo - 3| h(x) =_lo| - 3 4. As before, we set i(x) = 0 to find the zeros of i to get 4 - 2|3x +1|= 0. Once again, we need to isolate the absolute value expression before we apply Theorem 2.1. So 4 - 2|3x + 1|I= 0 becomes 2|3x + 1|I= 4 and hence, |3x + 1|I= 2. Applying Theorem 2.1, we get 3x + 1 = 2 or 3x + 1 = -2, from which we get x= } or x= -1. Our x-intercepts are (1, 0) and (-1,0). Substituting x = 0 gives y = i(0) = 4 - 2|3(0) + 1| = 2, for a y-intercept of (0, 2). Rewriting the formula for i(x) without absolute values gives . 4-2(-(3x+ 1)), if 3x + 1 < 0 _ 6x+6, if z < -} 4 - 2(3x + 1), if 3x + 1 > 0 -6x+2, if > -} The usual analysis near the trouble spot, x -} gives the 'corner' of this graph is (-} 4), and we get the distinctive 'V' shape:  2.2 ABSOLUTE VALUE FUNCTIONS 121 y 3 2 - 1 x -1- -2- -3- i(x) 4 - 2|3x + 2| The domain of i is (-oc, o) while the range is (-oc, 4]. The function i is increasing on (-oc, -fj] and decreasing on [-j, oc). The relative maximum occurs at the point (-i, 4) and the relative and absolute maximum value of i is 4. Since the graph of i extends downwards forever more, there is no absolute minimum value. As we can see from the graph, there is no relative minimum, either. D Note that all of the functions in the previous example bear the characteristic 'V' shape as the graph of y = . In fact, we could have graphed all of the functions g, h, and i in Example 2.2.1 starting with the graph of f(x) = x and applying transformations as in Section 1.8. For example, the for the function g, we have g(x) =I - 3| = f(x - 3). Theorem 1.3 tells us this causes the graph of f to be shifted to the right 3 units. Choosing three representative points on the graph of f, (-1, 1), (0, 0) and (1, 1), we can graph g as follows. y y 4 4 3 3 2 2 (-1, 1) 1 (1, 1) 1..(2,1)(4,)1) -3 -2 -1 (0,0)1 2 3 4 5 x shift right 3 unit -3 -2 -1 1 2 (3,0)4 5 x f (x) =I| add 3 to each x-coordinate g(x) = f (x - 3) =|x - 3 Similarly, the graph of h in Example 2.2.1 can be understood via Theorem 1.2 as a vertical shift down 3 units. The function i can be graphed using Theorem 1.7 by finding the final destinations of the three points (-1, 1), (0, 0) and (1, 1) and connecting them in the characteristic 'V' fashion. While the methods in Section 1.8 can be used to graph an entire family of absolute value functions, not all functions involving absolute values posses the characteristic 'V' shape, as the next example illustrates. EXAMPLE 2.2.2. Graph each of the following functions. Find the zeros of each function and the x- and y-intercepts of each graph, if any exist. From the graph, determine the domain and range  122 LINEAR AND QUADRATIC FUNCTIONS of each function, list the intervals on which the function is increasing, decreasing, or constant, and find the relative and absolute extrema, if they exist. 1. f(x) I2. g(x)_ =Ix+2| - Ix-3 +1 SOLUTION. 1. We first note that, due to the fraction in the formula of f(x), x - 0. Thus the domain is (-oc, 0) U (0, oc). To find the zeros of f, we set f(x) =_f = 0. This last equation implies Ix= 0, which, from Theorem 2.1, implies x= 0. However, x= 0 is not in the domain of f, which means we have, in fact, no x-intercepts. For the same reason, we have no y-intercepts, since f(0) is undefined. Re-writing the absolute value in the function gives ff, if x< 0 f-1, if x<0 -, ifx>0 1, ifx>0 To graph this function, we graph two horizontal lines: y = -1 for x < 0 and y = 1 for x > 0. We have open circles at (0, -1) and (0, 1) (Can you explain why?) so we get f(x)= $ As we found earlier, the domain is (-oc, 0) U (0, oc). The range consists of just 2 y values: {-1, 1}.2 The function f is constant on (-oc, 0) and (0, oc). The local minimum value of f is the absolute minimum value of f, namely -1; the local maximum and absolute maximum values for f also coincide - they both are 1. Every point on the graph of f is simultaneously a relative maximum and a relative minimum. (Can you see why in light of Definition 1.9? This was explored in the exercises in Section 1.7.2.) 2. To find the zeros of g, we set g(x) = 0. The result is |x + 2| - Ix - 3|+ 1 = 0. Attempting to isolate the absolute value term is complicated by the fact that there are two terms with absolute values. In this case, it easier to proceed using cases by re-writing the function g with two separate applications of Definition 2.4 to remove each instance of the absolute values, one at a time. In the first round we get -(+2)- I-3 3|+1, if z+ 2 <0 f --1 - I -3|, if z0 x+ 3-x|z-3|, if x> -2 2These are set braces, not parentheses or brackets. We used this same 'set builder' notation in Section 1.4.  2.2 ABSOLUTE VALUE FUNCTIONS 123 Given that ix - 3|= {l -(x - 3), if x x - 3, if x 3<0 3>0 { -x+3, if x<3 x-3, if x>3 we need to break up replace Ix - 3| with - form of g yields the domain again at x= 3. Note that if x < -2, then x < 3, so we -x + 3 for that part of the domain, too. Our completed revision of the g(x) = x - 1 - (-x + 3), if x< -2 x+3-(-x+3), if x>-2 andx<3 x+3-(x-3), if x>3 {-4, if x< -2 2x, if -2 3 To solve g(x) = 0, we see that the only piece which contains a variable is g(x) = 2x for -2 3, and the point (3, 6) fills in the open circle left by the previous part of the graph. The finished graph is x g(x) = x +21 - lo- 3|+1 The domain of g is all real numbers, (-oc, oc), and the range of g is all real numbers between -4 and 6 inclusive, [-4, 6]. The function is increasing on [-2, 3] and constant on (-oc, -2] and [3, oo). The relative minimum value of f is 4 which matches the absolute minimum. The relative and absolute maximum values also coincide at 6. Every point on the graph of y = g(x) for x < -2  124 LINEAR AND QUADRATIC FUNCTIONS and x > 3 yields both a relative minimum and relative maximum. The point (-2, -4), however, gives only a relative minimum and the point (3,6) yields only a relative maximum. (Recall the exercises in Section 1.7.2 which dealt with constant functions.) D Many of the applications that the authors are aware of involving absolute values also involve absolute value inequalities. For that reason, we save our discussion of applications for Section 2.4.  2.2 ABSOLUTE VALUE FUNCTIONS 125 2.2.1 EXERCISES 1. Graph each of the following functions using transformations or the definition of absolute value, as appropriate. Find the zeros of each function and the x- and y-intercepts of each graph, if any exist. From the graph, determine the domain and range of each function, list the intervals on which the function is increasing, decreasing, or constant, and find the relative and absolute extrema, if they exist. (a) f (x) = Ix + 4| (b) f(x)x +4 (c) f(x) =|4x (d) f (x) = -3x (e) f (x) = 3|z + 4| - 4 (f) f ) |+4| (g) f (z) = i2 i 2 -yx (h) f (x) = -2x - 1| 3 (i) fz)~ =xz+4 4+ I- 2| 2. With the help of your classmates, prove the second, third and fifth properties listed in The- orem 2.. 3. With the help of your classmates, find a function involving absolute values whose graph is given below. 4 3 2 1 -8-7 -6 -5-4 -3-2 -1 y 1 2 3 4 5 6 7 8 x 4. With the help of your classmates, prove the following two properties of absolute value. (a) (The Triangle Inequality) For all real numbers a and b, la + b <; |a + |bl. (b) If f(x) =_Ig(x)| then either f () = g(x) or f () = -g(x). 5. Use the result from Exercise 4b above to solve the following equations. Interpret your results graphically. (a) |3x - 21 = |2x + 7| (b) |3x +1|=|4x| (c) 1 (d) 2 2x| 5x| |x+1| 5|x+1|  126 LINEAR AND QUADRATIC FUNCTIONS 2.2.2 ANSWERS 1. (a) f() x+4 f(-4) = 0 x-intercept (-4, 0) y-intercept (0, 4) Domain (-oc, oc) Range [0,oo) Decreasing on (-oc, -4] Increasing on [-4, ox) Relative and absolute min. at (-4, 0) No relative or absolute maximum (b) f (x) |j| + 4 No zeros No x-intercepts y-intercept (0, 4) Domain (-oc, o) Range [4,oo) Decreasing on (-oc,0] Increasing on [0, oo) Relative and absolute minimum at (0, 4) No relative or absolute maximum 4 3 2 -8 -7 -6 -5 -4 -3 -2 -1 i 1 8- 7- 6- 5- 4 3- 2- 1- I I I I I I I I I -4 -3 -2 -1 1 2 3 4 x (c) f (x) =4x| f (0) = 0 x-intercepts (0, 0) y-intercept (0,0) Domain (-oc, oc) Range [0,oo) Decreasing on (-oc, 0] Increasing on [0, oo) Relative and absolute minimum at (0, 0) No relative or absolute maximum  2.2 ABSOLUTE VALUE FUNCTIONS 127 (d) f (x)= -3|z| f(0) = 0 x-intercepts (0, 0) y-intercept (0,0) Domain (-oc, oc) Range (-oc, 0] Increasing on (-oc, 0] Decreasing on [0, oo) Relative and absolute maximum at (0, 0) No relative or absolute minimum (e) f (x) = 3|x + 4| - 4 f (-L6) = 0, f (-j) = 0 x-intercepts (-9,0), (-),0) y-intercept (0,8) Domain (-oc, oc) Range [-4,oo) Decreasing on (-oc, -4] Increasing on [-4, oc) Relative and absolute min. at (-4, -4) No relative or absolute maximum S+ 4| (f) f )-x + 41 No zeros No x-intercept y-intercept (0, 1) Domain (-oo,-4) U(-4,oo) Range {-1, 1} Constant on (-oc, -4] Constant on [-4, oc) Absolute minimum at every point (x, -1) (g) f v 2 - x No zeros No x-intercept y-intercept (0, 1) Domain (-oo,2) U(2,oo) Range {-1, 1} Constant on (-oo, 2] Constant on [2, oo) Absolute minimum at every point (x, -1) x where x < -4 Absolute maximum at every point (x, 1) where x> -4 Relative maximum AND minimum at ev- ery point on the graph -8 -7 -6 -5 -4 -3 -2 -1 1 x where x> 2 Absolute maximum at every point (x, 1) where x < 2 Relative maximum AND minimum at ev- ery point on the graph y -3-2-1 1 2 3 4 5 x -1  128 (h) f (x) = }|12x - 1| f (j) 0 x-intercepts (2, 0) y-intercept (0, j) Domain (-oc, o) Range [0,oo) Decreasing on (-oc, f] LINEAR AND QUADRATIC FUNCTIONS Increasing on [2, o) Relative and absolute min. at (i, 0) No relative or absolute maximum 2 -3 -2 -1 1 2 3 4 X (i) Re-write f (x)_ =Ix + 4| + |x- 2| as -2x - 2 if x< -4 f (z) = 6 if -4 < z < 2 2x + 2 if x > 2 No zeros No x-intercept y-intercept (0,6) Domain (-oc, oc) Range [6,oo) Decreasing on (-oc, -4] Constant on [-4, 2] Increasing on [2, oc) Absolute minimum at every point (x, -1) where -4 < x < 2 No absolute maximum Relative minimum where -4 < x < 2 Relative maximum where -4 < x < 2 8 7- 5-- 4- 3-- 2-- 1-- at every point (x, 1) at every point (x, 1) y i i i i i i i----- i-----i----> -5 -4 -3 -2 -1 1 2 3 x 3. f (x)_ = |x| - 4| 5. (a) x= -1 or, (b) x= 1 or x x=9 _ 1 -7 (c) x= 0 or x= 2 (d) x=-3  2.3 QUADRATIC FUNCTIONS 129 2.3 QUADRATIC FUNCTIONS You may recall studying quadratic equations in Intermediate Algebra. In this section, we review those equations in the context of our next family of functions: the quadratic functions. DEFINITION 2.5. A quadratic function is a function of the form f (x) = ax2 + bx + c, where a, b, and c are real numbers with a - 0. The domain of a quadratic function is (-oc, oc). EXAMPLE 2.3.1. Graph each of the following quadratic functions. Find the zeros of each function and the x- and y-intercepts of each graph, if any exist. From the graph, determine the domain and range of each function, list the intervals on which the function is increasing, decreasing, or constant and find the relative and absolute extrema, if they exist. 1. f(X) = x2 - 4x + 3. 2. g(x) = -2(x - 3)2 + 1. SOLUTION. 1. To find the zeros of f, we set f(x) = 0 and solve the equation 2 - 4x + 3 = 0. Factoring gives us (x- 3)(x- 1) = 0 so that x= 3 or x= 1. The x-intercepts are then (1, 0) and (3, 0). To find the y-intercept, we set x= 0 and find that y = f(0) = 3. Hence, the y-intercept is (0, 3). Plotting additional points, we get y x f(x) = 2 - 4x + 3 From the graph, we see the domain is (-oc, oc) and the range is [-1, oc). The function f is increasing on [2, oc) and decreasing on (-oc, 2]. A relative minimum occurs at the point (2, -1) and the value -1 is both the relative and absolute minimum of f.  130 LINEAR AND QUADRATIC FUNCTIONS 2. Note that the formula for g(x) doesn't match the form given in Definition 2.5. However, if we took the time to expand g(x) = -2(x - 3)2 + 1, we would get g(x) = -2x2 + 12x - 17 which does match with Definition 2.5. When we find the zeros of g, we can use either formula, since both are equivalent. Using the formula which was given to us, we get g(x) -2(x - 3)2 + 1 -2(x - 3)2 0 0 -1 1 (x - 3)2 _1 2 2 X = 3f i 2 z = 2 divide by -2 extract square roots rationalize the denominator get a common denominator Hence, we have two x-intercepts: (6+2, 0) and (62, 0). (The inquisitive reader may wonder what we would have done had we chosen to set the expanded form of g(x) equal to zero. Since -2x2 + 12x - 17 does not factor nicely, we would have had to resort to other methods, which are reviewed later in this section, to solve -2x2 + 12x - 17 = 0.) To find the y-intercept, we set x= 0 and get g(0) = -17. Our y-intercept is then (0, -17). Plotting some additional points, we get -1. g(x) = -2(x - 3)2 + 1  2.3 QUADRATIC FUNCTIONS 131 The domain of g is (-oc, o) and the range is (-oc, 1]. The function g is increasing on (-oc, 3] and decreasing on [3, oc). The relative maximum occurs at the point (3, 1) with 1 being both the relative and absolute maximum value of g. D Hopefully the previous examples have reminded you of some of the basic characteristics of the graphs of quadratic equations. First and foremost, the graph of y = ax2 + bx + c where a, b, and c are real numbers with a # 0 is called a parabola. If the coefficient of x2, a, is positive, the parabola opens upwards; if a is negative, it opens downwards, as illustrated below.1 vertex vertex a>0 a<0 Graphs of y = ax2 + bx+ c. The point at which the relative minimum (if a > 0) or relative maximum (if a < 0) occurs is called the vertex of the parabola. Note that each of the parabolas above is symmetric about the dashed vertical line which contains its vertex. This line is called the axis of symmetry of the parabola. As you may recall, there are two ways to quickly find the vertex of a parabola, depending on which form we are given. The results are summarized below. EQUATION 2.4. Vertex Formulas for Quadratic Functions: Suppose a, b, c, h, and k are real numbers with a # 0. " If f (x) = a(x - h)2 + k, the vertex of the graph of y = f(x) is the point (h, k). " If f(x) = ax2 + bx + c, the vertex of the graph of y = f(x) is the point -+, f- . 2a, 2a EXAMPLE 2.3.2. Use Equation 2.4 to find the vertex of the graphs in Example 2.3.1. SOLUTION. 1. The formula f(x) =9x2 -4x+3 is in the form f(x) =ax2+ bx+ c. We identify a = 1, b and c = 3, so that -4, b -4 2a 2(1) 2' and f( ) f(2) 1, so the vertex is (2, -1) as previously stated. 'We will justify the role of a in the behavior of the parabola later in the section.  132 LINEAR AND QUADRATIC FUNCTIONS 2. We see that the formula g(x) = -2(x - 3)2+1 is in the form g(x) = a(x - h)2 + k. We identify a = -2, xz- h as x - 3 (so h = 3), and k = 1 and get the vertex (3, 1), as required. D The formula f(x) = a(x - h)2 + k, a # 0 in Equation 2.4 is sometimes called the standard form of a quadratic function; the formula f(x) = ax2 + bx + c, a # 0 is sometimes called the general form of a quadratic function. To see why the formulas in Equation 2.4 produce the vertex, let us first consider a quadratic function in standard form. If we consider the graph of the equation y = a(x - h)2 + k we see that when x = h, we get y = k, so (h, k) is on the graph. If x # h, then x - h # 0 and so (x - h)2 is a positive number. If a > 0, then a(x - h)2 is positive, and so y = a(x - h)2 + k is always a number larger than k. That means that when a > 0, (h, k) is the lowest point on the graph and thus the parabola must open upwards, making (h, k) the vertex. A similar argument shows that if a < 0, (h, k) is the highest point on the graph, so the parabola opens downwards, and (h, k) is also the vertex in this case. Alternatively, we can apply the machinery in Section 1.8. The vertex of the parabola y =x2 is easily seen to be the origin, (0, 0). We leave it to the reader to convince oneself that if we apply any of the transformations in Section 1.8 (shifts, reflections, and/or scalings) to y = x2, the vertex of the resulting parabola will always be the point the graph corresponding to (0,0). To obtain the formula f (x) = a(x - h)2 + k, we start with g(x) =9x2 and first define gi(x) = ag(x) = ax2. This is results in a vertical scaling and/or reflection.2 Since we multiply the output by a, we multiply the y-coordinates on the graph of g by a, so the point (0, 0) remains (0, 0) and remains the vertex. Next, we define g2(X) = g1(x - h) = a(x - h)2. This induces a horizontal shift right or left h units? moves the vertex, in either case, to (h, 0). Finally, f(x) = g2(X) + k = a(x - h)2 + k which effects a vertical shift up or down k units4 resulting in the vertex moving from (h, 0) to (h, k). To verify the vertex formula for a quadratic function in general form, we complete the square to convert the general form into the standard form.5 f(x)= ax2+bx+c - a z2+-z)+c - 2 b b2 b2 - az2+-x+ $) + c - a ()complete the square a 4a24a2 b 2 4ac -b2 = a x++ +4factor; get a common denominator 2a 4a Comparing this last expression with the standard form, we identify (x - h) as (x + a) so that h - . Instead of memorizing the value k = 4ac-b2 we see that f (--) = 4ab2 As such, 2ag4a 2a 4 2Just a scaling if a > 0. If a < 0, there is a reflection involved. 3Right if h > 0, left if h < 0. 4Up if k > 0, down if k < 0 5Actually, we could also take the standard form, f(x) =a(x - h)2 + k, expand it, and compare the coefficients of it and the general form to deduce the result. However, we will have another use for the completed square form of the general form of a quadratic, so we'll proceed with the conversion.  2.3 QUADRATIC FUNCTIONS 133 we have derived the vertex formula for the general form as well. Note that the value a plays the exact same role in both the standard and general equations of a quadratic function - it is the coefficient of 2 in each. No matter what the form, if a > 0, the parabola opens upwards; if a < 0, the parabola opens downwards. Now that we have the completed square form of the general form of a quadratic function, it is time to remind ourselves of the quadratic formula. In a function context, it gives us a means to find the zeros of a quadratic function in general form. EQUATION 2.5. The Quadratic Formula: If a, b, c are real numbers with a # 0, then the solutions to ax2 + bx + c = 0 are -b+ kb2 - 4ac 2a Assuming the conditions of Equation 2.5, the solutions to ax2 + bx + c = 0 are precisely the zeros of f (x) =ax2 + bx + c. We have shown an equivalent formula for f is b \2 Aac - b2 f(x) -a + + 4a. b Hence, an equation equivalent to ax2 + bx + c = 0 is 2 4ac-b2 a\+ a + 4a = 0. Solving gives ( b \ 2 4ac -b2 ayz+y) + 4a = 0 / b 2 4ac - b2 2a)4a 1 b 1 b2-4ac 2a a 4a b2 b2 - 4ac b -4aac 2a2a -b+ b2-4ac z =  134 LINEAR AND QUADRATIC FUNCTIONS In our discussions of domain, we were warned against having negative numbers underneath the square root. Given that b2 - 4ac is part of the Quadratic Formula, we will need to pay special attention to the radicand b2 - 4ac. It turns out that the quantity b2 - 4ac plays a critical role in determining the nature of the solutions to a quadratic equation. It is given a special name and is discussed below. DEFINITION 2.6. If a, b, c are real numbers with a # 0, then the discriminant of the quadratic equation ax2 + bx + c = 0 is the quantity b2 - 4ac. THEOREM 2.2. Discriminant Trichotomy: Let a, b, and c be real numbers with a # 0. " If b2 - 4ac < 0, the equation ax2 + bx + c = 0 has no real solutions. " If b2 - 4ac = 0, the equation ax2 + bx + c = 0 has exactly one real solution. " If b2 - 4ac > 0, the equation ax2 + bx + c = 0 has exactly two real solutions. The proof of Theorem 2.2 stems from the position of the discriminant in the quadratic equation, and is left as a good mental exercise for the reader. The next example exploits the fruits of all of our labor in this section thus far. EXAMPLE 2.3.3. The profit function for a product is defined by the equation Profit = Revenue - Cost, or P(x) = R(x) - C(x). Recall from Example 2.1.7 that the weekly revenue, in dollars, made by selling x PortaBoy Game Systems is given by R(x) = -1.5x2 + 250x. The cost, in dollars, to produce x PortaBoy Game Systems is given in Example 2.1.5 as C(x) = 80x + 150, x > 0. 1. Determine the weekly profit function, P(x). 2. Graph y = P(x). Include the x- and y-intercepts as well as the vertex and axis of symmetry. 3. Interpret the zeros of P. 4. Interpret the vertex of the graph of y = P(x). 5. Recall the weekly price-demand equation for PortaBoys is: p(x) = -1.5x + 250, where p(x) is the price per PortaBoy, in dollars, and x is the weekly sales. What should the price per system be in order to maximize profit? SOLUTION. 1. To find the profit function P(x), we subtract P(x) = R(x) - C(x) = (-1.5x2 + 250x) - (80x + 150) = -1.5x2 + 170x - 150. 2. To find the x-intercepts, we set P(x) = 0 and solve -1.5x2 + 170x - 150 = 0. The mere thought of trying to factor the left hand side of this equation could do serious psychological damage, so we resort to the quadratic formula, Equation 2.5i. Identifying a =-1.5, b =170, and c =-150, we obtain  2.3 QUADRATIC FUNCTIONS 135 -b+ b2 - 4ac X 2a -170 + 11702 - 4(-1.5)(-150) 2(-1.5) -170+ 28000 -3 170 +20v70 3 We get two x-intercepts: (170-270, 0) and (170+2a7,0). To find the y-intercept, we set x = 0 and find y = P(0) =-150 for a y-intercept of (0, -150). To find the vertex, we use the fact that P(x) =-1.5x2 + 170x - 150 is in the general form of a quadratic function and appeal to Equation 2.4. Substituting a = -1.5 and b = 170, we get x= - 170 170 2(-1.5) 3 To find the y-coordinate of the vertex, we compute P (170) = 14000 and find our vertex is (1, 14P00) The axis of symmetry is the vertical line passing through the vertex so it is the 3. line x = 13. To sketch a reasonable graph, we approximate the x-intercepts, (0.89, 0) and (112.44, 0), and the vertex, (56.67, 4666.67). (Note that in order to get the x-intercepts and the vertex to show up in the same picture, we had to scale the x-axis differently than the y-axis. This results in the left-hand x-intercept and the y-intercept being uncomfortably close to each other and to the origin in the picture.) y 4000-- 3000-- 2000-- 1000-- 10 20 30 40 50 160 70 80 90 100 11 120 x 3. The zeros of P are the solutions to P(x) = 0, which we have found to be approximately 0.89 and 112.44. Since P represents the weekly profit, P(x) = 0 means the weekly profit is $0. Sometimes, these values of x are called the 'break-even' points of the profit function, since these are places where the revenue equals the cost; in other words we gave sold enough product to recover the cost spent to make the product. More importantly, we see from the graph that as long as x is between 0.89 and 112.44, the graph y = P(x) is above the x-axis, meaning y = P(x) > 0 there. This means that for these values of x, a profit is being made. Since x represents the weekly sales of PortaBoy Game Systems, we round the zeros to positive integers and have that as long as 1, but no more than 112 game systems are sold weekly, the retailer will make a profit.  136 LINEAR AND QUADRATIC FUNCTIONS 4. From the graph, we see the maximum value of P occurs at the vertex, which is approximately (56.67, 4666.67). As above, x represents the weekly sales of PortaBoy systems, so we can't sell 56.67 game systems. Comparing P(56) = 4666 and P(57) = 4666.5, we conclude we will make a maximum profit of $4666.50 if we sell 57 game systems. 5. In the previous part, we found we need to sell 57 PortaBoys per week to maximize profit. To find the price per PortaBoy, we substitute x= 57 into the price-demand function to get p(57) = -1.5(57) + 250 = 164.5. The price should be set at $164.50. D We conclude this section with a more complicated absolute value function. EXAMPLE 2.3.4. Graph f (x) =9Iz2 - x - 6. SOLUTION. Using the definition of absolute value, Definition 2.4, we have {(z2 -x-6)if 2-x-6<0 f)2 -x-6, if 9 -x-6>0 The trouble is that we have yet to develop any analytic techniques to solve nonlinear inequalities such as 2 - x - 6 < 0. You won't have to wait long; this is one of the main topics of Section 2.4. Nevertheless, we can attack this problem graphically. To that end, we graph y = g(x) = z2 - x - 6 using the intercepts and the vertex. To find the x-intercepts, we solve 2 - x - 6 = 0. Factoring gives (x - 3)(x + 2) = 0 so x= -2 or x= 3. Hence, (-2, 0) and (3, 0) are x-intercepts. The y- intercept is found by setting x= 0, (0, -6). To find the vertex, we find x =2a =-2() =, and y (1)2 - ( 6) -6 - = -6.25. Plotting, we get the parabola seen below on the left. To obtain points on the graph of y = f(x) =I2 - x -6|, we can take points on the graph of g(x) =9x2 - x -6 and apply the absolute value to each of the y values on the parabola. We see from the graph of g that for x < -2 or x > 3, the y values on the parabola are greater than or equal to zero (since the graph is on or above the x-axis), so the absolute value leaves this portion of the graph alone. For x between -2 and 3, however, the y values on the parabola are negative. For example, the point (0, -6) on y =9z - x - 6 would result in the point (0, l- 6|) =(0, -(-6)) =(0, 6) on the graph of f (x) = I2 - x - 6|. Proceeding in this manner for all points with x-coordinates between -2 and 3 results in the graph seen above on the right.  2.3 QUADRATIC FUNCTIONS 137 y y=g(x)-=x2 x -6 y f(x)= -x - 6 If we take a step back and look at the graphs of g and f in the last example, we notice that to obtain the graph of f from the graph of g, we reflect a portion of the graph of g about the x-axis. We can see this analytically by substituting g(x) = 2 - x - 6 into the formula for f(x) and calling to mind Theorem 1.4 from Section 1.8. f ) -g(x), if g(x) < 0 g(x), if g(x) > 0 The function f is defined so that when g(x) is negative (i.e., when its graph is below the x-axis), the graph of f is its refection across the x-axis. This is a general template to graph functions of the form f(x) = Ig(x)|. From this perspective, the graph of f(x) =_Ix can be obtained by reflection the portion of the line g(x) = x which is below the x-axis back above the x-axis creating the characteristic 'V' shape.  138 LINEAR AND QUADRATIC FUNCTIONS 2.3.1 EXERCISES 1. Graph each of the following quadratic functions. Find the x- and y-intercepts of each graph, if any exist. If it is given in the general form, convert it into standard form. Find the domain and range of each function and list the intervals on which the function is increasing or decreasing. Identify the vertex and the axis of symmetry and determine whether the vertex yields a relative and absolute maximum or minimum. (a) f (x) = x2 + 2 (d) f (x) = -2(x+ 1)2 +4 (b) f (x) = -(x+ 2)2 (e) f (x) = -3x2 + 4x - 7 (c) f (X) = X2 - 2x - 8 (f) f (X) = zX2- X - 16 100 2. Graph f (x) =1-9x2 3. Find all of the points on the line y = 1 - x which are 2 units from (1, -1) 4. With the help of your classmates, show that if a quadratic function f(x) =ax2 + bx + c has two real zeros then the x-coordinate of the vertex is the midpoint of the zeros. 5. Assuming no air resistance or forces other than the Earth's gravity, the height above the ground at time t of a falling object is given by s(t) = -4.9t2 + vot + so where h is in meters, t is in seconds, vo is the object's initial velocity in meters per second and so is its initial position in meters. (a) What is the applied domain of this function? (b) Discuss with your classmates what each of vo > 0, vo = 0 and vo < 0 would mean. (c) Come up with a scenario in which so < 0. (d) Let's say a slingshot is used to shoot a marble straight up from the ground (so = 0) with an initial velocity of 15 meters per second. What is the marble's maximum height above the ground? At what time will it hit the ground? (e) Now shoot the marble from the top of a tower which is 25 meters tall. When does it hit the ground? (f) What would the height function be if instead of shooting the marble up off of the tower, you were to shoot it straight DOWN from the top of the tower? 6. The International Silver Strings Submarine Band holds a bake sale each year to fund their trip to the National Sasquatch Convention. It has been determined that the cost in dollars of baking x cookies is C(x) = 0.1x + 25 and that the demand function for their cookies is p =10 - .01z. How many cookies should they bake in order to maximize their profit? 6We have already seen the graph of this function. It was used as an example in Section 1.7 to show how the graphing calculator can be misleading.  2.3 QUADRATIC FUNCTIONS 139 7. The two towers of a suspension bridge are 400 feet apart. The parabolic cable7 attached to the tops of the towers is 10 feet above the point on the bridge deck that is midway between the towers. If the towers are 100 feet tall, find the height of the cable directly above a point of the bridge deck that is 50 feet to the right of the left-hand tower. 8. What is the largest rectangular area one can enclose with 14 inches of string? 9. Solve the following quadratic equations for the indicated variable. (a) 2 - 10y2 = 0 for xv (b) y2 - 4y = 2 - 4 for x (c) 2 - mx= 1 for xv (d) -gt2+ vet + so = 0 for t (Assume g #0.) (e) y2 - 3y = 4x for y (f) y2-4y=x2 4fory 7The weight of the bridge deck forces the bridge cable into a parabola and a free hanging cable such as a power line does not form a parabola. We shall see in Exercise 12 in Section 6.5 what shape a free hanging cable makes.  140 LINEAR AND QUADRATIC FUNCTIONS 2.3.2 ANSWERS 1. (a) f(x)=2+2 No x-intercepts y-intercept (0, 2) Domain: (-oo,oo) Range: [2,oo) Decreasing on (-oc, 0] Increasing on [0, oo) Vertex (0, 2) is a minimum Axis of symmetry x= 0 (b) f (x) = -(x + 2)2 x-intercept (-2, 0) y-intercept (0, -4) Domain: (-oo,oo) Range: (-oo,0] Increasing on (-oc, -2] Decreasing on [-2, oc) Vertex (-2, 0) is a maximum Axis of symmetry x= -2 (c) f (X) = Xz2 - 2x - 8 =(X - 1)2_9 x-intercepts (-2, 0) and (4, 0) y-intercept (0, -8) Domain: (-oo,oo) Range: [-8,oo) Decreasing on (-oo,1] Increasing on [1, oo) Vertex (1, -9) is a minimum Axis of symmetry x= 1 (d) f (x) = -2(x+ 1)2 +4 x-intercepts (-1 - v2,0) and (-1 + v2,0) y-intercept (0, 2) Domain: (-oo,oo) Range: (-oo,4]) Increasing on (-oc, -1] Decreasing on [-1, oo) Vertex (-1, 4) is a maximum Axis of symmetry zx -1 y  2.3 QUADRATIC FUNCTIONS 141 (e) f (x) = -3x2 + 4x - 7 = - 3(x - 3)2 17 No x-intercepts y-intercept (0, -7) Domain: (-oo,oo) Range: (-oo, - ] Increasing on (-oc, 4] Decreasing on [4, oo) Vertex ( , - 3) is a maximum Axis of symmetry z-= (f) f(xv) = 2 - bz c- 1=(z -2 )2- c(-intercepts 1+ 2000 and -200 y-intercept (0, -1) Domain: (-oo,oo) Range: [- ,oo) Decreasing on (-o, 200 Increasing on oo) Vertex ( ,200 -40000is a minimum8 Axis of symmetry z =200 -40001 40000 )001) 2. y=|1-z2 3- -1+ 3+ -1- 3. 2 ' 2 2 ' 2 5. (a) The applied domain is [0, oc). (d) The height function is this case is s(t) = -4.9t2 + 15t. The vertex of this parabola is approximately (1.53, 11.48) so the maximum height reached by the marble is 11.48 meters. It hits the ground again when t 3.06 seconds. 8You'll need to use your calculator to zoom in far enough to see that the vertex is not the y-intercept.  142 LINEAR AND QUADRATIC FUNCTIONS (e) The revised height function is s(t) = -4.9t2 + 15t+25 which has zeros at t -1.20 and t 4.26. We ignore the negative value and claim that the marble will hit the ground after 4.26 seconds. (f) Shooting down means the initial velocity is negative so the height functions becomes s(t) = -4.9t2 - 15t + 25. 6. 500 cookies 7. Make the vertex of the parabola (0, 10) so that the point on the top of the left-hand tower where the cable connects is (-200,100) and the point on the top of the right-hand tower is (200, 100). Then the parabola is given by p(x) = 402 +10. Standing 50 feet to the right of the left-hand tower means you're standing at x= -150 and p(-150) = 60.625. So the cable is 60.625 feet above the bridge deck there. 8. The largest rectangle has area 12.25in2. 9. (a) x = tyV1/0 (d) t = vo i/v2 + 4gso 2g (b) =t(y - 2)e) 3+ /16x+9 mi m2+4 2 (c) = 2 (f) y = 2+ t  2.4 INEQUALITIES 143 2.4 INEQUALITIES In this section, not only do we develop techniques for solving various classes of inequalities analyt- ically, we also look at them graphically. The next example motivates the core ideas. EXAMPLE 2.4.1. Let f (x) = 2x - 1 and g(x) = 5. 1. Solve f(x) = g(x). 2. Solve f () < g(x). 3. Solve f () > g(x). 4. Graph y = f(x) and y = g(x) on the same set of axes and interpret your solutions to parts 1 through 3 above. SOLUTION. 1. To solve f(x) = g(x), we replace f(x) with 2x - 1 and g(x) with 5 to get 2x - 1 = 5. Solving for x, we get x = 3. 2. The inequality f(x) < g(x) is equivalent to 2x - 1 < 5. Solving gives x < 3 or (-oc, 3). 3. To find where f (x) > g(x), we solve 2x - 1 > 5. We get x> 3, or (3, oo). 4. To graph y = f(x), we graph y = 2x - 1, which is a line with a y-intercept of (0, -1) and a slope of 2. The graph of y = g(x) is y = 5 which is a horizontal line through (0, 5). x To see the connection between the graph and the algebra, we recall the Fundamental Graphing Principle for Functions in Section 1.7: the point (a, b) is on the graph of f if and only if f(a) = b. In other words, a generic point on the graph of y = f(x) is (x, f(x)), and a generic  144 LINEAR AND QUADRATIC FUNCTIONS point on the graph of y = g(x) is (x, g(x)). When we seek solutions to f(x) = g(x), we are looking for values x whose y values on the graphs of f and g are the same. In part 1, we found x = 3 is the solution to f(x) = g(x). Sure enough, f(3) = 5 and g(3) = 5 so that the point (3, 5) is on both graphs. We say the graphs of f and g intersect at (3, 5). In part 2, we set f (x) < g(x) and solved to find x < 3. For x < 3, the point (x, f (x)) is below (x, g(x)) since the y values on the graph of f are less than the y values on the graph of g there. Analogously, in part 3, we solved f(x) > g(x) and found x > 3. For x > 3, note that the graph of f is above the graph of g, since the y values on the graph of f are greater than the y values on the graph of g for those values of x. y 8- 7- 6- 5- 4- y gyfWx g =g(x) g(x) 4- 3- 2- 3 2- 1- x 1 2 3 4 -1 f(x)g(x) The preceding example demonstrates the following, which is a consequence of the Fundamental Graphing Principle for Functions. Graphical Interpretation of Equations and Inequalities Suppose f and g are functions. " The solutions to f(x) = g(x) are precisely the x values where the graphs of y = f(x) and y = g(x) intersect. " The solutions to f(x) < g(x) are precisely the x values where the graph of y = f(x) is below the graph of y = g(x). " The solutions to f(x) > g(x) are precisely the x values where the graph of y = f(x) above the graph of y = g(x). The next example turns the tables and furnishes the graphs of two functions and asks for solutions to equations and inequalities.  2.4 INEQUALITIES 145 EXAMPLE 2.4.2. The graphs of f and g are below. The graph of y = f(x) resembles the upside down V shape of an absolute value function while the graph of y = g(x) resembles a parabola. Use these graphs to answer the following questions. 1,Y g~g(x) x 1. Solve f () = g(x). 2. Solve f(x) < g(x). 3. Solve f () g(x). SOLUTION. 1. To solve f(x) = g(x), we look for where the graphs of f and g intersect. These appear to be at the points (-1, 2) and (1, 2), so our solutions to f(x) = g(x) are x= -1 and x= 1. 2. To solve f(x) < g(x), we look for where the graph of f is below the graph of g. This appears to happen for the x values less than -1 and greater than 1. Our solution is (-oc, -1)u(1, oc). 3. To solve f(x) g(x), we look for solutions to f(x) = g(x) as well as f(x) > g(x). We solved the former equation and found x = t1. To solve f(x) > g(x), we look for where the graph of f is above the graph of g. This appears to happen between x= -1 and x= 1, on the interval (-1, 1). Hence, our solution to f(x) g(x) is [-1, 1]. 4 4 / y =g(x) 3-\ y =g(x) x fWx < g(x) f(x) >_ g(x) El  146 LINEAR AND QUADRATIC FUNCTIONS We now turn our attention to solving inequalities involving the absolute value. We have the following theorem from Intermediate Algebra to help us. THEOREM 2.3. Inequalities Involving the Absolute Value: Let c be a real number. " For c > 0, x < c is equivalent to -c < x< c. " For c < 0, Ix < c has no solution. " For c > 0, x > c is equivalent to x< -c or x> c. " For c < 0, x > c is true for all real numbers. As with Theorem 2.1 in Section 2.2, we could argue Theorem 2.3 using cases. However, in light of what we have developed in this section, we can understand these statements graphically. For instance, if c > 0, the graph of y = c is a horizontal line which lies above the x-axis through (0, c). To solve Ix < c, we are looking for the x values where the graph of y = x is below the graph of y = c. We know the graphs intersect when Ix|= c, which, from Section 2.2, we know happens when x= c or x= -c. Graphing, we get We see the graph of y =Iz| is below y = c for x between -c and c, and hence we get Ix < c is equivalent to -c < x < c. The other properties in Theorem 2.3 can be shown similarly. EXAMPLE 2.4.3. Solve the following inequalities analytically; check your answers graphically. 1. Ix-11 >3 3. 2< Ix-11 <5 4. Ix + 1| > 2 2. 4-312x+11| >-2 SOLUTION. 1. To solve Ix - 1| > 3, we seek solutions to ix - 1| > 3 as well as solutions to ix - 1| = 3. From Theorem 2.3, cv - 1| > 3 is equivalent to c - 1 < -3 or x - 1 > 3. From Theorem 2.1, ix - 1| = 3 is equivalent to c - 1 = -3 or c - 1 = 3. Combining these equations with the inequalities, we solve c - 1 < -3 or c - 1 > 3. Our answer is c < -2 or c > 4, which, in interval notation is (-oc, -2] U [4, oc). Graphically, we have  2.4 INEQUALITIES 147 y -I I "" z i 6 ., - i 16 r -4 -3 -2 -1 -F i 2 3 4 5 x We see the graph of y = Ix - 1| (the V) is above the horizontal line y = 3 for x < x > 4, and, hence, this is where Ix - 1| > 3. The two graphs intersect when x= x = 4, and so we have graphical confirmation of our analytic solution. -2 and -2 and 2. To solve 4 - 312x + 1| > -2 analytically, we first isolate the absolute value before applying Theorem 2.3. To that end, we get -312x + 1| > -6 or |2x + 1| < 2. Rewriting, we now have -2 < 2x + 1 < 2 so that -2 < x < 2. In interval notation, we write (- , 1). Graphically we see the graph of y = 4 - 312x + 1| is above y = -2 for x values between -2 and 1. y 3. Rewriting the compound inequality 2 < ix - 1| < 5 as '2 < ix - 1| and Ix - 1| < 5' allows us to solve each piece using Theorem 2.3. The first inequality, 2 < Ix - 1| can be re-written as |x - 1| > 2 and so x - 1 < -2 or x - 1 > 2. We get x< -1or x> 3. Our solution to the first inequality is then (-oc, -1) U (3, oc). For Ix - 1| < 5, we combine results in Theorems 2.1 and 2.3 to get -5 x4. When variables are both inside and outside of the absolute value, it's usually best to refer to the definition of absolute value, Definition 2.4, to remove the absolute values and proceed from there. To that end, we have |x+ 1|I= -(x+ 1) if x < -1and Ix + 1| = x + 1 if x> -1. We break the inequality into cases, the first case being when x < -1. For these values of x, our inequality becomes -(x + 1)x 24. Solving, we get -2x - 2 x + 4, so that -3x > 6, which means x < -2. Since all of these solutions fall into the category x < -1, we keep them all. For the second case, we assume x> -1. Our inequality becomes x+1 2x4, which gives 2x +2 >x+ 4 or x > 2. Since all of these values of x are greater than or equal to -1, we accept all of these solutions as well. Our final answer is (-oc, -2] U [2, oc). 4.-- 3.3 -4 -3 -2 -1 1 2 3 4 We now turn our attention to quadratic inequalities. In the last example of Section 2.3, we needed to determine the solution to 2 - x - 6 < 0. We will now re-visit this problem using some of the techniques developed in this section not only to reinforce our solution in Section 2.3, but to also help formulate a general analytic procedure for solving all quadratic inequalities. If we consider f (x) = 2- x - 6 and g(x) = 0, then solving 2 - x - 6 < 0 corresponds graphically to finding the values of x for which the graph of y = f(x) =9x2 - x - 6 (the parabola) is below the graph of y = g(x) = 0 (the x-axis.) We've provided the graph again for reference.  2.4 INEQUALITIES 149 y y =2 -x-6 We can see that the graph of f does dip below the x-axis between its two x-intercepts. The zeros of f are x= -2 and x= 3 in this case and they divide the domain (the x-axis) into three intervals: (-oc, -2), (-2, 3), and (3, oc). For every number in (-oc, -2), the graph of f is above the x-axis; in other words, f (x) > 0 for all x in (-oc, -2). Similarly, f (x) < 0 for all x in (-2, 3), and f (x) > 0 for all x in (3, oo). We can schematically represent this with the sign diagram below. (+) 0 (-) 0 (+) C I I )~ -2 3 Here, the (+) above a portion of the number line indicates f(x) > 0 for those values of x; the (-) indicates f(x) < 0 there. The numbers labeled on the number line are the zeros of f, so we place 0 above them. We see at once that the solution to f(x) <0 is (-2, 3). Our next goal is to establish a procedure by which we can generate the sign diagram without graphing the function. An important property' of quadratic functions is that if the function is positive at one point and negative at another, the function must have at least one zero in between. Graphically, this means that a parabola can't be above the x-axis at one point and below the x-axis at another point without crossing the x-axis. This allows us to determine the sign of all of the function values on a given interval by testing the function at just one value in the interval. This gives us the following. 'We will give this property a name in Chapter 3 and revisit this concept then.  150 LINEAR AND QUADRATIC FUNCTIONS Steps for Solving a Quadratic Inequality 1. Rewrite the inequality, if necessary, as a quadratic function f (x) on one side of the inequality and 0 on the other. 2. Find the zeros of f and place them on the number line with the number 0 above them. 3. Choose a real number, called a test value, in each of the intervals determined in step 2. 4. Determine the sign of f(x) for each test value in step 3, and write that sign above the corresponding interval. 5. Choose the intervals which correspond to the correct sign to solve the inequality. EXAMPLE 2.4.4. Solve the following inequalities analytically using sign diagrams. Verify your answer graphically. 1. 2x2<3-x 3.92+4<12x 2. 2> 2x +1 4. 2x -2> I -1| -1 SOLUTION. 1. To solve 2x2 < 3 - x, we first get 0 on one side of the inequality which yields 2x2+ cx - 3 < 0. We find the zeros of f(x) = 2x2 + x - 3 by solving 2x2 + x - 3 = 0 for x. Factoring gives (2x + 3)(x - 1) = 0, so xv= -2,or xc= 1. We place these values on the number line with 0 above them and choose test values in the intervals (-oc, -(), (- , 1), and (1, oc). For the interval (-oo, -2), we choose2 x = -2; for (-2,1), we pick xc= 0; and for (1,oo), xc= 2. Evaluating the function at the three test values gives us f(-2) = 3 > 0 (so we place (+) above (-oo,- ); f(0) -3 < 0 (so (-) goes above the interval (- 2,1)); and, f(2) = 7 (which means (+) is placed above (1, oc)). Since we are solving 2x2 + x - 3 < 0, we look for solutions to 2x2 + x - 3 < 0 as well as solutions for 2x2 + x - 3 =0. For 2x2 + x - 3 < 0, we need the intervals which we have a (-). Checking the sign diagram, we see this is (-2, 1). We know 2x2+ c - 3 = 0 when x = - 2 and xc= 1, so or final answer is [- , 1]. To check our solution graphically, we refer to the original inequality, 2x2 < 3 - c. We let g(x) = 2x2 and h(x) = 3 - c. We are looking for the x values where the graph of g is below that of h (the solution to g(x) < h(x)) as well as the two graphs intersect (the solutions to g(x) = h(x).) The graphs of g and h are given on the right with the sign chart on the left. 2We have to choose something in each interval. If you don't like our choices, please feel free to choose different numbers. You'll get the same sign chart.  2.4 INEQUALITIES 151 (+) 0 (-) 0 (+) 4 C I I ) t -i t I1t3. -2 0 2 2.. -2 -1 1 2 x 2. Once again, we re-write 2 > 2x + 1 as 2 - 2x - 1 > 0 and we identify f(x) =9x2 - 2x - 1. When we go to find the zeros of f, we find, to our chagrin, that the quadratic 2 - 2x - 1 doesn't factor nicely. Hence, we resort to the quadratic formula to solve 2 - 2x - 1 =0, and arrive at x= 1 i 2. As before, these zeros divide the number line into three pieces. To help us decide on test values, we approximate 1 - v2~ -0.4 and 1+ v2~ 2.4. We choose x =-1, x = 0, and x = 3 as our test values and find f(-1) = 2, which is (+); f(0) =--1 which is (-); and f(3) = 2 which is (+) again. Our solution to 2 - 2x - 1 > 0 is where we have (+), so, in interval notation (-oc, 1 - 2) U (1 + 2, oo). To check the inequality 2 > 2x + 1 graphically, we set g(x) = x2 and h(x) = 2x + 1. We are looking for the x values where the graph of g is above the graph of h. As before we present the graphs on the right and the sign chart on the left. y 8-- 7- 6-- (+) 0 (-) 0 (+) CI I )4 - t 1 - 2 t 1 + 2 t 3- -1 0 3 2 -1 1 2 3 x 3. To solve 92+4 < 12x, as before, we solve 9x2-12x+4 < 0. Setting f (x) = 9X2-12x+4 = 0, we find the only one zero of f, x = . This one x value divides the number line into two intervals, from which we choose x= 0 and x= 1 as test values. We find f (0) = 4 > 0 and f(1) = 1 > 0. Since we are looking for solutions to 9x2 - 12x + 4 < 0, we are looking for  152 LINEAR AND QUADRATIC FUNCTIONS x values where 92 - 12x + 4 < 0 as well as where 92 - 12x + 4 = 0. Looking at our sign diagram, there are no places where 92 - 12x + 4 < 0 (there are no (-)), so our solution is only x = (where 9x2 - 12x + 4 = 0). We write this as {3}. Graphically, we solve 92 + 4 < 12x by graphing g(x) = 92 + 4 and h(x) = 12x. We are looking for the x values where the graph of g is below the graph of h (for 92 + 4 < 12x) and where the two graphs intersect (92+ 4 = 12x). We see the line and the parabola touch at (3, 8), but the parabola is always above the line otherwise.3 13-- 12- 11-- 10-- 9- (+) 0 (+)8 I 7 t 3 t 6 O 1 4 3-- 2-- 1-- -1 1 X 4. To solve our last inequality, 2x - 2 > x - 1 - 1, we re-write the absolute value using cases. For x < 1, Iz- 1|= -(x -1) = 1-x, so we get 2x -z2 1 - x-l1, or x2 -3x <0. Finding the zeros of f(x) =9x2 -3x, we get x= 0 and x= 3. However, we are only concerned with the portion of the number line where x < 1, so the only zero that we concern ourselves with is x= 0. This divides the interval x < 1 into two intervals: (-oo, 0) and (0, 1). We choose x = -1 and x= 2 as our test values. We find f(-1) = 4 and f (1) =_-4. Solving 2 - 3x < 0 for x < 1 gives us [0, 1). Next, we turn our attention to the case x > 1. Here, x - 1| = x- 1, so our original inequality becomes 2x - 22 x - 1 - 1, or2 - x - 2 < 0. Setting g(x) =9x2 - x - 2, we find the zeros of g to be x= -1 and x= 2. Of these, only x= 2 lies in the region x 1, so we ignore x= -1. Our test intervals are now [1,2) and (2,oo). We choose x= 1 and x= 3 as our test values and find g(1) = -2 and g(3) = 4. To solve g(x) < 0, we have [1, 2). (+) 0 (-) (-) 0 (+) t 06t 1 2 -1 1 3 Combining these into one sign diagram, we get our solution is [0, 2]. Graphically, to check 2x-9>2 x z-1| - 1, we set h(x) = 2x -9z and i(x) =xz-1| - 1 and look for thexzvalues 3In this case, we say the line y =12x is tangent to y =9x2 + 4 at (j, 8). Finding tangent lines to arbitrary functions is the stuff of legends, I mean, Calculus.  2.4 INEQUALITIES 153 where the graph of h is above the the graph of i (the solution of h(x) i-coordinates of the intersection points of both graphs (where h(x) = sign chart is given on the left and the graphs are on the right. > i(x)) as well as the i(x)). The combined y (+) 0 (-) 0 (+) 0 2 -1 0 3 x H It is quite possible to encounter inequalities where the analytical methods developed so far will fail us. In this case, we resort to using the graphing calculator to approximate the solution, as the next example illustrates. EXAMPLE 2.4.5. Suppose the revenue R, in thousands of dollars, from producing and selling x hundred LCD TVs is given by R(x) = -5x3 + 35x2 + 155x for i > 0, while the cost, in thousands of dollars, to produce i hundred LCD TVs is given by C(x) = 200x + 25 for i > 0. How many TVs, to the nearest TV, should be produced to make a profit? SOLUTION. Recall that profit = revenue - cost. If we let P denote the profit, in thousands of dollars, which results from producing and selling x hundred TVs then P(x) = R(x) - (x) = (-5x3 + 35x2 + 155 ) - (200x + 25) -5x3 + 35x2 - 451 - 25, where i > 0. If we want to make a profit, then we need to solve P(x) > 0; in other words, -5x3 + 35x2 - 451 - 25 > 0. We have yet to discuss how to go about finding the zeros of P, let alone making a sign diagram for such an animal,4 as such we resort to the graphing calculator. After finding a suitable window, we get WINDOW Xmin=0 Xmax=6 Xsc1=1 Ymin= -100 Ymax=100 Ysc1=100 Xres=1 We are looking for the i values for which P(x) > 0, that is, where the graph of P is above the i-axis. We make use of the 'Zero' command and find two i-intercepts. 4The procedure, as we shall see in Chapter 3 is identical to what we have developed here.  154 LINEAR AND QUADRATIC FUNCTIONS H=2.41423ii Y=1E -11 X=5 Y=3E -11 We remember that x denotes the number of TVs in hundreds, so if we are to find our solution using the calculator, we need our answer to two decimal places. The zero5 2.414 ... corresponds to 241.4... TVs. Since we can't make a fractional part of a TV, we round this up to 242 TVs.7 The other zero seems dead on at 5, which corresponds to 500 TVs. Hence to make a profit, we should produce (and sell) between 242 and 499 TVs, inclusive. Q Our last example in the section demonstrates how inequalities can be used to describe regions in the plane, as we saw earlier in Section 1.2. EXAMPLE 2.4.6. Sketch the following relations. 1. R={(x,y):y> x}. 2. S={(x,y):y<2-x}. 3. T={(x,y): x 0 so we need to round up to 242 in order to make a profit.  2.4 INEQUALITIES 155 3. Finally, the relation T takes the points whose y-coordinates satisfy both the conditions in R and S. So we shade the region between y = x and y = 2 -9x2, keeping those points on the parabola, but not the points on y z. To get an accurate graph, we need to find where these two graphs intersect, so we set |xl= 2 - 2. Proceeding as before, breaking this equation into cases, we get x= -1, 1. Graphing yields y ;jjjj... F, 1,: \ ,' -2 -1 I 2 x -1- The graph of T El  156 LINEAR AND QUADRATIC FUNCTIONS 2.4.1 EXERCISES 1. Solve the inequality. Express your answer in interval form. (a) |3x-51 <4 (b) |7x + 2| > 10 (c)G < 2x - 9 <3 (d) | - 2x + 1| > z + 5 (e) |x + 3| < |6x + 9| (f) 2 + 2x - 3 > 0 (g) 2 + 1 <0 (h) 3x2 < llx + 4 (i) x > 2 (j) 2 < 92 - 9 <9 (k) 2 < 4x - 3| (1) 2 + z + 1 > 0 2. Prove the second, third and fourth parts of Theorem 2.3. 3. If a slingshot is used to shoot a marble straight up into the air from 2 meters above the ground with an initial velocity of 30 meters per second, for what values of time t will the marble be over 35 meters above the ground? (Refer to Exercise 5 in Section 2.3 for assistance if needed.) Round your answers to two decimal places. 4. What temperature values in degrees Celsius are equivalent to the temperature range 50°F to 95 F? (Refer to Exercise 3 in Section 2.1 for assistance if needed.) 5. The surface area S of a cube with edge length x is given by S(x) = 6x2 for x > 0. Suppose the cubes your company manufactures are supposed to have a surface area of exactly 42 square centimeters, but the machines you own are old and cannot always make a cube with the precise surface area desired. Write an inequality using absolute value that says the surface area of a given cube is no more than 3 square centimeters away (high or low) from the target of 42 square centimeters. Solve the inequality and express your answer in interval form. 6. Sketch the following relations. (a) R ={(x,y):y < -1} (b) R = {(x,y): y > x2+ 1} (c) R ={(x,y):-1 1 is a natural number.a The domain of a polynomial function is (-oc, oc). aRecall this means n is a 'counting number' n = 1, 2, 3, ... There are several things about Definition 3.1 that may be off-putting or downright frightening. The best thing to do is look at an example. Consider f(x) = 4x5 - 3x2 + 2x - 5. Is this a polynomial function? We can re-write the formula for f as f (x) = 4x5 + Ox4 + Ox3 + (-3)2 + 2x + (-5). Comparing this with Definition 3.1, we identify n = 5, a5 = 4, a4 = 0, a3= 0, a2 = -3, a1= 2, and ao = -5. In other words, a5 is the coefficient of x5, a4 is the coefficient of x4, and so forth; the subscript on the a's merely indicates to which power of x the coefficient belongs. The business of restricting n to be a natural number lets us focus on well-behaved algebraic animals.1 EXAMPLE 3.1.1. Determine if the following functions are polynomials. Explain your reasoning. 1. g(x) 4+x3 x 2. p(x) 4x+x3 x 4x + x3 3. q(x) = 'Enjoy this while it lasts. Before we're through with the book, you'll have been exposed to the most terrible of algebraic beasts. We will tame them all, in time.  170 POLYNOMIAL FUNCTIONS 4. f(x)=3x 5. h(x)= x 6. z(x)=0 SOLUTION. 1. We note directly that the domain of g(x) 23+4 is x # 0. By definition, a polynomial has all real numbers as its domain. Hence, g can't be a polynomial. 2. Even though p(x) = x3+4x simplifies to p(x) =9x2 + 4, which certainly looks like the form given in Definition 3.1, the domain of p, which, as you may recall, we determine before we simplify, excludes 0. Alas, p is not a polynomial function for the same reason g isn't. 3. After what happened with p in the previous part, you may be a little shy about simplifying q(x) = "2+4 to q(x) = z, which certainly fits Definition 3.1. If we look at the domain of q before we simplified, we see that it is, indeed, all real numbers. A function which can be written in the form of Definition 3.1 whose domain is all real numbers is, in fact, a polynomial. 4. We can rewrite f(x) = cv as f(x) = xz. Since } is not a natural number, f is not a polynomial. 5. The function h(x) =_Iz| isn't a polynomial, since it can't be written as a combination of powers of x (even though it can be written as a piecewise function involving polynomials.) As we shall see in this section, graphs of polynomials possess a quality2 that the graph of h does not. 6. There's nothing in Definition 3.1 which prevents all the coefficients an, etc., from being 0. Hence, z(x) = 0, is an honest-to-goodness polynomial. DEFINITION 3.2. Suppose f is a polynomial function. " Given f (x) = anz" + an_1x"-1 + ... a2x2 + a1x + ao with an # 0, we say - The natural number n is called the degree of the polynomial f. - The term anz" is called the leading term of the polynomial f. - The real number an is called the leading coefficient of the polynomial f. - The real number ao is called the constant term of the polynomial f. " If f () = a0, and ao # 0, we say f has degree 0. " If f () = 0, we say f has no degree.a "Some authors say f(x) =0 has degree -oo for reasons not even we will go into. 2One which really relies on Calculus to verify.  3.1 GRAPHS OF POLYNOMIALS 171 The reader may well wonder why we have chosen to separate off constant functions from the other polynomials in Definition 3.2. Why not just lump them all together and, instead of forcing n to be a natural number, n = 1, 2, ..., let n be a whole number, n = 0, 1, 2,.... We could unify all the cases, since, after all, isn't ao0 = ao? The answer is 'yes, as long as x $ 0.' The function f(x) = 3 and g(x) = 3x° are different, because their domains are different. The number f(0) = 3 is defined, whereas g(0) = 3(0)0 is not.3 Indeed, much of the theory we will develop in this chapter doesn't include the constant functions, so we might as well treat them as outsiders from the start. One good thing that comes from Definition 3.2 is that we can now think of linear functions as degree 1 (or 'first degree') polynomial functions and quadratic functions as degree 2 (or 'second degree') polynomial functions. EXAMPLE 3.1.2. Find the degree, leading term, leading coefficient and constant term of the following polynomial functions. 1. f(x) =4x5-3x2+2x-5 3. h(x) = 4-x 5 2. g(x) = 12x + x3 4. p(x) (2x - 1)3(x - 2)(3x + 2) SOLUTION. 1. There are no surprises with f (x) = 4x5 - 3x2 + 2x - 5. It is written in the form of Definition 3.2, and we see the degree is 5, the leading term is 4x5, the leading coefficient is 4 and the constant term is -5. 2. The form given in Definition 3.2 has the highest power of 3 first. To that end, we re-write g(x) = 12x + -3 = x3 + 12x, and see the degree of g is 3, the leading term is x3, the leading coefficient is 1 and the constant term is 0. 3. We need to rewrite the formula for h so that it resembles the form given in Definition 3.2: h(x) = 4-x -=5 - 5=-}z + j. We see the degree of h is 1, the leading term is -}z:, the leading coefficient is -1 and the constant term is 5. 4. It may seem that we have some work ahead of us to get p in the form of Definition 3.2. However, it is possible to glean the information requested about p without multiplying out the entire expression (2x - 1)3(x - 2)(3x + 2). The leading term of p will be the term which has the highest power of 3. The way to get this term is to multiply the terms with the highest power of 3 from each factor together - in other words, the leading term of p(x) is the product of the leading terms of the factors of p(x). Hence, the leading term of p is (2x)3(x)(3x) = 24x5. This means the degree of p is 5 and the leading coefficient is 24. As for the constant term, we can perform a similar trick. The constant term is obtained by multiplying the constant terms from each of the factors (-1)3(-2)(2) =4. 3Technically, 00 is an indeterminant form, which is a special case of being undefined. The authors realize this is beyond pedantry, but we wouldn't mention it if we didn't feel it was neccessary.  172 POLYNOMIAL FUNCTIONS Our next example shows how polynomials of higher degree arise 'naturally'4 in even the most basic geometric applications. EXAMPLE 3.1.3. A box with no top is to be fashioned from a 10 inch x 12 inch piece of cardboard by cutting out congruent squares from each corner of the cardboard and then folding the resulting tabs. Let x denote the length of the side of the square which is removed from each corner. .I I 12 in - height . depth ---- --- --- w idth 10 in 1. Find the volume V of the box as a function of x. Include an appropriate applied domain. 2. Use a graphing calculator to graph y = V(x) on the domain you found in part 1 and approx- imate the dimensions of the box with maximum volume to two decimal places. What is the maximum volume? SOLUTION. 1. From Geometry, we know Volume = width x height x depth. The key is to now find each of these quantities in terms of x. From the figure, we see the height of the box is x itself. The cardboard piece is initially 10 inches wide. Removing squares with a side length of x inches from each corner leaves 10 - 2x inches for the width.5 As for the depth, the cardboard is initially 12 inches long, so after cutting out x inches from each side, we would have 12 - 2x inches remaining. As a function6 of x, the volume is V(x) = x(10 - 2x)(12 - 2x) = 4x3 44x2 + 120x To find a suitable applied domain, we note that to make a box at all we need x > 0. Also the shorter of the two dimensions of the cardboard is 10 inches, and since we are removing 2x inches from this dimension, we also require 10 - 2x > 0 or x < 5. Hence, our applied domain is 0 0 a<0 We now turn our attention to functions of the form f (x) = x where n > 3 is an odd natural number.'3 Below we have graphed y = x3, y =x5, and y = x7. The 'flattening' and 'steepening' that we saw with the even powers presents itself here as well, and, it should come as no surprise that all of these functions are odd.14 The end behavior of these functions is all the same, with f(x) - -oo as x - -oo and f(x) - oo as x - oo. =X =X =X As with the even degreed functions we studied earlier, we can generalize their end behavior. End Behavior of functions f(x)= ax", rn odd. Suppose f(x) = ax" where a # 0 is a real number and n> 3 is an odd natural number. The end behavior of the graph of y = f(x) matches one of the following: a>0 a<0 Despite having different end behavior, all functions of the form f(x) = ax" for natural numbers n share two properties which help distinguish them from other animals in the algebra zoo: they are continuous and smooth. While these concepts are formally defined using Calculus,-j informally, graphs of continuous functions have no 'breaks' or 'holes' in their graphs, and smooth functions have no 'sharp turns.' It turns out that these traits are preserved when functions are added together, so general polynomial functions inherit these qualities. Below we find the graph of a function which is 13We ignore the case when n = 1, since the graph of f(x) = x is a line and doesn't fit the general pattern of higher-degree odd polynomials. 14And are, perhaps, the inspiration for the moniker 'odd function'. 15In fact, if you take Calculus, you'll find that smooth functions are automatically continuous, so that saying 'polynomials are continuous and smooth' is redundant.  3.1 GRAPHS OF POLYNOMIALS 175 neither smooth nor continuous, and to its right we have a graph of a polynomial, for comparison. The function whose graph appears on the left fails to be continuous where it has a 'break' or 'hole' in the graph; everywhere else, the function is continuous. The function is continuous at the 'corner' and the 'cusp', but we consider these 'sharp turns', so these are places where the function fails to be smooth. Apart from these four places, the function is smooth and continuous. Polynomial functions are smooth and continuous everywhere, as exhibited in graph on the right. 'hole' 'corner' | 'cusp' 'break' Pathologies not found on graphs of polynomials The graph of a polynomial The notion of smoothness is what tells us graphically that, for example, f(x) = Ix, whose graph is the characteric 'V' shape, cannot be a polynomial. The notion of continuity is what allowed us to construct the sign diagram for quadratic inequalities as we did in Section 2.4. This last result is formalized in the following theorem. THEOREM 3.1. The Intermediate Value Theorem (Polynomial Zero Version): If f is a polynomial where f(a) and f(b) have different signs, then f has at least one zero between x= a and x= b; that is, for at least one real number c such that a < c < b, we have f(c) = 0. The Intermediate Value Theorem is extremely profound; it gets to the heart of what it means to be a real number, and is one of the most oft used and under appreciated theorems in Mathematics. With that being said, most students see the result as common sense, since it says, geometrically, that the graph of a polynomial function cannot be above the x-axis at one point and below the x-axis at another point without crossing the x-axis somewhere in between. The following example uses the Intermediate Value Theorem to establish a fact that that most students take for granted. Many students, and sadly some instructors, will find it silly. EXAMPLE 3.1.4. Use the Intermediate Value Theorem to establish that 2 is a real number. SOLUTION. Consider the polynomial function f (x) 9= 2 - 2. Then f (1) = -1 and f (3) = 9. Since f (1) and f(3) have different signs, the Intermediate Value Theorem guarantees us a real number c between 1 and 3 with f(c) = 0. If c2 - 2 = 0 then c = i2. Since c is between 1 and 3, c is positive, so c =V2. Our primary use of the Intermediate Value Theorem is in the construction of sign diagrams, as in Section 2.4, since it guarantees us that polynomial functions are always positive (+) or always neg- ative (-) on intervals which do not contain any of its zeros. The general algorithm for polynomials is given below.  176 POLYNOMIAL FUNCTIONS Steps for Constructing a Sign Diagram for a Polynomial Function Suppose f is a polynomial function. 1. Find the zeros of f and place them on the number line with the number 0 above them. 2. Choose a real number, called a test value, in each of the intervals determined in step 1. 3. Determine the sign of f(x) for each test value in step 2, and write that sign above the corresponding interval. EXAMPLE 3.1.5. Construct a sign diagram for f(x) =x3( - 3)2(X + 2) (2 + 1). Use it to give a rough sketch of the graph of y = f(x). SOLUTION. First, we find the zeros of f by solving x3(x - 3)2(X + 2) (2 + 1) = 0. We get x= 0, x = 3, and x =-2. (The equation 2 + 1 = 0 produces no real solutions.) These three points divide the real number line into four intervals: (-oc, -2), (-2, 0), (0, 3) and (3, oc). We select the test values x= -3, x= -1, x= 1, and x= 4. We find f (-3) is (+), f (-1) is (-) and f (1) is (+) as is f(4). Wherever f is (+), its graph is above the x-axis; wherever f is (-), its graph is below the x-axis. The x-intercepts of the graph of f are (-2, 0), (0, 0) and (3, 0). Knowing f is smooth and continuous allows us to sketch its graph. y (+) 0 (-) 0 (+) 0 (+) -2 0 3 t -3 -1 1 4 Asketch ofy =f(x) A couple of notes about the Example 3.1.5 are in order. First, note that we purposefully did not label the y-axis in the sketch of the graph of y= f (x). This is because the sign diagram gives us the zeros and the relative position of the graph - it doesn't give us any information as to how high or low the graph strays from the x-axis. Furthermore, as we have mentioned earlier in the text, without Calculus, the values of the relative maximum and minimum can only be found approximately using a calculator. If we took the time to find the leading term of f, we would find it to be x8. Looking at the end behavior of f, we notice it matches the end behavior of y = x8. This is no accident, as we find out in the next theorem. THEOREM 3.2. End Behavior for Polynomial Functions: The end behavior of a polynomial f (x) = anx + an_1x--1 + . . . a2x2 + a1x + ao with an - 0 matches the end behavior of y = anz" To see why Theorem 3.2 is true, let's first look at a specific example. Consider f(x) =4x3 - -+,5. If we wish to examine end behavior, we look to see the behavior of f as x -~ io. Since we're concerned with z's far down the x-axis, we are far away from x 0 and so can rewrite f~x) for these values of x as  3.1 GRAPHS OF POLYNOMIALS 177 f(x) = 4x3 1 1 5 4x2 +4x3 As x becomes unbounded (in either direction), the terms { and 43 0, as the table below indicates. become closer and closer to 1 5 4x2 4x3 -1000 0.00000025 -0.00000000125 -100 0.000025 -0.00000125 -10 0.0025 -0.00125 10 0.0025 0.00125 100 0.000025 0.00000125 1000 0.00000025 0.00000000125 In other words, as x - too, f(x) ~ 4x3 (1 - 0 +0) = 4x3, which is the leading term of f. The formal proof of Theorem 3.2 works in much the same way. Factoring out the leading term leaves fa(x) = anz 1 + an-1 a2 a1 ao anx anxn-2 anxn-1 anxn As x - too, any term with an x in the denominator becomes closer and closer to 0, and we have f(x) ~ anz". Geometrically, Theorem 3.2 says that if we graph y = f(x), say, using a graphing calculator, and continue to 'zoom out,' the graph of it and its leading term become indistinguishable. Below are the graphs of y = 4x3 - x + 5 (the thicker line) and y = 4x3 (the thinner line) in two different windows. A view 'close' to the origin. A 'zoomed out' view. Let's return to the function in Example 3.1.5, f(x) = x3(x -3)2(x+2) (x2 + 1), whose sign diagram and graph are reproduced below for reference. Theorem 3.2 tells us that the end behavior is the same as that of its leading term, x8. This tells us that the graph of y = f(x) starts and ends above the i-axis. In other words, f(x) is (+) as x - too, and as a result, we no longer need to evaluate f at the test values x = -3 and x = 4. Is there a way to eliminate the need to evaluate f at the other test values? What we would really need to know is how the function behaves near its zeros - does it cross through the i-axis at these points, as it does at x = -2 and x = 0, or does it simply touch and rebound like it does at x = 3. From the sign diagram, the graph of f will cross the  178 POLYNOMIAL FUNCTIONS x-axis whenever the signs on either side of the zero switch (like they do at x= -2 and x= 0); it will touch when the signs are the same on either side of the zero (as is the case with x= 3). What we need to determine is the reason behind whether or not the sign change occurs. y (+) 0 (-) 0 (+) 0 (+) CII IIF -2 0 3 -3 -1 1 4 A sketch of y = f(x) Fortunately, f was given to us in factored form: f(x) = x3(x - 3)2(x + 2). When we attempt to determine the sign of f(-4), we are attempting to find the sign of the number (-4)3(_7)2(_2), which works out to be (-)(+)(-) which is (+). If we move to the other side of x= -2, and find the sign of f(-1), we are determining the sign of (-1)3(_4)2(+1), which is (-)(+)(+) which gives us the (-). Notice that signs of the first two factors in both expressions are the same in f(-4) and f(-1). The only factor which switches sign is the third factor, (x + 2), precisely the factor which gave us the zero x= -2. If we move to the other side of 0 and look closely at f(1), we get the sign pattern (+1)3(_2)2(+3) or (+)(+)(+) and we note that, once again, going from f(-1) to f(1), the only factor which changed sign was the first factor, x3, which corresponds to the zero x= 0. Finally, to find f(4), we substitute to get (+4)3(+2)2(+5) which is (+)(+)(+) or (+). The sign didn't change for the middle factor (x - 3)2. Even though this is the factor which corresponds to the zero x = 3, the fact that the quantity is squared kept the sign of the middle factor the same on either side of 3. If we look back at the exponents on the factors (x + 2) and x3, we note they are both odd - so as we substitute values to the left and right of the corresponding zeros, the signs of the corresponding factors change which results in the sign of the function value changing. This is the key to the behavior of the function near the zeros. We need a definition and then a theorem. DEFINITION 3.3. Suppose f is a polynomial function and m is a natural number. If (x - c)m is a factor of f(x) but (x - c)m+l is not, then we say x= c is a zero of multiplicity m. Hence, rewriting f(x) =sx3(z - 3)2(x+ 2) as f(x) = (x - 0)3(x - 3)2(X - (-2))1, we see that x= 0 is a zero of multiplicity 3, x = 3 is a zero of multiplicity 2, and x =-2 is a zero of multiplicity 1. THEOREM 3.3. The Role of Multiplicity: Suppose f is a polynomial function and x= c is a zero of multiplicity m. * If m is even, the graph of y =f(x) touches and rebounds from the x-axis as (c, 0). * If m is odd, the graph of y =f(x) crosses through the x-axis as (c, 0).  3.1 GRAPHS OF POLYNOMIALS 179 Our last example shows how end behavior and multiplicity allow us to sketch a decent graph without appealing to a sign diagram. EXAMPLE 3.1.6. Sketch the graph of f(x) -3(2x - 1)(x + 1)2 using end behavior and the multiplicity of its zeros. SOLUTION. The end behavior of the graph of f will match that of its leading term. To find the leading term, we multiply by the leading terms of each factor to get (-3)(2x)(x)2 = -6x3. This tells us the graph will start above the x-axis, in Quadrant II, and finish below the x-axis, in Quadrant IV. Next, we find the zeros of f. Fortunately for us, f is factored.16 Setting each factor equal to zero gives is x 1= and x= -1 as zeros. To find the multiplicity of x 1= we note that it corresponds to the factor (2x - 1). This isn't strictly in the form required in Definition 3.3. If we factor out the 2, however, we get (2x - 1) = 2 (x - 2), and we see the multiplicity of x= 2 is 1. Since 1 is an odd number, we know from Theorem 3.3 that the graph of f will cross through the x-axis at (,0). Since the zero x= -1 corresponds to the factor (x + 1)2 = (x - (-1))2, we see its multiplicity is 2 which is an even number. As such, the graph of f will touch and rebound from the x-axis at (-1, 0). Though we're not asked to, we can find the y-intercept by finding f(0) = -3(2(0) - 1)(0 + 1)2 = 3. Thus (0, 3) is an additional point on the graph. Putting this together gives us the graph below. 1 $ 1D 16Obtaining the factored form of a polynomial is the main focus of the next few sections.  180 POLYNOMIAL FUNCTIONS 3.1.1 EXERCISES 1. For each polynomial given below, find the degree, the leading term, the leading coefficient, the constant term and the end behavior. (a) f(x) 3v57 + 22.5x10 - 7x7 + 3 (d) s(t) -4.9t2 + vet + so (b) p(t) -t2(3 - 5t)(t2 + t + 4) (e) P(x) = (x - 1)( - 2)(x - 3)(x - 4) (c) Z(b) = 42b - b3 (f) q(r) = 1 - 16r4 2. For each polynomial given below, find its real zeros and their corresponding multiplicities. Use this information along with a sign chart to provide a rough sketch of the graph of the polynomial. (a) a(x) x= z( + 2)2 (d) Z(b) = b(42 - b2) (b) F(x) x=z3 -+ 2)2 (e) Q(x) (x + 5)2(x - 3)4 (c) P(x) (x - 1)(x - 2)(x - 3)(x - 4) (f) g(x) =x(x + 2)3 3. According to US Postal regulations, a rectangular shipping box must satisfy the inequality "Length + Girth < 130 inches" for Parcel Post and "Length + Girth < 108 inches" for other services.17 Let's assume we have a closed rectangular box with a square face of side length x as drawn below. The length is the longest side and is clearly labeled. The girth is the distance around the box in the other two dimensions so in our case it is the sum of the four sides of the square, 4x. (a) Assuming that we'll be mailing the box Parcel Post, express the length of the box in terms of x and then express the volume, V, of the box in terms of x. (b) Find the dimensions of the box with maximum volume that can be shipped via Parcel Post. (c) Repeat parts 3a and 3b above assuming that we'll be shipping the box using "other services". 17See here for details.  3.1 GRAPHS OF POLYNOMIALS 181 4. Use transformations to sketch the graphs of the following polynomials. (a) f (x) = (x+ 2)3 + 1 (c) h(x) = -zX5-3 (b) g(x) = (x + 2)4 + 1 (d) j(x) = 2 - 3(x - 1)4 5. Use the Intermediate Value Theorem to find intervals of length 1 which contain the real zeros of f (X) = z3 - 9x + 5. 6. The original function used to model the cost of producing x PortaBoys Game Systems given in Example 2.1.5 was C(x) = 80x+150. While developing their newest game, Sasquatch Attack!, the makers of the PortaBoy revised their cost function using a cubic polynomial. The new cost of producing x PortaBoys is given by C() =.03x3 - 4.52 + 225x + 250. Market research indicates that the demand function p(x) -1.5x + 250 remains unchanged. Find the production level x that maximizes the profit made by producing and selling x PortaBoys. 7. We now revisit the data set from Exercise 4b in Section 2.5. In that exercise, you were given a chart of the number of hours of daylight they get on the 21st of each month in Fairbanks, Alaska based on the 2009 sunrise and sunset data found on the U.S. Naval Observatory web- site. We let x= 1 represent January 21, 2009, x= 2 represent February 21, 2009, and so on. The chart is given again for reference. Month Number 1 2 3 4 5 6 7 8 9 10 11 12 Hours of Daylight 5.8 9.3 12.4 15.9 19.4 21.8 19.4 15.6 12.4 9.1 5.6 3.3 Find cubic (third degree) and quartic (fourth degree) polynomials which model this data and comment on the goodness of fit for each. What can we say about using either model to make predictions about the year 2020? (Hint: Think about the end behavior of polynomials.) Use the models to see how many hours of daylight they got on your birthday and then check the website to see how accurate the models are. Knowing that Sasquatch are largely nocturnal, what days of the year according to your models are going to allow for at least 14 hours of darkness for field research on the elusive creatures? 8. An electric circuit is built with a variable resistor installed. For each of the following resis- tance values (measured in kilo-ohms, kQ), the corresponding power to the load (measured in milliwatts, mW) is given in the table below. 18 Resistance: (kQ) 1.012 2.199 3.275 4.676 6.805 9.975 Power: (mW) 1.063 1.496 1.610 1.613 1.505 1.314 (a) Make a scatter diagram of the data using the Resistance as the independent variable and Power as the dependent variable. 18The authors wish to thank Don Anthan and Ken White of Lakeland Community College for devising this problem and generating the accompanying data set.  182 POLYNOMIAL FUNCTIONS (b) Use your calculator to find quadratic (2nd degree), cubic (3rd degree) and quartic (4th degree) regression models for the data and judge the reasonableness of each. (c) For each of the models found above, find the predicted maximum power that can be delivered to the load. What is the corresponding resistance value? (d) Discuss with your classmates the limitations of these models - in particular, discuss the end behavior of each. 9. Show that the end behavior of a linear function f(x) = mx + b is as it should be according to the results we've established in the section for polynomials of odd degree. (That is, show that the graph of a linear function is "up on one side and down on the other" just like the graph of y = anx for odd numbers n.) 10. There is one subtlety about the role of multiplicity that we need to discuss further; specifically we need to see 'how' the graph crosses the x-axis at a zero of odd multiplicity. In the section, we deliberately excluded the function f(x) = x from the discussion of the end behavior of f (x) =x" for odd numbers n and we said at the time that it was due to the fact that f (x) = x didn't fit the pattern we were trying to establish. You just showed in the previous exercise that the end behavior of a linear function behaves like every other polynomial of odd degree, so what doesn't f (x) = x do that g(x) = x3 does? It's the 'flattening' for values of x near zero. It is this local behavior that will distinguish between a zero of multiplicity 1 and one of higher odd multiplicity. Look again closely at the graphs of a(x) = x(x + 2)2 and F(x)= x3(x + 2)2 from Exercise 2. Discuss with your classmates how the graphs are fundamentally different at the origin. It might help to use a graphing calculator to zoom in on the origin to see the different crossing behavior. Also compare the behavior of a(x) =x(x + 2)2 to that of g(x) = x(x + 2)3 near the point (-2, 0). What do you predict will happen at the zeros of f(x) =_(x - 1)(x - 2)2(x - 3)3(x - 4)4(x -5)5? 11. Here are a few other questions for you to discuss with your classmates. (a) How many local extrema could a polynomial of degree n have? How few local extrema can it have? (b) Could a polynomial have two local maxima but no local minima? (c) If a polynomial has two local maxima and two local minima, can it be of odd degree? Can it be of even degree? (d) Can a polynomial have local extrema without having any real zeros? (e) Why must every polynomial of odd degree have at least one real zero? (f) Can a polynomial have two distinct real zeros and no local extrema? (g) Can an x-intercept yield a local extrema? Can it yield an absolute extrema? (h) If the y-intercept yields an absolute minimum, what can we say about the degree of the polynomial and the sign of the leading coefficient?  3.1 GRAPHS OF POLYNOMIALS 183 3.1.2 ANSWERS 1. (a) f (x) =3x17 + 22.5x10 - 7x7 + 3 Degree 17 Leading term 517 Leading coefficient 3 Constant term j As x - -oo, f(x) -- -oo As x oo, f (x) -- oo (b) p(t) = -t2(3 - 5t)(t2 + t + 4) Degree 5 Leading term 5t5 Leading coefficient 5 Constant term 0 As t - -oo, p(t) - -oo As t - oo, p(t) -- oo (d) s(t) = -4.9t2 + vet + so Degree 2 Leading term -4.9t2 Leading coefficient -4.9 Constant term so As t - -oo, s(t) - -o As t - oo, s(t) - -o (e) P(x) = (x - 1)(x - 2)(x - 3)(x - 4) Degree 4 Leading term x4 Leading coefficient 1 Constant term 24 As x - -oo, P(x) - o0 As x - oo, P(x) - o0 (c) Z(b) = 42b - b3 Degree 3 Leading term -b3 Leading coefficient -1 Constant term 0 As b- -oo, Z(b) --coo As b- oo, Z(b) ---oo 2. (a) a(x) - x(z + 2)2 x = 0 multiplicity 1 x =-2 multiplicity 2 (f) q(r) = 1 - 16r4 Degree 4 Leading term -16r4 Leading coefficient -16 Constant term 1 As r - -oo, q(r) - -oo As r oo, q(r) -- -oo (b) F(x)=x3(x + 2)2 x = 0 multiplicity 3 x =-2 multiplicity 2  184 POLYNOMIAL FUNCTIONS (c) P(x) = (x - 1)(x - 2)(x - 3)(x - 4) x = 1 multiplicity 1 x = 2 multiplicity 1 x = 3 multiplicity 1 x = 4 multiplicity 1 (e) Q(x) = (xz+ 5)2(x - 3)4 x = -5 multiplicity 2 x = 3 multiplicity 4 y (f) g(x) -x(x + 2)3 x - 0 multiplicity 1 x = -2 multiplicity 3 (d) Z(b) = b(42 - b2) b = -v 42multiplicity 1 b = 0 multiplicity 1 b = 42 multiplicity 1 y Z x 3. (a) Our ultimate goal is to maximize the volume, so we'll start with the maximum Length + Girth of 130. This means the length is 130 - 4x. The volume of a rectangular box is always length x width x height so we get V(x) = 2(130 - 4x) = -4x3 + 130x2. (b) Graphing y = V(x) on [0, 33] x [0, 21000] shows a maximum at (21.67, 20342.59) so the dimensions of the box with maximum volume are 21.67in. x 21.67in. x 43.32in. for a volume of 20342.59in.3. (c) If we start with Length + Girth = 108 then the length is 108 - 4x and the volume is V(x) = -4x3 + 108x2. Graphing y = V(x) on [0, 27] x [0,11700] shows a max- imum at (18.00, 11664.00) so the dimensions of the box with maximum volume are 18.OOin. x 18.OOin. x 36in. for a volume of 11664.OOin.3. (Calculus will confirm that the measurements which maximize the volume are exactly 18in. by 18in. by 36in., however, as I'm sure you are aware by now, we treat all calculator results as approximations and list them as such.)  3.1 GRAPHS OF POLYNOMIALS 185 4. (a) f (x) (x + 2)3 + 1 (c) h(x) = -x5 -3 y y 12 10 11- -9 10- 8-- 9 7- 8 6-- 5-- 4-- 5-- 3 4 2 31 2 1 1 . 1.X -2 -4 3 -2 -1 -1 _ -2 -4 -3 -5 -4 -6 -5 -7 -6 -8 -7- -9. -8- -10. -9- -10. (b) g(x) =(x + 2)4 + 1 (d) j(x) = 2 - 3(x - 1)4 21 2 20- 1 19 18. _ 1 X 17 -2 16--3 15._ 14 13._ 1 --8 1 ..9.. 8 - 1 7 - 6-- 3 5 4.. 3.. 2.. -4 -3 -2 -1 X 5. We have f(-4) =-23, f(-3) =5, f(0) = 5, f(1) =-3, f(2) =-5 and f(3) - 5 so the Intermediate Value Theorem tells us that f(x) = z3 - 9x + 5 has real zeros in the intervals [-4, -3], [0,1] and [2,3]. 6. Making and selling 71 PortaBoys yields a maximized profit of $5910.67. 7. The cubic regression model is p3(x)= 0.0226x3 - 0.950892 + 8.615x - 3.446. It has R2 0.93765 which isn't bad. The graph of y= p3(x) in the viewing window [-1, 13] x [0, 24] along with the scatter plot is shown below on the left. Notice that p3 hits the x-axis at about x 12.45 making this a bad model for future predictions. To use the model to approximate the number of hours of sunlight on your birthday, you'll have to figure out what decimal value  186 POLYNOMIAL FUNCTIONS of x is close enough to your birthday and then plug it into the model. My (Jeff's) birthday is July 31 which is 10 days after July 21 (x = 7). Assuming 30 days in a month, I think x = 7.33 should work for my birthday and p3(7.33) - 17.5. The website says there will be about 18.25 hours of daylight that day. To have 14 hours of darkness we need 10 hours of daylight. We see that p3(1.96) - 10 and p3(10.05) - 10 so it seems reasonable to say that we'll have at least 14 hours of darkness from December 21, 2008 (x = 0) to February 21, 2009 (x = 2) and then again from October 21,2009 (x = 10) to December 21, 2009 (x = 12). The quartic regression model is p4(x) = 0.0144x4 -0.3507x3+2.259x2 -1.571x+5.513. It has R2 = 0.98594 which is good. The graph of y = p4(x) in the viewing window [-1, 15] x [0, 35] along with the scatter plot is shown below on the right. Notice that p4(15) is above 24 making this a bad model as well for future predictions. However, p4(7.33) - 18.71 making it much better at predicting the hours of daylight on July 31 (my birthday). This model says we'll have at least 14 hours of darkness from December 21, 2008 (x = 0) to about March 1, 2009 (x = 2.30) and then again from October 10, 2009 (x = 9.667) to December 21, 2009 (x = 12). 0 y=p3(x) Y=p4(x) 8. (a) The scatter plot is shown below with each of the three regression models. (b) The quadratic model is P2(x) = -0.02x2 + 0.241x + 0.956 with R2 = 0.77708. The cubic model is P3 (x) = 0.005x3 - 0.103x2 + 0.602x + 0.573 with R2 = 0.98153. The quartic model is P4(x) = -0.000969x4 + 0.0253x3 - 0.240x2 + 0.944x + 0.330 with R2 = 0.99929. (c) The maximums predicted by the three models are P2(5.737) - 1.648, P3(4.232) - 1.657 and P4(3.784) - 1.630, respectively. a y P2(x) yy P3(x)y P(x P4(x) P2(x)  3.2 THE FACTOR THEOREM AND THE REMAINDER THEOREM 187 3.2 THE FACTOR THEOREM AND THE REMAINDER THEOREM Suppose we wish to find the zeros of f(x) = x3 + 4x2 - 5x - 14. Setting f(x) = 0 results in the polynomial equation x3 + 4x2 - 5x - 14 = 0. Despite all of the factoring techniques we learned' in Intermediate Algebra, this equation foils2 us at every turn. If we graph f using the graphing calculator, we get The graph suggests that x = 2 is a zero, and we can verify f(2) = 0. The other two zeros seem to be less friendly, and, even though we could use the 'Zero' command to find decimal approximations for these, we seek a method to find the remaining zeros exactly. Based on our experience, if x = 2 is a zero, it seems that there should be a factor of (x - 2) lurking around in the factorization of f (x). In other words, it seems reasonable to expect that x3 + 4x2 - 5x - 14 = (x - 2) q(x), where q(x) is some other polynomial. How could we find such a q(x), if it even exists? The answer comes from our old friend, polynomial division. Dividing x3 + 4x2 - 5x - 14 by x - 2 gives x2 + 6x +7 x-2 x3 + 4x2 - 5x - 14 - (x3 - 2x2) 6x2 - 5x -(6x2 -12x) 7x - 14 - (7x -14) 0 As you may recall, this means x3 + 4x2 - 5x - 14 = (x - 2) (x2 + 6x + 7), and so to find the zeros of f, we now can solve (x - 2) (x2 + 6x + 7) = 0. We get x - 2 = 0 (which gives us our known zero, x = 2) as well as x2 + 6x + 7 = 0. The latter doesn't factor nicely, so we apply the Quadratic Formula to get x = -3 + /2. The point of this section is to generalize the technique applied here. First up is a friendly reminder of what we can expect when we divide polynomials. THEOREM 3.4. Polynomial Division: Suppose d(x) and p(x) are nonzero polynomials where the degree of p is greater than or equal to the degree of d. There exist two unique polynomials, q(x) and r(x), such that p(x) = d(x) q(x) + r(x), where either r(x) = 0 or the degree of r is strictly less than the degree of d. 1and probably forgot 2pun intended  188 POLYNOMIAL FUNCTIONS As you may recall, all of the polynomials in Theorem 3.4 have special names. The polynomial p is called the dividend; d is the divisor; q is the quotient; r is the remainder. If r(x) = 0 then d is called a factor of p. The proof of Theorem 3.4 is usually relegated to a course in Abstract Algebra,3 but we will use the result to establish two important facts which are the basis of the rest of the chapter. THEOREM 3.5. The Remainder Theorem: Suppose p is a polynomial of degree at least 1 and c is a real number. When p(x) is divided by x - c the remainder is p(c). The proof of Theorem 3.5 is a direct consequence of Theorem 3.4. When a polynomial is divided by x - c, the remainder is either 0 or has degree less than the degree of x - c. Since x - c is degree 1, this means the degree of the remainder must be 0, which means the remainder is a constant. Hence, in either case, p(x) =_(x - c) q(x) + r, where r, the remainder, is a real number, possibly 0. It follows that p(c) = (c - c) q(c) + r = 0 - q(c) + r = r, and so we get r = p(c), as required. There is one last 'low hanging fruit'4 to collect - it is an immediate consequence of The Remainder Theorem. THEOREM 3.6. The Factor Theorem: Suppose p is a nonzero polynomial. The real number c is a zero of p if and only if (x - c) is a factor of p(x). The proof of The Factor Theorem is a consequence of what we already know. If (x - c) is a factor of p(x), this means p(x) - (x - c) q(x) for some polynomial q. Hence, p(c) - (c - c) q(c) = 0, and so c is a zero of p. Conversely, if c is a zero of p, then p(c) = 0. In this case, The Remainder Theorem tells us the remainder when p(x) is divided by (x - c), namely p(c), is 0, which means (x - c) is a factor of p. What we have established is the fundamental connection between zeros of polynomials and factors of polynomials. Of the things The Factor Theorem tells us, the most pragmatic is that we had better find a more efficient way to divide polynomials by quantities of the form x - c. Fortunately, people like Ruffini and Horner have already blazed this trail. Let's take a closer look at the long division we performed at the beginning of the section and try to streamline it. First off, let's change all of the subtractions into additions by distributing through the -1s. 2 + 6x +7 x-2 x3 + 42 - 5x -14 -x3+ 2x2 6x2 - 5x -6x2+12x 7x -14 -7x+14 0 Next, observe that the terms -x3, -6x2, and -7x are the exact opposite of the terms above them. The algorithm we use ensures this is always the case, so we can omit them without losing any 3Yes, Virginia, there are algebra courses more abstract than this one. 4Jeff hates this expression and Carl included it just to annoy him.  3.2 THE FACTOR THEOREM AND THE REMAINDER THEOREM 189 information. Also note that the terms we 'bring down' (namely the -5x and -14) aren't really necessary to recopy, and so we omit them, too. z2+6x+7 x-2 3-+42- 5x -14 2x2 6x2 12x 7z 14 0 Now, let's move things up a bit and, for reasons which will become clear in a moment, copy the x3 into the last row. z2+6x+7 x-2 3-+42- 5x -14 2x2 12x 14 33 6x2 7x 0 Note that by arranging things in this manner, each term in the last row is obtained by adding the two terms above it. Notice also that the quotient polynomial can be obtained by dividing each of the first three terms in the last row by 3 and adding the results. If you take the time to work back through the original division problem, you will find that this is exactly the way we determined the quotient polynomial. This means that we no longer need to write the quotient polynomial down, nor the 3 in the divisor, to determine our answer. -2 z3-+43:2- 5x -14 2x2 12x 14 x3 6x2 7x 0 We've streamlined things quite a bit so far, but we can still do more. Let's take a moment to remind ourselves where the 2x2, 12x, and 14 came from in the second row. Each of these terms was obtained by multiplying the terms in the quotient, x2, 6x and 7, respectively, by the -2 in x - 2, then by -1 when we changed the subtraction to addition. Multiplying by -2 then by -1 is the same as multiplying by 2, and so we replace the -2 in the divisor by 2. Furthermore, the coefficients of the quotient polynomial match the coefficients of the first three terms in the last row, so we now take the plunge and write only the coefficients of the terms to get 2| 1 4 -5 -14 2 12 14 1 6 7 0  190 POLYNOMIAL FUNCTIONS We have constructed is the synthetic division tableau for this polynomial division problem. Let's re-work our division problem using this tableau to see how it greatly streamlines the division process. To divide x3 + 42 - 5x - 14 by x - 2, we write 2 in the place of the divisor and the coefficients of x3 + 42 - 5x - 14 in for the dividend. Then 'bring down' the first coefficient of the dividend. 2 1 4 -5 -14 2 1 4 -5 -14 1 Next, take the 2 from the divisor and multiply by the 1 that was 'brought down' to get 2. Write this underneath the 4, then add to get 6. 2 1 4 -5 -14 2 1 4 -5 -14 1,2 1,2 1 1 6 Now take the 2 from the divisor times the 6 to get 12, and add it to the -5 to get 7. 2 1 4 -5 -14 2 1 4 -5 -14 1,2 12 1,2 12 1 6 1 6 7 Finally, take the 2 in the divisor times the 7 to get 14, and add it to the -14 to get 0. 2 1 4 -5 -14 2 1 4 -5 -14 1 2 12 14 1 2 12 14 1 6 7 1 6 7 10 The first three numbers in the last row of our tableau are the coefficients of the quotient polynomial. Remember, we started with a third degree polynomial and divided by a first degree polynomial, so the quotient is a second degree polynomial. Hence the quotient is 2 + 6x + 7. The number in the box is the remainder. Synthetic division is our tool of choice for dividing polynomials by divisors of the form x - c. It is important to note that it works only for these kinds of divisors.5 Also take note that when a polynomial (of degree at least 1) is divided by x - c, the result will be a polynomial of exactly one less degree. Finally, it is worth the time to trace each step in synthetic division back to its corresponding step in long division. While the authors have done their best to indicate where the algorithm comes from, there is no substitute for working through it yourself. EXAMPLE 3.2.1. Use synthetic division to perform the following polynomial divisions. Find the quotient and the remainder polynomials, then write the dividend, quotient and remainder in the form given in Theorem 3.4. 5You'll need to use good old-fashioned polynomial long division for divisors of degree larger than 1.  3.2 THE FACTOR THEOREM AND THE REMAINDER THEOREM 191 1. (5x3 - 2x2 + 1) +±(x-3) 3.4 8x - 12x2 2x - 3 2. (X3 + 8) + (x + 2) SOLUTION. 1. When setting up the synthetic division tableau, we need to enter 0 for the coefficient of x in the dividend. Doing so gives 3 5 -2 0 1 1f 15 39 117 5 13 39 118 Since the dividend was a third degree polynomial, the quotient is a quadratic polynomial with coefficients 5, 13 and 39. Our quotient is q(x) = 5x2 + 13x + 39 and the remainder is r (x) = 118. According to Theorem 3.4, we have 5x3 -2x2+1 = (x-3) (52 + 13x + 39) +118. 2. For this division, we rewrite x + 2 as x - (-2) and proceed as before -2 1 0 0 8 1 -2 4 -8 1 -2 4 0 We get the quotient q(x) =9x2 - 2x + 4 and the remainder r(x) = 0. Relating the dividend, quotient and remainder gives x3 + 8 = (x + 2) (2 - 2x + 4). 3. To divide 4 - 8x - 12x2 by 2x - 3, two things must be done. First, we write the dividend in descending powers of x as -12x2 - 8x + 4. Second, since synthetic division works only for factors of the form x - c, we factor 2x - 3 as 2 (x - 2). Our strategy is to first divide -12x2 - 8x + 4 by 2, to get -6x2 - 4x +2. Next, we divide by (x - 2). The tableau becomes 3 -6 -4 2 1.-9 39 -6 -13 -5 From this, we get -6x2 - 4x + 2 =(x - 2) (-6x - 13) - 2. Multiplying both sides by 2 and distributing gives -12x2 - 8x + 4 = (2x - 3) (-6x - 13) - 35. At this stage, we have written -12x2 -8x + 4 in the form (2x - 3)q(x) + r(z), but how can we be sure the quotient polynomial is -6x -13 and the remainder is -35? The answer is the word 'unique' in Theorem 3.4. The theorem states that there is only one way to decompose -1292 - 8x + 4 into a multiple of (2x - 3) plus a constant term. Since we have found such a way, we can be sure it is the only way. D The next example pulls together all of the concepts discussed in this section.  192 POLYNOMIAL FUNCTIONS EXAMPLE 3.2.2. Let p(x) = 2x3 - 5x + 3. 1. Find p(-2) using The Remainder Theorem. Check your answer by substitution. 2. Use the fact that x = 1 is a zero of p to factor p(x) and find all of the real zeros of p. SOLUTION. 1. The Remainder Theorem states p(-2) is the remainder when p(x) is divided by x - (-2). We set up our synthetic division tableau below. We are careful to record the coefficient of 2 as 0, and proceed as above. -2 2 0 -5 3 1 -4 8 -6 2 -4 3 -3 According to the Remainder Theorem, p(-2) = -3. We can check this by direct substitution into the formula for p(x): p(-2) = 2(-2)3 - 5(-2) + 3 = -16 + 10 + 3 = -3. 2. The Factor Theorem tells us that since x= 1 is a zero of p, x -1 is a factor of p(x). To factor p(x), we divide 1 2 0 -5 3 1f 2 2 -3 2 2 -3 0 We get a remainder of 0 which verifies that, indeed, p(l) = 0. Our quotient polynomial is a second degree polynomial with coefficients 2, 2, and -3. So q(x) = 2x2 + 2x - 3. Theorem 3.4 tells us p(x) = (x - 1) (2x2 + 2x - 3). To find the remaining real zeros of p, we need to solve 2x2+2x - 3 = 0 for x. Since this doesn't factor nicely, we use the quadratic formula to find that the remaining zeros are x =-12 . 2 In Section 3.1, we discussed the notion of the multiplicity of a zero. Roughly speaking, a zero with multiplicity 2 can be divided twice into a polynomial; multiplicity 3, three times and so on. This is illustrated in the next example. EXAMPLE 3.2.3. Let p(x) = 4x4 -4x3 - 11x2 + 12x - 3. Given that x= 2 is a zero of multiplicity 2, find all of the real zeros of p. SOLUTION. We set up for synthetic division. Since we are told the multiplicity of 2 is two, we continue our tableau and divide 1 into the quotient polynomial 1l4 -4 -11 12 -3 2 1J 2 -1 -6 3 1 4 -2 -12 6 %j 1. 2 0 -6 4 0 -12 0  3.2 THE FACTOR THEOREM AND THE REMAINDER THEOREM 193 From the first division, we get 4x4 - 4x3 - 11x2 + 12x - 3 = (x - 2) (4x3 - 2x2 - 12x + 6). The second division tells us 4x3 - 2x2 - 12x + 6 = (- 2) (4x2 - 12). Combining these results, we have 4x4 - 4x3 - 11x2+ 12x - 3 = (x - 2)2 (4x2 - 12). To find the remaining zeros of p, we set 4x2-12=0andgetx = 3. D A couple of things about the last example are worth mentioning. First, the extension of the synthetic division tableau for repeated divisions will be a common site in the sections to come. Typically, we will start with a higher order polynomial and peel off one zero at a time until we are left with a quadratic, whose roots can always be found using the Quadratic Formula. Secondly, we found x = t3 are zeros of p. The Factor Theorem guarantees (x - 3) and (x - (-v3)) are both factors of p. We can certainly put the Factor Theorem to the test and continue the synthetic division tableau from above to see what happens. 1l4 -4 -11 12 -3 2 t 2 -1 -6 3 1 4 -2 -12 6 0_ J, 2 0 -6 4 0 -12[0_ S44 3 12 -3 4 43 0 1.-4v3 4 0_ This gives us 4x4 - 4x3 - 11x2+ 12x - 3 = (x- j)2 (x- 3) (z- (-3)) (4), or, when written with the constant in front p(x) = 4 x - (x - V3) x - -) We have shown that p is a product of its leading term times linear factors of the form (x - c) where c are zeros of p. It may surprise and delight the reader that, in theory, all polynomials can be reduced to this kind of factorization. We leave that discussion to Section 3.4, because the zeros may not be real numbers. Our final theorem in the section gives us an upper bound on the number of real zeros. THEOREM 3.7. Suppose f is a polynomial of degree n, n > 1. Then f has at most n real zeros, counting multiplicities. Theorem 3.7 is a consequence of the Factor Theorem and polynomial multiplication. Every zero c of f gives us a factor of the form (x - c) for f(x). Since f has degree n, there can be at most n of these factors. The next section provides us some tools which not only help us determine where the real zeros are to be found, but which real numbers they may be. We close this section with a summary of several concepts previously presented. You should take the time to look back through the text to see where each concept was first introduced and where each connection to the other concepts was made.  194 POLYNOMIAL FUNCTIONS Connections Between Zeros, Factors and Graphs of Polynomial Functions Suppose p is a polynomial function of degree n > 1. The following statements are equivalent: " The real number c is a zero of p " p(c) = 0 " x= c is a solution to the polynomial equation p(x) = 0 " (x - c) is a factor of p(x) " The point (c, 0) is an x-intercept of the graph of y = p(x)  3.2 THE FACTOR THEOREM AND THE REMAINDER THEOREM 195 3.2.1 EXERCISES 1. (An Intermediate Algebra review exercise) Use polynomial long division to perform the indi- cated division. Write the polynomial in the form p(x) = d(x)q(x) + r(x). (a) (5x4 - 3x3 +2x2 - 1)±+ (x2 + 4) (c) (9x3+ 5)±+ (2x - 3) (b) (-X5 + 73-x).(x3 -2 + 1) (d) (4x2 - x - 23)±+ (x2 -1) 2. Use synthetic division and the Remainder Theorem to test whether or not the given number is a zero of the polynomial p(x) = 15x5 - 121x4 + 17x3 - 73x2 + 2x + 48. (a) c = -1 (d) c = 3 (b) c = 8 (e) c = 0 (c) c = I (f) c = -3 1 3 3. For each polynomial given below, you are given one of its zeros. Use the techniques in this section to find the rest of the real zeros and factor the polynomial. (a) x3-6x2+11x-6, c=1 (b) x3 - 24x2 + 192x - 512, c = 8 (c) 4x4 - 28x3 + 61x2 -42x+9, c = (d) 3x3 +4x2 -x-2, c=-3 (e) x4 -z2, c = 0 (f) 92 -2x-2, c=1- v/3 (g) 125x5 - 275x4 - 2265x3 - 3213x2 - 1728x - 324, c = -j 4. Create a polynomial p with the following attributes. " As x -- -oc, p(x) -- oc. " The point (-2, 0) yields a local maximum. " The degree of p is 5. " The point (3, 0) is one of the x-intercepts of the graph of p. 3 29 5. Find a quadratic polynomial with integer coefficients which has x= - i as its real zeros. 5 5  196 196 POLYNOMIAL FUNCTIONS 3.2.2 ANSWERS 1. (a) 5x4 -3x3 +2x2 -1 =(:2 + 4)(5x:2- 3x- 18) +(12x +71) (b) -x5 +7x - x =(X3 X2 + 1)(-:2 -X + 6) +(7X2 6) (d) 492 - x - 23 =(92 - 1)(4) + (-x - 19) 2. (a) p(-1) - 100 (b) p(8) =0 (C) p( 2) - 825 (d) p(3) =0 (e) p(O)= 48 (f) P(- ) 0 3. (a) x3-6x2+11x-6=(x-1)(x-2)(x-3) (b) x3 - 2492 + 192x - 512= (x -8) (c) 4x4 - 28x3 + 6192 - 42x + 9= 4(x - 2)2(X - 3)2 (d) 33+42- ( -3( ) (e) X2 - 9x 2 =(x 1(I-+v'1))(-(I+v3) (g) 125x5 - 275x4 - 2265x3 - 321392 - 1728x - 324= 125(x + 5)3(X - 6)(x + 2) 4. Something like p~x) -(x + 2)2(x - 3) (x + 3) (x - 4) will work. 5. q(x) = 9x - 6x - 4  3.3 REAL ZEROS OF POLYNOMIALS 197 3.3 REAL ZEROS OF POLYNOMIALS In Section 3.2, we found that we can use synthetic division to determine if a given real number is a zero of a polynomial function. This section presents results which will help us determine good candidates to test using synthetic division. There are two approaches to the topic of finding the real zeros of a polynomial. The first approach (which is gaining popularity) is to use a little bit of mathematics followed by a good use of technology like graphing calculators. The second approach (for purists) makes good use of mathematical machinery (theorems) only. For completeness, we include the two approaches but in separate subsections.1 Both approaches benefit from the following two theorems, the first of which is due to the famous mathematician Augustin Cauchy. It gives us an interval on which all of the real zeros of a polynomial can be found. THEOREM 3.8. Cauchy's Bound: Suppose f (x) = anx+an_1x"-1+. . .+aiz+a° is a polynomial of degree n with n> 1. Let M be the largest of the numbers: a, a . a., . Then all the real zeros of f lie in in the interval [-(M + 1), M + 1]. The proof of this fact is not easily explained within the confines of this text. This paper contains the result and gives references to its proof. Like many of the results in this section, Cauchy's Bound is best understood with an example. EXAMPLE 3.3.1. Let f(x) = 2x4 + 4x3 - x2 - 6x - 3. Determine an interval which contains all of the real zeros of f. SOLUTION. Cauchy's Bound says to take the absolute value of each of the non-leading coefficients of f, namely, 4, 1, 6 and 3, and divide them by the absolute value of the leading coefficient, 2. Doing so produces the list of numbers 2, 1, 3, and 2. Next, we take the largest of these values, 3, as our value M in the theorem and add one to it to get 4. The real zeros of f are guaranteed to lie in the interval [-4, 4]. Whereas the previous result tells us where we can find the real zeros of a polynomial, the next theorem gives us a list of possible real zeros. THEOREM 3.9. Rational Zeros Theorem: Suppose f (x) = anx + an_1x"-1 + ... + a1x + a° is a polynomial of degree n with n > 1, and a0, a1, ... an are integers. If r is a rational zero of f, then r is of the form t, where p is a factor of the constant term a0, and q is a factor of the leading coefficient an. The rational zeros theorem gives us a list of numbers to try in our synthetic division and that is a lot nicer than simply guessing. If none of the numbers in the list are zeros, then either the polynomial has no real zeros at all, or all of the real zeros are irrational numbers. To see why the Rational Zeros Theorem works, suppose c is a zero of f and c = in lowest terms. This means p and q have no common factors. Since f(c) = 0, we have an -P + an_1 -P) + .. . + ai -P + a° = 0. Carl is the purist and is responsible for all of the theorems in this section. Jeff, on the other hand, has spent too much time in school politics and has been polluted with notions of 'compromise.' You can blame the slow decline of civilization on him and those like him who mingle mathematics with technology.  198 POLYNOMIAL FUNCTIONS Multiplying both sides of this equation by q", we clear the denominators to get anp" + an_1p"- q +... + a1pq"-1 + aoq" = 0 Rearranging this equation, we get anp" = -an_1p"- q - ... - aipq" - - aoq" Now, the left hand side is an integer multiple of p, and the right hand side is an integer multiple of q. (Can you see why?) This means anp" is both a multiple of p and a multiple of q. Since p and q have no common factors, an must be a multiple of q. If we rearrange the equation anp" + an_1p"-1q +... + a1pq"-1 + aoq" = 0 as aoq" = -anp" - an_1p"-lq - ... - a1pq"-1 we can play the same game and conclude ao is a multiple of p, and we have the result. EXAMPLE 3.3.2. Let f(x) = 2x4 + 4x3 - x2 - 6x - 3. Use the Rational Zeros Theorem to list all possible rational zeros of f. SOLUTION. To generate a complete list of rational zeros, we need to take each of the factors of constant term, ao = -3, and divide them by each of the factors of the leading coefficient a4 = 2. The factors of -3 are t 1 and +3. Since the Rational Zeros Theorem tacks on a + anyway, for the moment, we consider only the positive factors 1 and 3. The factors of 2 are 1 and 2, so the Rational Zeros Theorem gives the list {+ }, + 1, + j, +2} or {+ 1, + 1, + 2, + 3}. D Our discussion now diverges between those who wish to use technology and those who do not. 3.3.1 FOR THOSE WISHING TO USE A GRAPHING CALCULATOR At this stage, we know not only the interval in which all of the zeros of f (x) = 2x4+4x3-z2 -6x -3 are located, but we also know some potential candidates. We can now use our calculator to help us determine all of the real zeros of f, as illustrated in the next example. EXAMPLE 3.3.3. Let f (x) = 2x4 + 4x3 - 2 - 6x - 3. 1. Graph y = f(x) on the calculator using the interval obtained in Example 3.3.1 as a guide. 2. Use the graph to help narrow down the list of candidates for rational zeros you obtained in Example 3.3.2. 3. Use synthetic division to find the real zeros of f, and state their multiplicities. SOLUTION. 1. In Example 3.3.1, we determined all of the real zeros of f lie in the interval [-4, 4] We set our window accordingly and get  3.3 REAL ZEROS OF POLYNOMIALS 199 WINDOW Xmin= -4 Xmax=4 Xsc1=1 Ymin= -4 Ymax=4 Ysc1=1 Xr es=1 2. In Example 3.3.2, we learned that any rational zero of f must be in the list {+ j, +1, + 9, + 3}. From the graph, it looks as if we can rule out any of the positive rational zeros, since the graph seems to cross the x-axis at a value just a little greater than 1. On the negative side, -1 looks good, so we try that for our synthetic division. -1 2 4 -1 -6 -3 1 -2 -2 3 3 2 2 -3 -3 0 We have a winner! Remembering that f was a fourth degree polynomial, we now know our quotient is a third degree polynomial. If we can do one more successful division, we will have knocked the quotient down to a quadratic, and, if all else fails, we can use the quadratic formula to find the last two zeros. Since there seems to be no other rational zeros to try, we continue with -1. Also, the shape of the crossing at x = -1 leads us to wonder if the zero x = -1 has multiplicity 3. -1 2 4 -1 -6 -3 1 -2 -2 3 3 -1 2 2 -3 -3 0 { -2 0 3 2 0 -3 0 Success! Our quotient polynomial is now 2x2 - 3. Setting this to zero gives 2x2 - 3 = 0, or x2 = 2, which gives us x =+ -. Concerning multiplicities, based on our division, we have that -1 has a multiplicity of at least 2. The Factor Theorem tells us our remaining zeros, + -Q, each have multiplicity at least 1. However, Theorem 3.7 tells us f can have at most 4 real zeros, counting multiplicity, and so we conclude that -1 is of multiplicity exactly 2 and + each has multiplicity 1. (Thus, we were wrong to think that -1 had multiplicity 3.) E It is interesting to note that we could greatly improve on the graph of y = f(x) in the previous example given to us by the calculator. For instance, from our determination of the zeros of f and their multiplicities, we know the graph crosses at x = --2 ~ -1.22 then turns back upwards to touch the x-axis at x = -1. This tells us that, despite what the calculator showed us the first time, there is a relative maximum occurring at x = -1 and not a 'flattened crossing' as we originally  200 POLYNOMIAL FUNCTIONS believed. After resizing the window, we see not only the relative maximum but also a relative minimum just to the left of x = technology, instead of vice-versa. 1 which shows us, once again, that Mathematics enhances the WINDOW Min= -1.5 Xmax= -. 5 Xsc1=1 Ymin= -. 01 Ymax=. 01 Ysc=1 Xres=1 Our next example shows how even a mild-mannered polynomial can cause problems. EXAMPLE 3.3.4. Let f(x) = x4 + 2 - 12. 1. Use Cauchy's Bound to determine an interval in which all of the real zeros of f lie. 2. Use the Rational Zeros Theorem to determine a list of possible rational zeros of f. 3. Graph y = f(x) using your graphing calculator. 4. Find all of the real zeros of f and their multiplicities. SOLUTION. 1. Applying Cauchy's Bound, we find M = 12, so all of the real zeros lie in the interval [ 13,13]. 2. Applying the Rational Zeros Theorem with constant term ao a4 = 1, we get the list {+1, +2, +3, +4, +6, +12}. 12 and leading coefficient 3. Graphing y = f(x) on the interval [ 13,13] produces the graph below on the left. Zooming in a bit gives the graph below on the right. Based on the graph, none of our rational zeros will work. (Do you see why not?) ~I. . . . . . 1 i ). . . I  3.3 REAL ZEROS OF POLYNOMIALS 201 4. From the graph, we know f has two real zeros, one positive, and one negative. Our only hope at this point is to try and find the zeros of f by setting f(x) = 0 and solving. Doing so results in the equation z4 + 2 - 12 = 0. If we stare at this equation long enough, we may recognize it as a 'quadratic in disguise'.2 In other words, we have three terms: 34, 92, and 12, and the exponent on the first term, 34, is exactly twice that of the second term, 2. We may rewrite this as (2)2+ (2) - 12 0. To better see the forest for the trees, let us momentarily replace 2 with the variable u. In terms of u, our equation becomes u2 + u - 12 = 0. We can readily factor the left hand side of this equation as (a +4)(u - 3) = 0, which means we have factored the left hand side of x4 +9x2 -12 = 0 as (2 - 3) (x2 + 4) =0. We get 2 = 3, which gives us x = i3, or 2 = -4, which admits no real solutions. Since V3~ 1.73, the two zeros match what we expected from the graph. In terms of multiplicity, the Factor Theorem guarantees (x- 3) and (x + 3) are factors of f(x). We note that our work for finding the zeros of f shows f(x) can be factored as f(x) = (2 - 3) (x2 + 4). Since 2 + 4 has no real zeros, the quantities (x - 3) and (x + 3) must both be factors of 2 - 3. According to Theorem 3.7, 2 - 3 can have at most 2 zeros, counting multiplicity, hence each of t 3is a zero of f of multiplicity 1. Q The technique used to factor f(x) in Example 3.3.4 is called u-substitution. We shall see more of this technique in Section 5.3. In general, substitution can help us identify a 'quadratic in disguise' provided that there are exactly three terms and the exponent of the first term is exactly twice that of the second. It is entirely possible that a polynomial has no real roots at all, or worse, it has real roots but none of the techniques discussed in this section can help us find them exactly. In the latter case, we are forced to approximate, which in this subsection means we use the 'Zero' command on the graphing calculator. We now present other theorems and discuss how to find zeros of polynomials when we do not have access to a graphing calculator. 3.3.2 FOR THOSE WISHING NOT TO USE A GRAPHING CALCULATOR Suppose we wish to find the zeros of f(x) = 2x4 + 4x3 - 2 - 6x - 3 without using the calculator. In this subsection, we present some more mathematical tools to help us. Our first result is due to Rene Descartes and gives us an estimate of how many positive and how many negative real zeros are to be found. The theorem requires us to discuss what is meant by the variations in sign of a polynomial function. For example, consider f(x) = 2x + 4x3 - 2 - 6x - 3. If we focus on only the signs of the coefficients, we start with a (+), followed by another (+), then switch to (-), and stay (-) for the remaining two coefficients. Since the signs of the coefficients switched once, we say f (x) has one variation in sign. When we speak of the variations in sign of a polynomial function, we assume the formula for f(x) is written with descending powers of x, as in Definition 3.1, and concern ourselves only with the nonzero coefficients. 2More appropriately, this equation is 'quadratic in form.' Carl likes to call it a 'quadratic in disguise' because it reminds him of The Transformers.  202 POLYNOMIAL FUNCTIONS THEOREM 3.10. Descartes' Rule of Signs: Suppose f(x) is the formula for a polynomial function written with descending powers of x. " If P denotes the number of variations of sign in the formula for f(x), then the number of positive real zeros (counting multiplicity) is one of the numbers {P, P - 2, P - 4, ... }. " If N denotes the number of variations of sign in the formula for f(-x), then the number of negative real zeros (counting multiplicity) is one of the numbers {N, N - 2, N - 4, ... }. A couple of remarks about Descartes' Rule of Signs are in order. First, Descartes' Rule of Signs gives us an estimate to the number of real zeros, not the actual value of the zeros. Second, Descartes' Rule of Signs counts multiplicities. This means that, for example, if one of the zeros has multiplicity 2, Descsartes' Rule of Signs would count this as two zeros. Lastly, note that the number of positive or negative real zeros always starts with the number of sign changes and decreases by an even number. For example, if f(x) has 7 changes in sign, then, counting multplicities, f has either 7, 5, 3, or 1 positive real zero. This implies that the graph of y = f(x) crosses the positive x-axis at least once. If f(-x) results in 4 sign changes, then, counting multiplicities, f has 4, 2, or 0 negative real zeros; hence, the graph of y = f(x) may not cross the negative x-axis at all. The proof of Descartes' Rule of Signs is a bit technical, and can be found here. EXAMPLE 3.3.5. Let f(x) = 2x4 + 4x3 - 2 - 6x - 3. Use Descartes' Rule of Signs to determine the possible number and location of the real zeros of f. SOLUTION: As noted above, the variations of sign of f(x) is 1. This means, counting multiplicities, f has exactly 1 positive real zero. To find the possible number of negative real zeros, we consider f(-x) = 2(-x)4 + 4(-X)3 - (-x)2 - 6(-x) - 3 which simplifies to 2x4 - 4x3 - 2 + 6x - 3. This has 3 variations in sign, hence f has either 3 negative real zeros or 1 negative real zero, counting multiplicities. D Cauchy's Bound gives us a general bound on the zeros of a polynomial function. Our next result helps us determine bounds on the real zeros of a polynomial as we synthetically divide which are often sharper3 bounds than Cauchy's Bound. THEOREM 3.11. Upper and Lower Bounds: Suppose f is a polynomial of degree n with n> 1. " If c > 0 is synthetically divided into f and all of the numbers in the final line of the division tableau have the same signs, then c is an upper bound for the real zeros of f. That is, there are no real zeros greater than c. " If c < 0 is synthetically divided into f and the numbers in the final line of the division tableau alternate signs, then c is a lower bound for the real zeros of f. That is, there are no real zeros less than c. NOTE: If the number 0 occurs in the final line of the division tableau in either of the above cases, it can be treated as (+) or (-) as needed. 3That is, better, or more accurate.  3.3 REAL ZEROS OF POLYNOMIALS 203 The Upper and Lower Bounds Theorem works because of Theorem 3.4. For the upper bound part of the theorem, suppose c > 0 is divided into f and the resulting line in the division tableau contains, for example, all nonnegative numbers. This means f(x) = (x - c)q(x) + r, where the coefficients of the quotient polynomial and the remainder are nonnegative. (Note the leading coefficient of q is the same as f and so q(x) is not the zero polynomial.) If b > c, then f(b) - (b - c)q(b) + r, where (b - c) and q(b) are both positive and r > 0. Hence f(b) > 0 which shows b cannot be a zero of f. Thus no real number b > c can be a zero of f, as required. A similar argument proves f(b) <0 if all the numbers in the final line of the synthetic division tableau are non-positive. To prove the lower bound part of the theorem, we note that a lower bound for the negative real zeros of f(x) is an upper bound for the positive real zeros of f(-x). Applying the upper bound portion to f(-x) gives the result. (Do you see where the alternating signs come in?) With the additional mathematical machinery of Descartes' Rule of Signs and the Upper and Lower Bounds Theorem, we can find the real zeros of f(x) = 2x4 + 4x3 -2 - 6x -3 without the use of a graphing calculator. EXAMPLE 3.3.6. Let f (x) = 2x4 + 4x3 - 2 - 6x - 3. 1. Find all real zeros of f and their multiplicities. 2. Sketch the graph of y = f(x). SOLUTION. 1. We know from Cauchy's Bound that all of the real zeros lie in the interval [-4, 4] and that our possible rational zeros are i, ,i1, , i ±, and t+3. Descartes' Rule of Signs guarantees us at least one negative real zero and exactly one positive real zero, counting multiplicity. We try our positive rational zeros, starting with the smallest, 2. Since the remainder isn't zero, we know 2 isn't a zero. Sadly, the final line in the division tableau has both positive and negative numbers, so j is not an upper bound. The only information we get from this division is courtesy of the Remainder Theorem which tells us f (j) =--8 so the point (j, -8) is on the graph of f. We continue to our next possible zero, 1. As before, the only information we can glean from this is that (1, -4) is on the graph of f. When we try our next possible zero, 2, we get that it is not a zero, and we also see that it is an upper bound on the zeros of f, since all of the numbers in the final line of the division tableau are positive. This means there is no point trying our last possible rational zero, 3. Descartes' Rule of Signs guaranteed us a positive real zero, and at this point we have shown this zero is irrational. Furthermore, the Intermediate Value Theorem, Theorem 3.1, tells us the zero lies between 1 and , since f(1) < 0 and f ( ) > 0. 2 4 -1 -6 -3 1 2 4 -1 -6 -3 2 4 -1 -6 -3 S1 5 3 21 13 2 7 9 1,1 2 4 8ta2 6 5 -1 3 2 4 8 295 3 21 - 2 6 5 -1-4 27 12337 We now turn our attention to negative real zeros. We try the largest possible zero, -. Synthetic division shows us it is not a zero, nor is it a lower bound (since the numbers in  204 POLYNOMIAL FUNCTIONS the final line of the division tableau do not alternate), so we proceed to -1. This division shows -1 is a zero. Descartes' Rule of Signs told us that we may have up to three negative real zeros, counting multiplicity, so we try -1 again, and it works once more. At this point, we have taken f, a fourth degree polynomial, and performed two successful divisions. Our quotient polynomial is quadratic, so we look at it to find the remaining zeros. -1 2 4 -1 -6 -3 -i 2 4 -1 -6 -3 1 1 3 5 19 -2 -2 3 2 4 8 -1 2 2 -3 -3[0_ 2 3 5 19 5 J,-2 0 3 2 4 8 2 0 -3 0 Setting the quotient polynomial equal to zero yields 2x2-3 = 0, so that 2 = 3, or x = i . Descartes' Rule of Signs tells us that the positive real zero we found, has multiplicity 1. Descartes also tells us the total multiplicity of negative real zeros is 3, which forces -1to be a zero of multiplicity 2 and -4 to have multiplicity 1. 2. We know the end behavior of y = f(x) resembles its leading term, y = 2x4. This means the graph enters the scene in Quadrant II and exits in Quadrant I. Since +4are zeros of odd multiplicity, we know the graph crosses through the x-axis at the points (-4 o) and ( , 0). Since -1 is a zero of multiplicity 2, we know the graph of y = f(x) touches and rebounds off the x-axis at (-1, 0). Putting this together, we get y You can see why the 'no calculator' approach is not very popular these days. It requires more computation and more theorems than the alternative.4 In general, no matter how many theorems 4This is apparently a bad thing.  3.3 REAL ZEROS OF POLYNOMIALS 205 you throw at a polynomial, it may well be impossible5 to find their zeros exactly. The polynomial f (x) x5 - x - 1 is one such beast.6 According to Descartes' Rule of Signs, f has exactly one positive real zero, and it could have two negative real zeros, or none at all. The Rational Zeros Test gives us +1 as rational zeros to try but neither of these work since f(1) = f(-1) = -1. If we try the substitution technique we used in Example 3.3.4, we find f(x) has three terms, but the exponent on the x5 isn't exactly twice the exponent on x. How could we go about approximating this zero without resorting to the 'Zero' command of a graphing calculator? We use the Bisection Method. The first step in the Bisection Method is to find an interval on which f changes sign. We know f(1) = -1 and we find f(2) = 29. By the Intermediate Value Theorem, we know the zero of f lies in the interval [1, 2]. Next, we 'bisect' this interval and find the midpoint, 1.5. We find f(1.5) 5.09. This means our zero is between 1 and 1.5, since f changes sign on this interval. Now, we 'bisect' the interval [1, 1.5] and find f(1.25) 0.08, so now we have the zero between 1 and 1.25. Bisecting [1, 1.25], we find f(1.125) -0.32, which means the zero of f is between 1.125 and 1.25. We continue in this fashion until we have 'sandwiched' the zero between two numbers which differ by no more than a desired accuracy. You can think of the Bisection Method as reversing the sign diagram process: instead of finding the zeros and checking the sign of f using test values, we are using test values to determine where the signs switch to find the zeros. It is a slow and tedious, yet fool-proof, method for determining an approximation of a real zero. Our last example reminds us that finding the zeros of polynomials is a critical step in solving polynomial equations and inequalities. EXAMPLE 3.3.7. 1. Find all of the real solutions to the equation 2x5 + 6x3 + 3 = 3x4 + 8x2. 2. Solve the inequality 2x5 + 6x3 + 3 < 3x4 + 8x2. 3. Interpret your answer to part 2 graphically, and verify using a graphing calculator. SOLUTION. 1. Finding the real solutions to 2x5 + 6x3 + 3 = 3x4 + 8x2 is the same as finding the real solutions to 2x5 - 3x4 + 6x3 - 8x2 + 3 = 0. In other words, we are looking for the real zeros of p(x) = 2x5 - 3x4 + 6x3 -8x2+3. Using the techniques developed in this section, we divide as follows. 5We don't use this word lightly; it can be proven that the zeros of some polynomials cannot be expressed using the usual algebraic symbols. 6See this page.  206 POLYNOMIAL FUNCTIONS 1 2 -3 6 -8 0 3 1 2 -1 5 -3 -3 1 2 -1 5 -3 -3 0 1 2 1 6 3 -1 2 1 6 3 0 2 4 -1 0 -3 2 0 6 0 The quotient polynomial is 2x2 + 6 which has no real zeros so we get x = - 2 and x = 1. 2. To solve this nonlinear inequality, we follow the same guidelines set forth in Section 2.4: we get 0 on one side of the inequality and construct a sign diagram. Our original inequality can be rewritten as 2x5-3x4+6x3-8x2+3 < 0. We found the zeros of p(x) = 2x5-3x4+6x3-8x2+3 in part 1 to be x = - and x = 1. We construct our sign diagram as before. (-) 0 (+) 0 (+) C ~ I ' I t -2 t 1 t -1 0 2 The solution to p(x) < 0 is (-oc, -2, and we know p(x) = 0 at x = -j and x = 1. Hence, the solution to p(x) < 0 is (-oc, -2] U {1}. 3. To interpret this solution graphically, we set f(x) = 2x5 + 6x3 +3 and g(x) = 3x4 + 8x2. We recall that the solution to f(x) < g(x) is the set of x values for which the graph of f is below the graph of g (where f(x) < g(x)) along with the x values where the two graphs intersect (f(x) = g(x)). Graphing f and g on the calculator produces the picture on the lower left. (The end behavior should tell you which is which.) We see the graph of f is below the graph of g on (-oo, -i). However, it is difficult to see what is happening near x = 1. Zooming in (and making the graph of g thicker), we see that the graphs of f and g do intersect at x = 1, but the graph of g remains below the graph of f on either side of x = 1. D  3.3 REAL ZEROS OF POLYNOMIALS 207 3.3.3 EXERCISES 1. Find the real zeros of the polynomial using the techniques specified by your instructor. State the multiplicity of each real zero. (a) p(x) = x3 - 2x2 - 5x + 6 (b) p(x) =_-2x3 + 19x2 - 49x + 20 (c) p(x) = x4 - 92 - 4x + 12 (d) p(x) = x3 +42 - 11x + 6 (e) p(x) = 3x3 + 3x2 - 11x - 10 (f) p(x) = x4 + 2x3 - 12x2 - 40x - 32 (g) p(x) = 6x4 - 5x3 - 92 (h) p(x) = 36x4 - 12x3 - 11x2 + 2x + 1 (i) p(x3) =-17x3 + 52 + 34x - 10 (j) p(x) = 25x5 - 105x4 + 174x3 - 142x2 + 57x - 9 (k) p(x) = x5 - 60x3 - 80x2 + 960x + 2304 (1) p(x) =3:3 - 72 + - 7 (m) p(x) = 90x4 - 399x3 + 622x2 - 399x + 90 (n) p(x) = 9x3 - 52 - x 2. Find the real zeros of f(x) = x3 - 122 - 2x + 7 by first finding a polynomial q(x) with integer coefficients such that q(x) = N . f((x) for some integer N. (Recall that the Rational Zeros Theorem required the polynomial in question to have integer coefficients.) Show that f and q have the same real zeros. 3. Solve the polynomial inequality and give your answer in interval form. (a) -2x3 + 192 - 49x + 20 > 0 (b) x4 - 9x2 <4x - 12 (c) ( + - 1)2>4 (d) 7 -5x3 + 35x2 - 45x - 25 > 0 (e) 4x3 > 3x + 1 (f) x322x2 < x + 2 (g) x4 < 16 + 4x - x3 (h) 2x2 + 3x < 34 4. Let f (x) = 5x7 - 33x6 + 3x5 - 71zX4 - 597x3 + 2097x2 - 1971x + 567. With the help of your classmates, find the 3- and y- intercepts of the graph of f. Find the intervals on which the function is increasing, the intervals on which it is decreasing and the local extrema. Sketch the graph of f, using more than one picture if necessary to show all of the important features of the graph. 5. With the help of your classmates, create a list of five polynomials with different degrees whose real zeros cannot be found using any of the techniques in this section. 7This inequality was solved using a graphing calculator in Example 2.4.5. Compare your answer now to what was given at the time.  208 208 POLYNOMIAL FUNCTIONS 3.3.4 ANSWERS 1. (a) p(cv) cx3 -22 -5x + 6 x = -2 (mult. 1), xcv 1 (mult. 1), xv =3 (mult. 1) (b) p(v) -2xv3 + 1992 - 49xv + 20 x =~ (mult. 1), x = 4 (mult. 1), x = 5 (mult. 1) (c) p(cv) cx4 - 992 - 4xv + 12 x = -2 (mult. 2), x = 1 (mult. 1), x = 3 (mult. 1) (d) p(v) -=cx3 +492 - 11cv + 6 x = -6 (mult. 1), xcv 1 (mult. 2) (e) p(cv) =3x3 + 392 - 11xv - 10 xc= -2 (mult. 1), cv= 3±6 69 (each has mult. 1) (f) p(v) -=cx4 + 2xv3 - 12xv2 - 40xv - 32 xc= -2 (mult. 3), xcv 4 (mult. 1) (g) p(cv) =6c4 - 5cv3 - 9cv2 x = 0 (mult. 2), xc 5±2 41 (each has mult. 1) (h) p(cv)= 36cv4 - 12cv3 - 1192 + 2cv + 1 cv= 2 (mult. 2), xcv - (mult. 2) (i) p(cv) -17xv3 + 5xv2 + 34cv - 10 xcv 7 (mult. 1), xc v'2 (each mult. 1) (j) p(cv)= 25cv5- 105cv4 + 174cv3 - 142cv2 + 57xv - 9 cv= 5 (mult. 2), xcv 1 (mult. 3) (k) p(cv) cx5 - 60cv3 - 80cv2 + 960xv + 2304 cv= -4 (mult. 3), xcv 6 (mult. 2) (1) p(cv) cx3 - 7cv2 + cv - 7 cv - 7 (mult. 1) (in) p(cv)= 90cv4 - 399cv3 + 622cv2 - 399cv + 90 cv= 4 (mult. 1), xcv 2 (mult. 1), xcv 3 (mult. 1), cv (n) p(cv) =9x3 - 5cv2 - cv cv - 0 (mult. 1), cv 5±1861 (each has mult. 1) 5(mult. 1) 2. We choose q(cv) - 72xv3 - 6cv2 - 7xv + 1 - 72 "f so they have the same real zeros. In this case, of both f and q. 3. (a) (-oo,1) U (4,5) (b) f{-2} U[1, 3] (c) (-oo, -1] U [3, oo) (d) (-1,0)U(1,oo) (cv). Clearly f (cv) 0 if and only if q(cv) =0 x x = 6 andcxv= 4are the real zeros 2(f) (- c ,c - 2 ) U (- v 2 ,) v 2) (g) [-2,2] (h) (- oc, - /3) U(v/3, oo)  3.4 COMPLEX ZEROS AND THE FUNDAMENTAL THEOREM OF ALGEBRA 209 3.4 COMPLEX ZEROS AND THE FUNDAMENTAL THEOREM OF ALGEBRA In Section 3.3, we were focused on finding the real zeros of a polynomial function. In this section, we expand our horizons and look for the non-real zeros as well. Consider the polynomial p(x) = 2+1. The zeros of p are the solutions to 2+ 1 = 0, or 2 = -1. This equation has no real solutions, but you may recall from Intermediate Algebra that we can formally extract the square roots of both sides to get x = iv . The quantity -1 is usually re-labeled i, the so-called imaginary unit.1 The number i, while not a real number, plays along well with real numbers, and acts very much like any other radical expression. For instance, 3(2i) = 6i, 7i - 3i = 4i, (2 - 7i) + (3 + 4i) = 5 - 3i, and so forth. The key properties which distinguish i from the real numbers are listed below. DEFINITION 3.4. The imaginary unit i satisfies the two following properties 1. i2 _1 2. If c is a real number with c > 0 then / = iv/c Property 1 in Definition 3.4 establishes that i does act as a square root2 of -1, and property 2 establishes what we mean by the 'principal square root' of a negative real number. In property 2, it is important to remember the restriction on c. For example, it is perfectly acceptable to say -4 v/4= i(2) = 2i. However, -(-4) - iv , otherwise, we'd get 2 = v4 =_ -(-4) = i i = i(2i) = 2i2 = 2(-1) = -2, which is unacceptable.3 We are now in the position to define the complex numbers. DEFINITION 3.5. A complex number is a number of the form a + bi, where a and b are real numbers and i is the imaginary unit. Complex numbers include things you'd normally expect, like 3 + 2i and 5 - iv/3. However, don't forget that a or b could be zero, which means numbers like 3i and 6 are also complex numbers. In other words, don't forget that the complex numbers include the real numbers, so 0 and r - v21 are both considered complex numbers. The arithmetic of complex numbers is as you would expect. The only thing you need to remember are the two properties in Definition 3.4. The next example should help recall how these animals behave. EXAMPLE 3.4.1. Perform the indicated operations and simplify. Write your final answer in the form4 a + bi. Some technical mathematics textbooks label it 'j'. 2Note the use of the indefinite article 'a'. Whatever beast is chosen to be i, -i is the other square root of -1. 3We want to enlarge the number system so we can solve things like x2 = -1, but not at the cost of the established rules already set in place. For that reason, the general properties of radicals simply do not apply for even roots of negative quantities. 4We'll accept an answer of say 3 - 2i, although, technically, we should write this as 3 + (-2)i. Even we pedants have our limits.  210 POLYNOMIAL FUNCTIONS 1. (1 - 2i) - (3 + 4i) 2. (1 - 2i)(3 + 4i) 1 - 2i 3. 3 - 4i 4. v-3 -12 5. (-3)(-12) 6. (x -[1+2i])(-[1 -2i]) SOLUTION. 1. As mentioned earlier, we treat expressions involving i as we would any other radical. We combine like terms to get (1 - 2i) - (3 + 4i) = 1 - 2i - 3 - 4i = -2 - 6i. 2. Using the distributive property, we get (1 - 2i) (3+ 4i)= 3 + 4i - 6i - 8i2. Recalling i2 =-1, we get 3 + 4i - 6i (1)(3) + (1)(4i) - (2i)(3) - (21)(41) -8i2=3-2i-(-8)=11-2i. 3. How in the world are we supposed to simplify }-? Well, we deal with the denominator 3 - 4i as we would any other denominator containing a radical, and multiply both numerator and denominator by 3 + 4i (the conjugate of 3 - 4i).5 Doing so produces 1-2i 3+4i 3-4i 3+4i (1-2i)(3+4i) (3-4i)(3+4i) 11-2i 11 25 25 2 25 4. We use property 2 of Definition 3.4 first, then apply the rules of radicals applicable to real radicals to get -3 -12= (iv/) (iv/12)= i2/3 - 12 = -v/36=-6. 5. We adhere to the order of operations here and perform the multiplication before the radical to get (-3)(-12) =v36= 6. 6. We can brute force multiply using the distributive property and see that ( -[1+2i])(x-[1-2i]) 2 - x[1 - 2i] - x[1 + 2i] + [1 - 2i][1 + 2i] x2 -x+2ix-x-2ix+1-2i+2i-4i2 2 -2x + 5 El A couple of remarks about the last example are in order. First, the conjugate of a complex number a + bi is the number a - bi. The notation commonly used for conjugation is a 'bar': a + bi = a - bi. For example, 3 + 2i = 3 - 2i, 3 - 2i = 3+2i, 6= 6, 4i = -4i, and 3+ v/5 =3+ v/5. The properties of the conjugate are summarized in the following theorem. 5We will talk more about this in a moment.  3.4 COMPLEX ZEROS AND THE FUNDAMENTAL THEOREM OF ALGEBRA 211 THEOREM 3.12. Suppose z and w are complex numbers. e z+=z w * z+W =z+W " zw = zw " (2) 7= , for any natural number n = 1, 2, 3, .. . " z is a real number if and only if z = z. Essentially, Theorem 3.12 says that complex conjugation works well with addition, multiplication, and powers. The proof of these properties can best be achieved by writing out z = a + bi and w = c + di for real numbers a, b, c, and d. Next, we compute the left and right hand side of each equation and check to see that they are the same. The proof of the first property is a very quick exercise. To prove the second property, we compare z + w and z + w. We have z+ = a + bi+ c+ di = a - bi+ c - di. To find z + w, we first compute z+w = (a+bi) + (c+di) = (a+c) + (b+d)i so z +w = (a+ c) + (b+ d)i = (a+ c) - (b+ d)i = a - bi + c - di As such, we have established z+ w = z + w. The proof for multiplication works similarly. The proof that the conjugate works well with powers can be viewed as a repeated application of the product rule, and is best proved using a technique called Mathematical Induction.7 The last property is a characterization of real numbers. If z is real, then z = a + Oi, so z = a - Oi = a = z. On the other hand, if z = z, then a+ bi = a - bi which means b= -b sob= 0. Hence, z = a+Oi = a and is real. We now return to the business of zeros. Suppose we wish to find the zeros of f(x) =9x2 - 2x + 5. To solve the equation 2 - 2x + 5 = 0, we note the quadratic doesn't factor nicely, so we resort to the Quadratic Formula, Equation 2.5 and obtain -(-2)1 t (-2)2 - 4(1)(5)_ 21 t -16 21t 4i 2(1) 2 2 Two things are important to note. First, the zeros, 1 + 2i and 1 - 2i are complex conjugates. If ever we obtain non-real zeros to a quadratic function with real coefficients, the zeros will be a complex conjugate pair. (Do you see why?) Next, we note that in Example 3.4.1, part 6, we found (x - [1 + 2i])(x - [1 - 2i]) =9x2 - 2x + 5. This demonstrates that the factor theorem holds even for non-real zeros, i.e, x= 1 + 2i is a zero of f, and, sure enough, (x - [1 + 2i]) is a factor of f(x). It turns out that polynomial division works the same way for all complex numbers, real and non-real alike, and so the Factor and Remainder Theorems hold as well. But how do we know if a general polynomial has any complex zeros at all? We have many examples of polynomials with no real 6TTust us on this. 7See Section 9.3.  212 POLYNOMIAL FUNCTIONS zeros. Can there be polynomials with no zeros whatsoever? The answer to that last question is "No." and the theorem which provides that answer is The Fundamental Theorem of Algebra. THEOREM 3.13. The Fundamental Theorem of Algebra: Suppose f is a polynomial function with complex number coefficients of degree n> 1, then f has least one complex zero. The Fundamental Theorem of Algebra is an example of an 'existence' theorem in mathematics. Like the Intermediate Value Theorem, Theorem 3.1, the Fundamental Theorem of Algebra guarantees the existence of at least one zero, but gives us no algorithm to use in finding it. In fact, as we mentioned in Section 3.3, there are polynomials whose real zeros, though they exist, cannot be expressed using the 'usual' combinations of arithmetic symbols, and must be approximated. The authors are fully aware that the full impact and profound nature of the Fundamental Theorem of Algebra is lost on most students this level, and that's fine. It took mathematicians literally hundreds of years to prove the theorem in its full generality, and some of that history is recorded here. Note that the Fundamental Theorem of Algebra applies to polynomial functions with not only real coefficients, but, those with complex number coefficients as well. Suppose f is a polynomial of degree n with n> 1. The Fundamental Theorem of Algebra guarantees us at least one complex zero, z1, and, as such, the Factor Theorem guarantees that f (x) factors as f(x) =_(x - z1) qi(x) for a polynomial function qi, of degree exactly n - 1. If n - 1 > 1, then the Fundamental Theorem of Algebra guarantees a complex zero of qi as well, say z2, and so the Factor Theorem gives us qi(x) = (x - z2) q2(x), and hence f(x) =_(x - z1) (x - z2) q2(x). We can continue this process exactly n times, at which point our quotient polynomial q, has degree 0 so it's a constant. This argument gives us the following factorization theorem. THEOREM 3.14. Complex Factorization Theorem: Suppose f is a polynomial function with complex number coefficients. If the degree of f is n and n > 1, then f has exactly n complex zeros, counting multiplicity. If z1, z2, ..., zk are the distinct zeros of f, with multiplicities mi, m2, ... , mk, respectively, then f (x) = a (x - zi)ml (x - z2)m2 ...-(x - zk)mk. Note that the value a in Theorem 3.14 is the leading coefficient of f(x) (Can you see why?) and as such, we see that a polynomial is completely determined by its zeros, their multiplicities, and its leading coefficient. We put this theorem to good use in the next example. EXAMPLE 3.4.2. Let f (x) = 12x5 - 20x4 + 19x3 - 6x2 - 2x + 1. 1. Find all complex zeros of f and state their multiplicities. 2. Factor f(x) using Theorem 3.14 SOLUTION. 1. Since f is a fifth degree polynomial, we know we need to perform at least three successful divisions to get the quotient down to a quadratic function. At that point, we can find the remaining zeros using the Quadratic Formula, if necessary. Using the techniques developed in Section 3.3, we get  3.4 COMPLEX ZEROS AND THE FUNDAMENTAL THEOREM OF ALGEBRA 213 12 -20 19 -6 -2 1 2 1 6 -7 6 0 -1 12 -14 12 0 -2 0 2 1 6 -4 4 2 -i 12 -8 8 4 j 1 -4 4 -4 12 -12 12 0 Our quotient is 12x2 - 12x + 12, whose zeros we find to be 1i. From Theorem 3.14, we 21 know f has exactly 5 zeros, counting multiplicities, and as such we have the zero j with multiplicity 2, and the zeros - j1, v and 1-/, each of multiplicity 1. 3, 2 2 eahomutpiiy1 2. Applying Theorem 3.14, we are guaranteed that f factors as f(x) =12 x - -2x + - x - 1 _ x -1-i 2 3 2 2 A true test of Theorem 3.14 (and a student's mettle!) would be to take the factored form of f(x) in the previous example and multiply it out8 to see that it really does reduce to the formula f(x) = 12x5 - 20x4 + 19x3 - 6x2 - 2x + 1. When factoring a polynomial using Theorem 3.14, we say that it is factored completely over the complex numbers, meaning that it is impossible to factor the polynomial any further using complex numbers. If we wanted to completely factor f(x) over the real numbers then we would have stopped short of finding the nonreal zeros of f and factored f using our work from the synthetic division to write f (x) (x - )2 (x + }) (12x - 12x + 12), or f(x) = 12 (x- )2 (x+ ) ( - x + 1). Since the zeros of x2 - x + 1 are nonreal, we call x2 - x + 1 an irreducible quadratic meaning it is impossible to break it down any further using real numbers. The last two results of the section show us that, at least in theory, if we have a polynomial function with real coefficients, we can always factor it down enough so that any nonreal zeros come from irreducible quadratics. THEOREM 3.15. Conjugate Pairs Theorem: If f is a polynomial function with real number coefficients and z is a zero of f, then so is z. To prove the theorem, suppose f is a polynomial with real number coefficients. Specifically, let f (x) = anx + an_1x"-1 + .. . a2x2 + a1x + ao. If z is a zero of f, then f (z) = 0, which means anz" + an iz"- + . . . a2z2 + aiz + a= 0. Next, we consider f () and apply Theorem 3.12 below. 8You really should do this once in your life to convince yourself that all of the theory actually does work!  214 POLYNOMIAL FUNCTIONS f(Z) = an ()Th + an1 (z)" 1 + ...a2(z)2 + aiz + ao = anz+ an_1zn-1 + ... a2z2 + a1z + ao since ()Th = 7 = an-71+ a_1z-1+ ... -z+ a-+ a- since the coefficients are real = anz" + an_1z"-1 + ...a2z2 + a1z + a- since z -w = zw Sanz"+an_1z"-1 +...a2z2+a1z+ao sincez+ w = z+w = f(z) =0 This shows that z is a zero of f. So, if f is a polynomial function with real number coefficients, Theorem 3.15 tells us if a + bi is a nonreal zero of f, then so is a - bi. In other words, nonreal zeros of f come in conjugate pairs. The Factor Theorem kicks in to give us both (x - [a + bi]) and (x - [a - bi]) as factors of f (x) which means (x - [a + bi])(x - [a - bi]) =9x2 + 2ax + (a2 + b2) is an irreducible quadratic factor of f. As a result, we have our last result of the section. THEOREM 3.16. Real Factorization Theorem: Suppose f is a polynomial function with real number coefficients. Then f(x) can be factored into a product of linear factors corresponding to the real zeros of f and irreducible quadratic factors which give the nonreal zeros of f. We now present an example which pulls together all of the major ideas of this section. EXAMPLE 3.4.3. Let f (x) = z4 + 64. 1. Use synthetic division to show x = 2 + 2i is a zero of f. 2. Find the remaining complex zeros of f. 3. Completely factor f(x) over the complex numbers. 4. Completely factor f(x) over the real numbers. SOLUTION. 1. Remembering to insert the 0's in the synthetic division tableau we have 2-+-2i 1 0 0 0 64 1 2 + 2i 8i -16 + 16i -64 1 2+2i 8i -16+16i j 2. Since f is a fourth degree polynomial, we need to make two successful divisions to get a quadratic quotient. Since 2 + 2i is a zero, we know from Theorem 3.15 that 2 - 2i is also a zero. We continue our synthetic division tableau.  3.4 COMPLEX ZEROS AND THE FUNDAMENTAL THEOREM OF ALGEBRA 215 2+2i 1 0 0 0 64 1t 2+2i 8i -16+16i -64 2 - 2i 1 2 + 2i 8i -16 + 16i 0_ t 2-2i 8-8i 16-16i 1 4 8 Our quotient polynomial is 2+4x +8. Using the quadratic formula, we obtain the remaining zeros -2 + 2i and -2 - 2i. 3. Using Theorem 3.14, we get f (x) = (x - [2 - 2i])(x - [2 + 2i])(x - [-2 + 2i])(x - [-2 - 2i]). 4. We multiply the linear factors of f(x) which correspond to complex conjugate pairs. We find (x - [2 - 2i])(x - [2 + 2i]) =9x2 - 4x + 8, and (x - [-2 + 2i])(x - [-2 - 2i]) =9x2 + 4x + 8. Our final answer f (x) =_(x2- 4x + 8) (x2 + 4x + 8). D Our last example turns the tables and asks us to manufacture a polynomial with certain properties of its graph and zeros. EXAMPLE 3.4.4. Find a polynomial p of lowest degree that has integer coefficients and satisfies all of the following criteria: " the graph of y = p(x) touches the x-axis at (j, 0) " x= 3i is a zero of p. " as x - -oo, p(x) - -oo e as x - oo, p(x) -+ -oo SOLUTION. To solve this problem, we will need a good understanding of the relationship between the x-intercepts of the graph of a function and the zeros of a function, the Factor Theorem, the role of multiplicity, complex conjugates, the Complex Factorization Theorem, and end behavior of polynomial functions. (In short, you'll need most of the major concepts of this chapter.) Since the graph of p touches the x-axis at (., 0), we know x = is a zero of even multiplicity. Since we are after a polynomial of lowest degree, we need x to have multiplicity exactly 2. The Factor Theorem now tells us (x - 1)2 is a factor of p(x). Since x= 3i is a zero and our final answer is to have integer (real) coefficients, x= -3i is also a zero. The Factor Theorem kicks in again to give us (x -3i) and (x+3i) as factors of p(x). We are given no further information about zeros or intercepts so we conclude, by the Complex Factorization Theorem that p(x) = a (x - })2 (x - 3i)(x + 3i) for some real number a. Expanding this, we get p(x) = ax4 - 3 x3 + 8a92 - 6ax +a. In order to obtain integer coefficients, we know a must be an integer multiple of 9. Our last concern is end behavior. Since the leading term of p(x) is az4, we need a < 0 to get p(x) - -oC as x - too. Hence, if we choose x -9, we get p(x) =-9x4 + 6x3 - 8292 + 54x - 9. We can verify our handiwork using the techniques developed in this chapter. D  216 POLYNOMIAL FUNCTIONS This example concludes our study of polynomial functions.9 The last few sections have contained what is considered by many to be 'heavy' mathematics. Like a heavy meal, heavy mathematics takes time to digest. Don't be overly concerned if it doesn't seem to sink in all at once, and pace yourself on the exercises or you're liable to get mental cramps. But before we get to the exercises, we'd like to offer a bit of an epilogue. Our main goal in presenting the material on the complex zeros of a polynomial was to give the chapter a sense of completeness. Given that it can be shown that some polynomials have real zeros which cannot be expressed using the usual algebraic operations, and still others have no real zeros at all, it was nice to discover that every polynomial of degree n > 1 has n complex zeros. So like we said, it gives us a sense of closure. But the observant reader will note that we did not give any examples of applications which involve complex numbers. Students often wonder when complex numbers will be used in 'real-world' applications. After all, didn't we call i the imaginary unit? How can imaginary things be used in reality? It turns out that complex numbers are very useful in many applied fields such as fluid dynamics, electromagnetism and quantum mechanics, but most of the applications require Mathematics well beyond College Algebra to fully understand them. That does not mean you'll never be be able to understand them; in fact, it is the authors' sincere hope that all of you will reach a point in your studies when the glory, awe and splendor of complex numbers are revealed to you. For now, however, the really good stuff is beyond the scope of this text. We invite you and your classmates to find a few examples of complex number applications and see what you can make of them. A simple Internet search with the phrase 'complex numbers in real life' should get you started. Basic electronics classes are another place to look, but remember, they might use the letter j where we have used i. For the remainder of the text, we will restrict our attention to real numbers. We do this primarily because the first Calculus sequence you will take, ostensibly the one that this text is preparing you for, studies only functions of real variables. Also, lots of really cool scientific things don't require any deep understanding of complex numbers to study them, but they do need more Mathematics like exponential, logarithmic and trigonometric functions. We believe it makes more sense pedagogically for you to learn about those functions now and then, after you've completed the Calculus sequence, take a course in Complex Function Theory in your junior or senior year. It is in that course that the true power of the complex numbers is released. But for now, in order to fully prepare you for 1 life immediately after College Algebra, we will say that functions like f (x) = 1 have a domain of all real numbers, even though we know 2 + 1 = 0 has two complex solutions, namely x = ti. Because 2 + 1 > 0 for all real numbers x, the fraction X2+1 is never undefined in our real variable setting. 9With the exception of the exercises on the next page, of course.  3.4 COMPLEX ZEROS AND THE FUNDAMENTAL THEOREM OF ALGEBRA 217 3.4.1 EXERCISES 1. We know i2 = -1 which meansi3 _i2 - i = (-1) - i Use this information to simplify the following. i and i4 (-i2.i2 _)(_l) 1. (a) is (b) i304 (c) (2i)3 (d) (-i)23 2. Let z = 3+ 4i and w = 2 - i. Compute the following and express your answer in a + bi form. (a) z+w (b) w - z (c) z - w (d)z w (e)w z (f) w 3. Simplify the following. (a) -49 (b) v/ v-9 -16 (c) (-9)(-16) (d) 49 4 4. Find the complex solutions of the following quadratic equations. (a) 2 - 4x + 13 = 0 (b) 3x2 + 2x + 10 = 0 5. For each polynomial given below find all of its zeros, completely factor it over the real numbers and completely factor it over the complex numbers. (a) 2 - 2x + 5 (b) x3 - 2x2 + 9x - 18 (c) x3 + 6x2 + 6x + 5 (d) 3x3 - 13x2 + 43x - 13 (e) x4 + 92 + 20 (f) 4x4 - 4x3 + 13x2 - 12x + 3 (g) x4 - 2x3 + 272 - 2x + 26 (Hint: x= i is one of the zeros.) (h) 2x4 + 5x3 + 13x2 +7x+5 (Hint: x= -1+2i is a zero.) 6. Let z and w be arbitrary complex numbers. Show that i = zw and z = z. 7. With the help of your classmates, build a polynomial p with integer coefficients such that x =-2 - i is a zero of p, p has a local maximum at the point (4, 0) and p(x) -- -oc as 3 00  218 218 POLYNOMIAL FUNCTIONS 3.4.2 ANSWERS 1.(a) (b) -30 (l/41)76 =17 (c) (2i)3 - 8i3= -8i 2 2(a) z+w=5+3i (b) w - z-1 -5i (c) z~w=10+5i (d) z 2 +11.z w 5 5~ (ew 2 11. ()z 25 25Z (f) w3-2 - li (C) -V(-9)(-16) 3. (a) -49 - 7i (b) -9 -16 V Th4 12 =14i (3i) "(4i) =1212 12 (d) 49v/--4 =7 2i 4. (a) xv=2+3i 5. (a) x2- 2x + 5 =(x -(1 +2i))(v Zeros: xc= 1 1 2i - 1 29 fib 3 32 (1 -2) (b) x3~ - 292 + 9xv - 18 Zeros: xcv 2, ±3i (cv - 2) (92 + 9) _(xv - 2)(xv - 3i)(cv + 3i) (c) c3+6xv2+6xv+5 =(cv+5)(cv2+cv+1) (xv+5) (xv C 1 2i) (xc (. 1 1 3 Zeros: x= -5, x=--2f1 (d) 3c3 -132 + 43x- 13 (3x- 1)(2- 4xv+ 13) (3x- 1)(xv- (2 +3i)) (x 1 Zeros: xc= 3, xcv 2 1 3i - 2i) (2 -3i)) (e) xv4 + 9cv2 + 20 - (cv2 + 4) (cv2 + 5) Zeros: xcv ±2i, Ii'5 (cv - 2i) (x + 2i) (xv i v/5)(vi ) (f) 4cv4 - 4cv3 + 13cv2 - 12xv + 3= (2xv - 1)2(xv2 + 3)_ (2xv - 1)2(xv + v3i) (x v3iZ) Zeros:cv -,c2 x 3i (g) c4-2cv3+27c2-2x+26=(cv2 -2xv+ 26)(v2 +1) _(xv-(1+5i))(v (1-5i)(xv+ i) (xv-i) Zeros: xcv 1 1 Si, xcv +i (h) 24 + 53 +132 + 7x+ 5 =(2 + 2xv+5)(22 +cv+ 1) 2(x- (-1 +2i))(c- (-1 -2i)) (cv- (4 + 4))(cv '4 4 C 1 24) )  CHAPTER 4 RATIONAL FUNCTIONS 4.1 INTRODUCTION TO RATIONAL FUNCTIONS If we add, subtract or multiply polynomial functions according to the function arithmetic rules defined in Section 1.6, we will produce another polynomial function. If, on the other hand, we divide two polynomial functions, the result may not be a polynomial. In this chapter we study rational functions - functions which are ratios of polynomials. DEFINITION 4.1. A rational function is a function which is the ratio of polynomial functions. Said differently, r is a rational function if it is of the form p(x) where p and q are polynomial functions' "According to this definition, all polynomial functions are also rational functions. (Take q(x) = 1). As we recall from Section 1.5, we have domain issues anytime the denominator of a fraction is zero. In the example below, we review this concept as well as some of the arithmetic of rational expressions. EXAMPLE 4.1.1. Find the domain of the following rational functions. Write them in the form p(X) q(x) for polynomial functions p and q and simplify. 1. f (x) 2= - x+1 3 2. g(x) 2 - z+1 2x2-1 3. h(x) = 3x - 2 x2-1 2x24) -1 3x-2 .r(z) X2+ SOLUTION. 1. To find the domain of f, we proceed as we did in Section 1.5: we find the zeros of the denominator and exclude them from the domain. Setting x+1= 0 results in x= -1. Hence,  220 RATIONAL FUNCTIONS our domain is (-oc, -1) U (-1, oc). The expression f(x) is already in the form requested and when we check for common factors among the numerator and denominator we find none, so we are done. 2. Proceeding as before, we determine the domain of g by solving x +1 = 0. As before, we find the domain of g is (-oc, -1) U (-1, oc). To write g(x) in the form requested, we need to get a common denominator 3 g9x) = 2- x+1 2 3 1 x+1 (2)(x+1) 3 (1)(x + 1) - + 1 (2x + 2) - 3 x+1 2x - 1 x+1 This formula is also completely simplified. 3. The denominators in the formula for h(x) are both 2 - 1 whose zeros are x =il. As a result, the domain of h is (-oc, -1) U (-1, 1) U (1, oc). We now proceed to simplify h(x). Since we have the same denominator in both terms, we subtract the numerators. We then factor the resulting numerator and denominator, and cancel out the common factor. h(x) 2x2-1 3x-2 x2-1 x2-1 (2x2 - 1) - (3x - 2) x2-1 2x2 - 1 - 3x + 2 x2-1 2x2 - 3x + 1 x-1 (2x - 1)(x - 1) (x+1)(x-1) (2x - 1 2x + 1  4.1 INTRODUCTION TO RATIONAL FUNCTIONS 221 4. To find the domain of r, it may help to temporarily rewrite r(x) as 2x2 1 1 r(z) = x2 -1 F(~z3x-2 x21 We need to set all of the denominators equal to zero which means we need to solve not only x2 - 1 = 0, but also X-- = 0. We find x +1 for the former and x = 2 for the latter. Our domain is (-oc, -1) U (-1, 2) U (2, 1) U (1, oc). We simplify r(x) by rewriting the division as multiplication by the reciprocal and then simplifying r(x) 2x2 -1 3x-2 2x2-1 x2-1 2x2 -1 x2-_1 x2-1 3x-1 (2x2 - 1) (x2 _ 1) (x2 - 1) (3x - 1) (2x2 - 1) 2 (3x - 1) 2x2 1 3x - 1 A few remarks about Example 4.1.1 are in order. Note that the expressions for f(x), g(x) and h(x) work out to be the same. However, only two of these functions are actually equal. Recall that functions are ultimately sets of ordered pairs,1 and so for two functions to be equal, they need, among other things, to have the same domain. Since f(x) = g(x) and f and g have the same domain, they are equal functions. Even though the formula h(x) is the same as f(x), the domain of h is different than the domain of f, and thus they are different functions. We now turn our attention to the graphs of rational functions. Consider the function f(x) = 2x-- from Example 4.1.1. Using a graphing calculator, we obtain You should review Sections 1.2 and 1.4 if this statement caught you off guard.  222 RATIONAL FUNCTIONS Two behaviors of the graph are worthy of further discussion. First, note that the graph appears to 'break' at xc= -1. We know from our last example that xc= -1 is not in the domain of f which means f(-1) is undefined. When we make a table of values to study the behavior of f near x = -1 we see that we can get 'near' x = -1 from two directions. We can choose values a little less than -1, for example xc= -1.1, xc= -1.01, xc= -1.001, and so on. These values are said to 'approach -1 from the left.' Similarly, the values xc= -0.9, xc= -0.99, xc= -0.999, etc., are said to 'approach -1 from the right.' If we make two tables, we find that the numerical results confirm what we see graphically. v f(cv) (X, f ()) -1.1 32 (-1.1, 32) -1.01 302 (-1.01, 302) -1.001 3002 (-1.001, 3002) -1.0001 30002 (-1.001, 30002) v f () (X, f ()) -0.9 -28 (-0.9, -28) -0.99 -298 (-0.99, -298) -0.999 -2998 (-0.999, -2998) -0.9999 -29998 (-0.9999, -29998) As the x values approach -1 from the left, the function values become larger and larger positive numbers.2 We express this symbolically by stating as x - -1-, f(x) - oc. Similarly, using analogous notation, we conclude from the table that as x - -1+, f(x) - -oc. For this type of unbounded behavior, we say the graph of y= f(x) has a vertical asymptote of xc= -1. Roughly speaking, this means that near xc= -1, the graph looks very much like the vertical line xc= -1. Another feature worthy of note about the graph of y= f(cx) is it seems to 'level off' on the left and right hand sides of the screen. This is a statement about the end behavior of the function. As we discussed in Section 3.1, the end behavior of a function is its behavior as x as x attains larger3 and larger negative values without bound, x - -oc, and as x becomes large without bound, x - oo. Making tables of values, we find S f () (x,f(cv)) -10 i 2.3333 (-10, 2.3333) -100 i 2.0303 m (-100, 2.0303) -1000 i 2.0030 (-1000, 2.0030) -10000 i 2.0003 (-10000, 2.0003) c f () (x,f(cv)) 10 i 1.7273 (10, 1.7273) 100 i 1.9703 (100, 1.9703) 1000 i 1.9970 (1000,1.9970) 10000 i 1.9997 (10000,1.9997) From the tables, we see as v - -oc, f(x) 2+ and as v - oc, f(x) - 2-. Here the '+' means 'from above' and the '-' means 'from below'. In this case, we say the graph of y= f(cx) has a horizontal asymptote of y = 2. This means that the end behavior of f resembles the horizontal line y = 2, which explains the 'leveling off' behavior we see in the calculator's graph. We formalize the concepts of vertical and horizontal asymptotes in the following definitions. DEFINITION 4.2. The line xc= c is called a vertical asymptote of the graph of a function y f(x) if as x -> c or as x > c+, either f(x) - oo or f(x) - -oo. 2We would need Calculus to confirm this analytically. 3Here, the word 'larger' means larger in absolute value.  4.1 INTRODUCTION TO RATIONAL FUNCTIONS 223 DEFINITION 4.3. The line y = c is called a horizontal asymptote of the graph of a function y = f(x) if as x -- -ox or as x -- oc, either f(x) -- c or f(x) - c+. In our discussion following Example 4.1.1, we determined that, despite the fact that the formula for h(x) reduced to the same formula as f(x), the functions f and h are different, since x= 1 is in the domain of f, but x= 1 is not in the domain of h. If we graph h(x) = -21 -2 using a graphing calculator, we are surprised to find that the graph looks identical to the graph of y = f(x). There is a vertical asymptote at x= -1, but near x= 1, everything seem fine. Tables of values provide numerical evidence which supports the graphical observation. x h(x) (x,h(x)) 0.9 0.4210 (0.9, 0.4210) 0.99 0.4925 (0.99, 0.4925) 0.999 0.4992 (0.999, 0.4992) 0.9999 0.4999 (0.9999, 0.4999) x h(x) (x,h(x)) 1.1 ~ 0.5714 ~ (1.1, 0.5714) 1.01 ~ 0.5075 (1.01,0.5075) 1.001 ~R0.5007 R(1.001, 0.5007) 1.0001 ~ 0.5001 (1.0001, 0.5001) We see that as x - 1-, h(x) - 0.5- and as x - 1+, h(x) - 0.5+. In other words, the points on the graph of y = h(x) are approaching (1, 0.5), but since x= 1 is not in the domain of h, it would be inaccurate to fill in a point at (1,0.5). As we've done in past sections when something like this occurs,4 we put an open circle (also called a 'hole' in this case) at (1, 0.5). Below is a detailed graph of y = h(x), with the vertical and horizontal asymptotes as dashed lines. y 4For instance, graphing piecewise defined functions in Section 1.7. 5Stay tuned. In Calculus, we will see how these 'holes' can be 'plugged' when embarking on a more advanced study of contintuity.  224 RATIONAL FUNCTIONS Neither x= -1 nor x= 1 are in the domain of h, yet we see the behavior of the graph of y = h(x) is drastically different near these points. The reason for this lies in the second to last step when we simplified the formula for h(x) in Example 4.1.1. We had h(x) =(2x.1)(x-1) The reason x= -1 is not in the domain of h is because the factor (x + 1) appears in the denominator of h(x); similarly, x = 1 is not in the domain of h because of the factor (x - 1) in the denominator of h(x). The major difference between these two factors is that (x - 1) cancels with a factor in the numerator whereas (x + 1) does not. Loosely speaking, the trouble caused by (x - 1) in the denominator is canceled away while the factor (x + 1) remains to cause mischief. This is why the graph of y = h(x) has a vertical asymptote at x= -1 but only a hole at x= 1. These observations are generalized and summarized in the theorem below, whose proof is found in Calculus. THEOREM 4.1. Location of Vertical Asymptotes and Holes:' Suppose r is a rational function which can be written as r(x) - where p and q have no common zeros. b Let c be a real number q(x) which is not in the domain of r. " If q(c) $ 0, then the graph of y = r(x) has a hole at (c, P()). " If q(c) = 0, then the the line x= c is a vertical asymptote of the graph of y = r(z). aOr, 'How to tell your asymptote from a hole in the graph.' bIn other words, r(x) is in lowest terms. In English, Theorem 4.1 says if x= c is not in the domain of r but, when we simplify r(z), it no longer makes the denominator 0, then we have a hole at x= c. Otherwise, we have a vertical asymptote. EXAMPLE 4.1.2. Find the vertical asymptotes of, and/or holes in, the graphs of the following rational functions. Verify your answers using a graphing calculator. 2x 2 -x-6 1. f () 23 3. h) 2 +9 2 -x-6 4x2-x-6 2. g(x) -=(x 4. ) = x2_9 x2+4x+4 SOLUTION. 1. To use Theorem 4.1, we first find all of the real numbers which aren't in the domain of f. To do so, we solve 2 - 3 = 0 and get x = v5. Since the expression f(x) is in lowest terms, there is no cancellation possible, and we conclude that the lines x= -v/ and x = are vertical asymptotes to the graph of y = f(x). The calculator verifies this claim. 2. Solving 2 - 9 = 0 gives x 1=i3. In lowest terms g(x) =29-6 = -.2) x±2 Since z -3 continues to make trouble in the denominator, we know the line z -3 is a vertical asymptote of the graph of y =g(x). Since z 3 no longer produces a 0 in the denominator, we have a hole at z 3. To find the y-coordinate of the hole, we substitute z 3 into x+  4.1 INTRODUCTION TO RATIONAL FUNCTIONS and find the hole is at (3, ). When we graph y 225 g(x) using a calculator, we clearly see the vertical asymptote at x -3, but everything seems calm near x 3. J The graph of y f(x) The graph of y g(x) 3. The domain of h is all real numbers, since x2 + 9 0 has no real solutions. Accordingly, the graph of y = h(x) is devoid of both vertical asymptotes and holes. 4. Setting x2 + 4x + 4 = 0 gives us x = - we see r(z) =f +4x+4 (x2) denominator of the reduced function, we which the calculator confirms. 2 as the only real number of concern. Simplifying, -. Since x = -2 continues to produce a 0 in the 2 is a vertical asymptote to the graph, The graph of y h(x) The graph of y r(x) H Our next example gives us a physical interpretation of a vertical asymptote. This type of model arises from a family of equations cheerily named 'doomsday' equations.6 The unfortunate name will make sense shortly. EXAMPLE 4.1.3. A mathematical model for the population P, in thousands, of a certain species of bacteria, t days after it is introduced to an environment is given by P(t) 1. Find and interpret P(0). 2. When will the population reach 100,000? 6This is a class of Calculus equations in which a population grows very rapidly. 10 0 m, and hence n - m is a positive number and r(c) ax m, which becomes unbounded as v - boo. As we said before, if a rational function has a horizontal asymptote, then it will have only one. (This is not true for other types of functions we shall see in later chapters.) EXAMPLE 4.1.4. Determine the horizontal asymptotes, if any, of the graphs of the following func- tions. Verify your answers using a graphing calculator. 5xv c2-4 6x3-3x+1 1. f (c) =2 2. g(x) = 3. h(x) = X+1x+1 5 - 2x3 SOLUTION. 1. The numerator of f(x) is 5x, which is degree 1. The denominator of f(x) is x2 + 1, which is degree 2. Applying Theorem 4.2, y = 0 is the horizontal asymptote. Sure enough, as c - boo, the graph of y= f(cx) gets closer and closer to the c-axis. 2. The numerator of g(x), x2 - 4, is degree 2, but the degree of the denominator, xv+ 1, is degree 1. By Theorem 4.2, there is no horizontal asymptote. From the graph, we see the graph of 7Note that as x -~ -oo, f(x) -~ 2±, whereas as x -e oo, f(x) -~ 2-. We write f(x) -~ 2 if we are unconcerned from which direction the function values f(x) approach the number 2.  228 RATIONAL FUNCTIONS y = g(x) doesn't appear to level off to a constant value, so there is no horizontal asymptote.8 3. The degrees of the numerator and denominator of h(x) are both three, so Theorem 4.2 tells us y = = -3 is the horizontal asymptote. The calculator confirms this. J I The graph of y f(x) The graph of y g(x) The graph of y h(x) H Our last example of the section gives us a real-world application of a horizontal asymptote. Though the population below is more accurately modeled with the functions in Chapter 6, we approximate it9 using a rational function. EXAMPLE 4.1.5. The number of students, N, at local college who have had the flu t months after the semester begins can be modeled by the formula N(t) 500 - 45 fort > 0. 1. Find and interpret N(0). 2. How long will it take until 300 students will have had the flu? 3. Determine the behavior of N as t - 0c. context of the problem. Interpret this result graphically and within the SOLUTION. 1. N(0) = 500 - 450 1+3(0) have had the flu. 50. This means that at the beginning of the semester, 50 students 2. We set N(t) = 300 to get 500 - +3 300 and solve. Isolating the fraction gives 450 = 200. Clearing denominators gives 450 = 200(1 + 3t). Finally, we get t = 12. This means it will take 2 months, or about 13 days, for 300 students to have had the flu. 3. To determine the behavior of N as t - 0c, we can use a table. 8The graph does, however, seem to resemble a non-constant line as x -> ±oo. We will discuss this phenomenon in the next section. 9Using techniques you'll see in Calculus.  4.1 INTRODUCTION TO RATIONAL FUNCTIONS 229 t N(t) 10 485.48 100 498.50 1000 ~ 499.85 10000 ~ 499.98 The table suggests that as t - oc, N(t) - 500. (More specifically, 500-.) This means as time goes by, only a total of 500 students will have ever had the flu. D  230 RATIONAL FUNCTIONS 4.1.1 EXERCISES 1. For each rational function f given below: " Find the domain of f. " Identify any vertical asymptotes of the graph of y = f(x) and describe the behavior of the graph near them using proper notation. " Identify any holes in the graph. " Find the horizontal asymptote, if it exists, and describe the end behavior of f using proper notation. (a) f (x) 3xx6 (d) f (x)=2x1 _3+ 7z z+7 (b) f (x) = 5-2x(e) f () ( +3)2 sXx3 +1 (c) f() 2+X-12(f) f (z) = 2. In Exercise 9 in Section 1.5, the population of Sasquatch in Portage County was modeled by the function P(t) = 50, where t = 0 represents the year 1803. Find the horizontal asymptote of the graph of y = P(t) and explain what it means. 3. In Exercise 8 in Section 3.1, we fit a few polynomial models to the following electric circuit data. (The circuit was built with a variable resistor. For each of the following resistance values (measured in kilo-ohms, kQ), the corresponding power to the load (measured in milliwatts, mW) is given in the table below.)10 Resistance: (kQ) 1.012 2.199 3.275 4.676 6.805 9.975 Power: (mW) 1.063 1.496 1.610 1.613 1.505 1.314 Using some fundamental laws of circuit analysis mixed with a healthy dose of algebra, we can derive the actual formula relating power to resistance. For this circuit, it is P(x) (x±3 9)2, where x is the resistance value, x > 0. (a) Graph the data along with the function y = P(x) on your calculator. (b) Approximate the maximum power that can be delivered to the load. What is the corre- sponding resistance value? (c) Find and interpret the end behavior of P(x) as x - xoo. 4. In his now famous 1919 dissertation The Learning Curve Equation, Louis Leon Thurstone presents a rational function which models the number of words a person can type in four ioThe authors wish to thank Don Anthan and Ken White of Lakeland Community College for devising this problem and generating the accompanying data set.  4.1 INTRODUCTION TO RATIONAL FUNCTIONS 231 minutes as a function of the number of pages of practice one has completed. (This paper, which is now in the public domain and can be found here, is from a bygone era when students at business schools took typing classes on manual typewriters.) Using his original notation and original language, we have Y (XP)where L is the predicted practice limit in terms of speed units, X is pages written, Y is writing speed in terms of words in four minutes, P is equivalent previous practice in terms of pages and R is the rate of learning. In Figure 5 of the paper, he graphs a scatter plot and the curve Y = 216(X+19) Discuss this equation with your X+148.Dicsthseutowihyr classmates. How would you update the notation? Explain what the horizontal asymptote of the graph means. You should take some time to look at the original paper. Skip over the computations you don't understand yet and try to get a sense of the time and place in which the study was conducted.  232 RATIONAL FUNCTIONS 4.1.2 ANSWERS 1. (a) f ()3j6 Domain: (-oo,2) U (2,oo) Vertical asymptote: x = 2 As x - 2-, f(x) - -oo As x - 2+, f(x) - oo No holes in the graph Horizontal asymptote: y = As x - -oo, f(x) -} As x - oo, f(x) + (b) f (x) = 2 7 Domain: (-oc, 2) U (2, oo) Vertical asymptote: x- 2 As x - , f(z) - oo As x - + f (X) - -oo No holes in the graph Horizontal asymptote: y As x - -oo, f(x) --+ As x - oo, f(x) - -2 (d) f() =21 Domain: (-oo,oo) No vertical asymptotes No holes in the graph Horizontal asymptote: y = 0 As x - -oc, f(x) - 0- As x - oo, f (x) -- 0+ 1 (e) f (x) =(x +32 Domain: (-oc, -3) U (-3, o) Vertical asymptote: x =-3 As x - -3-, f(x) - oo As x -3+, f(x) - oo No holes in the graph Horizontal asymptote: y = 0 As x - -oc, f (x) - 0- As x - oo, f(z) - 0+ 7 (c) f X 2 + - 12 (x +4)(x - 3) Domain: (-oc, -4) U (-4, 3) U (3, ox) Vertical asymptotes: x = -4, x= 3 As x - -4-, f () - -oo As x -4+, f (X) - oc As x - 3-, f () - -oo As x > 3+, f(x) - oo No holes in the graph Horizontal asymptote: y = 0 As x - -oc, f () - 0- As x oc, f (x) - 0+ 2. The horizontal asymptote of the graph of P(t) predicts the population of Sasquatch in Portage 3+1 (f) f(X) = Domain: (-oc, -1) U (-1, 1) U (1, o) Vertical asymptote: x = 1 As x - 1-, f(x) - -oc As x - 1+, f (x) - o0 Hole at (-1, -3) No horizontal asymptote As x - -oo, f(x) -- -oo As x oo, f(z) - oc 150t is y 150 and it means that the model t+15 County will never exceed 150. "This is hard to see on the calculator, but trust me, the graph is below the x-axis to the left of x 7.  4.1 INTRODUCTION TO RATIONAL FUNCTIONS 233 3. (a) (b) The maximum power is approximately 1.603 mW which corresponds to 3.9 kf2. (c) As x - 0c, P(x) -* 0+ which means as the resistance increases without bound, the power diminishes to zero.  234 RATIONAL FUNCTIONS 4.2 GRAPHS OF RATIONAL FUNCTIONS In this section, we take a closer look at graphing rational functions. In Section 4. 1, we learned that the graphs of rational functions may include vertical asymptotes, holes in the graph, and horizontal asymptotes. Theorems 4.1 and 4.2 tell us exactly when and where these behaviors will occur, and if we combine these results with what we already know about graphing functions, we will quickly be able to generate reasonable graphs of rational functions. One of the standard tools we will use is the sign diagram which was first introduced in Section 2.4, and then revisited in Section 3.1. In those sections, we operated under the belief that a function couldn't change its sign without its graph crossing through the x-axis. The major theorem we used to justify this belief was the Intermediate Value Theorem, Theorem 3.1. It turns out the Intermediate Value Theorem applies to all continuous functions,' not just polynomials. Although rational functions are continuous on their domains,2 Theorem 4.1 tells us vertical asymptotes and holes occur at the values excluded from their domains. In other words, rational functions aren't continuous at these excluded values which leaves open the possibility that the function could change sign without crossing through the x-axis. Consider the graph of y = h(x) from Example 4.1.1, recorded below for convenience. We have added its x-intercept at (2, 0) for the discussion that follows. Suppose we wish to construct a sign diagram for h(x). Recall that the intervals where h(x) > 0, or (+), correspond to the x-values where the graph of y = h(x) is above the x-axis; the intervals on which h(x) < 0, or (-) correspond to where the graph is below the x-axis. Y 7-- 6-- 5-- 4-- 3-- - --------I --------- (+) T (-) 0 (+) T (+) I 1 C | | -1 1 1 1 1 1 11 2 -4 -3 -2 1 1 2 3 4 x -1 2 14 1 5.. 15 1 6 As we examine the graph of y= h(x), reading from left to right, we note that from (-oo, -1), 1Recall that, for our purposes, this means the graphs are devoid of any breaks, jumps or holes 2Another result from Calculus.  4.2 GRAPHS OF RATIONAL FUNCTIONS 235 the graph is above the x-axis, so h(x) is (+) there. At x= -1, we have a vertical asymptote, at which point the graph 'jumps' across the x-axis. On the interval (-1, j), the graph is below the x-axis, so h(x) is (-) there. The graph crosses through the x-axis at (1, 0) and remains above the x-axis until x= 1, where we have a 'hole' in the graph. Since h(1) is undefined, there is no sign here. So we have h(x) as (+) on the interval (j, 1). Continuing, we see that on (1, oc), the graph of y = h(x) is above the x-axis, and so we mark (+) there. To construct a sign diagram from this information, we not only need to denote the zero of h, but also the places not in the domain of h. As is our custom, we write '0' above 1 on the sign diagram to remind us that it is a zero of h. We need a different notation for -1 and 1, and we have chosen to use '?' - a nonstandard symbol called the interrobang. We use this symbol to convey a sense of surprise, caution, and wonderment - an appropriate attitude to take when approaching these points. The moral of the story is that when constructing sign diagrams for rational functions, we include the zeros as well as the values excluded from the domain. Steps for Constructing a Sign Diagram for a Rational Function Suppose r is a rational function. 1. Place any values excluded from the domain of r on the number line with an '?' above them. 2. Find the zeros of r and place them on the number line with the number 0 above them. 3. Choose a test value in each of the intervals determined in steps 1 and 2. 4. Determine the sign of r(z) for each test value in step 3, and write that sign above the corresponding interval. We now present our procedure for graphing rational functions and apply it to a few exhaustive examples. Please note that we decrease the amount of detail given in the explanations as we move through the examples. The reader should be able to fill in any details in those steps which we have abbreviated. Steps for Graphing Rational Functions Suppose r is a rational function. 1. Find the domain of r. 2. Reduce r(z) to lowest terms, if applicable. 3. Find the x- and y-intercepts of the graph of y = r(x), if they exist. 4. Determine the location of any vertical asymptotes or holes in the graph, if they exist. Analyze the behavior of r on either side of the vertical asymptotes, if applicable. 5. Analyze the end behavior of r. Use long division, as needed. 6. Use a sign diagram and plot additional points, as needed, to sketch the graph of y =r(x).  236 RATIONAL FUNCTIONS 3xv EXAMPLE 4.2.1. Sketch a detailed graph of f(x) = 2- x - 4 SOLUTION. We follow the six step procedure outlined above. 1. As usual, we set the denominator equal to zero to get 2 - 4 = 0. We find x = t2, so our domain is (-oc, -2) U (-2, 2) U (2, oc). 2. To reduce f(x) to lowest terms, we factor the numerator and denominator which yields f (x) = .2x+2 There are no common factors which means f(x) is already in lowest terms. 3. To find the x-intercepts of the graph of y = f (x), we set y = f (x) = 0. Solving x_2)x±2)= 0 results in xc= 0. Since xc= 0 is in our domain, (0, 0) is the x-intercept. To find the y-intercept, we set xc= 0 and find y = f(0) = 0, so that (0,0) is our y-intercept as well.? 4. The two numbers excluded from the domain of f are xc= -2 and xc= 2. Since f(x) didn't reduce at all, both of these values of x still cause trouble in the denominator, and so, by Theorem 4.1, cx= -2 and xc= 2 are vertical asymptotes of the graph. We can actually go a step farther at this point and determine exactly how the graph approaches the asymptote near each of these values. Though not absolutely necessary,4 it is good practice for those heading off to Calculus. For the discussion that follows, it is best to use the factored form of f(=) = 3x " The behavior of y= f(cx) as v - -2: Suppose x - -2-. If we were to build a table of values, we'd use x-values a little less than -2, say -2.1, -2.01 and -2.001. While there is no harm in actually building a table like we did in Section 4.1, we want to develop a 'number sense' here. Let's think about each factor in the formula of f(x) as we imagine substituting a number like xc= -2.000001 into f(x). The quantity 3x would be very close to -6, the quantity (xv -2) would be very close to -4, and the factor (xv+2) would be very close to 0. More specifically, (x + 2) would be a little less than 0, in this case, -0.000001. We will call such a number a 'very small (-)', 'very small' meaning close to zero in absolute value. So, mentally, as x - -2-, we estimate 3x -6 3 (x - 2)(x + 2) (-4) (very small (-)) 2 (very small (-)) Now, the closer x gets to -2, the smaller (x + 2) will become, and so even though we are multiplying our 'very small (-)' by 2, the denominator will continue to get smaller and smaller, and remain negative. The result is a fraction whose numerator is positive, but whose denominator is very small and negative. Mentally, _3 _ 3 f(x)t 3 _______(_ very big (-) 2 (very small (-)) very small (-) 3As we mentioned at least once earlier, since functions can have at most one y-intercept, once we find (0, 0) is on the graph, we know it is the y-intercept. 4The sign diagram in step 6 will also determine the behavior near the vertical asymptotes.  4.2 GRAPHS OF RATIONAL FUNCTIONS 237 The term 'very big (-)' means a number with a large absolute value which is negative.5 What all of this means is that as x - -2-, f(x) - -oc. Now suppose we wanted to determine the behavior of f(x) as x - -2+. If we imagine substituting something a little larger than -2 in for x, say -1.999999, we mentally estimate -6 (-4) (very small (+)) 33very big (+) 2 (very small (+)) very small (+) We conclude that as x - +-2+, f(x) -- oxc. " The behavior of y = f(x) as x -- 2: Consider x - 2-. We imagine substituting x = 1.999999. Approximating f(x) as we did above, we get 6 3 3 f lx) very big (-) (very small (-)) (4) 2 (very small (-)) very small (-) We conclude that as x -2-, x = 2.000001, we get f(x) Graphically, we have that near xz f (x) - -oc. Similarly, as x - 2+, we imagine substituting very smal (±) very big (+). So as x - 2+, f(x) - oc. = -2 and x = 2 the graph of y= f(x) looks like6 y I I I I I I I I I _3 I I I I I I I I -1 ii i ;p 3 x 5. Next, we determine the end behavior of the graph of y = f(x). Since the degree of the numerator is 1, and the degree of the denominator is 2, Theorem 4.2 tells us that y = 0 is the horizontal asymptote. As with the vertical asymptotes, we can glean more detailed information using 'number sense'. For the discussion below, we use the formula f(x) = X24. * The behavior of y = f(x) as x - -oc: If we were to make a table of values to discuss the behavior of f as x - -oc, we would substitute very 'large' negative numbers in for x, say, for example, x = -1 billion. The numerator 3x would then be -3 billion, whereas 5The actual retail value of f(-2.000001) is approximately -1,500,000. 6We have deliberately left off the labels on the y-axis because we know only the behavior near x= +2, not the actual function values.  238 RATIONAL FUNCTIONS the denominator 2 - 4 would be (-1 billion)2 - 4, which is pretty much the same as 1(billion)2. Hence, -3 billion 3 f (-1 billion) - very small (-) 1(billion)2 billion Notice that if we substituted in x = -1 trillion, essentially the same kind of cancellation would happen, and we would be left with an even 'smaller' negative number. This not only confirms the fact that as x - -oo, f (x) - 0, it tells us that f (x) - 0-. In other words, the graph of y= f(x) is a little bit below the x-axis as we move to the far left. " The behavior of y = f(x) as x -oc: On the flip side, we can imagine substituting very large positive numbers in for x and looking at the behavior of f(x). For example, let x = 1 billion. Proceeding as before, we get 3 billion 3 f (1 billion) 3 i ~o~ very small (+) 1(billion)2 billion The larger the number we put in, the smaller the positive number we would get out. In other words, as x - oo, f(x) - 0+, so the graph of y= f(x) is a little bit above the x-axis as we look toward the far right. Graphically, we have? -1 6. Lastly, we construct a sign diagram for f(x). The x-values excluded from the domain of f are +=i2, and the only zero of f is x = 0. Displaying these appropriately on the number line gives us four test intervals, and we choose the test values8 we x = -3, x = -1, x = 1, and x = 3. We find f(-3) is (-), f(-1) is (+), f(1) is (-), and f(3) is (+). Combining this with our previous work, we get the graph of y= f(x) below. 7As with the vertical asymptotes in the previous step, we know only the behavior of the graph as x - too. For that reason, we provide no x-axis labels. 8In this particular case, we can eschew test values, since our analysis of the behavior of f near the vertical asymptotes and our end behavior analysis have given us the signs on each of the test intervals. In general, however, this won't always be the case, so for demonstration purposes, we continue with our usual construction.  4.2 GRAPHS OF RATIONAL FUNCTIONS 239 | 3 | (-) T (+) 0 (-) T (+) i 1-- i -3 -1 1 3 1 -1- 1 I-3-- A couple of notes are in order. First, the graph of y = f(x) certainly seems to possess symmetry with respect to the origin. In fact, we can check f(-x) = -f(x) to see that f is an odd function. In some textbooks, checking for symmetry is part of the standard procedure for graphing rational functions; but since it happens comparatively rarely9 we'll just point it out when we see it. Also note that while y = 0 is the horizontal asymptote, the graph of f nevertheless crosses the x-axis at (0,0). The myth that graphs of rational functions can't cross their horizontal asymptotes is completely false, as we shall see again in our next example. 292 - 3x - 5 EXAMPLE 4.2.2. Sketch a detailed graph of g(x) = 2- x -x-6 SOLUTION. 1. Setting 2 - x - 6 = 0 gives x =-2 and x = 3. Our domain is (-oo, -2) U (-2,3) U (3, oo). 2. Factoring g(x) gives g(x) >=(2-5)x). There is no cancellation, so g(x) is in lowest terms. 3. To find the x-intercept we set y = g(x) = 0. Using the factored form of g(x) above, we find the zeros to be the solutions of (2x - 5)(x + 1) = 0. We obtain x = 2 and x = -1. Since both of these numbers are in the domain of g, we have two x-intercepts, (2, 0) and (-1,0). To find the y-intercept, we set x= 0 and find y = g(0) = , so our y-intercept is (0, 6). 4. Since g(x) was given to us in lowest terms, we have, once again by Theorem 4.1 vertical asymptotes x = -2 and x= 3. Keeping in mind g(x) (x-_)(x±), we proceed to our analysis near each of these values. " The behavior of y = g(x) asx- + -2: As x- -2-, we imagine substituting a number a little bit less than -2. We have g~x) (-)-1 ______ very big (+) (-5)(very small (-)) very small (+) 9And Jeff doesn't think much of it to begin with...  240 RATIONAL FUNCTIONS so as x - -2-, g(x) -- oo. On the flip side, as x - -2+, we get 9 g~x) very big (-) very small (-) so g(x ) - -oo. " The behavior of y = g(x) as x -- 3: As x - 3-, we imagine plugging in a number just shy of 3. We have (1)(4) 4. gWx) ( )_ very big (-) (very small (-))(5) very small (-) Hence, as x -- 3-, g(x) - -oo. As x - 3+, we get 4 g()~ very big (+) very small (+) so g(x) -- oo. Graphically, we have (again, without labels on the y-axis) y II -3 I-1 KI 1 2 I I I I I I I 5. Since the degrees of the numerator and denominator of g(x) are the same, we know from Theorem 4.2 that we can find the horizontal asymptote of the graph of g by taking the ratio of the leading terms coefficients, y = 2 = 2. However, if we take the time to do a more detailed analysis, we will be able to reveal some 'hidden' behavior which would be lost otherwise.-m As in the discussion following Theorem 4.2, we use the result of the long division (2x2 - 3x - 5)+ (2 - x - 6) to rewrite g(x) = -- as g(x) 2 - We focus our attention on the term 2x-7 ioThat is, if you use a calculator to graph. Once again, Calculus is the ultimate graphing power tool.  4.2 GRAPHS OF RATIONAL FUNCTIONS 241 " The behavior of y = g(x) as x - -oc: If imagine substituting x= -1 billion into x2--6, we estimate 2-67 ibillion very small (-).11 Hence, g(x) = 2 - x 6 2 - very small (-) = 2 +very small (+) In other words, as x - -oo, the graph of y = g(x) is a little bit above the line y = 2. " The behavior of y = g(x) as x -oo. To consider 2x--6 as x - oo, we imagine substituting x = 1 billion and, going through the usual mental routine, find 2 x very small (+) X -x-6 Hence, g(x) 2 - very small (+), in other words, the graph of y = g(x) is just below the line y =2 as x - oo. On y = g(x), we have (again, without labels on the x-axis) I:1 . 6. Finally we construct our sign diagram. We place an T above x= -2 and x= 3, and a '0' above x = and x= -1. Choosing test values in the test intervals gives us f(x) is (+) on the intervals (-oo, -2), (-1,j), and (3, oo), and (-) on the intervals (-2, -1) and (2, 3). As we piece together all of the information, we note that the graph must cross the horizontal asymptote at some point after x= 3 in order for it to approach y = 2 from underneath. This is the subtlety that we would have missed had we skipped the long division and subsequent end behavior analysis. We can, in fact, find exactly when the graph crosses y = 2. As a result of the long division, we have g(x) = 2 - x2 _6. For g(x) = 2, we would need x27_6 = 0. This gives x - 7 = 0, or x= 7. Note that x - 7 is the remainder when 2x2 - 3x - 5 is divided by 2 - x -6, and so it makes sense that for g(x) to equal the quotient 2, the remainder from the division must be 0. Sure enough, we find g(7) = 2. Moreover, it stands to reason that g must attain a relative minimum at some point past x= 7. Calculus verifies that at x= 13, we have such a minimum at exactly (13, 1.96). The reader is challenged to find calculator windows which show the graph crossing its horizontal asymptote on one window, and the relative minimum in the other. 11I1 the denominator, we would have (1billion)2 - 1billion - 6. It's easy to see why the 6 is insignificant, but to ignore the 1 billion seems criminal. However, compared to (1 billion)2, it's on the insignificant side; it's 1018 versus 109. We are once again using the fact that for polynomials, end behavior is determined by the leading term, so in the denominator, the x2 term wins out over the x term.  242 RATIONAL FUNCTIONS (+) ? (-) 0 (+) 0 (-) T (+) C I I I I ) -2 -1 5 3 2 t x Our next example gives us not only a hole in the graph, but also some slightly different end behavior. 2x3 + 592 + 4x + 1 EXAMPLE 4.2.3. Sketch a detailed graph of h(x) =32 T + 3x + 2 SOLUTION. 1. For domain, you know the drill. Solving 2 + 3x + 2 = 0 gives x answer is (-oc, -2) U (-2, -1) U (-1, oc). -2 and x= -1. Our 2. To reduce h(x), we need to factor the numerator and denominator. To factor the numerator, we use the techniques12 set forth in Section 3.3 and we get 2x3 + 5x2 + 4x + 1 x2+3x+2 (2x + 1)(x + 1)2 (x + 2)(x + 1) (2x + 1)(x + 1)/ ( + 2)fzMT (2x + 1)(x + 1) x+2 We will use this reduced formula for h(x) as long as we're not substituting x this exclusion specific, we write h(x) (2x+1)(x+) , z #-1. x+2 -1. To make 3. To find the x-intercepts, as usual, we set h(x) = 0 and solve. Solving (2x+1)(x+)= 0 yields x+2 x = -j and x= -1. The latter isn't in the domain of h, so we exclude it. Our only x- intercept is (-i, 0). To find the y-intercept, we set x= 0. Since 0 # -1, we can use the reduced formula for h(x) and we get h(0) = for a y-intercept of (0, 2). 4. From Theorem 4.1, we know that since x= -2 still poses a threat in the denominator of the reduced function, we have a vertical asymptote there. As for x= -1, we note the factor (x + 1) was canceled from the denominator when we reduced h(x), and so it no longer causes trouble there. This means we get a hole when x -1. To find the y-coordinate of the hole, we substitute x = -1 into (2x+) +), per Theorem 4.1 and get 0. Hence, we have a hole on 12Bet you never thought you'd never see that stuff again before the Final Exam!  4.2 GRAPHS OF RATIONAL FUNCTIONS 243 the x-axis at (-1, 0). It should make you uncomfortable plugging x = -1 into the reduced formula for h(x), especially since we've made such a big deal concerning the stipulation about not letting x= -1 for that formula. What we are really doing is taking a Calculus short-cut to the more detailed kind of analysis near x- -1 which we will show below. Speaking of which, for the discussion that follows, we will use the formula h(x) _(2x+1)(x+1) -1. x+2 " The behavior of y = h(x) as x - -2: As x - -2-,we imagine substituting a number a little bit less than -2. We have h(x) (very sal (-)) 3(very sman ( very big (-) and so as x - -2-, h(x) - -oc. On the other side of -2, as x - -2+, we find that h(x) very sa (+) very big (+), so h(x) - oc. " The behavior of y = h(x) as x - -1. As x - -1-, we imagine plugging in a number a bit less than x= -1. We have h(x) (-1)(very small (-)) = very small (+) Hence, as 1 x - -1-, h(x) - 0+. This means, as x - -1-, the graph is a bit above the point (-1, 0). As x - -1+, we get h(x) (-l)(very smal (±)) =very small (-). This gives us that as x - -1+, h(x) - 0-, so the graph is a little bit lower than (-1, 0) here. Graphically, we have y -3 x 5. For end behavior, we note that the degree of the numerator of h(x), 2x3 + 52 + 4x + 1 is 3, and the degree of the denominator, 92+3x +2, is 2. Theorem 4.2 is of no help here, since the degree of the numerator is greater than the degree of the denominator. That won't stop us, however, in our analysis. Since for end behavior we are considering values of x as x -too, we are far enough away from x -1 to use the reduced formula, h(x) _ (2x+)+(xl), # -1. To perform long division, we multiply out the numerator and get h(x) = 2x2+32+1, z # -1, and, as a result, we rewrite h(x) = 2x - 1 + x+2 , # -1. As in the previous example, we focus our attention on the term generated from the remainder, x2 * The behavior of y =h(x) as x -~ -coo: Substituting x =-1 billion into x+, we get the estimate -1 binion very small (-). Hence, h(x) =2z-1+< ~ 2x-1+very small (-). This means the graph of y =h~x) is a little bit below the line y =2x - 1 as x -~ -oo.  244 RATIONAL FUNCTIONS * The behavior of y = h(x) asx-oo: If x oo, then x+2 very small (+). This means h(x) 2x - 1 + very small (+), or that the graph of y = h(x) is a little bit above the line y = 2x - 1 as x - oc. This is end behavior unlike any we've ever seen. Instead of approaching a horizontal line, the graph is approaching a slanted line. For this reason, y = 2x - 1 is called a slant asymptote13 of the graph of y = h(x). A slant asymptote will always arise when the degree of the numerator is exactly one more than the degree of the denominator, and there's no way to determine exactly what it is without going through the long division. Graphically we have y 4 3 2 1 I / / I S /' ' x -1 /-3 -4 6. To make our sign diagram, we place an '?' above x= -2 and x= -1 and a '0' above x = - . On our four test intervals, we find h(x) is (+) on (-2, -1) and (--j, o) and h(x) is (-) on (-oc, -2) and (-1, -i). Putting all of our work together yields the graph below. 1. AYEM (-) ? (+) T (-) 0 (+) CI II) -2 -1 _1 2 13Also called an 'oblique' asymptote in some texts.  4.2 GRAPHS OF RATIONAL FUNCTIONS 245 We could ask whether the graph of y = h(x) crosses its slant asymptote. From the formula h(x) = 2x -1+ 2 x -x -1, we see that if h(x) = 2x - 1, we would have +2= 0. Since this will never happen, we conclude the graph never crosses its slant asymptote.14 D We end this section with an example that shows it's not all pathological weirdness when it comes to rational functions and technology still has a role to play in studying their graphs at this level. z1 + 1 EXAMPLE 4.2.4. Sketch the graph of r(x) = +. z2 +1 SOLUTION. 1. The denominator 2 + 1 is never zero so the domain is (-oc, ox). 2. With no real zeros in the denominator, 2 + 1 is an irreducible quadratic. Our only hope of reducing r(x) is if 2 + 1 is a factor of 34 + 1. Performing long division gives us 31+1 +2 2 x2+1 x2+1 The remainder is not zero so r(x) is already reduced. 3. To find the x-intercept, we'd set r(x) = 0. Since there are no real solutions to +1 = 0, we x +l have no x-intercepts. Since r(0) = 1, so we get (0, 1) for the y-intercept. 4. This step doesn't apply to r, since its domain is all real numbers. 5. For end behavior, once again, since the degree of the numerator is greater than that of the denominator, Theorem 4.2 doesn't apply. We know from our attempt to reduce r(x) that we can rewrite r () = 2_ 1 + X2+1, and so we focus our attention on the term corresponding to the remainder, X2+1 It should be clear that as x t- oo,7 ~ very small (+), which means r(x) -2 1 + very small (+). So the graph y = r(x) is a little bit above the graph of the parabola y =9x2 - 1 as x - +oo. Graphically, \l / 5-- 4- \ 2. . I \ 11/ \ / \ / x \ / 14But rest assured, some graphs do!  246 RATIONAL FUNCTIONS 6. There isn't much work to do for a sign diagram for r(x), since its domain is all real numbers and it has no zeros. Our sole test interval is (-oc, c), and since we know r(0) = 1, we conclude r(x) is (+) for all real numbers. At this point, we don't have much to go on for a graph. Below is a comparison of what we have determined analytically versus what the calculator shows us. We have no way to detect the relative extrema analytically15 apart from brute force plotting of points, which is done more efficiently by the calculator. 6 Y 5 I 4 1 3I1 \ / 2 i \ , >H 15Without appealing to Calculus, of course.  4.2 GRAPHS OF RATIONAL FUNCTIONS 4.2.1 EXERCISES 1. Find the slant asymptote of the graph of the rational function. 247 3:3 - 3x + 1 (a) f(x) 32-+1 2x2 + 5x - 3 (b) f() = 3x+2 _C f -5x4 - 3x3 -+ z2 - 10 x3 - 3x2 + 3x - 1 -za3-+ 4x (d) f (x) =22 3:29 2. Use the six-step procedure to graph each rational function given. Be sure to draw any asymptotes as dashed lines. 4 (a) f (x) = (b) f (x)=652x (c) f 1)= 2 (e) f (x) = 2x-i -2x2 - 5x +3 (f) f(3:)= 2 x x2 +z-12 -x3 +4x (g) f (z) =2 - (h) f x(z) = :22x+116 f 3+:- 2x (d) f(z) 1 32+: -12 3. Example 4.2.4 showed us that the six-step procedure cannot tell us everything of importance about the graph of a rational function. Without Calculus, we need to use our graphing calculators to reveal the hidden mysteries of rational function behavior. Working with your classmates, use a graphing calculator to examine the graphs of the following rational functions. Compare and contrast their features. Which features can the six-step process reveal and which features cannot be detected by it? 1 (a) f(z)=21 (b) f (x) =2 + 1 2 1 2 (c) f () = 2 + 3 (d) f (X) =2 + 1 4. Graph the following rational functions by applying transformations to the graph of y =-. 1 -2x + 1 (a) f(z)- (c) h(3) =(Hint: Divide) 3 3x- 7 (b) g(x)=1-- (d) j(x) = - (Hint: Long division) X x-2 az: + b Discuss with your classmates how you would graph f((x) C + d What restrictions must 1 be placed on a, b, c and d so that the graph is indeed a transformation of y -? 16Once you've done the six-step procedure, use your calculator to graph this function on the viewing window [0, 12] x [0, 0.25]. What do you see?  248 RATIONAL FUNCTIONS 5. In Example 3.1.1 in Section 3.1 we showed that p(x) = 4xx3 is not a polynomial even though its formula reduced to 4 + 2 for x # 0. However, it is a rational function similar to those studied in the section. With the help of your classmates, graph p(x). x4 - 8x3 -+ 24x2 - 72x + 135 6. Let g(x) = .x3-+19 - 7 1 With the help of your classmates, find the x- and y- intercepts of the graph of g. Find the intervals on which the function is increasing, the intervals on which it is decreasing and the local extrema. Find all of the asymptotes of the graph of g and any holes in the graph, if they exist. Be sure to show all of your work including any polynomial or synthetic division. Sketch the graph of g, using more than one picture if necessary to show all of the important features of the graph.  4.2 GRAPHS OF RATIONAL FUNCTIONS 4.2.2 ANSWERS 249 1. (a) (b) 2 11 5 : + 9~ (c) y (d) y -5x - 18 -x 4 2. (a) f(z)= Domain: (-oc, -2) U (-2, No 3-intercepts y-intercept: (0, 0) Vertical asymptote: x = - As 3: -~ -2-- f (3) -~ -Xo As : - -2+, f(X) > oo Horizontal asymptote: y = As : - -oo, f(x) -0 As 3 x o, f (x) -- 0+ (b) f (x) 6 = 2 Domain: (-oo,3) U (3,oo) 3-intercept: (0, 0) y-intercept: (0, 0) Vertical asymptote: x = 3 As x3 -, 3 f(x) - 00 As : - 3+, f(X) -oo Horizontal asymptote: y = As 3: - -0, f (x) - - + As 3 oo, f(z) - - oo) 2 0 5. 4- 2. 1- y -6-5-4---1 2345x y 3 2 1 -i2- -3- -4- -5- -6- -7- [ 2 34 5 3 7 8 Vx - I - I _ I _ I 5 (c) f ( =2 Domain: (-oc, 0) U (0, o) No 3-intercepts No y-intercepts Vertical asymptote: x = 0 As 3: - 0-, f() - 00 As : - 0+, f (x) 00 Horizontal asymptote: y = 0 As : - -o, f (X) - 0+ As 3 x o, f (x) - 0+ y x  250 250 RATIONAL FUNCTIONS (d) f(x)-= 1 x2 +X - 12 Domain: (-oc, -4) U (-4, 3) U (3, oc) No x-intercepts Vertical asymptotes: x= -4 and x =3 As x - -4-, f (x) - o0 As x - 4+ f (X) -C -c As x - 3-, f (x) - -oc As x - 3+, f(x)- oc Horizontal asymptote: y =0 As x - -oc, f (X) - As x - oc, f (x) -- 0 (e) f(x) = 2x - 1 (e)f~v --2x2-5x + 3 Domain: (-oc, -3) U (-3, ~ U (2, oc) No x-intercepts Hole in the graph at (2) - 7 Vertical asymptote: x= -3 As x - -3-, f (x) - o0 As x > 3+, f (X) -C -c Horizontal asymptote: y =0 As x - -oc, f (X) - As x - oc, f (x) -- 0 (f) f3X:x x -1 Domain: (-oc, -4) U (-4, 3) U (3, oc) x-intercept: (0, 0) y-intercept: (0, 0) Vertical asymptotes: x= -4 and x =3 As x - -4-, f (x) - -oc As x >-4+, f (X) -00 As x - 3-, f (x) - -oc As x 3+, f(x)- oc Horizontal asymptote: y =0 y y i i i i i i i i  ND 8 II6 ° IImQ ztw zII + NDND(J~ 8 8'Ii 8 S. II " II IN .. C ^8z H 0 ii 0 C 8 cz 8 I I I i i i I I I I I II -I C~ Q~ / / / / / / - --- A---- / ?~ ? '. - - - - - - - - tP Cn I I I I I I I I I I I / / / ~Q / - -1---~-------- / / / / / / LIZ  252 RATIONAL FUNCTIONS 3 (b) g(x) = 1 - - Vertically stretch the graph of y by a factor of 3. 3 Reflect the graph of y =- about the x-axis. 3 Shift the graph of y -- up 1 unit. 1 - -2. -3. -4. -5. (c) h(x) 2x+1 Shift the graph of y: down 2 units. 1 -2+- 3T 1 =x 1. IN I I I I 4 I I I -3 -2 -1 -1. -3. 4. 1 2 3 3z-7 1 (d)j(x)= 3- x-2 x-2 1 Shift the graph of y = - to the right 2 units. 1 Reflect the graph of y = about the x-axis. 1 Shift the graph of y = - 2 up 3 units. 7- 6- 5- 4- 2- 1- y - I - I - I - - - - - - - . I I I , I , I -3 -2 -1 -1- 1 3 I I 4 5 X  4.3 RATIONAL INEQUALITIES AND APPLICATIONS 253 4.3 RATIONAL INEQUALITIES AND APPLICATIONS In this section, we use sign diagrams to solve rational inequalities including some that arise from real-world applications. Our first example showcases the critical difference in procedure between solving a rational equation and a rational inequality. EXAMPLE 4.3.1. z3 - 2x +1 1 1. Solve = x--1. x- 1 2 z3 -2x+1 1 2. Solve > -x - 1. x-1 2 3. Use your calculator to graphically check your answers to 1 and 2. SOLUTION. 1. To solve the equation, we clear denominators -1 - z- X1 2x1 _ x1 x-1 2 ( x3 1+ 1 2(-1) ( x 1).2(x 1) 2x3 - 4x - 2 = 2 - 3x + 2 expand 2x3 - 2 - x = 0 x(2x + 1)(x- 1) = 0 factor x = -2, 0, 1 Since we cleared denominators, we need to check for extraneous solutions. Sure enough, we see that x = 1 does not satisfy the original equation and must be discarded. Our solutions are xz = -jand z = 0. 2. To solve the inequality, it may be tempting to begin as we did with the equation - namely by multiplying both sides by the quantity (cv - 1). The problem is that, depending on cv, (cv - 1) may be positive (which doesn't affect the inequality) or (cv - 1) could be negative (which would reverse the inequality). Instead of working by cases, we collect all of the terms on one side of the inequality with 0 on the other and make a sign diagram using the technique given on page 235 in Section 4.2.  254 RATIONAL FUNCTIONS x3-2x-+1 z-1 x3-2x-+1 1 x~ x~ -2-+111 x-1 2 2(x3-2 + 1) -x~x 1) + 1(2(x7 1)) 2(x - 1) 2x3 - x2 - x 2x -2 1 > - -1 --2 > 0 > 0 get a common denominator > 0 expand Viewing the left hand side as a rational function r"() we make a sign diagram. The only value excluded from the domain of r is x = 1 which is the solution to 2x - 2 = 0. The zeros of r are the solutions to 2x3 - x2 - x = 0, which we have already found to be x = 0, x = - and x = 1, the latter was discounted as a zero because it is not in the domain. Choosing test values in each test interval, we construct the sign diagram below. (+) 0 (-) 0 (+) t (+) 1 2 0 1 We are interested in where r"() > 0. We find r"() > 0, or (+), on the intervals (- (0, 1) and (1, c). We add to these intervals the zeros of r, -2 and 0, to get our final (-oo, -4] U[0, 1) U(1, oo). solutI), solution: 3. Geometrically, if we set f(x) = x-2x--1 and g(x) = jx - 1, the solutions to f(x) = g(x) are the i-coordinates of the points where the graphs of y = f(x) and y = g(x) intersect. The solution to f(x) > g(x) represents not only where the graphs meet, but the intervals over which the graph of y = f(x) is above (>) the graph of g(x). We obtain the graphs below. The 'Intersect' command confirms that the graphs cross when x = -j and x = 0. It is clear from the calculator that the graph of y = f(x) is above the graph of y = g(x) on (-oc, -i) as well as on (0, c). According to the calculator, our solution is then (-oc, -j] U [0, cc)  4.3 RATIONAL INEQUALITIES AND APPLICATIONS 255 which almost matches the answer we found analytically. We have to remember that f is not defined at x= 1, and, even though it isn't shown on the calculator, there is a hole1 in the graph of y = f(x) when x= 1 which is why x= 1 needs to be excluded from our final answer. D Our next example deals with the average cost function of PortaBoy Game systems from Example 2.1.5 in Section 2.1. EXAMPLE 4.3.2. Given a cost function C(x), which returns the total cost of producing x products, the average cost function, AC(x) = C(x), computes the cost per item. Recall that the cost C, in dollars, to produce x PortaBoy game systems for a local retailer is C(x) = 80x + 150, x > 0. 1. Find an expression for the average cost function AC(x). Determine an appropriate applied domain for AC. 2. Find and interpret AC(10). 3. Solve AC(x) < 100 and interpret. 4. Determine the behavior of AC(x) as x - oc and interpret. SOLUTION. 1. From AC(x) C= 07), we obtain AC(x) = 80x+150 The domain of C is x > 0, but since x= 0 causes problems for AC(x), we get our domain to be x > 0, or (0, oc). 2. We find AC(10) = 80(10)+150 = 95, so the average cost to produce 10 game systems is $95 10 per system. 3. Solving AC(x) < 100 means we solve 80x+150 < 100. We proceed as in the previous example. 80x + 150 < 100 80x + 150100<0 80x + 150 - 100x < 0 common denominator 150 - 20x < 0 If we take the left hand side to be a rational function r(x), we need to keep in mind the the applied domain of the problem is x > 0. This means we consider only the positive half of the number line for our sign diagram. On (0, oc), r is defined everywhere so we need only look for zeros of r. Setting r(x) =0 gives 150 - 20x = 0, so that x = = 7.5. The test intervals on our domain are (0, 7.5) and (7.5, ooD). We find r(x) < 0 on (7.5, ooD). 1There is no asymptote at x 1 since the graph is well behaved near x 1. According to Theorem 4.1, there must be a hole there.  256 RATIONAL FUNCTIONS T (+) 0 (-) I I >~ 0 7.5 In the context of the problem, x represents the number of PortaBoy games systems produced and AC(x) is the average cost to produce each system. Solving AC(x) < 100 means we are trying to find how many systems we need to produce so that the average cost is less than $100 per system. Our solution, (7.5, o) tells us that we need to produce more than 7.5 systems to achieve this. Since it doesn't make sense to produce half a system, our final answer is [8, oo). 4. We can apply Theorem 4.2 to AC(x) and we find y = 80 is a horizontal asymptote to the graph of y = AC(x). To more precisely determine the behavior of AC(x) as x - oc, we first use long division2 and rewrite AC(x) = 80 + 150. As x - +o, 10 > 0+, which means AC(x) 80 + very small (+). Thus the average cost per system is getting closer to $80 per system. If we set AC(x) = 80, we get l°= 0, which is impossible, so we conclude that AC(x) > 80 for all x > 0. This means the average cost per system is always greater than $80 per system, but the average cost is approaching this amount as more and more systems are produced. Looking back at Example 2.1.5, we realize $80 is the variable cost per system - the cost per system above and beyond the fixed initial cost of $150. Another way to interpret our answer is that 'infinitely' many systems would need to be produced to effectively counterbalance the fixed cost. D Our next example is another classic 'box with no top' problem. EXAMPLE 4.3.3. A box with a square base and no top is to be constructed so that it has a volume of 1000 cubic centimeters. Let x denote the width of the box, in centimeters. Refer to the figure below. height depth width, x 1. Express the height h in centimeters as a function of the width x and state the applied domain. 2. Solve h(x) > x and interpret. 3. Find and interpret the behavior of h(x) as x - 0+ and as x - oc. 4. Express the surface area S of the box as a function of x and state the applied domain. 5. Use a calculator to approximate (to two decimal places) the dimensions of the box which minimize the surface area. 2In this case, long division amounts to term-by-term division.  4.3 RATIONAL INEQUALITIES AND APPLICATIONS 257 SOLUTION. 1. We are told the volume of the box is 1000 cubic centimeters and that x represents the width, in centimeters. From geometry, we know Volume = width x height x depth. Since the base of the box is to be a square, the width and the depth are both x centimeters. Using h for the height, we have 1000 =9z2h, so that h 1=10. Using function notation, h(x) =O10 As for the applied domain, in order for there to be a box at all, x > 0, and since every such choice of x will return a positive number for the height h we have no other restrictions and conclude our domain is (0, oc). 2. To solve h(x) > x, we proceed as before and collect all nonzero terms on one side of the inequality and use a sign diagram. h(x) > x 1000 1000 -2 x > 0 1000 -3 > 0 common denominator We consider the left hand side of the inequality as our rational function r(x). We see r is undefined at x= 0, but, as in the previous example, the applied domain of the problem is x > 0, so we are considering only the behavior of r on (0, oc). The sole zero of r comes when 1000 - z3 = 0, which is x= 10. Choosing test values in the intervals (0, 10) and (10, oc) gives the following diagram. T (+) 0 (-) I I >~ 0 10 We see r(x) > 0 on (0, 10), and since r(x) = 0 at x= 10, our solution is (0, 10]. In the context of the problem, h represents the height of the box while x represents the width (and depth) of the box. Solving h(x) > x is tantamount to finding the values of x which result in a box where the height is at least as big as the width (and, in this case, depth.) Our answer tells us the width of the box can be at most 10 centimeters for this to happen. 3. As x - 0+, h(x) 1=O1 00 -oc. This means the smaller the width x (and, in this case, depth), the larger the height h has to be in order to maintain a volume of 1000 cubic centimeters. As x -e oc, we find h(x) -~ 0+, which means to maintain a volume of 1000 cubic centimeters, the width and depth must get bigger the smaller the height becomes. 3That is, h(x) means 'h of x', not 'h times x' here.  258 RATIONAL FUNCTIONS 4. Since the box has no top, the surface area can be found by adding the area of each of the sides to the area of the base. The base is a square of dimensions x by x, and each side has dimensions x by h. We get the surface area, S = x2 + 4xh. To get S as a function of x, we substitute h 1000 to obtain S = x2 + 4x (1000) Hence, as a function of x, S(x) = x2+ 4000. The domain of S is the same as h, namely (0, oc), for the same reasons as above. 5. A first attempt at the graph of y = S(x) on the calculator may lead to frustration. Chances are good that the first window chosen to view the graph will suggest y = S(x) has the x-axis as a horizontal asymptote. From the formula S(x) = x2 + 4000 however, we get S(x) - x2 as x - 00, so S(x) - oc. Readjusting the window, we find S does possess a relative minimum at x - 12.60. As far as we can tell,4 this is the only relative extremum, and so it is the absolute minimum as well. This means the width and depth of the box should each measure approximately 12.60 centimeters. To determine the height, we find h(12.60) - 6.30, so the height of the box should be approximately 6.30 centimeters. W I NDOW Xmin=O Xmax=50 Xsc1=10 Ym in=0 Ymax=10000 Ysc 1=1000 Xres=1 =12. 211 .Y=476.22032 In many instances in the sciences, rational functions are encountered as a result of fundamental natural laws which are typically a result of assuming certain basic relationships between variables. These basic relationships are summarized in the definition below. DEFINITION 4.4. Suppose x, y, and z are variable quantities. We say " y varies directly with (or is directly proportional to) x if there is a constant k such that y = kx. " y varies inversely with (or is inversely proportional to) x if there is a constant k such thaty = " z varies jointly with (or is jointly proportional to) x and y if there is a constant k such that z = kxy. The constant k in the above definitions is called the constant of proportionality. EXAMPLE 4.3.4. Translate the following into mathematical equations using Definition 4.4. 4without Calculus, that is...  4.3 RATIONAL INEQUALITIES AND APPLICATIONS 259 1. Hooke's Law The force F exerted on a spring is directly proportional the extension x of the spring. 2. Boyle's Law: At a constant temperature, the pressure P of an ideal gas is inversely propor- tional to its volume V. 3. The volume V of a right circular cone varies jointly with the height h of the cone and the square of the radius r of the base. 4. Ohm's Law: The current I through a conductor between two points is directly proportional to the voltage V between the two points and inversely proportional to the resistance R between the two points. 5. Newton's Law of Universal Gravitation: Suppose two objects, one of mass m and one of mass M, are positioned so that the distance between their centers of mass is r. The gravitational force F exerted on the two objects varies directly with the product of the two masses and inversely with the square of the distance between their centers of mass. SOLUTION. 1. Applying the definition of direct variation, we get F = kx for some constant k. 2. Since P and V are inversely proportional, we write p k= . 3. There is a bit of ambiguity here. It's clear the volume and height of the cone is represented by the quantities V and h, respectively, but does r represent the radius of the base or the square of the radius of the base? It is the former. Usually, if an algebraic operation is specified (like squaring), it is meant to be expressed in the formula. We apply Definition 4.4 to get V = khr2. 4. Even though the problem doesn't use the phrase 'varies jointly', the fact that the current I is given as relating to two different quantities implies this. Since I varies directly with V but inversely with R, we write I = k 5. We write the product of the masses mM and the square of the distance as r2. We have F varies directly with mM and inversely with r2, so that F = kmM In many of the formulas in the previous example, more than two varying quantities are related. In practice, however, usually all but two quantities are held constant in an experiment and the data collected is used to relate just two of the variables. Comparing just two varying quantities allows us to view the relationship between them as functional, as the next example illustrates. EXAMPLE 4.3.5. According to this website the actual data relating the volume V of a gas and its pressure P used by Boyle and his assistant in 1662 to verify the gas law that bears his name is given below.  260 RATIONAL FUNCTIONS V 48 46 44 42 40 38 36 34 32 30 28 26 24 P 29.13 30.56 31.94 33.5 35.31 37 39.31 41.63 44.19 47.06 50.31 54.31 58.81 V 23 22 21 20 19 18 17 16 15 14 13 12 P 61.31 64.06 67.06 70.69 74.13 77.88 82.75 87.88 93.06 100.44 107.81 117.56 1. Use your calculator to generate a scatter diagram for these data using V as the independent variable and P as the dependent variable. Does it appear from the graph that P is inversely proportional to V? Explain. 2. Assuming P and V do vary inversely, use the data to approximate the constant of propor- tionality. 3. Use your calculator to determine a 'Power Regression' to this data5 and use it verify your results in 1 and 2. SOLUTION. 1. If P really does vary inversely with V, then P = k for some constant k. From the data plot, the points do seem to like along a curve like y = k. 2. To determine the constant of proportionality, we note that from P , Vwe get k PV. Multiplying each of the volume numbers times each of the pressure numbers,6 we produce a number which is always approximately 1400. We suspect that P = 10. Graphing y = 1400 along with the data gives us good reason to believe our hypotheses that P and V are, in fact, inversely related. a an The graph of the data The data with y = 1400 3. After performing a 'Power Regression', the calculator fits the data to the curve y = azS where a ~ 1400 and b ~ -1 with a correlation coefficient which is darned near perfect7. In other words, y = 1400--1 or Y 140,as we guessed. 'We will talk more about this in the coming chapters. 6You can use tell the calculator to do this algebra on the lists and save yourself some time. 7We will revisit this example once we have developed logarithms in Chapter 6 to see how we can actually 'linearize' this data and do a linear regression to obtain the same result.  4.3 RATIONAL INEQUALITIES AND APPLICATIONS 261 Pwr Reg 'j=a*x~b a=1400.874407 b=-.9984033692 r2=.9999320488 r=-.9999660238  262 RATIONAL FUNCTIONS 4.3.1 EXERCISES 1. Solve each rational equation. Be sure to check for extraneous solutions. (a) =3 5x + 4- 3x - 1 _ (b) 2 - 1 x2+1 1 1 (c) + z+3 z-3 (d) 2x +17 x+1 (e) x22x +1 x3+x2 -2x z2 -3 -x3 + 4 x239 (f) =94x z2_9 2_9 2. Solve each rational inequality. Express your answer using interval notation. 1 (a) > 0 x+2 - z:-3 (b) < 0 x+2 (g) x+ 4x x2_9 -4 1 (h) x2+1< 0 (c) 2 x1< 0 (d) > 0 x2-1 3x - 1 (e) 2 <1i x2+1 2x+17 x+1 z4x3+x2-2x-15 X3 -_4x2- 5( 3 - 12x2 + 9x + 10 )> 3x - X2_-1 1 3. Another Classic Problem: A can is made in the shape of a right circular cylinder and is to hold one pint. (For dry goods, one pint is equal to 33.6 cubic inches.)8 (a) Find an expression for the volume V of the can based on the height h and the base radius r. (b) Find an expression for the surface area S of the can based on the height h and the base radius r. (Hint: The top and bottom of the can are circles of radius r and the side of the can is really just a rectangle that has been bent into a cylinder.) (c) Using the fact that V = 33.6, write S as a function of r and state its applied domain. (d) Use your graphing calculator to find the dimensions of the can which has minimal surface area. 4. In Exercise 9 in Section 1.5, the population of Sasquatch in Portage County was modeled by the function P(t) =5t1,where t = 0 represents the year 1803. When were there fewer than 100 Sasquatch in the county? 5. The cost C in dollars to remove p% of the invasive species of Ippizuti fish from Sasquatch Pond is given by C(p) = JJ where 0 p <100. 8According to www.dictionary.coim, there are different values given for this conversion. We will stick with 33.6in3 for this problem.  4.3 RATIONAL INEQUALITIES AND APPLICATIONS 263 (a) Find and interpret C(25) and C(95). (b) What does the vertical asymptote at x= 100 mean within the context of the problem? (c) If you have $40000 to spend for Ippizuti removal, what percentage of the fish can you remove? 6. Translate the following into mathematical equations. (a) At a constant pressure, the temperature T of an ideal gas is directly proportional to its volume V. (This is Charles's Law) (b) The frequency of a wave f is inversely proportional to the wavelength of the wave A. (c) The density d of a material is directly proportional to the mass of the object m and inversely proportional to its volume V. (d) The square of the orbital period of a planet P is directly proportional to the cube of the semi-major axis of its orbit a. (This is Kepler's Third Law of Planetary Motion ) (e) The drag of an object traveling through a fluid D varies jointly with the density of the fluid p and the square of the velocity of the object v. (f) Suppose two electric point charges, one with charge q and one with charge Q, are posi- tioned r units apart. The electrostatic force F exerted on the charges varies directly with the product of the two charges and inversely with the square of the distance between the charges. (This is Coulomb's Law) 1 T 7. According to this webpage, the frequency f of a vibrating string is given by f = 2L y where T is the tension, ,p is the linear mass of the string and L is the length of the vibrating part of the string. Express this relationship using the language of variation. 8. According to the Centers for Disease Control and Prevention www.cde.gov, a person's Body Mass Index B is directly proportional to his weight W in pounds and inversely proportional to the square of his height h in inches. (a) Express this relationship as a mathematical equation. (b) If a person who was 5 feet, 10 inches tall weighed 235 pounds had a Body Mass Index of 33.7, what is the value of the constant of proportionality? (c) Rewrite the mathematical equation found in part 8a to include the value of the constant found in part 8b and then find your Body Mass Index. 9. We know that the circumference of a circle varies directly with its radius with 27r as the constant of proportionality. (That is, we know C = 27rr.) With the help of your classmates, compile a list of other basic geometric relationships which can be seen as variations. 9Also known as the linear density. It is simply a measure of mass per unit length.  264 RATIONAL FUNCTIONS 4.3.2 ANSWERS 1. (a) x=- (b) x =1, x= 2 (c) x= -1 (d) x=-6, x=2 (e) No solution (f) x=0, x =i2 2. (a) (-2, oo) (b) (-2,3] (c) (-1,0)U(1,oo) (d) (-oc, -3) U (-2, -1) U (1, x) (e) (-oo, 1] U [2, oo) (f) (-oo, -6) U (-1, 2) (g) (-oo,-3) U (-v8,0) U (v8,3) (h) (i) (j) No solution [-3,0) U (0,4) U [5,oo) (-1, - 1 ] U (1, oo)) 3. (a) V = wr2h. (b) S = 27r2 + 2wrh (c) S(r) = 27r2 + 62, Domain r > 0 (d) r ~ 1.749 in. and h 3.498 in. 4. P(30) = 100 so before 1903 there were fewer than 100 Sasquatch in Portage County. 5. (a) C(25) = 590 means it costs $590 to remove 25% of the fish and and C(95) = 33630 means it would cost $33630 to remove 95% of the fish from the pond. (b) The vertical asymptote at x= 100 means that as we try to remove 100% of the fish from the pond, the cost increases without bound. Thus it is impossible to remove 100% of the fish. (c) For $40000 you could remove about 95.76% of the fish. 6. (a) T = kV k (b) m / - A (c) d =-km (d) P2 = ka3 (e) " D = kpv2. (f) 12 F = kqQ 1T 7. Rewriting f =2 - as f square root of the tension and linear mass. 2v varies we see that the frequency f varies directly with the inversely with the length and the square root of the kW 8. (a) B =Ih2 (b) k = 702.68 (The CDC uses 703.) 703W (c) B =7h2 1oThe character A is the lower case Greek letter 'lambda.' "Note: The characters p and v are the lower case Greek letters 'rho' and 'nu,' respectively. 12Note the similarity to this formula and Newton's Law of Universal Gravitation as discussed in Example 5.  CHAPTER 5 FURTHER Tonics IN FUNCTIONS 5.1 FUNCTION COMPOSITION Before we embark upon any further adventures with functions, we need to take some time to gather our thoughts and gain some perspective. Chapter 1 first introduced us to functions in Section 1.4. At that time, functions were specific kinds of relations - sets of points in the plane which passed the Vertical Line Test, Theorem 1.1. In Section 1.5, we developed the idea that functions are processes - rules which match inputs to outputs - and this gave rise to the concepts of domain and range. We spoke about how functions could be combined in Section 1.6 using the four basic arithmetic operations, took a more detailed look at their graphs in Section 1.7 and studied how their graphs behaved under certain classes of transformations in Section 1.8. In Chapter 2, we took a closer look at three families of functions: linear functions (Section 2.1), absolute value functions1 (Section 2.2), and quadratic functions (Section 2.3). Linear and quadratic functions were special cases of polynomial functions, which we studied in generality in Chapter 3. Chapter 3 culminated with the Real Factorization Theorem, Theorem 3.16, which says that all polynomial functions with real coefficients can be thought of as products of linear and quadratic functions. Our next step was to enlarge our field2 of study to rational functions in Chapter 4. Being quotients of polynomials, we can ultimately view this family of functions as being built up of linear and quadratic functions as well. So in some sense, Chapters 2, 3, and 4 can be thought of as an exhaustive study of linear and quadratic3 functions and their arithmetic combinations as described in Section 1.6. We now wish to study other algebraic functions, such as f(x) =/ and g(x) = 2/3, and the purpose of the first two sections of this chapter is to see how these kinds of functions arise from polynomial and rational functions. To that end, we first study a new way to combine functions as defined below. DEFINITION 5.1. Suppose f and g are two functions. The composite of g with f, denoted g o f, is defined by the formula ((g o f)(x) = g(f(x)), provided x is an element of the domain of f and f (x) is an element of the domain of g. 'These were introduced, as you may recall, as piecewise-defined linear functions. 2This is a really bad math pun. 31f we broaden our concept of functions to allow for complex valued coefficients, the Complex Factorization Theorem, Theorem 3.14, tells us every function we have studied thus far is a combination of linear functions.  266 FURTHER TOPICS IN FUNCTIONS The quantity g o f is also read 'g composed with f' or, more simply 'g of f.' At its most basic level, Definition 5.1 tells us to obtain the formula for (g o f) (x), we replace every occurrence of x in the formula for g(x) with the formula we have for f(x). If we take a step back and look at this from a procedural, 'inputs and outputs' perspective, Defintion 5.1 tells us the output from g o f is found by taking the output from f, f(x), and then making that the input to g. The result, g(f(x)), is the output from g o f. From this perspective, we see g o f as a two step process taking an input x and first applying the procedure f then applying the procedure g. Abstractly, we have f g gof In the expression g(f (x)), the function f is often called the 'inside' function while g is often called the 'outside' function. There are two ways to go about evaluating composite functions - 'inside out' and 'outside in' - depending on which function we replace with its formula first. Both ways are demonstrated in the following example. 2x EXAMPLE 5.1.1. Let f(x) 9 z2 - 4x, g(x) = 2 - x + 3, and h(x) = . Find and simplify the indicated composite functions. State the domain of each. 1. (go f)(x) 2. (f o g)(x) 3. (g o h)(x) 4. (ho g)(x) 5. (h o h)(x) 6. (ho(gof))(z) 7. ((h o g) o f)(z) SOLUTION. 1. By definition, (g o f)(x) = g(f(x)). We now illustrate the two ways to evaluate this. * inside out: We insert the expression f(x) into g first to get (go f)(x) =g(f(x)) = g (x2 - 4x) = 2 - (x2 - 4x) + 3 = 2 - x2 - 4x + 3 Hence, (gof)(x)=2- x2-4x+3.  5.1 FUNCTION COMPOSITION 267 " outside in: We use the formula for g first to get (g o f)(z)= g(f(x))= 2 - ff(x)+ 3 =2 - (x2 - 4x) + 3= 2 - Vz2 - 4 We get the same answer as before, (g o f)(z) = 2 - x2 - 4x + 3. To find the domain of g o f, we need to find the elements in the domain of f whose outputs f(x) are in the domain of g. We accomplish this by following the rule set forth in Section 15, that is, we find the domain before we simplify. To that end, we examine (g o f)(x) 2 - (x2 - 4x) + 3. To keep the square root happy, we solve the inequality 2 - 4x + 3 > 0 by creating a sign diagram. If we let r(x) 9= 2 - 4x + 3, we find the zeros of r to be x = 1 and x= 3. We obtain (+) 0 (-) 0 (+) E I I 1 3 Our solution to 2 - 4x + 3 > 0, and hence the domain of g o f, is (-oc, 1] U [3, oc). 2. To find (f o g)(x), we find f(g(x)). " inside out: We insert the expression g(x) into f first to get (fo0g)() =f(g(x)) = f(2- x+3) = (2- x+3)2-4(2- x+3) = 4-4 x+3+( x+3)2-8+4 x+3 = 4+x+3-8 = x-1 " outside in: We use the formula for f(x) first to get (fo0g)() =f(g(x)) - (g(x))2-4(g(x)) - (2- x+3)2-4(2- x+3) = x - 1 same algebra as before Thus we get (f o g)(x) = x - 1. To find the domain of (f o g), we look to the step before we did any simplification and find (f o g)(x) = (2 - Vx + 3)2 - 4 (2 - Vx + 3). To keep the square root happy, we set x + 3 ;> 0 and find our domain to be [-3, oo). 3. To find (g o h)(x), we compute g(h(x)).  268 FURTHER TOPICS IN FUNCTIONS " inside out: We insert the expression h(x) into g first to get (go h)(x) = g(h(x)) 2x - 2- (j+1)+3 2_ 2x 3+ (x + 1) get common denominators x+l x+l 2- 5x+3 x+1 " outside in: We use the formula for g(x) first to get (go h)(x) = g(h(x)) - 2-- h(x)+ 3 = 2- (2)+3 = 2 - +1 get common denominators as before Hence, (g o h)(x) = 2 - . To find the domain, we look to the step before we began to simplify: (g o h)(x) = 2 - (+1) + 3. To avoid division by zero, we need x # -1. To keep the radical happy, we need to solve + + 3 > 0. Getting common denominators as before, this reduces to S+3 > 0. Defining r(x) = ±3, we have that r is undefined at x= -1 and r(x)=0 at x= -5.We get (+) T (-) 0 (+) C I I ) - 1_3 Our domain is (-oo, -1) U [-5, oo). 4. We find (h o g)(x) by finding h(g(x)). * inside out: We insert the expression g(x) into h first to get (h o g)(z) =h(g(x))  5.1 FUNCTION COMPOSITION 269 = h(2- x+3) 2(2- x+3) (2- xz+ 3) +1 4-2 xz+ 3 3- x+3 " outside in: We use the formula for h(x) first to get (ho g)(x) = h(g(x)) 2 (g(x)) (g x))+1 2(2- x+3) (2- x+ 3) +1 4-2 x+3 3- x+3 Hence, (h o g)(x) =4-2 x+3 To find the domain of h o g, we look to the step before any 3- x+3 simplification: (hog)(x) =2(2-.x3 To keep the square root happy, we require x + 3 > 0 or x > -3. Setting the denominator equal to zero gives (2 - x + 3) + 1 = 0 or x + 3 = 3. Squaring both sides gives us x+ 3 = 9, or x= 6. Since x= 6 checks in the original equation, (2 - x + 3) +1 = 0, we know x= 6 is the only zero of the denominator. Hence, the domain of h o g is [-3, 6) U (6,oo). 5. To find (h o h)(z), we substitute the function h into itself, h(h(x)). * inside out: We insert the expression h(x) into h to get (ho h)(x) = h(h(x)) h 2x x +1 4x 2 2 x +l x1 2x ++1 (xl+ 1 4x  270 FURTHER TOPICS IN FUNCTIONS 4x - .(Xz+1 _ z+1 2x 4x 3x +1 " outside in: This approach yields (ho h)(x) = h(h(x)) 2(h(x)) h(x) + 1 2 2x +1 x+ 4x same algebra as before 3x + 1 2x To find the domain of h o h, we analyze (h o h)(x) = 1'-. To keep the denominator x + 1 happy, we need 3 # -1. Setting the denominator 1+ 1 = 0 gives x= -}. Our domain is (-oc, -1) U (-1, -}) U (-}, oo). 6. The expression (h o (g o f))(x) indicates that we first find the composite, g o f and compose the function h with the result. We know from number 1 that (g o f)(z) = 2 - 3z2 - 4x + 3. We now proceed as usual. * inside out: We insert the expression (g o f)(z) into h first to get (ho(gof))(z) = h((g o f)(z)) = h (2 - 3z2-4x + 3) 2(- :2-43+ 3)+1  5.1 FUNCTION COMPOSITION 271 4-2x2-4x+3 3- /2-4x+3 " outside in: We use the formula for h(x) first to get (ho (go f))(x) = h((g o f)(x)) 2 ((g o f)(z)) ((g o0f)(x))+1 2 (2 - x2 - 4x + 3) (2- x2-4x+3)+1 4-2 x2-4x+3 3- /2-4x+3 So we get (h o (g o f))(x) = 4-2 X2-4x+3 To find the domain, we look at the step before 3- x2-4x+3 ' we began to simplify, (h o (g o f))(z) =2(2 X24X+3) For the square root, we need (2- X2_4X+3)+1 x2 - 4x + 3 > 0, which we determined in number 1 to be (-oc, 1] U [3, oc). Next, we set the denominator to zero and solve: (2 - /2 - 4x + 3) + 1 = 0. We get x2 - 4x + 3 = 3, and, after squaring both sides, we have 2 - 4x + 3 = 9. To solve 2 - 4x - 6 = 0, we use the quadratic formula and get x= 2 t 0IU. The reader is encouraged to check that both of these numbers satisfy the original equation, (2 - /2 - 4x + 3) + 1 = 0. Hence we must exclude these numbers from the domain of h o (g o f). Our final domain for h o (f o g) is (-oo,2- /I)U(2- /I,1]U [3,2+ 1/i) U (2+ /I,coo). 7. The expression ((hog) of)(x) indicates that we first find the composite hog and then compose that with f. From number 4, we gave (ho g)(x) = 4-2 x+3 We now proceed as before. 3- x+3 * inside out: We insert the expression f(x) into h o g first to get ((hog)of)(x) (h o g)(f (x)) (hog)(z2-4x) 4-2 (x2-4x)+3 3 - (x2 - 4x) + 3 4-2 x2-4x+3 3 -/x2-4x+3 * outside in: We use the formula for (h o g)(x) first to get ((hog) o f)(x) = (ho g)(f(x))  272 FURTHER TOPICS IN FUNCTIONS 4 - 2 (f (x)) + 3 3- f(x))+3 4 - 2 (x2 - 4x) + 3 3 - (x2 - 4x) + 3 4-2 x2-4x+3 3 - 2-4x+3 We note that the formula for ((h o g) o f)(x) before simplification is identical to that of (h o (g o f))(x) before we simplified it. Hence, the two functions have the same domain, h o (fog) is (-oo, 2 - /I0) U (2 - l0, 1] U [3, 2 + 10) U (2+ /10, oo). It should be clear from Example 5.1.1 that, in general, when you compose two functions, such as f and g above, the order matters.4 We found that the functions f o g and g o f were different as were g o h and h o g. Thinking of functions as processes, this isn't all that surprising. If we think of one process as putting on our socks, and the other as putting on our shoes, the order in which we do these two tasks does matter.5 Also note the importance of finding the domain of the composite function before simplifying. For instance, the domain of f o g is much different than its simplified formula would indicate. Composing a function with itself, as in the case of h o h, may seem odd. Looking at this from a procedural perspective, however, this merely indicates performing a task h and then doing it again - like setting the washing machine to do a 'double rinse'. Composing a function with itself is called 'iterating' the function, and we could easily spend an entire course on just that. The last two problems in Example 5.1.1 serve to demonstrate the associative property of functions. That is, when composing three (or more) functions, as long as we keep the order the same, it doesn't matter which two functions we compose first. This property as well as another important property are listed in the theorem below. THEOREM 5.1. Properties of Function Composition: Suppose f, g, and h are functions. " h o (g o f) - (h o g) o f, provided the composite functions are defined. " If I is defined as I(x) = x for all real numbers x, then I o f = f o I = f. By repeated applications of Definition 5.1, we find (h o (g o f))(x) = h((g o f)(x)) = h(g(f(x))). Similarly, ((h o g) o f)(x) = (h o g)(f(x)) = h(g(f(x))). This establishes that the formulas for the two functions are the same. We leave it to the reader to think about why the domains of these two functions are identical, too. These two facts establish the equality h o (g o f) = (h o g) o f. A consequence of the associativity of function composition is that there is no need for parentheses 4This shows us function composition isn't commutative. An example of an operation we perform on two functions which is commutative is function addition, which we defined in Section 1.6. In other words, the functions f + g and g + f are always equal. Which of the remaining operations on functions we have discussed are commutative? 5Amore mathematical example in which the order of two processes matters can be found in Section 1.8. In fact, all of the transformations in that section can be viewed in terms of composing functions with linear functions.  5.1 FUNCTION COMPOSITION 273 when we write h o g o f. The second property can also be verified using Definition 5.1. Recall that the function I(x) = x is called the identity function and was introduced in Exercise 12 in Section 2.1. If we compose the function I with a function f, then we have (I o f)(x) =_I(f(x)) = f(x), and a similar computation shows (f o I)() = ff(x). This establishes that we have an identity for function composition much in the same way the real number 1 is an identity for real number multiplication. That is, just as for any real number x, 1 - x = x - 1 = x , we have for any function f, I o f = f o I = f. We shall see the concept of an identity take on great significance in the next section. Out in the wild, function composition is often used to relate two quantities which may not be directly related, but have a variable in common, as illustrated in our next example. EXAMPLE 5.1.2. The surface area S of a sphere is a function of its radius r and is given by the formula S(r) = 47r2. Suppose the sphere is being inflated so that the radius of the sphere is increasing according to the formula r(t) = 3t2, where t is measured in seconds, t > 0, and r is measured in inches. Find and interpret (S o r)(t). SOLUTION. If we look at the functions S(r) and r(t) individually, we see the former gives the surface area of a sphere of a given radius while the latter gives the radius at a given time. So, given a specific time, t, we could find the radius at that time, r(t) and feed that into S(r) to find the surface area at that time. From this we see that the surface area S is ultimately a function of time t and we find (S o r)(t) S(r(t)) 4F(r(t))2 - 4- (3t2)2 36wt4. This formula allows us to compute the surface area directly given the time without going through the 'middle man' r. D A useful skill in Calculus is to be able to take a complicated function and break it down into a composition of easier functions which our last example illustrates. EXAMPLE 5.1.3. Write each of the following functions as a composition of two or more (non-identity) functions. Check your answer by performing the function composition. 1. F(x) =13x - 1| 2 2. G(x) 2 X2 +1 c+ 1 3. H(x)x c- 1 SOLUTION. There are many approaches to this kind of problem, and we showcase a different methodology in each of the solutions below. 1. Our goal is to express the function F as F = g o f for functions g and f. From Definition 5.1, we know F(x) = g(f(x)), and we can think of f(x) as being the 'inside' function and g as being the 'outside' function. Looking at F(x) =_|3x - 1| from an 'inside versus outside' perspective, we can think of 3x - 1 being inside the absolute value symbols. Taking this cue, we define f (x) =3cv - 1. At this point, we have F(xv) = f(x)|. What is the outside function? The function which takes the absolute value of its input, g(xv) = Iz. Sure enough, (g o f)(cv) = g(f(xv)) = If ()| = |3cv- 1| =F(xv), so we are done.  274 FURTHER TOPICS IN FUNCTIONS 2. We attack deconstructing G from an operational approach. Given an input x, the first step is to square x, then add 1, then divide the result into 2. We will assign each of these steps a function so as to write G as a composite of three functions: f, g and h. Our first function, f, is the function that squares its input, f(x) = x2. The next function is the function that adds 1 to its input, g(x) = x + 1. Our last function takes its input and divides it into 2, h(x) = j. The claim is that G = h o g o f. We find (hog o f)() = h(g(f (x))) = h(g (x2)) h (2 + 1) = 2 = G(x). 3. If we look H(x) = 1with an eye towards building a complicated function from simpler functions, we see the expression V1 is a simple piece of the larger function. If we define f (x) =xfd, we have H(x) f(x)± If we want to decompose H = g o f, then we can glean the formula from g(x) by looking at what is being done to f(x). We find g(x) = . We check (go f)(x) g(f(x)) = x= H(x), as required. D  5.1 FUNCTION COMPOSITION 275 5.1.1 EXERCISES 1 1. Let f(x) = 3x - 6, g(x) =_Ix, h(x) = xV1 and k(x) =-. Find and simplify the indicated composite functions. State the domain of each. (a) (f o g)(x) (h) (k o f)(x) (b) (gof)(x) (i) (h o k)(x) (c) (foh)(x) (j) (koh)(x) (d) (e) (f) (g) (h o f)(x) (g o h)(x) (h o g)(x) (f o k)(x) (k) (1) (m) (n) (fogoh)(x) (ho gok)(x) (kohof)(x) (hokogof)(x) 2. Let f(x)=2x+ 1and g(x) = x2 - x - 6 and h(x) composite functions. Find the domain of each. = +. Find and simplify the indicated x-6 (a) (g o f)() (b) (h o f)() (c) (h o g) (x) (d) (h o h) (x) 3. Let f(x) = 9 - x2 and g(x) =9x2 - 9. Find and simplify the indicated composite functions. State the domain of each. (a) (f o f)(x) (b) (g o g)(x) (c) (g o f)() (d) (f o g)(x) 4. Let f be the function defined by f = {(-3, 4), (-2, 2), (-1, 0), (0, 1), (1, 3), (2, 4), (3,-1)} and let g be the function defined g = {(-3, -2), (-2, 0), (-1, -4), (0, 0), (1, -3), (2, 1), (3, 2)}. Find the each of the following values if it exists. (a) (b) (c) (d) (e) (f) (g) (f o g)(3) f (g (-1)) (f o f)(0) (f o g)(-3) (g o f)(3) g(f(-3)) (g o g)(-2) (h) (go f)(-2) (i) g(f(g(0))) (j) f(f(f(-1))) (k) f(f(f(f(f(1))))) n times (1) '(g ogo...og)(0) 5. Let g(x) = -x, h(x) = x+2, j(x) = 3x and k(x) = x - 4. In what order must these functions be composed with f (x) =xV1 to create F(x) = 3 -x + 2 - 4?  276 FURTHER TOPICS IN FUNCTIONS 6. What linear functions could be used to transform f(x) = x3 into F(x) What is the proper order of composition? 7. Write the following as a composition of two or more non-identity functions. '(2x - 7)3 + 1? (a) h(x) =v2x - 1 (b) r () = 2 5x + 1 8. Write the function F(x) (c) F(x) =(92 _ 13 (d) R(x) 2x + 1 x3 E_ -i x3 + 6 -3_ as a composition of three or more non-identity functions. 9. The volume V of a cube is a function of its side length x. Let's assume that x= t + 1 is also a function of time t, where x is measured in inches and t is measured in minutes. Find a formula for V as a function of t. 10. Suppose a local vendor charges $2 per hot dog and that the number of hot dogs sold per hour x is given by x(t) = -4t2 + 20t + 92, where t is the number of hours since 10 AM, 0 < t < 4. (a) Find an expression for the revenue per hour R as a function of x. (b) Find and simplify (R o x) (t). What does this represent? (c) What is the revenue per hour at noon? 11. Discuss with your classmates how 'real-world' processes such as filling out federal income tax forms or computing your final course grade could be viewed as a use of function composition. Find a process for which composition with itself (iteration) makes sense.  5.1 FUNCTION COMPOSITION27 277 5.1.2 ANSWERS 1.(a) (f og)(3) =33: - 6 Domain: (-oo,oo) Db gomain:(x)_1-o6 Domain: ([0,oo) Dc (omain:[2,oo Domain: [0,oo) 1 (i) ( h o k ) ( x)_Dm i :) -o o Domain: (0,oo) (j) (kooh)() =3x1 Domain: (,oo) (k) (f oghoh)(3) -3x3- 6 Domain: [0,oo) (1) (hokgo)()/ 1 Domain: (o)(2,oo) D om ain: (- f o ) x D o m ain : ((h-ko ) x Domain: (- 3- -oo, 0) U (0),oo) 1 3xv - 6 -oo, 2) U (2),oo) 492+ 23:-6 -00, o0) 23x+ 7 23:- 5 -o U ° 2. (a) (g o f ) (:) D m i :(b(h of (xDo a n(3. (a) (fof)( :)~ D om ain: [- g o ) x Domain: (- (c) (ho g)(x:) Do a n ( ) ( )()Dom ain: (- 3:2-x-1 -oo, -3) U (-3, 4) U (4,oo) 7x - 30 5x -42 -x, 6) U (6,) 1U (,o 4. (a) (b) (c) (d) (e) (f) (g) (f og) (3)- f (9(- 1))- (fo0f)(0) (f o g) (-3) (go f)(3)- g(f (-3))- (g o g)(-2) -3, 3] 3:4-183:2+72 f (9(3)) -f (2) - 4 f (-4) which is undefined - f (g(-3)) =f (-2) =2 9(f (3)) - g(-1) =--4 g(4) which is undefined - g(g(-2)) - J(0) - 0 (C) (g o f)(3:) =3-x Domain: [-3, 3] (d) (f og) (x) = - Domain: [- 12, (h) (g o f)(-2)= g( Mi g (f (g(0))) =g (f 2 3:4 + 18:2 -72 V6]U [V16,1216 f(-2)) - g(2) - 1 g~) (1) _-3 (k) f (f(f (f (f(1))))) =f (f (f (f(3)))) f (f (f (-))) - f (f (0)) - f (1) - 3 rtimes 6The quantity -x4 + 18x2 - 72 is a 'quadratic in disguise' which factors nicely. See Example 3.3.4 is Section 3.3.  278 FURTHER TOPICS IN FUNCTIONS 5. F(a') = 3 -a' + 2 - 4 - k(J(f (h.(J(x))))) 6. One possible solution is F(x) = - 2(2 x - 7, j~x) =- jz and k~z) =x + G(x) = 2x - 7 and H(x) = -2x + 1. 7. (a) h(x) = (g o f) (x) where fv(x) = and g(x) =xV7. (b) r(x) (g o f) (x) where f (x) = 2 and g(x) =-. 3: 8. F(x) == (h(g(f(x))) where 9. V(x) 9=z3 so V(x(t)) = (t + 1)3 x - 7)3+1 = k(j(f(h(g(x))))) where g(x) = 2x, h(x) = 1. You could also have F(x) = H(f(G(x))) where 2x - 1 (c) F(x) = (g o f) (x) where f(x) and g(x) = x3. 5x + 1 (d) R(x) = (g o f) (x) where f (x) 2x + 1 g(x)= x - 1 2 -1 x3 and X3, g(x) x+ -and h(x) V/T. 10. (a) R(x) = 2x (b) (R ox) (t) of time. -8t2 + 40t + 184, 0 < t < 4. This gives the revenue per hour as a function (c) Noon corresponds to t = 2, so (R o x) (2) = 232. The hourly revenue at noon is $232 per hour.  5.2 INVERSE FUNCTIONS 279 5.2 INVERSE FUNCTIONS Thinking of a function as a process like we did in Section 1.5, in this section we seek another function which might reverse that process. As in real life, we will find that some processes (like putting on socks and shoes) are reversible while some (like cooking a steak) are not. We start by discussing a very basic function which is reversible, f(x) = 3x + 4. Thinking of f as a process, we start with an input x and apply two steps, as we saw in Section 1.5 1. multiply by 3 2. add 4 To reverse this process, we seek a function g which will undo each of these steps and take the output from f, 3x + 4, and return the input x. If we think of the real-world reversible two-step process of first putting on socks then putting on shoes, to reverse the process, we first take off the shoes, and then we take off the socks. In much the same way, the function g should undo the second step of f first. That is, the function g should 1. subtract 4 2. divide by 3 Following this procedure, we get g(x) =x3. Let's check to see if the function g does the job. If x = 5, then f(5) = 3(5) + 4 = 15 + 4 = 19. Taking the output 19 from f, we substitute it into g to get g(19) = 193 4 = 1 = 5, which is our original input to f. To check that g does the job for all x in the domain of f, we take the generic output from f, f(x) = 3x + 4, and substitute that into g. That is, g(f(x)) = g(3x + 4) = (3x+4)-4 = 33= x, which is our original input to f. If we carefully examine the arithmetic as we simplify g(f(x)), we actually see g first 'undoing' the addition of 4, and then 'undoing' the multiplication by 3. Not only does g undo f, but f also undoes g. That is, if we take the output from g, g(x) =x35, and put that into f, we get f(g(x)) = f (x ) = 3 (x) +5 = (x -5) +5 =x. Using the language of function composition developed in Section 5.1, the statements g(f(x)) = x and f(g(x)) = x can be written as (g o f)(=) =x and (f o g)(z) c=v, respectively. Abstractly, we can visualize the relationship between f and g in the diagram below. f g  280 FURTHER TOPICS IN FUNCTIONS The main idea to get from the diagram is that g takes the outputs from f and returns them to their respective inputs, and conversely, f takes outputs from g and returns them to their respective inputs. We now have enough background to state the central definition of the section. DEFINITION 5.2. Suppose f and g are two functions such that 1. (g o f)(x) = x for all x in the domain of f and 2. (f o g)(x) = x for all x in the domain of g. Then f and g are said to be inverses of each other. The functions f and g are said to be invertible. Our first result of the section formalizes the concepts that inverse functions exchange inputs and outputs and is a consequence of Definition 5.2 and the Fundamental Graphing Principle for Func- tions. THEOREM 5.2. Properties of Inverse Functions: Suppose f and g are inverse functions. * The rangea of f is the domain of g and the domain of f is the range of g * f(a) = b if and only if g(b) = a * (a, b) is on the graph of f if and only if (b, a) is on the graph of g aRecall this is the set of all outputs of a function. The third property in Theorem 5.2 tells us that the graphs of inverse functions are reflections about the line y = x. For a proof of this, we refer the reader to Example 1.1.6 in Section 1.1. A plot of the inverse functions f(x) = 3x + 4 and g(x) = confirms this to be the case. 2- y f(x) 1- -2 1 / -1- -2- y /y 1 / x 1 2 g =g(x) If we abstract one step further, we can express the sentiment in Definition 5.2 by saying that f and g are inverses if and only if g o f = I1 and f o g = I2 where I1 is the identity function restricted' to the domain of f and 12 is the identity function restricted to the domain of g. In other words, Ii(x) = x for all x in the domain of f and 12(x) = x for all x in the domain of g. Using this description of inverses along with the properties of function composition listed in Theorem 5.1, 'The identity function I, which was introduced in Section 2.1 and mentioned in Theorem 5.1, has a domain of all real numbers. In general, the domains of f and g are not all real numbers, which necessitates the restrictions listed here.  5.2 INVERSE FUNCTIONS 281 we can show that function inverses are unique.2 Suppose g and h are both inverses of a function f. By Theorem 5.2, the domain of g is equal to the domain of h, since both are the range of f. This means the identity function 12 applies both to the domain of h and the domain of g. Thus h = h o I2 = h o (f o g) = (h o f) o g = I1 o g = g, as required.3 We summarize the discussion of the last two paragraphs in the following theorem.4 THEOREM 5.3. Uniqueness of Inverse Functions and Their Graphs : Suppose f is an invertible function. " There is exactly one inverse function for f, denoted f-1 (read f-inverse) " The graph of y = f-1(x) is the reflection of the graph of y = f(x) across the line y = x. The notation f -1 is an unfortunate choice since you've been programmed since Elementary Algebra to think of this as }. This is most definitely not the case since, for instance, f(x) = 3x + 4 has as its inverse f-1(x) =x4, which is certainly different than 1 1= 3x+4. Why does this confusing notation persist? As we mentioned in Section 5., the identity function I is to function composition what the real number 1 is to real number multiplication. The choice of notation f-1 alludes to the property that f -1 o f = I1 and f o f-1 = 12, in much the same way as 3-1 . 3= 1 and 3 . 3-1 = 1. Let's turn our attention to the function f (x) 9=x2. Is f invertible? A likely candidate for the inverse is the function g(x) =_x/E. Checking the composition yields (go f)(x) = g(ff(x)) = /2 x= Iz|, which is not equal to x for all x in the domain (-oc, oc). For example, when x= -2, f(-2) =_(-2)2 4, but g(4) = 4= 2, which means g failed to return the input -2 from its output 4. What g did, however, is match the output 4 to a different input, namely 2, which satisfies f(2) = 4. This issue is presented schematically in the picture below. f 9 We see from the diagram that since both f(-2) and f(2) are 4, it is impossible to construct a function which takes 4 back to both x 2 and x -2. (By definition, a function matches 2In other words, invertible functions have exactly one inverse. 3It is an excellent exercise to explain each step in this string of equalities. 4In the interests of full disclosure, the authors would like to admit that much of the discussion in the previous paragraphs could have easily been avoided had we appealed to the description of a function as a set of ordered pairs. We make no apology for our discussion from a function composition standpoint, however, since it exposes the reader to more abstract ways of thinking of functions and inverses. We will revisit this concept again in Chapter 8.  282 FURTHER TOPICS IN FUNCTIONS a real number with exactly one other real number.) From a graphical standpoint, we know that if y = f-1(x) exists, its graph can be obtained by reflecting y =9x2 about the line y = x, in accordance with Theorem 5.3. Doing so produces y 7 y 6 (4 , -2 ) 5 (-2, 4) (2,T4)r 3 2 - 1 -2 -1 1 2 X reflect across y= x (4, -2) Y =f (x) =X9 switch x and y coordinates Y =f-1(x) ? We see that the line x= 4 intersects the graph of the supposed inverse twice - meaning the graph fails the Vertical Line Test, Theorem 1.1, and as such, does not represent y as a function of x. The vertical line x= 4 on the graph on the right corresponds to the horizontal line y = 4 on the graph of y = f(x). The fact that the horizontal line y = 4 intersects the graph of f twice means two different inputs, namely x= -2 and x= 2, are matched with the same output, 4, which is the cause of all of the trouble. In general, for a function to have an inverse, different inputs must go to different outputs, or else we will run into the same problem we did with f(x) = x2. We give this property a name. DEFINITION 5.3. A function f is said to be one-to-one if f matches different inputs to different outputs. Equivalently, f is one-to-one if and only if whenever f(c) = f(d), then c = d. Graphically, we detect one-to-one functions using the test below. THEOREM 5.4. The Horizontal Line Test: A function f is one-to-one if and only if no horizontal line intersects the graph of f more than once. We say that the graph of a function passes the Horizontal Line Test if no horizontal line intersects the graph more than once; otherwise, we say the graph of the function fails the Horizontal Line Test. We have argued that if f is invertible, then f must be one-to-one, otherwise the graph given by reflecting the graph of y = f(x) about the line y = x will fail the Vertical Line Test. It turns out that being one-to-one is also enough to guarantee invertibility. To see this, we think of f as the set of ordered pairs which constitute its graph. If switching the x- and y-coordinates of the points results in a function, then f is invertible and we have found f-1. This is precisely what the Horizontal Line Test does for us: it checks to see whether or not a set of points describes x as a function of y. We summarize these results below.  5.2 INVERSE FUNCTIONS 283 THEOREM 5.5. Equivalent Conditions for Invertibility: Suppose f is a function. The fol- lowing statements are equivalent. " f is invertible. " f is one-to-one. " The graph of f passes the Horizontal Line Test. We put this result to work in the next example. EXAMPLE 5.2.1. Determine if the following functions are one-to-one in two ways: (a) analytically using Definition 5.3 and (b) graphically using the Horizontal Line Test. 1. f (X) =1 - 2x 5 2x 2. g(x) 1 - x 3. h(x) =9x2 - 2x +4 4. F {(-1, 1), (0, 2), (2, 1)} SOLUTION. 1. (a) To determine if f is one-to-one analytically, we assume f (c) deduce that c = d. f(d) and attempt to f (c) f (d) 1-2c 1 -2d 5 5 1 - 2c = 1 - 2d -2c = -2d c = d/ Hence, f is one-to-one. (b) To check if f is one-to-one graphically, we look to see if the graph of y = f(x) passes the Horizontal Line Test. We have that f is a non-constant linear function, which means its graph is a non-horizontal line. Thus the graph of f passes the Horizontal Line Test as seen below. 2. (a) We begin with the assumption that g(c) = g(d) and try to show c = d. g(c) g(d) 2c 2d 1-c 1-d 2c(1 - d) = 2d(1 - c) 2c - 2cd = 2d - 2dc 2c = 2d c = d/ We have shown that g is one-to-one.  284 FURTHER TOPICS IN FUNCTIONS (b) We can graph g using the six step procedure outlined in Section 4.2. We get the sole intercept at (0, 0), a vertical asymptote x= 1 and a horizontal asymptote (which the graph never crosses) y = -2. We see from that the graph of g passes the Horizontal Line Test. 4 I 2 3-- 2 I 1_ 1 1-2 -1 1 r 2 -2 -1 2 _----- -1 -3 I -4 I -2 -5 -6 y =.f(x) y-g(x) 3. (a) We begin with h(c) = h(d). As we work our way through the problem, we encounter a nonlinear equation. We move the non-zero terms to the left, leave a 0 on the right and factor accordingly. h(c) = h(d) c2-2c+4 = d2-2d+4 c2-2c = d2-2d c2-d2-2c+2d = 0 (c+d)(c-d)-2(c-d) = 0 (c - d)((c + d) - 2) = 0 factor by grouping c-d=0 or c+d-2=0 c=d or c = 2 - d We get c = d as one possibility, but we also get the possibility that c = 2 - d. This suggests that f may not be one-to-one. Taking d = 0, we get c = 0 or c = 2. With f(0) = 4 and f(2) = 4, we have produced two different inputs with the same output meaning f is not one-to-one. (b) We note that h is a quadratic function and we graph y = h(x) using the techniques presented in Section 2.3. The vertex is (1, 3) and the parabola opens upwards. We see immediately from the graph that h is not one-to-one, since there are several horizontal lines which cross the graph more than once. 4. (a) The function F is given to us as a set of ordered pairs. The condition F(c) =F(d) means the outputs from the function (the y-coordinates of the ordered pairs) are the same. We see that the points (-1, 1) and (2, 1) are both elements of F with F(-1) - 1 and F(2) =1. Since -1 -f 2, we have established that F is not one-to-one. (b) Graphically, we see the horizontal line y =1 crosses the graph more than once. Hence, the graph of F fails the Horizontal Line Test.  5.2 INVERSE FUNCTIONS 285 6 Y - 4- - - - - 2 3- - - - -0-.- - - .- - 42-- 2||6 X -2 -1 1 2 y = F(x) 1 2 x -1 Q y =h(x) We have shown that the functions f and g in Example 5.2.1 are one-to-one. This means they are invertible, so it is natural to wonder what f1(x) and g-1(x) would be. For f(x) =152x, we can think our way through the inverse since there is only one occurrence of x. We can track step-by-step what is done to x and reverse those steps as we did at the beginning of the chapter. The function g(x) = 1 is a bit trickier since x occurs in two places. When one evaluates g(x) for a specific value of x, which is first, the 2x or the 1 - x? We can imagine functions more complicated than these so we need to develop a general methodology to attack this problem. Theorem 5.2 tells us equation y = f-1(x) is equivalent to f(y) = x and this is the basis of our algorithm. Steps for finding the Inverse of a One-to-one Function 1. Write y = f(x) 2. Interchange x and y 3. Solve x= f(y) for y to obtain y = f1(x) Note that we could have simply written 'Solve x= f(y) for y' and be done with it. The act of interchanging the x and y is there to remind us that we are finding the inverse function by switching the inputs and outputs. EXAMPLE 5.2.2. Find the inverse of the following one-to-one functions. Check your answers ana- lytically using function composition and graphically. 1 - 2x 2x 1. f (x) =52. g(x) = 5 1-S SOLUTION. 1. As we mentioned earlier, it is possible to think our way through the inverse of f by recording the steps we apply to x and the order in which we apply them and then reversing those steps in the reverse order. We encourage the reader to do this. We, on the other hand, will practice the algorithm. We write y =f(x) and proceed to switch x and y  286 FURTHER TOPICS IN FUNCTIONS y f(x) 1-2x y = 5 1-2y x = 5 switch x and y 5x 5x - 1 5x - 1 -2 1 -2y r-y y 2+ We have f1(x) =_- x+ -. To check this answer analytically, we first check that (f1 o f) (X) x for all x in the domain of f, which is all real numbers. (f1o f)Q() f -1 (f ()) -i f /) + i - 2.1- 2x) + _ 1 -21 We now check that (f o f-1) (x) = x for all x in the range of f which is also all real numbers. (Recall that the domain of f-1) is the range of f.) (fof-) (x) f (f -,(X)) 1 - 2f-1(x) 5 1-2(-zx+ ) 5 1+5x -1 53 5 5 3:V To check our answer graphically, we graph y = f (x) and y They appear to be reflections across the line y = x. f -1(x) on the same set of axes.5 5Note that if you perform your check on a calculator for more sophisticated functions, you'll need to take advantage of the 'ZoomSquare' feature to get the correct geometric perspective.  5.2 INVERSE FUNCTIONS28 287 y y~x x yf(x) y =f-1(x) 2. To find g-1(x), we start with y =g(x). We note that the domain of g is (-00, 1) U (1, c). y -=g9W 2x y 1-x x = 1-y x(1l-y) =-2 x -xy =2 x xy + 2y X Y(x +2) y x+2 switch x and y factor Weoti -()in the domain of g, +X To check this analytically, we first check (g-1 o g) (x)= x for all x that is, for all x : 1. (g-,o g)() g-1(W) ___x )  288 FURTHER TOPICS IN FUNCTIONS 2x 2x + 2(1 - x) 2x 2x + 2 - 2x 2x 2 Next, we check g (g1(x)) = x for all x in the range of g. From the graph of g in Example 5.2.1, we have that the range of g is (-oc, -2) U (-2, oc). This matches the domain we get from the formula g-1(x) =_+2,as it should. (gog 1) () g (g- gx + 2) 1 - (Xi2 2 x x+2 2(2 1- (x+2) (xz+2). clear denominators (x + 2) 2x (x + 2)- x 2x 2 Graphing y = g(x) and y = g-1(x) on the same set of axes is busy, but we can see the sym- metric relationship if we thicken the curve for y = g-1(x). Note that the vertical asymptote x = 1 of the graph of g corresponds to the horizontal asymptote y = 1 of the graph of g-1, as it should since x and y are switched. Similarly, the horizontal asymptote y = -2 of the graph of g corresponds to the vertical asymptote x= -2 of the graph of g-1.  5.2 INVERSE FUNCTIONS 289 y x y = g(x) and y = g-1(x) We now return to f(x) = x2. We know that f is not one-to-one, and thus, is not invertible. However, if we restrict the domain of f, we can produce a new function g which is one-to-one. If we define g(x) = 2, x > 0, then we have y y 7- 6- 5- 4- 3- 2- x -2 -1 I 1 2 restrict domain to x;> 0 y=g(x)=x2,x>O yf(x) =x2 The graph of g passes the Horizontal Line Test. To find an inverse of g, we proceed as usual y y 3: y y g(x) z2, x> 0 y2, y> 0 switch xand y / since y > 0  290 FURTHER TOPICS IN FUNCTIONS We get g-1(xv) =cV. At first it looks like we'll run into the same trouble as before, but when we check the composition, the domain restriction on g saves the day. We get (g-1 o g) (x) g-1(g(x)) = g-1 (2) v2 =I =x, since x > 0. Checking (gog-)l() = g (g-1(x)) g () /) =(1/)2 = x. Graphing6 g and g- on the same set of axes shows that they are reflections about the line y = x. y 8 y = g(X) 7 ~/ 6-- 5-- 4-- y =g-1W 3- 2 -- 1 -- Our next example continues the theme of domain restriction. EXAMPLE 5.2.3. Graph the following functions to show they are one-to-one and find their inverses. Check your answers analytically using function composition and graphically. 1. j(x) = x2 - 2x + 4, z < 1. 2. k(x) = cx/+2 - 1 SOLUTION. 1. The function j is a restriction of the function h from Example 5.2.1. Since the domain of j is restricted to xv < 1, we are selecting only the 'left half' of the parabola. We see that the graph of j passes the Horizontal Line Test and thus j is invertible. y 6 4 3 2 1 2 x -1 y =j(x) 6We graphed y = in Section 1.8.  5.2 INVERSE FUNCTIONS29 291 We now use our algorithm to find j-1(x). y - y 92 2x+4, x< o - y2-2y+4-x y - 2(1) 2fV y 2 2 2(1+ x-3) 2 y =1+ x-3 y= 1- x-3 switch x and y quadratic formula, c 4-x since y < 1. We have j-1(x)= 1 - x - 3. When we simplify (j o j) (x), we need to remember that the domain of j is x < 1. (j-, o j)() j -(j~v)) j- (2 -2x + 4), x <1 1- (x2-2x+4)-3 1- 3:l223+ 1 - (x - 1)2 1- x-i1 1(-(x-) 3:V since 3: < 1 Checking j o j-1, we get (jojl)(:) We can use what we know from to get j (j-l(3:)) (i1- x 31-3 ) 2 (1 - x 3 ) + 4 1-2 x-3+( x3) 2+2 3x-3+4 3+x:-3 3:V Section 1.8 to graph y - j-1(x:) on the same axes as y=j (x:)  292 FURTHER TOPICS IN FUNCTIONS 6- yj(x) 4- 3- 2- 1- / / y =x / / - / V / I I I I 2:4- 1 1 N / / '-1 1 2 3 4 6 x y__-1(x) 2. We graph y = k(x) x= x + 2-1 using what we learned in Section 1.8 and see k is one-to-one. y 2 1 -2 -1 1 2 x -2-. y =k(x) We now try to find k-1. y y x x+1 (x + 1)2 2 + 2x + 1 y k(x) x+2-1 y+2- 1 ( y+2)2 y + 2 x2 +2x- 1 switch x and y We have k-1(x) = z2+2x - 1. Based on our experience, we know something isn't quite right. We determined k-1 is a quadratic function, and we have seen several times in this section that these are not one-to-one unless their domains are suitably restricted. Theorem 5.2 tells us that the domain of k-1 is the range of k. From the graph of k, we see that the range is [-1, oc), which means we restrict the domain of k-1 to x > -1. We now check that this works in our compositions.  5.2 INVERSE FUNCTIONS 293 (k-1 o k) (x) (k o k-1) () k-1(k(x)) k-1( x+2-1), x>-2 ( x+2-1)2+2(vx+2-1)-1 (v/x+2)2-2vx+2+1+2vx+2 x+2-2 xV~ 2-1 and k (x2 +2x-1) x> (x2 + 2x - 1) + 2 - x2 + 2x + 1 - 1 ( + 1)2 - 1 x+1 -1 x +1-1 -1 1 since x > -1 _ Xv Graphically, everything checks out as well, provided that we remember the domain restriction on k-1 means we take the right half of the parabola. 2- y - k(x) 1- -2 -1 / /-1- - 2- y / / -/ I I2 1 2 x .y - k-1 (x El Our last example of the section gives an application of inverse functions. EXAMPLE 5.2.4. Recall from Section 2.1 that the price-demand equation for the PortaBoy game system is p(x) = -1.5x + 250 for 0 < x < 166, where x represents the number of systems sold weekly and p is the price per system in dollars. 1. Explain why p is one-to-one and find a formula for p-1(x). State the restricted domain. 2. Find and interpret p-1(220). 3. Recall from Section 2.3 that the weekly profit P, in dollars, as a result of selling x systems is given by P(x) = -1.5x2 + 170x - 150. Find and interpret (P o0p-1) (x).  294 FURTHER TOPICS IN FUNCTIONS 4. Use your answer to part 3 to determine the price per PortaBoy which would yield the maxi- mum profit. Compare with Example 2.3.3. SOLUTION. 1. We leave to the reader to show the graph of p(x) = -1.5x + 250, 0 < x < 166, is a line segment from (0, 250) to (166, 1), and as such passes the Horizontal Line Test. Hence, p is one-to-one. We find the expression for p-1(x) as usual and get p-1(x) - 503 2x. The domain of p-1 should match the range of p, which is [1, 250], and as such, we restrict the domain of p-1 to 1 5 (i) f (X) =3+ (e) f(z) =3( + 4)2 -5, z <; -4 (j) f (x) = -- 2 2. Show that the Fahrenheit to Celsius conversion function found in Exercise 3 in Section 2.1 is invertible and that its inverse is the Celsius to Fahrenheit conversion function found in that same exercise. 3. Analytically show that the function f(x)= x3 + 3x + 1 is one-to-one. Since finding a formula for its inverse is beyond the scope of this textbook, use Theorem 5.2 to help you compute f-1(1), f-1(5), and f-1(-3). 4. With the help of your classmates, find a formula for the inverse of the following. (a) f (x) ax + b, a # 0 (b) f (x)=a x-h + k, a#x0, z > h (c) f (x)=aj+, a # 0, b # 0, c # 0, d #0 (d) f(x) =ax2+bx+cwherea#0, x > -0. 5. Let f(x) = . Using the techniques in Section 4.2, graph y = f(x). Verify f is one-to-one on the interval (-1, 1). Use the procedure outlined on Page 5.2 to find the possibilities for f-1(x) and use your graphing calculator to help choose the correct formula for f-1(x). Note that since f(0) = 0, it should be the case that f-1(0) = 0. What goes wrong when you attempt to substitute x= 0 into f--1(x)? Discuss with your classmates how this problem arose algebraically, and possible remedies. 6. Suppose f is an invertible function. Prove that if graphs of y = f (x) and y = f -1(x) intersect at all, they do so on the line y = x. 7. With the help of your classmates, explain why a function which is either strictly increasing or strictly decreasing on its entire domain would have to be one-to-one, hence invertible. 8. Let f and g be invertible functions. With the help of your classmates show that (f o g) is one-to-one, hence invertible, and that (f o g)-i(z) =(g- o fr-i)(x). 9. What graphical feature must a function f possess for it to be its own inverse?  296 FURTHER TOPICS IN FUNCTIONS 5.2.2 ANSWERS 1. (a)f() (b)f() (d)f() (e)f() x+2 6 3x 1 (x - 2)2 + 5, x < 2 5+ x+25 x+s -4 3 (f f (X) ,- +x+ (g) f W)- x- (h) f x) 3x1 (f1(x) =6x±4 (i) f -(x) -3x+32 3. Given that f (O)= 1, we have f-1(1) - 0. Similarly f-1(5)= 1 and f-1(-3) -1  5.3 OTHER ALGEBRAIC FUNCTIONS 297 5.3 OTHER ALGEBRAIC FUNCTIONS This section serves as a watershed for functions which are combinations of polynomial, and more generally, rational functions, with the operations of radicals. It is business of Calculus to discuss these functions in all the detail they demand so our aim in this section is to help shore up the requisite skills needed so that the reader can answer Calculus's call when the time comes. We briefly recall the definition and some of the basic properties of radicals from Intermediate Algebra.1 DEFINITION 5.4. Let x be a real number and n a natural number.' If n is odd, the principal nth root of x, denoted xis the unique real number satisfying (n /)= x. If n is even, xis defined similarlyb provided x > 0 and x > 0. The index is the number n and the radicand is the number x. For n = 2, we write x instead of . "Recall this means n = 1, 2, 3.... bRecall both x= -2 and x= 2 satisfy x4 = 16, but 16 = 2, not -2. It is worth remarking that, in light of Section 5.2, we could define f(x) = x functionally as the inverse of g(x) =x" with the stipulation that when n is even, the domain of g is restricted to [0, oo). From what we know about g(x) = x from Section 3.1 along with Theorem 5.3, we can produce the graphs of f(x) = xby reflecting the graphs of g(x) = x across the line y = x. Below are the graphs of y =V, y = and y = x. The point (0,0) is indicated as a reference. The axes are hidden so we can see the vertical steepening near x= 0 and the horizontal flattening as x - oc. y=V x 4 a 6 a The odd-indexed radical functions also follow a predictable trend - steepening near x = 0 and flattening as x - too. In the exercises, you'll have a chance to graph some basic radical functions using the techniques presented in Section 1.8. 3Y 5Y 7Y We have used all of the following properties at some point in the textbook for the case n square root), but we list them here in generality for completeness. 2 (the 'Although we discussed imaginary numbers in Section 3.4, we restrict our attention to real numbers in this section. See the epilogue on page 216 for more details.  298 FURTHER TOPICS IN FUNCTIONS THEOREM 5.6. Properties of Radicals: Let x and y be real numbers and m and n be natural numbers. If Cx, g/y are real numbers, then " Product Rule: gxy = /z gy * Powers of Radicals: X = (n /)' zX g/: " Quotient Rule: n - = -, provided y # 0. y vy " If n is odd, nY= x; ifn nis even, n= Iz|. The proof of Theorem 5.6 is based on the definition of the principal roots and properties of expo- nents. To establish the product rule, consider the following. If n is odd, then by definition xy is the unique real number such that ( /y)Th =y. Given that (C / n/y)"= (n = /)" (y)"T = y, it must be the case that cvy = cv g/y. If n is even, then cvy is the unique non-negative real number such that (n g/y)"n =zy. Also note that since n is even, c and g/y are also non-negative and hence so is cv g/y. Proceeding as above, we find that vy = / /y. The quotient rule is proved similarly and is left as an exercise. The power rule results from repeated application of the product rule, so long as C is a real number to start with.2 The last property is an application of the power rule when n is odd, and the occurrence of the absolute value when n is even is due to the requirement that c > 0 in Definition 5.4. For instance, /(-2)4=4 16 = 2 = | - 2, not -2. It's this last property which makes compositions of roots and powers delicate. This is especially true when we use exponential notation for radicals. Recall the following definition. DEFINITION 5.5. Let x be a real number, m an integer" and n a natural number. " X = c and is defined whenever gc is defined. * = (n = /)m = Vm, whenever ( /)m is defined. "Recall this means m = 0, +1, +2, ... The rational exponents defined in Definition 5.5 behave very similarly to the usual integer exponents from Elementary Algebra with one critical exception. Consider the expression (2/3)3/2. Applying the usual laws of exponents, we'd be tempted to simplify this as (2/3)3/2 _ .2 -2l - X However, if we substitute x = -1 and apply Definition 5.5, we find (-1)2/3 ( 12)= (_1)2 1 so that ((-1)2/3)3/2 - 13/2 - ()3 13 = 1. We see in this case that (z2/3)3/2 # z. If we take the time to rewrite (z2/3)3/2 with radicals, we see 2/3)3/2 )23/2 2 3 3 3 In the play-by-play analysis, we see that when we canceled the 2's in multiplying - , we were, 2Otherwise we'd run into the same paradox we did in Section 3.4.  5.3 OTHER ALGEBRAIC FUNCTIONS 299 in fact, attempting to cancel a square with a square root. The fact that =x2 I| and not simply x is the root3 of the trouble. It may amuse the reader to know that (z3/2)2/3 = x, and this verification is left as an exercise. The moral of the story is that when simplifying fractional exponents, it's usually best to rewrite them as radicals.4 The last major property we will state, and leave to Calculus to prove, is that radical functions are continuous on their domains, so the Intermediate Value Theorem, Theorem 3.1, applies. This means that if we take combinations of radical functions with polynomial and rational functions to form what the authors consider the algebraic functions,5 we can make sign diagrams using the procedure set forth in Section 4.2. Steps for Constructing a Sign Diagram for an Algebraic Function Suppose f is an algebraic function. 1. Place any values excluded from the domain of f on the number line with an ' above them. 2. Find the zeros of f and place them on the number line with the number 0 above them. 3. Choose a test value in each of the intervals determined in steps 1 and 2. 4. Determine the sign of f(x) for each test value in step 3, and write that sign above the corresponding interval. Our next example reviews quite a bit of Intermediate Algebra and demonstrates some of the new features of these graphs. EXAMPLE 5.3.1. For the following functions, state their domains and create sign diagrams. Check your answer graphically using your calculator. 1. f(x)=3x32-x 2. g(x) = 2-- 4x +3 3. h(x) = 38x x+1 4. k(x) 2x SOLUTION. 1. As far as domain is concerned, f(x) has no denominators and no even roots, which means its domain is (-oc, oc). To create the sign diagram, we find the zeros of f. 3Did you like that pun? 4In most other cases, though, rational exponents are preferred. 5As mentioned in Section 2.2, f(x) x= x2=Iz so that absolute value is also considered an algebraic functions.  300 FURTHER TOPICS IN FUNCTIONS f(x) = 0 3x32-x = 0 3x=0 or 32-x=0 x=0 or (32-x)3=03 x=0 or 2-x=0 x=0 or x=2 The zeros 0 and 2 divide the real number line into three test intervals. The sign diagram and accompanying graph are below. Note that the intervals on which f is (+) correspond to where the graph of f is above the i-axis, and where the graph of f is below the i-axis we have that f is (-). The calculator suggests something mysterious happens near x = 2. Zooming in shows the graph becomes nearly vertical there. You'll have to wait until Calculus to fully understand this phenomenon. / 1 (-) 0 (+) 0 (-) I I 0 2 y=f(x) y= f(x)nearx=2. 2. In g(x) = 2 - fx + 3, we have two radicals both of which are even indexed. To satisfy 4x + 3, we require x + 3 > 0 or x > -3. To satisfy 2 - 4x + 3, we need 2 - -4x +3 0. While it may be tempting to write this as 2 > ' x + 3 and take both sides to the fourth power, there are times when this technique will produce erroneous results.6 Instead, we solve 2 - 4x +3 > 0 using a sign diagram. If we let r(x) = 2 - 4x + 3, we know i > -3, so we concern ourselves with only this portion of the number line. To find the zeros of r we set r(x) =0 and solve 2 - 4x +3 =0. We get 4x +3 =2 so that (4x +3)4 24 from which we obtain x + 3 = 16 or x = 13. Since we raised both sides of an equation to an even power, we need to check to see if x = 13 is an extraneous solution.7 We find x = 13 does check since 2 - x +3 = 2 - ' 13 + 3 = 2 - ' 16 = 2 - 2 = 0. Below is our sign diagram for r. (+) 0 (-) I I -3 13 We find 2 - fx + 3 > 0 on [-3, 13] so this is the domain of g. To find a sign diagram for g, we look for the zeros of g. Setting g(x) = 0 is equivalent to 2 - fx + 3 = 0. After squaring 6For instance, -2 > ' x + 3, which has no solution or -2 < ' x + 3 whose solution is [-3, oo). 7Recall, this means we have produced a candidate which doesn't satisfy the original equation. Do you remember how raising both sides of an equation to an even power could cause this?  5.3 OTHER ALGEBRAIC FUNCTIONS 301 both sides, we get 2 - fx + 3 = 0, whose solution we have found to be x = 13. Since we squared both sides, we double check and find g(13) is, in fact, 0. Our sign diagram and graph of g are below. Since the domain of g is [-3,13], what we have below is not just a portion of the graph of g, but the complete graph. It is always above or on the x-axis, which verifies our sign diagram. (+) -3 13 The complete graph of y = g(x). 3. The radical in h(x) is odd, so our only concern is the denominator. Setting x + 1 = 0 gives x -1, so our domain is (-oc, -1) U (-1, oc). To find the zeros of h, we set h(x) = 0. To solve 3 8X 0, we cube both sides to get x = 0. We get 8x = 0, or x 0. Below is the resulting sign diagram and corresponding graph. From the graph, it appears as though x = -1 is a vertical asymptote. Carrying out an analysis as x - -1 as in Section 4.2 confirms this. (We leave the details to the reader.) Near x = 0, we have a situation similar to x = 2 in the graph of f in number 1 above. Finally, it appears as if the graph of h has a horizontal asymptote y = 2. Using techniques from Section 4.2, we find as x -bo, x -- 8. From this, it is hardly surprising that as x - toc, h(x) 8 35 = 2. (+) ? (-) 0 (+) C I I > -1 0 y = h(x) 4. To find the domain of k, we have both an even root and a denominator to concern ourselves with. To satisfy the square root, x2 - 1 > 0. Setting r(x) = x2 - 1, we find the zeros of r to be x +1, and we find the sign diagram of r to be (+) 0 (-) 0 (+) - I I -1 1  302 FURTHER TOPICS IN FUNCTIONS We find x2 - 1 > 0 for (-oc, -1] U [1, oc). To keep the denominator of k(x) away from zero, we set x2 - 1 = 0. We leave it to the reader to verify the solutions are x +1, both of which must be excluded from the domain. Hence, the domain of k is (-oc, -1) U (1, oc). To build the sign diagram for k, we need the zeros of k. Setting k(x) = 0 results in 2 1 = 0. We get 2x = 0 or x = 0. However, x = 0 isn't in the domain of k, which means k has no zeros. We construct our sign diagram on the domain of k below alongside the graph of k. It appears that the graph of k has two vertical asymptotes, one at x = -1 and one at x = 1. The gap in the graph between the asymptotes is because of the gap in the domain of k. Concerning end behavior, there appear to be two horizontal asymptotes, y = 2 and y = -2. To see why this is the case, we think of x - toc. The radicand of the denominator x2 - 1 - x2, and as such, k(x) 2x x . As x - oc, we have x = x so k(x) - 2 = 2. On the other hand, as x - -o, x = -x, and as such k(x) - 2 = -2. Finally, it appears as though the graph of k passes the Horizontal Line Test which means k is one to one and k-1 exists. Computing k-1 is left as an exercise. -1 1 y = k(x) As the previous example illustrates, the graphs of general algebraic functions can have features we've seen before, like vertical and horizontal asymptotes, but they can occur in new and exciting ways. For example, k(x) had two distinct horizontal asymptotes. You'll recall that rational functions could have at most one horizontal asymptote. Also some new characteristics like 'unusual steepness'8 and cusps9 can appear in the graphs of arbitrary algebraic functions. Our next example first demonstrates how we can use sign diagrams to solve nonlinear inequalities. (Don't panic. The technique is very similar to the ones used in Chapters 2, 3 and 4.) We then check our answers graphically with a calculator and see some of the new graphical features of the functions in this extended family. EXAMPLE 5.3.2. Solve the following inequalities. Check your answers graphically using a calculator. 1. x2/3 < x43 - 6 2. 3(2 - <)1/3 x(2 - -2/3 8The proper Calculus term for this is 'vertical tangent', but for now we'll be okay calling it 'unusual steepness'. 9See page 175 for the first reference to this feature.  5.3 OTHER ALGEBRAIC FUNCTIONS 303 SOLUTION. 1. To solve x2/3 < x4/3 - 6, we get 0 on one side and attempt to solve x4/3 - x2/3 - 6 > 0. We set r(x) = x4/3 - x2/3 - 6 and note that since the denominators in the exponents are 3, they correspond to cube roots, which means the domain of r is (-oc, oc). To find the zeros for the sign diagram, we set r(x) = 0 and attempt to solve x4/3 - x2/3 - 6 = 0. At this point, it may be unclear how to proceed. We could always try as a last resort converting back to radical notation, but in this case we can take a cue from Example 3.3.4. Since there are three terms, and the exponent on one of the variable terms, x4/3, is exactly twice that of the other, x2/3, we have ourselves a 'quadratic in disguise' and we can rewrite x4/3 - x2/3 - 6 = 0 as (x2/3)2 - x2/3 - 6 = 0. If we let u = x2/3, then in terms of u, we get u2 - u - 6 = 0. Solving for u, we obtain u = -2 or u = 3. Replacing x2/3 back in for u, we get x2/3 - -2 or x2/3 = 3. To avoid the trouble we encountered in the discussion following Definition 5.5, we now convert back to radical notation. By interpreting x2/3 as 3x2 we have x = -2 or 3x2 = 3. Cubing both sides of these equations results in x2 = -8, which admits no real solution, or x2 = 27, which gives x = +33. We construct a sign diagram and find x4/3 _ 2/3 - 6 > 0 on (-oc, -3v3) U (3v3, oc). To check our answer graphically, we set f (x) = x2/3 and g(x) = x4/3 - 6. The solution to x2/3 < x4/3 - 6 corresponds to the inequality f(x) < g(x), which means we are looking for the x values for which the graph of f is below the graph of g. Using the 'Intersect' command we confirm10 that the graphs cross at x = +3/3. We see that the graph of f is below the graph of g (the thicker curve) on (-o, -3 ) U (3v, oc). (+) 0 (-) 0 (+) C I I a -335 3 /5 y =f(x) and y = g(x) As a point of interest, if we take a closer look at the graphs of f and g near x = 0 with the axes off, we see that despite the fact they both involve cube roots, they exhibit different behavior near x = 0. The graph of f has a sharp turn, or cusp, while g does not." 100r at least confirm to several decimal places "Again, we introduced this feature on page 175 as a feature which makes the graph of a function 'not smooth'.  304 FURTHER TOPICS IN FUNCTIONS y =f(x) near x= 0 y =g(x) near x= 0 2. To solve 3(2 - x)1/3 < x(2 - x)-2/3, we gather all the nonzero terms on one side and obtain 3(2 - ()1/3 - x2 - x)-2/3 0. We set r(x) = 3(2 - ()1/3 - x2 - x)-2/3. As in number 1, the denominators of the rational exponents are odd, which means there are no domain concerns there. However, the negative exponent on the second term indicates a denominator. Rewriting r(x) with positive exponents, we obtain r(x) = 3(2 - x)1/3 (2-x)2/3. Setting the denominator equal to zero we get (2 - z)2/3 = 0, or 3 (2- x)2 = 0. After cubing both sides, and subsequently taking square roots, we get 2 - x = 0, or x = 2. Hence, the domain of r is (-oc, 2) U (2, oc). To find the zeros of r, we set r(x) = 0. There are two school of thought on how to proceed and we demonstrate both. * Factoring Approach. From r(x) = 3(2 - )1/3 - x(2 - x-2/3, we note that the quantity (2- x) is common to both terms. When we factor out common factors, we factor out the quantity with the smaller exponent. In this case, since -3 < j, we factor (2 - z)-2/3 from both quantities. While it may seem odd to do so, we need to factor (2 - z)-2/3 from (2 - x)1/3, which results in subtracting the exponent - from j. We proceed using the usual properties of exponents and exercise special care when reducing the 3 power to 1. r(x) 3(2 (2 - (2 - (2 - (2 - (2 - - x(2 - -2/3 x)-2/3 [3(2 - x)3-(-3) x)-2/3 [3(2 - ])3/3 _ x)-2/3 [3(2 - )1 - x] x)-2/3 (6 - 4x) x)-2/3 (6 - 4x) ] since 3 3 = (3 /)3 _u To solve r(x) or x=2. 0, we set (2 - x)-2/3 (6 - 4x) = 0, or 6-4/ =0. We have 6 - 4x = 0  5.3 OTHER ALGEBRAIC FUNCTIONS 305 e Common Denominator Approach. We rewrite r(x) = 3(2 - ()1/3 - x2 - -2/3 3(2 - z)1/3 _ (2 - z)2/3 3(2 - x)1/3(2 - )23 x (2 -x)2-3 (2 x)2 common denominator (2 -)2/3 (2 -2/ 3(2 - x)3+3 x (2 - z)2/3 (2 - )2/3 3(2 - z)33 x (2 - z)2/3 (2 - 2/3 3(2 - x)1 x since Y U- = (3 u)3 _ (2-x)2/3 (2-x2/3 3(2 - x) - x (2 - z)2/3 6 -4x (2 - z)2/3 As before, when we set r(x) = 0 we obtain x = 2. We now create our sign diagram and find 3(2 - )1/3 -x(2 - x-2/3 < 0 on [2, 2) U (2, o). To check this graphically, we set f(x) = 3(2 - x)1/3 and g(x) = x(2 - x)-2/3 (the thicker curve). We confirm that the graphs intersect at x = 2 and the graph of f is below the graph of g for X > 2, with the exception of x = 2 where it appears the graph of g has a vertical asymptote. . .LN.-. ( 0 3 2 I 1 y = f(x) and y = g(x) One application of algebraic functions was given in Example 1.7.6 in Section 1.1. Our last example is a more sophisticated application of distance. EXAMPLE 5.3.3. Carl wishes to get high speed internet service installed in his remote Sasquatch observation post located 30 miles from Route 117. The nearest junction box is located 50 miles downroad from the post, as indicated in the diagram below. Suppose it costs $15 per mile to run cable along the road and $20 per mile to run cable off of the road.  306 FURTHER TOPICS IN FUNCTIONS 0 Outpost y X >1<> Junction Box Route 117 50 miles 1. Express the total cost C of connecting the Junction Box to the Outpost as a function of x, the number of miles the cable is run along Route 117 before heading off road directly towards the Outpost. Determine a reasonable applied domain for the problem. 2. Use your calculator to graph y = C(x) on its domain. What is the minimum cost? How far along Route 117 should the cable be run before turning off of the road? SOLUTION. 1. The cost is broken into two parts: the cost to run cable along Route 117 at $15 per mile, and the cost to run it off road at $20 per mile. Since x represents the miles of cable run along Route 117, the cost for that portion is 15x. From the diagram, we see that the number of miles the cable is run off road is z, so the cost of that portion is 20z. Hence, the total cost is C = 15x + 20z. Our next goal is to determine z as a function of x. The diagram suggests we can use the Pythagorean Theorem to get y2 + 302 = z2. But we also see x + y = 50 so that y = 50 - x. Hence, z2 = (50 - X)2 + 900. Solving for z, we obtain z = ±(50 - X)2 + 900. Since z represents a distance, we choose z = V(50 - x)2 + 900 so that our cost as a function of x only is given by C(x) = 15x + 20 (50-z-x)2+900 From the context of the problem, we have 0 < x < 50. 2. Graphing y = C(x) on a calculator in a suitable window produces the graph below. Using the 'Minimum' feature, we find the relative minimum (which is also the absolute minimum in this case) to two decimal places be (15.98, 1146.86). Here the x-coordinate tells us that in order to minimize cost, we should run 15.98 miles of cable along Route 117 and then turn off of the road and head towards the outpost. The y-coordinate tells us that the minimum cost, in dollars, to do so is $1146.86. The ability to stream live SasquatchCasts? Priceless. D  5.3 OTHER ALGEBRAIC FUNCTIONS 307 5.3.1 EXERCISES 1. For each function below " Find its domain. " Create a sign diagram. " Use your calculator to help you sketch its asymptotes, 'unusual steepness' or cusps. (a) f(x) = -2 (b) f() =4 16x9 (c) f (x) =3i(x - 7)3 graph and identify any vertical or horizontal (d) f (x) = 5z (e) f (x) = x2 (x - 7)3 (f) f (x) = y/xz- + 5)( - 4) (g) f () = 33 + 3X2 - 6x - 8 2. Use the transformations presented in Section 1.8 to graph the following functions. (a) f (x) (b) f (x) -23x + 1+4 34x-7-1 (c) f(x) = x +2 + 3 (d) f(z) =y - 2 3. Solve the following equations and inequalities. (a) 2x -1 = x -+3 ( b ) z) = 8 (c) zi =4 (d) 3 < x (e) vx-2+ fx-5=3 (f) 5 - (4- 2x)3 = 1 (g) 10- -x-2<11 (h) jxo(x - 3)-- + x-4(x - 3)3 <0 (i) 3x+3x2- -6x-(8 >x z+l1 (j) j(x+4)?(x -2)-- + (x+4)--i(x -2) ;>p0 4. Show that (xzj x for all x > 0. 5. Verify the Quotient Rule for Radicals in Theorem 5.6. 2x 6. Find the inverse of k(x) = . x-1 7. Show that 2 is an irrational number by first showing that it is a zero of p(x) = z3 - 2 and then showing p has no rational zeros. (You'll need the Rational Zeros Theorem, Theorem 3.9, in order to show this last part.) 8. With the help of your classmates, generalize the previous exercise to show that Vc/ is an irrational number provided that c" is not a rational number.  308 FURTHER TOPICS IN FUNCTIONS 9. The period of a pendulum in seconds is given by T = 2r -(for small displacements) where Vg L is the length of the pendulum in meters and g = 9.8 meters per second per second is the acceleration due to gravity. My Seth-Thomas antique schoolhouse clock needs T = j second and I can adjust the length of the pendulum via a small dial on the bottom of the bob. At what length should I set the pendulum? 10. The Cobb-Douglas production model states that the yearly total dollar value of the production output P in an economy is a function of labor x (the total number of hours worked in a year) and capital y (the total dollar value of all of the stuff purchased in order to make things). Specifically, P = axbyl-. By fixing P, we create what's known as an 'isoquant' and we can then solve for y as a function of x. Let's assume that the Cobb-Douglas production model for the country of Sasquatchia is P = 1.23x0.4y0.. (a) Let P = 300 and solve for y in terms of x. If x= 100, what is y? (b) Graph the isoquant 300 = 1.23x0.4y06. What information does an ordered pair (x, y) which makes P = 300 give you? With the help of your classmates, find several different combinations of labor and capital all of which yield P = 300. Discuss any patterns you may see. 11. According to Einstein's Theory of Special Relativity, the observed mass m of an object is a m function of how fast the object is traveling. Specifically, m(x) = Mr where m(0) = mr 12 is the mass of the object at rest, x is the speed of the object and c is the speed of light. (a) Find the applied domain of the function. (b) Compute m(.lc), m(.5c), m(.9c) and m(.999c). (c) As x -- c-, what happens to m(x)? (d) How slowly must the object be traveling so that the observed mass is no greater than 100 times its mass at rest? 12. Suppose Fritzy the Fox, positioned at a point (x, y) in the first quadrant, spots Chewbacca the Bunny at (0, 0). Chewbacca begins to run along a fence (the positive y-axis) towards his warren. Fritzy, of course, takes chase and constantly adjusts his direction so that he is always running directly at Chewbacca. If Chewbacca's speed is vi and Frity's speed is v2, the path Fritzy will take to intercept Chewbacca, provided v2 is directly proportional to, but not equal to, vi is modeled by y= - - x~1v)+  5.3 OTHER ALGEBRAIC FUNCTIONS 309 (a) Determine the path Fritzy will take if he runs exactly twice as fast as Chewbacca; that is, v2 = 2v1. Use your calculator to graph this path for x > 0. What is the significance of the y-intercept of the graph? (b) Determine the path Fritzy will take if Chewbacca runs exactly twice as fast as he does; that is, v1= 2v2. Use your calculator to graph this path for x > 0. Describe the behavior of y as x - 0+ and interpret this physically. (c) With the help of your classmates, generalize parts (a) and (b) to two cases: v2 > v1 and v2 < v1. We will discuss the case of v1= v2 in Exercise 9 in Section 6.5.  310 FURTHER TOPICS IN FUNCTIONS 5.3.2 ANSWERS 1. (a) f(x)= 1-2 Domain: [-1, 1] 0 (+) y 0 -1 1 1 (I No asymptotes Unusual steepness at x No cusps -1 and x= 1 ' 9 " . -1 1 'x (b) f () = 26 Domain: (-3,0] U (3,oo) T (+) 0 T (+) -3 0 Vertical asymptotes: x =-3 and x - 3 Horizontal asymptote: y = 0 Unusual steepness at x:= 0 No cusps I 5- I4- 3. I2- y ' i i i i i i i _13 -2 -1 1 2 $ >, i 5 6 7 (c) f (X) 3( -7)3 Domain: (-oo,oo) (-) 0 (-) o(+) 52 4- 3- 2- 1- y t I 'I ' 0 7 I I I 6 1 1 1 1 1 1 6 1 1 No vertical or horizontal asymptotes12 Unusual steepness at x:- 7 Cusp at 3:= 0 _ 5z: (d) f (x)-= Domain: (-oc, -2) U (-2, ox) (+) T (-) 0 (+) I I -2 0 -3-2-1 2- -3- -4- 1 2 3 4 5 6 8 9 Vertical asymptote 3 =: Horizontal asymptote y: No unusual steepness or -2 =5 cusps 12Using Calculus it can be shown that y = x - 3is a slant asymptote of this graph.  5.3 OTHER ALGEBRAIC FUNCTIONS 311 (e) f (x) 32z(x - 7)3 Domain: [O,oo) 0 (-) y 0(+) I I 0 7. No asymptotes Unusual steepness at 3: No cusps 7 x (f) f() =/x(+5)(x- 4) Domain: [-5, 0] U [4, o) 0 (+) 0 0 (+) -5 0 4 9- 8- 7- 6- 5- 4- 2 1- - No asymptotes Unusual steepness at x: x= 4 No cusps -5, x = 0 and -5-4-3-2-1 T 2 3 4 5 (g) f (x) = =z3 + 3x2 - 6x - 8 Domain: (-oo,oo) (-) 0 (+)0 (-)0 (+) -4 -1 2 6- 5- 4- 3- 2- 1- . . - No vertical or horizontal Unusual steepness at 3: No cusps asymptotes 1 = - 4, x = 1 and 2. (a) f(x) -23x+1+4 7- 6- 5- 4- 3- 1- v (b) f(x)=3fx- 7- 1 y 5 4 3 2 8123 -6-5-4-3-2-1 | 1 2 3 4 5 67 13Using Calculus it can be shown that y x= + 1 is a slant asymptote of this graph.  312 312 FURTHER TOPICS IN FUNCTIONS (c) f~)~ x+2+3 y 5- 4 -34 -2 30X f (x) = '---x- - y i i i i i Jo AA -I -20 -10 -2. x 3.(a) x (b) x (C) x (d) [ - (e) x 5 + 57 8 -4 (f) (g) (h) (i) (i) x= -2,6 [2, oo) (-oc, -4) U (-4, -i9] U (2, x)) -+f8 1, 0] U [1,oo) -6 6. k-1(x) IX 9. 9.8 r--- 0.062 meters or 6.2 centimeters (~47) 10. (a) First rewrite the model as P= 1.23x 5y 5. Then 300 =1.23X 5y 5 yields y:_ If x= 100 then y 441.93687. 5 (300 3 I 21 1.23x 5/ 1.(a) [0, c) (b) m(.lc) = 99 1.5m m(.5c) =mr5 1.155mr m(.9c) =m9 2.294mr m(999c) = M22.366mr .0.001999 (c) As x --+ c-, m(x) --+ o0 (d) If the object is traveling no faster than approximately 0.99995 times the speed of light, then its observed mass will be no greater than lO0mr.  5.3 OTHER ALGEBRAIC FUNCTIONS 313 12. (a) y = x32 vZ+ 2. The point (0, 3) is when Fritzy's path crosses Chewbacca's path - in other words, where Fritzy catches Chewbacca. (b) y = x3 + I - 2. Using the techniques from Chapter 4, we find as x - 0+, y oc which means, in this case, Fritzy's pursuit never ends; he never catches Chewbacca. This makes sense since Chewbacca has a head start and is running faster than Fritzy. y rx/ 13 1 _2 -fi +2- 3  314 FURTHER TOPICS IN FUNCTIONS  CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS 6.1 INTRODUCTION TO EXPONENTIAL AND LOGARITHMIC FUNCTIONS Of all of the functions we study in this text, exponential and logarithmic functions are possibly the ones which impact everyday life the most.1 This section will introduce us to these functions while the rest of the chapter will more thoroughly explore their properties. Up to this point, we have dealt with functions which involve terms like 2 or 2/3, in other words, terms of the form zp where the base of the term, x, varies but the exponent of each term, p, remains constant. In this chapter, we study functions of the form f(x) = bX where the base b is a constant and the exponent x is the variable. We start our exploration of these functions with f(x) = 2x. (Apparently this is a tradition. Every College Algebra book we have ever read starts with f(x) = 2x.) We make a table of values, plot the points and connect them in a pleasing fashion. x f ) (x, f )) -3 2-3 = j (-3, s) -2 2-2= j (-2, ) -1 2-1 (-1,) 0 20=1 (0,1) 1 21 = 2 (1, 2) 2 22 = 4 (2, 4) 3 23 = 8 (2, 8) y x y =f(x)=2x A few remarks about the graph of f(x) = 2x which we have constructed are in order. As x - -oc 'Take a class in Differential Equations and you'll see why.  316 EXPONENTIAL AND LOGARITHMIC FUNCTIONS and attains values like x= -100 or x= -1000, the function f(x) = 2X takes on values like f (-100) = 2-100 =100 or f(-1000) = 2-1000 =210. In other words, as x --oo, 1 2x 1 ~very small (+) very big (+) So as x - -oo, 2X - 0+. This is represented graphically using the x-axis (the line y = 0) as a horizontal asymptote. On the flip side, as x - oc, we find f(100) = 2100, f(1000) = 21000, and so on, thus 2X - oc. As a result, our graph suggests the range of f is (0, oo). The graph of f passes the Horizontal Line Test which means f is one-to-one and hence invertible. We also note that when we 'connected the dots in a pleasing fashion', we have made the implicit assumption that f(x) = 2x is continuous2 and has a domain of all real numbers. In particular, we have suggested that things like 25 exist as real numbers. We should take a moment to discuss what something like 2 might mean, and refer the interested reader to a solid course in Calculus for a more rigorous explanation. The number = 1.73205... is an irrational number3 and as such, its decimal representation neither repeats nor terminates. We can, however, approximate 3 by terminating decimals, and it stands to reason4 we can use these to approximate 24. For example, if we approximate v3 by 1.73, we can approximate 2 ~ 21.73 - 2ioo 100 2173. It is not, by any means, a pleasant number, but it is at least a number that we understand in terms of powers and roots. It also stands to reason that better and better approximations of 3 yield better and better approximations of 2v, so the value of 2v should be the result of this sequence of approximations.5 Suppose we wish to study the family of functions f (x) =bX. Which bases b make sense to study? We find that we run into difficulty if b < 0. For example, if b = -2, then the function f (x) =(-2)x has trouble, for instance, at x= 2 since (-2)1/2 = -2 is not a real number. In general, if x is any rational number with an even denominator, then (-2)x is not defined, so we must restrict our attention to bases b > 0. What about b = 0? The function f(x) = 0 is undefined for x < 0 because we cannot divide by 0 and 00 is an indeterminant form. For x > 0, 0 = 0 so the function f (x) = 0x is the same as the function f(x) = 0, x > 0. We know everything we can possibly know about this function, so we exclude it from our investigations. The only other base we exclude is b = 1, since the function f (x) = 1 = 1 is, once again, a function we have already studied. We are now ready for our definition of exponential functions. DEFINITION 6.1. A function of the form f(x) = bX where b is a fixed real number, b > 0, b # 1 is called a base b exponential function. We leave it to the reader to verify6 that if b > 1, then the exponential function f(x) = bX will share the same basic shape and characteristics as f(x) =2x. What if 0 < b < 1? Consider g(x) - j" We could certainly build a table of values and connect the points, or we could take a step back and note that g(x) = (1)x = (2-1)x = 2- = f(-x), where f(x) = 2x. Thinking back to Section 1.8, 2Recall that this means there are no holes or other kinds of breaks in the graph. 3You can actually prove this by considering the polynomial p(x) = x2 - 3 and showing it has no rational zeros by applying Theorem 3.9. 4This is where Calculus and continuity come into play. 5Want more information? Look up "convergent sequences" on the Internet. 6Meaning, graph some more examples on your own.  6.1 INTRODUCTION TO EXPONENTIAL AND LOGARITHMIC FUNCTIONS 317 the graph of f(-x) is obtained from the graph of f(x) by reflecting it across the y-axis. As such, we have y 8- 7- 6- 5- 4- 3- 2- y = f( y 8-- 6-- 5-- 4-- 3-- 2-- reflect across y-axis -3-2-17 123 x multiply each x-coordinate by -1 y = g(x) = 2-- = (})X :=2 x)=2 We see that the domain and range of g match that of f, namely (-oc, o) and (0, oo), respectively. Like f, g is also one-to-one. Whereas f is always increasing, g is always decreasing. As a result, as x - -oc, g(x) - oc, and on the flip side, as x - oc, g(x) - 0+. It shouldn't be too surprising that for all choices of the base 0 < b < 1, the graph of y = bX behaves similarly to the graph of g. We summarize these observations, and more, in the following theorem whose proof ultimately requires Calculus. THEOREM 6.1. Properties of Exponential Functions: Suppose f(x) =bx. * The domain of f is (-oc, oc) and the range of f is (0, oc). * (0, 1) is on the graph of f and y = 0 is a horizontal asymptote to the graph of f. * f is one-to-one, continuous and smootha * If b > 1: * If0 < b< 1: f is always increasing As x - -oc, f (X) - 0+ As x - oc, f (x) - oo The graph of f resembles: f is always decreasing As x - -oo, f (x) -+ oo As x - oo, f () - 0+ The graph of f resembles: y =b, b> 1 y =b, 0 < b < 1 "Recall that this means the graph of f has no sharp turns or corners.  318 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Of all of the bases for exponential functions, two occur the most often in scientific circles. The first, base 10, is often called the common base. The second base is an irrational number, e 2.718, called the natural base. We will more formally discuss the origins of this number in Section 6.5. For now, it is enough to know that since e > 1, f(x) = ex is an increasing exponential function. The following examples offer a glimpse as to the kind of real-world phenomena these functions can model. EXAMPLE 6.1.1. The value of a car can be modeled by V(t) = 25 (j)X, where x> 0 is age of the car in years and V(x) is the value in thousands of dollars. 1. Find and interpret V(0). 2. Sketch the graph of y = V(x) using transformations. 3. Find and interpret the horizontal asymptote of the graph you found in 2. SOLUTION. 1. To find V(0), we replace x with 0 to obtain V(0) = 25 ()= 25. Since x represents the age of the car in years, x = 0 corresponds to the car being brand new. Since V(x) is measured in thousands of dollars, V(0) = 25 corresponds to a value of $25, 000. Putting it all together, we interpret V(0) = 25 to mean the purchase price of the car was $25, 000. 2. To graph y = 25 (j)X, we start with the basic exponential function f(x) =_(j)X. Since the base b = 5 is between 0 and 1, the graph of y = f(x) is decreasing. We plot the y-intercept (0, 1) and two other points, (-1, j) and (1, 5), and label the horizontal asymptote y 0. To obtain V(x) 25 (j)X, x > 0, we multiply the output from f by 25, in other words, V(x) = 25f(x). In accordance with Theorem 1.5, this results in a vertical stretch by a factor of 25. We multiply all of the y values in the graph by 25 (including the y value of the horizontal asymptote) and obtain the points (-1, 145), (0, 25) and (1, 20). The horizontal asymptote remains y = 0. Finally, we restrict the domain to [0, o) to fit with the applied domain given to us. We have the result below. y y 2- 30- (0,25). (0, 1) 20 15 10 -3-2-1 1 2 3 5- H.A. y = 0 vertical scale by a factor of 25 HA. y 0 y = f(x) =) multiply each y-coordinate by 25 y = V(x) = 25f(x), x >0 3. We see from the graph of V that its horizontal asymptote is y =0. (We leave it to reader to verify this analytically by thinking about what happens as we take larger and larger powers of j.) This means as the car gets older, its value diminishes to 0.  6.1 INTRODUCTION TO EXPONENTIAL AND LOGARITHMIC FUNCTIONS 319 The function in the previous example is often called a 'decay curve'. Increasing exponential func- tions are used to model 'growth curves' and we shall see several different examples of those in Section 6.5. For now, we present another common decay curve which will serve as the basis for further study of exponential functions. Although it may look more complicated than the previ- ous example, it is actually just a basic exponential function which has been modified by a few transformations from Section 1.8. EXAMPLE 6.1.2. According to Netwon's Law of Cooling7 the temperature of coffee T (in degrees Fahrenheit) t minutes after it is served can be modeled by T(t) = 70 + 90e-01l. 1. Find and interpret T(0). 2. Sketch the graph of y = T(t) using transformations. 3. Find and interpret the horizontal asymptote of the graph. SOLUTION. 1. To find T(0), we replace every occurrence of the independent variable t with 0 to obtain T(0) = 70 + 90e-01'0) = 160. This means that the coffee was served at 160°F. 2. To graph y = T(t) using transformations, we start with the basic function, f(t) = et. As we have already remarked, e 2.718 > 1 so the graph of f is an increasing exponential with y-intercept (0, 1) and horizontal asymptote y = 0. The points (-1, e-1) (-1, 0.37) and (1, e) (1, 2.72) are also on the graph. Since the formula T(t) looks rather complicated, we rewrite T(t) in the form presented in Theorem 1.7 and use that result to track the changes to our three points and the horizontal asymptote. We have T(t) = 90e-01 +70 = 90f(-0.1t) + 70. Multiplication of the input to f, t, by -0.1 results in a horizontal expansion by a factor of 10 as well as a reflection about the y-axis. We divide each of the x values of our points by -0.1 (which amounts to multiplying them by -10) to obtain (10, e-1), (0, 1), and (-10, e). Since none of these changes affected the y values, the horizontal asymptote remains y = 0. Next, we see that the output from f is being multiplied by 90. This results in a vertical stretch by a factor of 90. We multiply the y-coordinates by 90 to obtain (10, 90e-1), (0, 90), and (-10, 90e). We also multiply the y value of the horizontal asymptote y = 0 by 90, and it remains y = 0. Finally, we add 70 to all of the y-coordinates, which shifts the graph upwards to obtain (10, 90e-1 + 70) (10,103.11), (0,160), and (-10, 90e + 70) (-10, 314.64). Adding 70 to the horizontal asymptote shifts it upwards as well to y = 70. We connect these three points using the same shape in the same direction as in the graph of f and, last but not least, we restrict the domain to match the applied domain [0, oo). The result is below. 7We will discuss this in greater detail in Section 6.5.  320 EXPONENTIAL AND LOGARITHMIC FUNCTIONS y 180 160-- 6-- 140- 6- 5 120-- 4 100- 4- -- - ---- 3 80-- 2 60 H.A. y = 70 (0,1 4 20 -3-2-1 1 2 3- H.A. y 0 2 4 6 8 10 1214 16 18 20 y=f(t)=et y=T(t) 3. From the graph, we see that the horizontal asymptote is y = 70. It is worth a moment or two of our time to see how this happens analytically and to review some of the 'number sense' developed in Chapter 4. As t - oo, We get T(t) = 70 + 90e-01lt 70 + 90every big (-. Since e > 1, every big (-) = 1eve 1i(+> 1verybi is(+> very small (+). The larger t becomes, the smaller e-0.lt becomes, so the term 90e-0.lt very small (+). Hence, T(t) 70 + very small (+) which means the graph is approaching the horizontal line y = 70 from above. This means that as time goes by, the temperature of the coffee is cooling to 70°F, presumably room temperature. D As we have already remarked, the graphs of f(x) = bX all pass the Horizontal Line Test. Thus the exponential functions are invertible. We now turn our attention to these inverses, the logarithmic functions, which are called 'logs' for short. DEFINITION 6.2. The inverse of the exponential function f (x) = bX is called the base b logarithm function, and is denoted f1(x) = logb(x) The expression logb(x) is read 'log base b of x.' We have special notations for the common base, b = 10, and the natural base, b =e. DEFINITION 6.3. The common logarithm of a real number x is logio(x) and is usually written log(x). The natural logarithm of a real number x is loge(x) and is usually written ln(x). Since logs are defined as the inverses of exponential functions, we can use Theorems 5.2 and 5.3 to tell us about logarithmic functions. For example, we know that the domain of a log function is the range of an exponential function, namely (0, oo), and that the range of a log function is the domain of an exponential function, namely (-oo, oo). Since we know the basic shapes of y =f(x) =bX for the different cases of b, we can obtain the graph of y =f-1(x) =logb~z) by reflecting the graph of f across the line y - x as shown below. The y-intercept (0, 1) on the graph of f corresponds to an x-intercept of (1, 0) on the graph of f-1. The horizontal asymptotes y =0 on the graphs of the exponential functions become vertical asymptotes x 0 on the log graphs.  6.1 INTRODUCTION TO EXPONENTIAL AND LOGARITHMIC FUNCTIONS 321 / / / / / / / / / / / / / / y= b7,b> 1 y = logo(x), b> 1 y= b,0 0 * If b > 1: * If0 < b< 1: f is always increasing As x - 0+, f (X) - -oc As x - oc, f(x) - oo The graph of f resembles: f is always decreasing As x - 0+, f (x) - oo As x - oc, f (x) - -oo The graph of f resembles: y = logo (x), b > 1 y =logb(x), 0 < b < 1  322 EXPONENTIAL AND LOGARITHMIC FUNCTIONS As we have mentioned, Theorem 6.2 is a consequence of Theorems 5.2 and 5.3. However, it is worth the reader's time to understand Theorem 6.2 from an exponential perspective. For instance, we know that the domain of g(x) = log2(x) is (0, oo). Why? Because the range of f(x) = 2x is (0, oc). In a way, this says everything, but at the same time, it doesn't. For example, if we try to find log2(-1), we are trying to find the exponent we put on 2 to give us -1. In other words, we are looking for x that satisfies 2x = -1. There is no such real number, since all powers of 2 are positive. While what we have said is exactly the same thing as saying 'the domain of g(x) = log2(x) is (0, oo) because the range of f (x) = 2 is (0, oc)', we feel it is in a student's best interest to understand the statements in Theorem 6.2 at this level instead of just merely memorizing the facts. EXAMPLE 6.1.3. Simplify the following. 1. log3(81) 4. In (3 e2) (1) 5. log(0.001) 2. log2 6. 21og2(8) 3. log,/5(25) 7. 117-1og117(6) SOLUTION. 1. The number log3(81) is the exponent we put on 3 to get 81. As such, we want to write 81 as a power of 3. We find 81 = 34, so that log3(81) = 4. 2. To find log2 (h), we need rewrite as a power of 2. We find = = 2-3, so log2 = -3. 3. To determine logyg(25), we need to express 25 as a power of 5. We know 25 = 52, and 5 = (d)2,so we have 25 = (()- =) (5)4. We get logg(25) = 4. 4. First, recall that the notation ln (3 e2) means loge (3 e2), so we are looking for the exponent to put on e to obtain Vce2. Rewriting V-= e2/3, we find ln (3'ce2) = ln (e2/3)-2 5. Rewriting log(0.001) as logio(0.001), we see that we need to write 0.001 as a power of 10. We have 0.001 =lni=10-3. Hence, log(0.001) =log (10-3) = -3. 6. We can use Theorem 6.2 directly to simplify 21og2(8) = 8. We can also understand this problem by first finding log2(8). By definition, log2(8) is the exponent we put on 2 to get 8. Since 8 = 23, we have log2(8) = 3. We now substitute to find 21og2(8) = 23 = 8. 7. We note that we cannot apply Theorem 6.2 directly to 117-1g117(6). (Why not?) We use 1 a property of exponents to rewrite 117- 1og117(6) as iio1 ).At this point, we can apply  6.1 INTRODUCTION TO EXPONENTIAL AND LOGARITHMIC FUNCTIONS 323 Theorem 6.2 to get 1171og117(6) = 6 and thus 117-1og117(6) - 1 -1 It is worth 1171og117(6) 6 a moment of your time to think your way through why 1171og117(6) = 6. By definition, 1og117(6) is the exponent we put on 117 to get 6. What are we doing with this exponent? We are putting it on 117. By definition we get 6. In other words, the exponential function f (x) = 117x undoes the logarithmic function g(x) = 1og117(X).D Up until this point, restrictions on the domains of functions came from avoiding division by zero and keeping negative numbers from beneath even radicals. With the introduction of logs, we now have another restriction. Since the domain of f(x) = logb(x) is (0, oc), the argument8 of the log must be strictly positive. EXAMPLE 6.1.4. Find the domain of the following functions. Check your answers graphically using the calculator. 1. f (x) = 2 log(3 - x) - 1 2. g(x) =In(x SOLUTION. 1. We set 3-x > 0 to obtain x < 3, or (-oc, 3). The graph from the calculator below verifies this. Note that we could have graphed f using transformations. Taking a cue from Theorem 1.7, we rewrite f (x) = 21og10(-x+3) -1 and find the main function involved is y = h(x) = logio(x). We select three points to track, ( 1, -1), (1, 0) and (10, 1), along with the vertical asymptote x= 0. Since f(x) = 2h(-x + 3) - 1, Theorem 1. (tells us that to obtain the destinations of these points, we first subtract 3 from the x-coordinates (shifting the graph left 3 units), then divide (multiply) by the x-coordinates by -1 (causing a reflection across the y-axis). These transformations apply to the vertical asymptote x= 0 as well. Subtracting 3 gives us x= -3 as our asymptote, then multplying by -1 gives us the vertical asymptote x= 3. Next, we multiply the y-coordinates by 2 which results in a vertical stretch by a factor of 2, then we finish by subtracting 1 from the y-coordinates which shifts the graph down 1 unit. We leave it to the reader to perform the indicated arithmetic on the points themselves and to verify the graph produced by the calculator below. 2. To find the domain of g, we set x-1 > 0 and use a sign diagram to solve this inequality. We define r(z) =x1 find its domain to be r is (-oc, 1) U (1, oo). Setting r(z) = 0 gives x= 0. (+) 0 (-) ? (+) C I I ) 0 1 8See page 41 if you've forgotten what this term means.  324 EXPONENTIAL AND LOGARITHMIC FUNCTIONS We find X1 > 0 on (-oc, 0) U (1, oc) to get the domain of g. The graph of y = g(x) confirms this. We can tell from the graph of g that it is not the result of Section 1.8 transformations being applied to the graph y = ln(x), so barring a more detailed analysis using Calculus, the calculator graph is the best we can do. One thing worthy of note, however, is the end behavior of g. The graph suggests that as x - toc, g(x) -* 0. We can verify this analytically. Using results from Chapter 4 and continuity, we know that as x -bo, 1 ~ 1. Hence, it makes sense that g(x) = ln (i ) ln(1) =0. y = f(x) = 2log(3 - x) - 1 y = g(x) = In ( ) z- 1 While logarithms have some interesting applications of their own which you'll explore in the exer- cises, their primary use to us will be to undo exponential functions. (This is, after all, how they were defined.) Our last example solidifies this and reviews all of the material in the section. EXAMPLE 6.1.5. Let f(x) = 2x-1 - 3. 1. Graph f using transformations and state the domain and range of f. 2. Explain why f is invertible and find a formula for f-1(x). 3. Graph f- using transformations and state the domain and range of f-1. 4. Verify (f-1 o f) (x) = x for all x in the domain of f and (f o f-1) (x) = x for all x in the domain of f-1. 5. Graph f and f- on the same set of axes and check the symmetry about the line y = x. SOLUTION. 1. If we identify g(x) = 2x, we see f(x) = g(x - 1) - 3. We pick the points (-1, ), (0, 1) and (1, 2) on the graph of g along with the horizontal asymptote y = 0 to track through the transformations. By Theorem 1.7 we first add 1 to the i-coordinates of the points on the graph of g (shifting g to the right 1 unit) to get (0, j), (1, 1) and (2, 2). The horizontal asymptote remains y = 0. Next, we subtract 3 from the y-coordinates, shifting the graph  6.1 INTRODUCTION TO EXPONENTIAL AND LOGARITHMIC FUNCTIONS 325 down 3 units. We get the points (0, -), (1, -2) and (2, -1) with the horizontal asymptote now at y = -3. Connecting the dots in the order and manner as they were on the graph of g, we get the graph below. We see that the domain of f is the same as g, namely (-oc, oc), but that the range of f is (-3, oc). y y 8. 8- 7--7- 6.. 6-- 5.. 5-- 4- 4 3. 3- 2. 2- 11 3 4 -3 2- 1 2 3 4 X -3-2- 1 1 3 4 X -2 -2 - 3 .- -- - - - - - - y =h(x) = 2x ) y =f(x) =2x-1 -_3 2. The graph of f passes the Horizontal Line Test so f is one-to-one, hence invertible. To find a formula for f-1(x), we normally set y = f(x), interchange the x and y, then proceed to solve for y. Doing so in this situation leads us to the equation x= 2Y-1 - 3. We have yet to discuss how to solve this kind of equation, so we will attempt to find the formula for f-- from a procedural perspective. If we break f (x) = 2x-1 - 3 into a series of steps, we find f takes an input x and applies the steps (a) subtract 1 (b) put as an exponent on 2 (c) subtract 3 Clearly, to undo subtracting 1, we will add 1, and similarly we undo subtracting 3 by adding 3. How do we undo the second step? The answer is we use the logarithm. By definition, log2(x) undoes exponentiation by 2. Hence, f-1 should (a) add 3 (b) take the logarithm base 2 (c) add 1 In symbols, f -1(x) =log2(z + 3) + 1. 3. To graph f -1(x) =log2(x + 3) + 1 using transformations, we start with j~x) =log2(x). We track the points (j, -1), (1, 0) and (2, 1) on the graph of j along with the vertical asymptote x 0 through the transformations using Theorem 1.7. Since f -1(x) =j(x + 3) + 1, we first subtract 3 from each of the x values (including the vertical asymptote) to obtain (-), -1),  326 EXPONENTIAL AND LOGARITHMIC FUNCTIONS (-2, 0) and (-1, 1) with a vertical asymptote x= -3. Next, we add 1 to the y values on the graph and get (- 2, 0), (-2,1) and (-1, 2). If you are experiencing djd vu, there is a good reason for it but we leave it to the reader to determine the source of this uncanny familiarity. We obtain the graph below. The domain of f-1 is (-3, oc), which matches the range of f, and the range of f-1 is (-oc, oo), which matches the domain of f. 4t 3- 2- 1- I 4 3 1 2-1 1 2 3 4 5 6 7 8 -2.. -3.. I-3-2-1 I I I I I -1- -2. -3- 1 2 3 4 5 6 y=j(x)= log2(x) = f (x) = log2(x + 3) + 1 4. We now verify that f(x) = 2x-1 - 3 and f1(x) = log2(z + 3) + 1 satisfy the composition requirement for inverses. For all real numbers x, (f o f)(x) f-I (f ()) f -1 (2- - 3) log2 ([2x-1 - 3] + 3) + 1 log2 (2x-1) + 1 (x -1) + 1 Since log2 (2") = u for all real numbers u Fxo For all real numbers x > -3, we have (f o f -1) f (f -I(X)) f(log2(x+3)+1) 21og2(x+3) - 3 (x + 3) - 3 37/ Since 21og2(u) = u for all real numbers a > 0 f (x) and y = f-1(x) on the same set of axes and 5. Last, but certainly not least, we graph y see the symmetry about the line y = x. 9Pay attention - can you spot in which step below we need x> -3?  6.1 INTRODUCTION TO EXPONENTIAL AND LOGARITHMIC FUNCTIONS 327 y / 8/ 7- 6- 1 // -~2-1/ 1 3 4 5 6 7 8X /2- Y= f(x) 2'-l 3 Y = f-1 (x) = log2(X+±3) ±+1  328 EXPONENTIAL AND LOGARITHMIC FUNCTIONS 6.1.1 EXERCISES 1. Evaluate the expression. (a) (b) (c) (d) (e) (f) log3 (27) log2(32) log8(4) logi (625) 5 log (0000) log4(8) (g) (h) (j) (k) (1) log6(1) log1s (\13) 71og73 ln (e5) log (91011) log5 (31og3 5) 2. Find the domain of the function. (a) f(x) = ln(2 + 1) (b) f(x) =log7(4x + 8) (c) f (z)=logQ72j ( ) f ( ) = l g (2(d ) f ( ) = ln ( 7 - ) + ln ( z (e) f(x) = log4x (f) f (x) = logg(|x + 3| - 4) (g) f (z) = ln( x -4 - 3) (h) f (x) 3 - logs x (*~\~ --1-z 4) (i) f (x) = logi x 2 (j) f (x)= ln(-2x3 - 92 + 13x - 6) 3. For each function given below " Find its inverse from the 'procedural perspective' discussed in Example 6.1.5. " Graph the function and its inverse on the same set of axes. (a) f (x) = 3x+2 - 4 (b) f (x) = log4(z - 1) (c) f (x) = -2-x + 1 (d) f (x) = 5 log(x) - 2 4. (The Logarithmic Scales) There are three widely used measurement scales which involve common logarithms: the Richter scale, the decibel scale and the pH scale. The computations involved in all three scales are nearly identical so pay close attention to the subtle differences. (a) Earthquakes are complicated events and it is not our intent to provide a complete dis- cussion of the science involved in them. Instead, we refer the interested reader to a solid course in Geology10 or the U.S. Geological Survey's Earthquake Hazards Program found here and present only a simplified version of the Richter scale. The Richter scale measures the magnitude of an earthquake by comparing the amplitude of the seismic ioRock-solid, perhaps?  6.1 INTRODUCTION TO EXPONENTIAL AND LOGARITHMIC FUNCTIONS 329 waves of the given earthquake to those of a "magnitude 0 event", which was chosen to be a seismograph reading of 0.001 millimeters recorded on a seismometer 100 kilometers from the earthquake's epicenter. Specifically, the magnitude of an earthquake is given by M(x) = log where x is the seismograph reading in millimeters of the earthquake 0.001 recorded 100 kilometers from the epicenter. i. Show that M(0.001) = 0. ii. Compute M(80, 000). iii. Show that an earthquake which registered 6.7 on the Richter scale had a seismograph reading ten times larger than one which measured 5.7. iv. Find two news stories about recent earthquakes which give their magnitudes on the Richter scale. How many times larger was the seismograph reading of the earthquake with larger magnitude? (b) While the decibel scale can be used in many disciplines," we shall restrict our attention to its use in acoustics, specifically its use in measuring the intensity level of sound.-2 The Sound Intensity Level L (measured in decibels) of a sound intensity I (measured in watts per square meter) is given by L(I) 10 log (10-12). Like the Richter scale, this scale compares I to baseline: 10-12 I2 is the threshold of human hearing. i. Compute L(10-6). ii. Damage to your hearing can start with short term exposure to sound levels around 115 decibels. What intensity I is needed to produce this level? iii. Compute L(1). How does this compare with the threshold of pain which is around 140 decibels? (c) The pH of a solution is a measure of its acidity or alkalinity. Specifically, pH - log[H+] where [H+] is the hydrogen ion concentration in moles per liter. A solution with a pH less than 7 is an acid, one with a pH greater than 7 is a base (alkaline) and a pH of 7 is regarded as neutral. i. The hydrogen ion concentration of pure water is [H+] = 10-7. Find its pH. ii. Find the pH of a solution with [H+] = 6.3 x 10-13. iii. The pH of gastric acid (the acid in your stomach) is about 0.7. What is the corre- sponding hydrogen ion concentration? 5. Show that logo1= 0 and log b = 1 for every b > 0, b # 1. 6. (Crazy bonus question) Without using your calculator, determine which is larger: er or re. "See this webpage for more information. "2As of the writing of this exercise, the Wikipedia page given here states that it may not meet the "general notability guideline" nor does it cite any references or sources. I find this odd because it is this very usage of the decibel scale which shows up in every College Algebra book I have read. Perhaps those other books have been wrong all along and we're just blindly following tradition.  330 330 EXPONENTIAL AND LOGARITHMIC FUNCTIONS 6.1.2 ANSWERS 1.(a) (b) (c) (d) (e) (f) 1093 27 = 3 log2 32=5 log8 4 =- logy 625= 5 lg 1 lg1000000 log4 8 =- (g) (h) (i) -6 (k) (1) log61 1= 0 log13 13 - 71og7 3 - 3 ln(e5) - 5 log (9 1011) logs (31°g3 5) 1 __11 9 2.()(-oo, oo) ()(- 2, -1) U(1, oo) (d 47) (f) (a) (h) (i) (i) (-oc, -7) U (1, oo) (13,oo) (0, 125) U (125, 00) No domain (-oc, -3) U (2, 2) 3.(a) f (x) = f-1&(X) 3x+2 - 4 - og3 (X + 4) -2 (C) f W)= -2-x + 1 _- 192 (1 I) y - -I ' - - - - -- y =f (x) =3x+2 -4 Y = f -(x) = log3(X+±4) -2 f (x) - log4(v - 1) f-1(x) 4X+ 1 / I / I Iy f (x) _ 2 x + 1 Y = f-1(x) = - log2(1 - x) (d) f (x) =-5log (x) - 2 f1(x) - 10 x5 (b) ky 6- 5- 4- 3- 2 -4- -2 - / / -2- yI I (I 1 2 34 56 X Tx / -2. y f (x) = 5log(x) - 2 y f-1(x) = 105 Y= f (x) =l104 (x - 1) y =f-1(x)=-4X+±1  6.1 INTRODUCTION TO EXPONENTIAL AND LOGARITHMIC FUNCTIONS 331 (0.001 ' 4. (a) i. M(0.001) =log ( 1= log(1) = 0. 0.001 / M(80000)log80 000) ii. M(80, 000) = log ' = log(80, 000, 000) 7.9. 0.001 (b) i. L(10-6) = 60 decibels. ii. I = 10-.5 ~ 0.316 watts per square meter. iii. Since L(1) = 120 decibels and L(100) = 140 decibels, a sound with intensity level 140 decibels has an intensity 100 times greater than a sound with intensity level 120 decibels. (c) i. The pH of pure water is 7. ii. If [H+] = 6.3 x 10-13 then the solution has a pH of 12.2. iii. [H+] = 10-0.7 ~ .1995 moles per liter.  332 EXPONENTIAL AND LOGARITHMIC FUNCTIONS 6.2 PROPERTIES OF LOGARITHMS In Section 6.1, we introduced the logarithmic functions as inverses of exponential functions and discussed a few of their functional properties from that perspective. In this section, we explore the algebraic properties of logarithms. Historically, these have played a huge rule in the scientific development of our society since, among other things, they were used to develop analog computing devices called slide rules which enabled scientists and engineers to perform accurate calculations leading to such things as space travel and the moon landing. As we shall see shortly, logs inherit analogs of all of the properties of exponents you learned in Elementary and Intermediate Algebra. We first extract two properties from Theorem 6.2 to remind us of the definition of a logarithm as the inverse of an exponential function. THEOREM 6.3. (Inverse Properties of Exponential and Log Functions) Let b > 0, b # 1. " bo*= c if and only if logb(c) = a " logo (bX) = x for all x and blgb (x) =x for all x > 0 Next, we spell out in more detail what it means for exponential and logarithmic functions to be one-to-one. THEOREM 6.4. (One-to-one Properties of Exponential and Log Functions) Let f(x) = b and g(x) = logb(x) where b > 0, b # 1. Then f and g are one-to-one. In other words: * b" = bW if and only if u = w for all real numbers u and w. " logb(u) = logo(w) if and only if u = w for all real numbers a > 0, w > 0. We now state the algebraic properties of exponential functions which will serve as a basis for the properties of logarithms. While these properties may look identical to the ones you learned in Elementary and Intermediate Algebra, they apply to real number exponents, not just rational exponents. Note that in the theorem that follows, we are interested in the properties of exponential functions, so the base b is restricted to b > 0, b # 1. An added benefit of this restriction is that it eliminates the pathologies discussed in Section 5.3 when, for example, we simplified (2/3)3/2 and obtained Ix instead of what we had expected from the arithmetic in the exponents, xi -X. THEOREM 6.5. (Algebraic Properties of Exponential Functions) Let f(x) = bX be an exponential function (b > 0, b # 1) and let u and w be real numbers. " Product Rule: f (a + w) = f(u)f(w). In other words, bu+w = bubW " Quotient Rule: f(u - w) f( . In other words, b"- = f(w ) bw * Power Rule: (f (a))W f (aw). In other words, (b")W b"W While the properties listed in Theorem 6.5 are certainly believable based on similar properties of integer and rational exponents, the full proofs require Calculus. To each of these properties of  6.2 PROPERTIES OF LOGARITHMS 333 exponential functions corresponds an analogous property of logarithmic functions. We list these below in our next theorem. THEOREM 6.6. (Algebraic Properties of Logarithm Functions) Let g(x) = logb(x) be a logarithmic function (b > 0, b - 1) and let u > 0 and w > 0 be real numbers. " Product Rule: g(uw) = g(u) + g(w). In other words, logb(uw) = logb(u) + logb(w) " Quotient Rule: g (-)= g(u) - g(w). In other words, log -()= logb(u) - logb(w) " Power Rule: g (u) = wg(u). In other words, logo (uW) = w logb(u) There are a couple of different ways to understand why Theorem 6.6 is true. Consider the product rule: logb(uw) = logb(u) + log(w). Let a = logb(uw), c = logb(u), and d =log(w). Then, by definition, ba = uw, bC = u and bd = w. Hence, b = uw = bcbd - bc+d so that bo = bc+d. By the one-to-one property of bx, we have a = c + d. In other words, logb(uw) = logb(u) + logb(w). The remaining properties are proved similarly. From a purely functional approach, we can see the properties in Theorem 6.6 as an example of how inverse functions interchange the roles of inputs in outputs. For instance, the Product Rule for exponential functions given in Theorem 6.5, f(a+ w) = f (u)f (w), says that adding inputs results in multiplying outputs. Hence, whatever f-- is, it must take the products of outputs from f and return them to the sum of their respective inputs. Since the outputs from f are the inputs to f-1 and vice-versa, we have that that f- must take products of its inputs to the sum of their respective outputs. This is precisely what the Product Rule for Logarithmic functions states in Theorem 6.6: g(uw) = g(u) +g(w). The reader is encouraged to view the remaining properties listed in Theorem 6.6 similarly. The following examples help build familiarity with these properties. In our first example, we are asked to 'expand' the logarithms. This means that we read the properties in Theorem 6.6 from left to right and rewrite products inside the log as sums outside the log, quotients inside the log as differences outside the log, and powers inside the log as factors outside the log. While it is the opposite process, which we will practice later, that is most useful in Algebra, the utility of expanding logarithms becomes apparent in Calculus. EXAMPLE 6.2.1. Expand the following using the properties of logarithms and simplify. Assume when necessary that all quantities represent positive real numbers. 1. log2 4. log3 100o2 2. log.1 (10x2) 3. In(+j) 5. log117 (2 -4) SOLUTION. 1. To expand log2 (i), we use the Quotient Rule identifying a 8 and w =x and simplify.  334 EXPONENTIAL AND LOGARITHMIC FUNCTIONS log2 (- =log2(8) - log2(z) Quotient Rule = 3- log2(x) Since 23 = 8 -log2(x)+3 2. In the expression logo.1 (10x2), we have a power (the 2) and a product. In order to use the Product Rule, the entire quantity inside the logarithm must be raised to the same exponent. Since the exponent 2 applies only to the x, we first apply the Product Rule with u = 10 and w =9x2. Once we get the 2 by itself inside the log, we may apply the Power Rule with u = x and w = 2 and simplify. logo.1 (10x2) = logo.1(10) + logo.1 (2) Product Rule = logo.(10) + 2logo.1(x) Power Rule -1 + 210go.1(x) Since (0.1)-1 = 10 2 logo.1(x) - 1 3. We have a power, quotient and product occurring in ln (A)2. Since the exponent 2 applies to the entire quantity inside the logarithm, we begin with the Power Rule with u = a and w = 2. Next, we see the Quotient Rule is applicable, with u = 3 and w = ex, so we replace ln (A) with the quantity ln(3) - ln(ex). Since ln (A) is being multiplied by 2, the entire quantity ln(3) - ln(ex) is multiplied by 2. Finally, we apply the Product Rule with u = e and w = x, and replace ln(ex) with the quantity ln(e) + ln(x), and simplify, keeping in mind that the natural log is log base e. InQ2 = 2lnQ ) Power Rule ex \ ex) = 2 [ln(3) - ln(ex)] Quotient Rule =2 ln(3) - 2 ln(ex) =2 ln(3) - 2 [ln(e) + ln(x)] Product Rule = 21n(3) - 21n(e) - 21n(x) = 21n(3) - 2 - 21n(x) Since el = e -21n(x) + 21n(3) - 2 4. In Theorem 6.6, there is no mention of how to deal with radicals. However, thinking back to Definition 5.5, we can rewrite the cube root as a j exponent. We begin by using the Power  6.2 PROPERTIES OF LOGARITHMS 335 Rule , and we keep in mind that the common log is log base 10. log31002 log 100x2 1/3 yz5 yz5 1 (010021 log 5Power Rule yz5 =3[log (100x2) - log (yz5)] Quotient Rule =log (100x2) - log (yz5) =3[log(100) + log (x2)1 - [log(y) + log (z5)] Product Rule = log(100) + 3 log (2) - i log(y) - I log (z5) = log(100) + 3 log(x) - } log(y) - 3 log(z) Power Rule = 2 1+log(x) - 1og(y) - j log(z) Since 102 = 100 = 3log(X) - j log(y) - 3 log(z) + 4 5. At first it seems as if we have no means of simplifying log117 (2 - 4), since none of the properties of logs addresses the issue of expanding a difference inside the logarithm. However, we may factor 2 - 4 = (x + 2)(x - 2) thereby introducing a product which gives us license to use the Product Rule. log117 (2 - 4) =log117 [(x + 2)(x - 2)] Factor =log117(X + 2) + 1og117(X - 2) Product Rule A couple of remarks about Example 6.2.1 are in order. First, while not explicitly stated in the above example, a general rule of thumb to determine which log property to apply first to a complicated problem is 'reverse order of operations.' For example, if we were to substitute a number for x into the expression logo.1 (10x2), we would first square the x, then multiply by 10. The last step is the multiplication, which tells us the first log property to apply is the Product Rule. In a multi-step problem, this rule can give the required guidance on which log property to apply at each step. The reader is encouraged to look through the solutions to Example 6.2.1 to see this rule in action. Second, while we were instructed to assume when necessary that all quantities represented positive real numbers, the authors would be committing a sin of omission if we failed to point out that, for instance, the functions f (x) =log117 (2 - 4) and g(x) = 1og117(x+2)+1og117(x -2) have different domains, and, hence, are different functions. We leave it to the reader to verify the domain of f is (-oo, -2) U (2, oo) whereas the domain of g is (2, oo). In general, when using log properties to 1At this point in the text, the reader is encouraged to carefully read through each step and think of which quantity is playing the role of a and which is playing the role of w as we apply each property.  336 EXPONENTIAL AND LOGARITHMIC FUNCTIONS expand a logarithm, we may very well be restricting the domain as we do so. One last comment before we move to reassembling logs from their various bits and pieces. The authors are well aware of the propensity for some students to become overexcited and invent their own properties of logs like log117 (2 - 4) =log117 (2) - 1og117(4), which simply isn't true, in general. The unwritten2 property of logarithms is that if it isn't written in a textbook, it probably isn't true. EXAMPLE 6.2.2. Use the properties of logarithms to write the following as a single logarithm. 1. log3(x - 1) - log3Q( + 1) 2. log(x) + 2 log(y) - log(z) 3. 4log2(z)+3 4. - ln(x) - 1 SOLUTION. Whereas in Example 6.2.1 we read the properties in Theorem 6.6 from left to right to expand logarithms, in this example we read them from right to left. 1. The difference of logarithms requires the Quotient Rule: log3(z-1)-log3(z+1) =log (). 2. In the expression, log(x) + 2log(y) - log(z), we have both a sum and difference of logarithms. However, before we use the product rule to combine log(x) + 2 log(y), we note that we need to somehow deal with the coefficient 2 on log(y). This can be handled using the Power Rule. We can then apply the Product and Quotient Rules as we move from left to right. Putting it all together, we have log(x) + 2 log(y) - log(z) log(x) + log (y2) - log(z) Power Rule log (xy2) - log(z) log (x 2 Product Rule Quotient Rule 3. We can certainly get started rewriting 4log2(x) + 3 by applying the Power Rule to 410g2(X) to obtain log2 (x4), but in order to use the Product Rule to handle the addition, we need to rewrite 3 as a logarithm base 2. From Theorem 6.3, we know 3 =log2 (23), so we get 41og2(x) + 3 log2 (x4) + 3 log2 (x4) + log2 (23) log2 (x4) + log2(8) log2 (8x4) Power Rule Since 3= log2 (23) Product Rule 2The authors relish the irony involved in writing what follows.  6.2 PROPERTIES OF LOGARITHMS 337 4. To get started with - ln(x) - 2, we rewrite - ln(x) as (-1) ln(x). We can then use the Power Rule to obtain (-1) ln(x) = in (x-). In order to use the Quotient Rule, we need to write 1 as a natural logarithm. Theorem 6.3 gives us i = n (e1/2) = in (e). We have - ln (x) - 2 = (-1) ln(x) - in (x-1) - 2 Power Rule in (x-1) - In (e1/2) Since 2 = In (e1/2) ln (-1) - In (e) X-1 In (Quotient Rule In1 As we would expect, the rule of thumb for re-assembling logarithms is the opposite of what it was for dismantling them. That is, if we are interested in rewriting an expression as a single logarithm, we apply log properties following the usual order of operations: deal with multiples of logs first with the Power Rule, then deal with addition and subtraction using the Product and Quotient Rules, respectively. Additionally, we find that using log properties in this fashion can increase the domain of the expression. For example, we leave it to the reader to verify the domain of f (x) = log3(Q-1)-log3(x+1) is (1, oc) but the domain of g(x) = log ()lis (-oc, -1)U(1, oc). We will need to keep this in mind when we solve equations involving logarithms in Section 6.4 - it is precisely for this reason we will have to check for extraneous solutions. The two logarithm buttons commonly found on calculators are the 'LOG' and 'LN' buttons which correspond to the common and natural logs, respectively. Suppose we wanted an approximation to log2(7). The answer should be a little less than 3, (Can you explain why?) but how do we coerce the calculator into telling us a more accurate answer? We need the following theorem. THEOREM 6.7. (Change of Base) Let a, b > 0, a, b - 1. " ax = blogb(a) for all real numbers x. " logo(x) =log ) for all real numbers x > 0. loge (a) The proofs of the Change of Base formulas are a result of the other properties studied in this section. If we start with bx loge (a) and use the Power Rule in the exponent to rewrite x logeI(a) as loge (ax) and then apply one of the Inverse Properties in Theorem 6.3, we get xlogb (a) = blog(ax)= ax,  338 EXPONENTIAL AND LOGARITHMIC FUNCTIONS as required. To verify the logarithmic form of the property, we also use the Power Rule and an Inverse Property. We note that log. (x) - logb(a) = logb (aloga (X)1 = logb (x) and we get the result by dividing through by logb(a). Of course, the authors can't help but point out the inverse relationship between these two change of base formulas. To change the base of an exponential expression, we multiply the input by the factor logb(a). To change the base of a logarithmic expression, we divide the output by the factor logb(a). While, in the grand scheme of things, both change of base formulas are really saying the same thing, the logarithmic form is the one usually encountered in Algebra while the exponential form isn't usually introduced until Calculus.3 What Theorem 6.7 really tells us is that all exponential and logarithmic functions are just scalings of one another. Not only does this explain why their graphs have similar shapes, but it also tells us that we could do all of mathematics with a single base - be it 10, e, 42, or 117. Your Calculus teacher will have more to say about this when the time comes. EXAMPLE 6.2.3. Use an appropriate change of base formula to convert the following expressions to ones with the indicated base. Verify your answers using a calculator, as appropriate. 1. 32 to base 10 2. 2x to base e 3. log4(5) to base e 4. ln(x) to base 10 SOLUTION. 1. We apply the Change of Base formula with a = 3 and b = 10 to obtain 32 the latter in the calculator produces an answer of 9 as required. 102iog(3) Typing 2. Here, a = 2 and b = e so we have 2x = ex in(2). f(x) = 2x and g(x) = exin(2). Their graphs that they are the same function. To verify this on our calculator, we can graph are indistinguishable which provides evidence 10^(21o9(3)) 9 y = f(x) = 2x and y = g(x) = ex in(2) 3The authors feel so strongly about showing students that every property of logarithms comes from and corresponds to a property of exponents that we have broken tradition with the vast majority of other authors in this field. This isn't the first time this happened, and it certainly won't be the last.  6.2 PROPERTIES OF LOGARITHMS 339 3. Applying the change of base with a = 4 and b = e leads us to write log4(5) 1n(5. Evaluating this in the calculator gives 1( ~ 1.16. How do we check this really is the value of log4(5)? By definition, log4(5) is the exponent we put on 4 to get 5. The calculator confirms this.4 4. We write ln(x) = log ( ) . We graph both f(x) both graphs appear to be identical. ln(x) and g(x) log() and find log(e) ln(5)/1n(4) 1. 160964047 4^Ans 5 y = f(x) = ln(x) and y g(x) = _____ 4Which means if it is lying to us about the first answer it gave us, at least it is being consistent.  340 EXPONENTIAL AND LOGARITHMIC FUNCTIONS 6.2.1 EXERCISES 1. Expand the following using the properties of logarithms and simplify. Assume when necessary that all quantities represent positive real numbers. (a) ln(x3y2) (e) ln(Z 128 zy (b) log2 z2 + 4 (f) logs (x2 - 25) (c) log5 (z)3 (g) log (4x3) (d) log(1.23 x 1037) (h) logi(9x(y3 - 8)) 2. Use the properties of logarithms to write the following as a single logarithm. (a) 41n(x) + 21n(y) (e) 2 log3(x) - 21og3(y) - log3(z) (b) 3 - log(x) (f) log2(x) + log.2 (x - 1) (c) log2() + log2(y) - log2(z) (g) log2() + log4(X - 1) (d) log3(x) - 2 log3(y) (h) log7(x) + log7(x - 3) - 2 3. Use an appropriate change of base formula to convert the following expressions to ones with the indicated base. (a) 7-1 to base e (c) 2xto base e (b) log3(x + 2) to base 10( \3 ) b (d) log(z2 + 1) to base e 4. Use the appropriate change of base formula to approximate the following logarithms. (a) log3(12) (c) log4 (1 (b) log5(80) '10/ (d) log3g(1000) 5. Compare and contrast the graphs of y = ln(z2) and y = 2ln(x). 6. Prove the Quotient Rule and Power Rule for Logarithms. 7. Give numerical examples to show that, in general, (a) logb(x + y) # logb(x) + logb(y) (b) loge(x - y) # loge(x) - loge(y)  6.2 PROPERTIES OF LOGARITHMS 341 8. The Henderson-Hasselbalch Equation: Suppose HA represents a weak acid. Then we have a reversible chemical reaction HA- H++ A-. The acid disassociation constant, Ka, is given by [H+][A -] [A-] [HA] [HA]' where the square brackets denote the concentrations just as they did in Exercise 4c in Section 6.1. The symbol pKa is defined similarly to pH in that pKa - log(Ka). Using the definition of pH from Exercise 4c and the properties of logarithms, derive the Henderson-Hasselbalch Equation which states [A-] pH = pKa + log [ HA] 9. Research the history of logarithms including the origin of the word 'logarithm' itself. Why is the abbreviation of natural log 'ln' and not 'nl'? 10. There is a scene in the movie 'Apollo 13' in which several people at Mission Control use slide rules to verify a computation. Was that scene accurate? Look for other pop culture references to logarithms and slide rules.  342 342 EXPONENTIAL AND LOGARITHMIC FUNCTIONS 6.2.2 ANSWERS 1.(a) (b) (c) (d) 3ln(x) + 2ln(y) 7 -12 (og2+ 4) 3 logs (z) - 6 log(1.23) + 37 (e) (f) (g) l n(z) - ln(x) - ln(y) 10g5 (X - 5) + 1og5Q(X + 5) 3logy/-Qv() + 8 -2 + logi (x) + logi (y - 2) + log i (y2 + 2y + 4) 3 3 3 2. (a) ln(x4y2) (b) log (1000) (c) loge XY (d) log3 (2X (e) log3 (2) (f) iogx ) (g) log2 (x x - i) (h) log7 ( 493) (c) ( )-ex1n() (d) log(x2 + 1) - ln~x2 + 1) ln(1O) 3. (a) (b) 4. (a) (b) 7X-1 - e(x-1)lIn(7) log (x+2 10g3(x++) _ 2)__ log3x + ) =log(3) log3(12) 2.26186 log5(80) 2.72271 5 1.66096 13.52273  6.3 EXPONENTIAL EQUATIONS AND INEQUALITIES 343 6.3 EXPONENTIAL EQUATIONS AND INEQUALITIES In this section we will develop techniques for solving equations involving exponential functions. Suppose, for instance, we wanted to solve the equation 2' = 128. After a moment's calculation, we find 128 = 27, so we have 2 = 27. The one-to-one property of exponential functions, detailed in Theorem 6.4, tells us that 2 = 27 if and only if x = 7. This means that not only is x = 7 a solution to 2X = 27, it is the only solution. Now suppose we change the problem ever so slightly to 2X = 129. We could use one of the inverse properties of exponentials and logarithms listed in Theorem 6.3 to write 129 = 21og2(129). We'd then have 2X = 21og2(129), which means our solution is c=v log2(129). This makes sense because, after all, the definition of log2(129) is 'the exponent we put on 2 to get 129.' Indeed we could have obtained this solution directly by rewriting the equation 2X = 129 in its logarithmic form log2(129) = x. Either way, in order to get a reasonable decimal approximation to this number, we'd use the change of base formula, Theorem 6.7, to give us something more calculator friendly,1 say log2(129) .n(129) Another way to arrive at this answer is as follows ln(2) 2X = 129 ln (2X) =ln(129) Take the natural log of both sides. x ln(2) = ln(129) Power Rule ln(129) X = ln(2) 'Taking the natural log' of both sides is akin to squaring both sides: since f (x) = ln(x) is a function, as long as two quantities are equal, their natural logs are equal.2 Also note that we treat ln(2) as any other non-zero real number and divide it through3 to isolate the variable x. We summarize below the two common ways to solve exponential equations, motivated by our examples. Steps for Solving an Equation involving Exponential Functions 1. Isolate the exponential function. 2. (a) If convenient, express both sides with a common base and equate the exponents. (b) Otherwise, take the natural log of both sides of the equation and use the Power Rule. EXAMPLE 6.3.1. Solve the following equations. Check your answer graphically using a calculator. 1. 23X = 161--x 3. 9 - 3X = 72X 5. 25X = 5X + 6 2. 2000 = 1000 -3-0-1 4. 75- =1+00_2 6. ex2x = 5 SOLUTION. 'You can use natural logs or common logs. We choose natural logs. (In Calculus, you'll learn these are the most 'mathy' of the logarithms.) 2This is also the 'if' part of the statement logb(u) =logb(w) if and only if a = w in Theorem 6.4. 3Please resist the temptation to divide both sides by 'ln' instead of ln(2). Just like it wouldn't make sense to divide both sides by the square root symbol 'I/' when solving zx/2 =5, it makes no sense to divide by 'ln'.  344 EXPONENTIAL AND LOGARITHMIC FUNCTIONS 1. Since 16 is a power of 2, we can exponents, we get 23x - 24(1-x) get 3x = 4(1-x) which gives x = and see that they intersect at x z rewrite 23x = 161-x as 23x - Using the one-to-one property 4. To check graphically, we set = 7 - 0.5714. (24) 1. Using properties of of exponential functions, we f(x) = 23x and g(x) = 161- 2. We begin solving 2000 which yields 3-0.1t 2 log to get ln (3-0.1t) divide both sides by f(x) = 2000 and g(x): = 1000 3-0-"t by dividing both sides by 2000 to isolate the exponential . Since it is inconvenient to write 2 as a power of 3, we use the natural = ln(2). Using the Power Rule, we get -0.1t ln(3) = ln(2), so we -0.11n(3) to get t = -l 1n2(3) _ -10 n(2). On the calculator, we graph = 1000 . 3-01x and find that they intersect at x = - ) -6.3093. n Inter Sction H=.57142D57 y=3.2DI3414I.I y = f(x) =23x and y = g(x) = 161- Intersection H:=5.30f96 .Y=2000... y = f (x) = 2000 and y = g(x) = 1000 . 3-O.1x 3. We first note that we can rewrite the equation 9-3x = 72x as 32.3x = 72x to obtain 3x+2 72x. Since it is not convenient to express both sides as a power of 3 (or 7 for that matter) we use the natural log: In (3x+2) = ln (72x). The power rule gives (x + 2) ln(3) = 2x ln(7). Even though this equation appears very complicated, keep in mind that ln(3) and ln(7) are just constants. The equation (x + 2) ln(3) = 2x ln(7) is actually a linear equation and as such we gather all of the terms with x on one side, and the constants on the other. We then divide both sides by the coefficient of x, which we obtain by factoring. (x + 2) ln(3) x ln(3) + 2 1n(3) 2 In(3) 2In(3) = 2x ln(7) = 2x ln(7) = 2x ln(7) - x ln(3) x(21n(7) - ln(3)) __ 2 in(3) 2ln(7)-ln(3) Factor. Graphing f(x) = 9-3x and g(x) = 72x on the calculator, we see that these two graphs intersect at x 2in( i(3) -i 0.7866. 21n(7)is to 4. Our objective in solving 75 = 13e_2t is to clear denominators and get 75 (1 + 3c-2t) which leads to 225e-2t = 25, and finally, e first isolate the exponential. To that end, we 100. From this we get 75 + 225e-2t = 100, -2t = 1 Taking the natural log of both sides  6.3 EXPONENTIAL EQUATIONS AND INEQUALITIES 345 gives In (e-2t) = In (j). Since natural log is log base e, In (e-2t) = -2t. We can also use the Power Rule to write In (j) = -ln(9). Putting these two steps together, we simplify In (e-2t) = In (4) to -2t = - In(9). We arrive at our solution, t = 1" which simplifies to t = In(3). (Can you explain why?) The calculator confirms the graphs of f(x) = 75 and g(x) = 1+3e°-2x intersect at x = In(3) t 1.099. Inter sc tion Inter s ction H=.7'656312 'Y=2.35D019 .H=i.0EID6123 .Y=75 y= f(x) =9.3x and y= f(x) =75 and Y = g(x) = 7x y = 100 Y = U =1+3e ax 5. We start solving 25x = 52 + 6 by rewriting 25 = 52 so that we have (52)x = 5x + 6, or 52x = 57 + 6. Even though we have a common base, having two terms on the right hand side of the equation foils our plan of equating exponents or taking logs. If we stare at this long enough, we notice that we have three terms with the exponent on one term exactly twice that of another. To our surprise and delight, we have a 'quadratic in disguise'. Letting u = 5x, we have u2 = (5x)2 = 52x so the equation 52x = 5 + 6 becomes u2 = + 6. Solving this as u2 - u-6 = 0 gives u = -2 or u = 3. Since u = 5x, we have 52 -2 or 5x = 3. Since 52 -2 has no real solution, (Why not?) we focus on 5 = 3. Since it isn't convenient to express 3 as a power of 5, we take natural logs and get In (52) = In(3) so that x in(5) = In(3) in(S or = 5) . When we graph f(x) = 252 and g(x) = 52 + 6, we see that they intersect at = 0.6826. 6. At first, it's unclear how to proceed with Cx2 r 5, besides clearing the denominator to obtain ex - e-x = 10. Of course, if we rewrite e- x e, we see we have another denominator lurking in the problem: ex - + = 10. Clearing this denominator gives us e2x - 1 = 10e, and once again, we have an equation with three terms where the exponent on one term is exactly twice that of another - a 'quadratic in disguise.' If we let u = ex, then u2 =e22 so the equation e2x - 1 = 10e can be viewed as u2 - 1 = 10u. Solving u2 - OU - 1 = 0, we obtain by the quadratic formula u = 5+ 26. From this, we have ex = 5+ 26. Since 5 - 26 < 0, we get no real solution to ex = 5 - 26, but for e = 5 + 26, we take natural logs to obtain x = In (5 + 26). If we graph f(x) = x 2 and g(x) = 5, we see that the graphs intersect at x = In (5 + 26) 2.312  346 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Intersection =.6260619 .Y=9 y = f(x) = 25x and y = g(x) = 5x + 6 Intersection H=2.3143r3 Y=5. y = f (x) = e- and y = g(x) = 5 H The authors would be remiss not to mention that Example 6.3.1 still holds value. Much can be learned about logarithms and exponentials by verifying the in Example 6.3.1 analytically. For example, to verify our solution to 2000 = substitute t = - 3 ) and obtain great educational solutions obtained 1000 - 3-0.1, we -0. - 1o in(2) 2000 1 1000 - 301("(3) In (2) 2000 - 1000-3n(3) 2000 ? 1000 - 31o93(2) 2000 = 1000 - 2 2000 ~ 2000 Change of Base Inverse Property The other solutions can be verified by using a combination of log and inverse properties. Some fall out quite quickly, while others are more involved. We leave them to the reader. Since exponential functions are continuous on their domains, the Intermediate Value Theorem 3.1 applies. As with the algebraic functions in Section 5.3, this allows us to solve inequalities using sign diagrams as demonstrated below. EXAMPLE 6.3.2. Solve the following inequalities. Check your answer graphically using a calculator. 1. 2x2-3x -16 > 0 2. <3 eX - 4 3. xe2x < 4x SOLUTION. 1. Since we already have 0 on one side of the inequality, we set r(x) = 2X2-3x - 16. The domain of r is all real numbers, so in order to construct our sign diagram, we seed to find the zeros of r. Setting r(x) = 0 gives 2x2-3x - 16 = 0 or 2X2-3x = 16. Since 16 = 24 we have 2X2-3x - 24, so by the one-to-one property of exponential functions, x2 - 3x = 4. Solving x2 - 3x - 4 = 0 gives x = 4 and x = -1. From the sign diagram, we see r(x) > 0 on (-oc, -1] U [4, c), which corresponds to where the graph of y = r(x) = 2X2-3x - 16, is on or above the i-axis.  6.3 EXPONENTIAL EQUATIONS AND INEQUALITIES 347 (+) 0 (-) 0 (+) I I4 -1 4 y = r(x) = 2X2-3x - 16 2. The first step we need to take to solve ex 4 < 3 is to get 0 on one side of that end, we subtract 3 from both sides and get a common denominator the inequality. To ex - 4 < 3 -3 < 0 ex - 4 ex 3 (ex - 4) ex - 4 ex - 4 12 - 2ex ex- 4 < 0 Common denomintors. < 0 We set r(x) = - and we note that r is undefined when its denominator ex - 4 = 0, or when ex = 4. Solving this gives x = ln(4), so the domain of r is (-oc, ln(4)) U (ln(4), oc). To find the zeros of r, we solve r(x) = 0 and obtain 12 - 2ex = 0. Solving for ex, we find ex = 6, or x = ln(6). When we build our sign diagram, finding test values may be a little tricky since we need to check values around ln(4) and ln(6). Recall that the function ln(x) is increasing4 which means ln(3) < ln(4) < ln(5) < ln(6) < ln(7). While the prospect of determining the sign of r (ln(3)) may be very unsettling, remember that ein(3) = 3, so r (l(3))12 - 2e1n(3) = e1'n(3) -4 12 - 2(3) 3-4 6 We determine the signs of r (ln(5)) and r (ln(7)) similarly.5 From the sign diagram, we find our answer to be (-oc, ln(4)) U [ln(6), oc). Using the calculator, we see the graph of f(x) = e e4 is below the graph of g(x) = 3 on (-oc, ln(4)) U (ln(6), oc), and they intersect at x = ln(6) ~ 1.792. 4This is because the base of ln(x) is e > 1. If the base b were in the interval 0 < b < 1, then logb(x) would decreasing. 5We could, of course, use the calculator, but what fun would that be?  348 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Intersection H=1.7917595 Y=3 (-+) 0 (-) ln(4) ln(6) y-rf(x)= 4 y = g(x) = 3 3. As before, we start solving xe2x < 4x by getting 0 on one side of the inequality, xe2x -4x < 0. We set r(x) = e2x - 4x and since there are no denominators, even-indexed radicals, or logs, the domain of r is all real numbers. Setting r(x) = 0 produces xe2x - 4x = 0. With x both in and out of the exponent, this could cause some difficulty. However, before panic sets in, we factor out the x to obtain x (e2x - 4) = 0 which gives x = 0 or e2x - 4 = 0. To solve the latter, we isolate the exponential and take logs to get 2x = ln(4), or x = n(4) = ln(2). (Can you explain the last equality using properties of logs?) As in the previous example, we need to be careful about choosing test values. Since ln(1) = 0, we choose ln (j), ln (2) and ln(3). Evaluating,6 we have r(ln (j)) = ln (Q) e21n() - 4n (2). Applying the Power Rule to the log in the exponent, we obtain ln (j) eln( Q) - 4 In (j)= ln (j) el1n() - 4 In (2). Using the inverse properties of logs, this reduces to 4 ln ( ) - 4ln (2) = - 4ln (2). Since j < 1, ln (j) < 0 and we get r(ln (j)) is (+). Continuing in this manner, we find r(x) < 0 on (0, ln(2)). The calculator confirms that the graph of f(x) = xe2x is below the graph of g(x) = 4x on this intervals.7 (+) 0 (-) 0 (+) 0 ln(2) Int ese tio H=.li5i7i10 '=2.??25DD7 y = f(x) = xe2x and y = g(x) = 4x 6A calculator can be used at this point. As usual, we proceed without apologies, with the analytical method. 7Note: ln(2) ~ 0.693.  6.3 EXPONENTIAL EQUATIONS AND INEQUALITIES 349 EXAMPLE 6.3.3. Recall from Example 6.1.2 that the temperature of coffee T (in degrees Fahrenheit) t minutes after it is served can be modeled by T(t) = 70 + 90e-0 l. When will the coffee be warmer than 100°F? SOLUTION. We need to find when T(t) > 100, or in other words, we need to solve the inequality 70 + 90e-0.lt > 100. Getting 0 on one side of the inequality, we have 90e-01lt - 30 > 0, and we set r(t) = 90e-0 lt - 30. The domain of r is artificially restricted due to the context of the problem to [0, oc), so we proceed to find the zeros of r. Solving 90e-01lt - 30 = 0 results in e-o.t = j so that t = -10 In ()3 which, after a quick application of the Power Rule leaves us with t = 10 ln(3). If we wish to avoid using the calculator to choose test values, we note that since 1 < 3, 0 = ln(1) < ln(3) so that 101ln(3) > 0. So we choose t = 0 as a test value in [0, 10ln(3)). Since 3 < 4, 10In(3) < 10In(4), so the latter is our choice of a test value for the interval (10 ln(3), oo). Our sign diagram is below, and next to it is our graph of t = T(t) from Example 6.1.2 with the horizontal line y = 100. y 180 160 140 (+) 0 (-) y =100 0 l0ln(3) 80 H.A. y = 70 40- 20- 2 4 6 8 10 12 14 16 18 20 t y =T(t) In order to interpret what this means in the context of the real world, we need a reasonable approximation of the number 10 In(3) 10.986. This means it takes approximately 11 minutes for the coffee to cool to 100°F. Until then, the coffee is warmer than that.8Q We close this section by finding the inverse of a function which is a composition of a rational function with an exponential function. 5ex EXAMPLE 6.3.4. The function f(x) = +5 1is one-to-one. Find a formula for f-1(x) and check e + 1 your answer graphically using your calculator. SOLUTION. We start by writing y = f(x), and interchange the roles of x and y. To solve for y, we first clear denominators and then isolate the exponential function. 8Critics may point out that since we needed to use the calculator to interpret our answer anyway, why not use it earlier to simplify the computations? It is a fair question which we answer unfairly: it's our book.  350 EXPONENTIAL AND LOGARITHMIC FUNCTIONS y 5e' ex+1 5ey = 1 x (ey + 1) = 5ey Switch x and y xeY + x 5ey x =5e - xey x =ey(5 - x) eY In (eY) X 5-x In 5 - x y = In5x 5- x We claim f-1(x) = In ( 5 X). To verify this analytically, we would need to (f-1 o f) (x) = x for all x in the domain of f and that (f o f-1) (x) = x of f-1. We leave this to the reader. To verify our solution graphically, we verify the compositions for all x in the domain graph y = f(x) ex SX+1 =X. and y = g(x) = In (5xx) on the same set of axes and Note the domain of f is the range of g and vice-versa. observe the symmetry about the line y y = f(x) and y = g(x) = In (x) D  6.3 EXPONENTIAL EQUATIONS AND INEQUALITIES 6.3.1 EXERCISES 1. Solve the following equations analytically. 351 (a) 3(x-1) (b) 3(x-1) (c) 3(x-) 27 29 2X (e) (1i+-) 01= 3 (f) e-5730k i 2 (g) 2(x3-x) -1 (h) e2x =-2ex (i) 70 + 90e-0 -= a*~ )+2150 so = 75 (k 1+29e-0-8t (k) 25 (4)X = 10 75 (d) 3(x-1) (1)(x+5) 2. Solve the following inequalities analytically. (a) ex > 53 (b) 1000 (1 + )10612t > 3000 (c) 2(x3-x) < 1 (d) 25 (5)x > 10 (e) 129e-O.8t < 130 (f) 70 + 90e-0 < 75 3. Use your calculator to help you solve the following equations and inequalities. (a) ex 0 (e) 3(-n) < 25 (f) ex = ln(x) + 5 4. Since f(x) = ln(x) is a strictly increasing function, if 0 < a < b then ln(a) < ln(b). Use this fact to solve the inequality e(3x-1) > 6 without a sign diagram. eCp-te-x 5. Compute the inverse of f(x) = 2 .State the domain and range of both f and f-1. 6. In Example 6.3.4, we found that the inverse of f(x) = +1was f-1(x) we left a few loose ends for you to tie up. ln (x\)but \5-zx (a) Show that (f-1 o f) (x) = x for all x in the domain of f and that (f o f-1) (x) = x for all x in the domain of f-1. (b) Find the range of f by finding the domain of f-1. (c) Let g(x)=5x and h(x) ex. Show that f = g o h and that (go h)-1 = h-1 o g-1. 9 x +1 7. With the help of your classmates, solve the inequality ex > x' for a variety of natural numbers n. What might you conjecture about the "speed" at which f(x) = ex grows versus any polynomial? 9We know this is true in general by Exercise 8 in Section 5.2, but it's nice to see a specific example of the property.  352 352 EXPONENTIAL AND LOGARITHMIC FUNCTIONS 6.3.2 ANSWERS 1. (a) x (b) x 4 in(29) + in(3) in(3) (f) k (g) x (h) x InQ(2) -5730 -1, 0, 1 :1, in(2) in(3) (c) x =in(3) - in(2) in(3) + 5in (CD (e) t 12in(3) (e) t= 1in(1.005) 2. (a) (in(53),oo) [b)121in(1.005)' cx) (c) (-oc, -1) U (0,1) (d) (i) t= lOin(18) (J) t =In(12) -0.8 (k) x = InQ 4) In 1 (e) -0.8)J (f) [In()) -(-00, -1n( [10 in(18), oc) 3. (a) (-oc, (b) x- (C) x-0 2.3217) U (4.3717, oc) 0.76666, x= 2, x =4 (d) (-oo,1] (e) (f ) (-oc, 2.7095) x 0.01866, x 1.7115 4. x > -'(In (6) + 1) 5. f-1 -in ( + x2 + I). Both f and f-1 have domain (-oc, oc) and range (-oc, oc).  6.4 LOGARITHMIC EQUATIONS AND INEQUALITIES 353 6.4 LOGARITHMIC EQUATIONS AND INEQUALITIES In Section 6.3 we solved equations and inequalities involving exponential functions using one of two basic strategies. We now turn our attention to equations and inequalities involving logarithmic functions, and not surprisingly, there are two basic strategies to choose from. For example, suppose we wish to solve log2(x) =log2(5). Theorem 6.4 tells us that the only solution to this equation is x = 5. Now suppose we wish to solve log2(x) = 3. If we want to use Theorem 6.4, we need to rewrite 3 as a logarithm base 2. We can use Theorem 6.3 to do just that: 3= log2 (23) = log2(8). Our equation then becomes log2(x) = log2(8) so that x= 8. However, we could have arrived at the same answer, in fewer steps, by using Theorem 6.3 to rewrite the equation log2(x) = 3 as 23 or x = 8. We summarize the two common ways to solve log equations below. Steps for Solving an Equation involving Logarithmic Fuctions 1. Isolate the logarithmic function. 2. (a) If convenient, express both sides as logs with the same base and equate the arguments of the log functions. (b) Otherwise, rewrite the log equation as an exponential equation. EXAMPLE 6.4.1. Solve the following equations. Check your solutions graphically using a calculator. 1. 1og117(1 - 3x) =log117 (2 - 3) 4. log7(1 - 2x) = 1 - log7(3 - x) 2. 2 - ln(xz- 3) = 1 5. log2(x + 3) =log2(6 - x) + 3 3. log6(X+4)+log6(3 - x) = 1 6. 1+2log4(x+1) = 2 log2(x) SOLUTION. 1. Since we have the same base on both sides of the equation 1og117(1 - 3x) =log117 (2 - 3), we equate what's inside the logs to get 1 - 3x =9x2 - 3. Solving 2 + 3x - 4 = 0 gives x= -4 and x= 1. To check these answers using the calculator, we make use of the change of base formula and graph f(x) ln(13x and g(x) nX23) and we see they intersect only gl ~x n(117) g l n(117) adw e hyitretol at x= -4. To see what happened to the solution x = 1, we substitute it into our original equation to obtain 1og117(-2) = log117(-2). While these expressions look identical, neither is a real number,l which means x= 1 is not in the domain of the original equation, and is not a solution. 2. Our first objective in solving 2-ln(x -3) = 1 is to isolate the logarithm. We get ln(z -3) = 1, which, as an exponential equation, is el = x - 3. We get our solution x= e + 3. On the calculator, we see the graph of f(x) = 2 - ln(z - 3) intersects the graph of g(x) = 1 at x =e + 3 ~5.718. 1They do, however, represent the same family of complex numbers. We stop ourselves at this point and refer the reader to a good course in Complex Variables.  354 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Intersection x= -4 Y=.5301 i0092 y = f (x) = og117(1 - 3x) and y = g(x) = log117 (2 - 3) Intersection K=5.7192H1 Y=1 y = f(x) = 2 -ln(x - y = g(x) = 1 3) and 3. We can start solving log6 (x + 4) +log6 (3 - x) = 1 rewrite the equation as log6 [(x + 4)(3 - x)] = 1. we get 61 = (x + 4)(3 - x). This reduces to x2 - by using the Product Rule for logarithms to Rewriting this as an exponential equation, x - 6 = 0, which gives x 2 and = 3. Graphing y x=-2 and f(x) n(6)+ n36x) and y g(x) = 1, we see they intersect twice, at x = 3. Intersection K=- - - Y=1 - t - - Intersection K=2 Y=1- -- y = f(x) = log6 (x + 4) + log6 (3 x) and y = g(x) = 1 4. Taking a cue from the previous problem, we begin solving log7(1 2x) = 1 - log7(3 - x) by - x) = 1, and then using first collecting the logarithms on the same side, log7(1 2x) +log7(3 the Product Rule to get logy[(1 2x)(3 - x)] = 1. Rewriting this as an exponential equation gives 71 = (1-2x)(3- x) which gives the quadratic equation 2x2 -7x -4 = 0. Solving, we find z = -1 and x = 4. Graphing, we find f(x) n(1-2x) nd y x 1in(3-x) intersect raig y l n(7) an =l() n(7)inesc only at x = - j. Checking x = 4 in the original equation produces log7( 7) = 1 - log7(-1), which is a clear domain violation. 5. Starting with log2(x + 3) = log2(6 - x) + 3, we gather the logarithms to one side and get log2(x + 3) - log2(6 - x) = 3, and then use the Quotient Rule to obtain log2 (6+) 3. Rewriting this as an exponential equation gives 23 =+X. This reduces to the linear equation 8(6 - x) = x + 3, which gives us x = 5. When we graph f(x) we find they intersect at x 5. ln(x+3) and g(x) ln(6-x) +3 ln(2)  6.4 LOGARITHMIC EQUATIONS AND INEQUALITIES 355 Intersection x-.5 L I Y=.35620719 y = f(x) = log7(1 - 2x) and y = g(x) = 1 - log7(3 - x) Intersection y = f(x) = log2(x + 3) and y = g(x) = log2(6 - x) + 3 6. Starting with 1 + 2 log4( + 1) = 2log2(x), we gather the logs to one side to get the equation 1 = 2logx2() - 2log4(x + 1). Before we can combine the logarithms, however, we need a common base. Since 4 is a power of 2, we use change of base to convert log4(x+1) =og2(x+1) Slog2(x + 1). Hence, our original equation becomes 1 = 2log2) - 2 (2 log(x + 1)) or 1 = 2log2(z) -log2(X + 1). Using the Power and Quotient Rules, we obtain 1 = log2 (-i1). Rewriting this in exponential form, we get x = 2 or x2 - 2x - 2 = 0. Using the quadratic formula, we get x = 1 + 3. Graphing f(x) = 1 + 21n (1) and g(x) = "t, we see the graphs intersect only at x = 1 + 3 ~ 2.732. The solution x = 1 - < 0, which means if substituted into the original equation, the term 2log2 (1 - 3) is undefined. Intersection N=2.7320509 Y=2.H99496 y = f(x) = 1+ 2log4(x +1) and y = g(x) = 21og2(X) If nothing else, Example 6.4.1 demonstrates the importance of checking for extraneous solutions2 when solving equations involving logarithms. Even though we checked our answers graphically, extraneous solutions are easy to spot - any supposed solution which causes a negative number inside a logarithm needs to be discarded. As with the equations in Example 6.3.1, much can be learned from checking all of the answers in Example 6.4.1 analytically. We leave this to the reader and turn our attention to inequalities involving logarithmic functions. Since logarithmic functions are continuous on their domains, we can use sign diagrams. EXAMPLE 6.4.2. Solve the following inequalities. Check your answer graphically using a calculator. 2Recall that an extraneous solution is an answer obtained analytically which does not satisfy the original equation.  356 EXPONENTIAL AND LOGARITHMIC FUNCTIONS 1. 1 < 1 2. (log2(X))2 < 21og2(x) + 3 3. xlog(x + 1) > x ln(x) + 1 SOLUTION. 1. We start solving in( )+1 < 1 by getting 0 on one side of the inequality: 1- 1 < 0. Getting a common denominator yields n(+1 - ln(x)+ < 0 which reduces to l < 0, or > 0. We define r(x) j and set about finding the domain and the zeros of r. Due to the appearance of the term ln(x), we require x > 0. In order to keep the denominator away from zero, we solve ln(x) + 1 = 0 so ln(x) = -1, so x = e- . Hence, the domain of r is (0, e) U (e, 00). To find the zeros of r, we set r(x) = " = 0 so that ln(x) = 0, and we find x e = 1. In order to determine test values for r without resorting to the calculator, we need to find numbers between 0, e, and 1 which have a base of e. Since e ~ 2.718 > 1, 0 < < < - -< 1 < e. To determine the sign of r () we use the fact that ln (+) = In (e-2) = -2, and find r (+) = _2+ = 2, which is (+). The rest of the test values are determined similarly. From our sign diagram, we find the solution to be (0, e) U [1, oc). Graphing f(x) = in( )+1 and g(x) = 1, we see the the graph of f is below the graph of g on the solution intervals, and that the graphs intersect at x = 1. LLLJ (+) T (-) 0 (+) I I I 0 A 1 Inr FS i~n x=i L7 Yi y = f(x) )n(j+1 and y = g(x) = 1 2. Moving all of the nonzero terms of (log2(W))2 < 2 log2 (x) + 3 to one side of the inequality, we have (log2(X))2 - 21og2(x) - 3 < 0. Defining r(x) = (log2(X))2 - 21og2(x) - 3, we get the domain of r is (0, oc), due to the presence of the logarithm. To find the zeros of r, we set r(x) = (log2 ())2 - 21og2(x) - 3 = 0 which results in a 'quadratic in disguise.' We set u = log2(x) so our equation becomes u2 - 2u - 3 = 0 which gives us u = -1 and u = 3. Since u = log2 (x), we get log2(x) = -1, which gives us x = 2-1 = 1, and log2 (x) = 3, which yields x = 23 = 8. We use test values which are powers of 2: 0 < < < 1 < 8 < 16, and from our sign diagram, we see r(x) < 0 on (2, 8). Geometrically, we see the graph of f(x) is below the graph of y =g(x) = " + 3 on the solution interval.  6.4 LOGARITHMIC EQUATIONS AND INEQUALITIES 357 (+) 0 (-) 0 (+) ! ! 1 3 0 1 8 y = f(x) = (log2())2 and y = g(x) = 21og2() + 3 3. We begin to solve x log(x+1) > x by subtracting x from both sides to get x log(x+1) -x > 0. We define r(x) = x log(x+1)-x and due to the presence of the logarithm, we require x+1 > 0, or x > -1. To find the zeros of r, we set r(x) = x log(x + 1) - x = 0. Factoring, we get x (log(x + 1) - 1) = 0, which gives x = 0 or log(x+1) -1 = 0. The latter gives log(x+1) = 1, or x + 1 = 101, which admits x = 9. We select test values x so that x + 1 is a power of 10, and we obtain -1 < -0.9 < 0 < 10 - 1 < 9 < 99. Our sign diagram gives the solution to be (-1, 0] U [9, c). The calculator indicates the graph of y = f(x) = x log(x + 1) is above y = g(x) = x on the solution intervals, and the graphs intersect at x = 0 and x = 9. (+) 0 (-) 0 (+) Intersection X=9 Y=9 -1 0 9 y= f(x) = xlog(x + 1) and y = g(x) = x Y1=Xloar+Dr Near x = 0 Intseacion X=9 / .Y=9. Near x= 9 H Our next example revisits the concept of pH as first introduced in the exercises in Section 6.1.  358 EXPONENTIAL AND LOGARITHMIC FUNCTIONS EXAMPLE 6.4.3. In order to successfully breed Ippizuti fish the pH of a freshwater tank must be at least 7.8 but can be no more than 8.5. Determine the corresponding range of hydrogen ion concentration. SOLUTION. Recall from Exercise 4c in Section 6.1 that pH = -log[H+] where [H+] is the hydrogen ion concentration in moles per liter. We require 7.8 < -log[H+] K 8.5 or -7.8 > log[H+] > -8.5. To solve this compound inequality we solve -7.8 > log[H+] and log[H+] > -8.5 and take the intersection of the solution sets.3 The former inequality yields 0 < [H+] K 10-7.8 and the latter yields [H+] 10-8.5. Taking the intersection gives us our final answer 10-8.5 [H+] 10-7.8. (Your Chemistry professor may want the answer written as 3.16 x 10-9 K [H+] < 1.58 x 10-8.) After carefully adjusting the viewing window on the graphing calculator we see that the graph of f(x) = - log(x) lies between the lines y = 7.8 and y = 8.5 on the interval [3.16 x 10-9, 1.58 x 10-8. Inters ctionN =3.ii23E -9 Y=.5 The graphs of y = f (x) Inters1E tion x=1.5049E -H Y=7.0 log(x), y = 7.8 and y = 8.5 H We close this section by finding an inverse of a one-to-one function which involves logarithms. EXAMPLE 6.4.4. The function f(x) = log(x) is one-to-one. Find a formula for ff-(x) and 1 - log(x) check your answer graphically using your calculator. SOLUTION. We first write y = f(x) then interchange the x and y and solve for y. y = f(x) log(x) 1 - log (x) log(y) 1 - log (y) x (1 - log(y)) = log(y) x - xlog(y) = log(y) X = x log(y) + log (y) X = (x +1)log(y) X + 1 = log(y) x +1 y = 10 ±1 Interchange x and y. Rewrite as an exponential equation. 3Refer to page 147 for a discussion of what this means.  6.4 LOGARITHMIC EQUATIONS AND INEQUALITIES 359 We have f--1(x) 10±1. Graphing f and f-1 on the same viewing window yields log(x) _ Y = f(x) = log(x) and y = g(x) = 10X1 1 - log(x) H  360 EXPONENTIAL AND LOGARITHMIC FUNCTIONS 6.4.1 EXERCISES 1. Solve the following equations analytically. (a) log x= -3 (b) ln(z2) = (ln(x))2 (c) log3(x-4)+ log3(+4) = 2 (d) log5(2x + 1) + log5(x + 2) 1 (e) log2(x3) = log2(z) 1 (f) 1og169(3x + 7) - log169(5x - 9) = 2. Solve the following inequalities analytically. (g) (h) (i) (j) log (10-3)4-7 - log(x) = 5.4 10 log (10- 12) = 150 log3(x)= logi(x) + 8 3 (a) x ln(x) - x > 0 (b) 5.6 < log (103) 7.1 (c) 10 log (10- 12) > 90 (d) 2.3 < - log(x) < 5.4 (e) (f) 1 - ln(z) <0 ln(2) < (ln(x))2 3. Use your calculator to help you solve the following equations and inequalities. (a) ln(x) = e-x (b) ln(z2 + 1) > 5 (c) ln(x) - =c (d) ln(-2x3 - 2 + 13x - 6) < 0 4. Since f(x) = ex is a strictly increasing function, if a < b then ea < eb. Use this fact to solve the inequality ln(2x + 1) <3 without a sign diagram. Also, compare this exercise to question 4 in Section 6.3. 5. Solve ln(3 - y) - ln(y) = 2x + ln(5) for y. 6. Let f(x) 1 (1+ x.f -ln' - . Compute f-1(x) and find its domain and range. log(x) 7. In Example 6.4.4 we found the inverse of f(x) 1 - log(x) to be f-1(x) = 10=+1. (a) Show that (f-1 o f) (x) = x for all x in the domain of f and that (f o f-1) (x) all x in the domain of f-1. x for (b) Find the range of f by finding the domain of f-1. (c) Let g(xc) =1 and h(x) = log(x). Show that f = g o h and (g o h)-1 = h-1 o g-1. (We know this is true in general by Exercise 8 in Section 5.2, but it's nice to see a specific example of the property.)  6.4 LOGARITHMIC EQUATIONS AND INEQUALITIES 361 8. Explain the equation in Exercise 1g and the inequality in Exercise 2b above in terms of the Richter scale for earthquake magnitude. (See Exercise 4a in Section 6.1.) 9. Explain the equation in Exercise li and the inequality in Exercise 2c above in terms of sound intensity level as measured in decibels. (See Exercise 4b in Section 6.1.) 10. Explain the equation in Exercise lh and the inequality in Exercise 2d above in terms of the pH of a solution. (See Exercise 4c in Section 6.1.) 11. With the help of your classmates, solve the inequality n x > ln(x) for a variety of natural numbers n. What might you conjecture about the "speed" at which f(x) = ln(x) grows versus any principal nth root function?  362 EXPONENTIAL AND LOGARITHMIC FUNCTIONS 6.4.2 ANSWERS 1. (a) x (b) x (c) x (d) x (e) x 8 1, x =e2 5 1 1 (f) (g) (h) (i) (j) (d) (e) (f) 3: 3: 3: 3: 3: 2 101.7 10-5.4 103 81 2. (a) (e,oo) (b) [1026, 104.1] (c) [10-3, oc) (10 ., 10-2.31 (e, oo) (0, 1] U [e2,ooe) 3. (a) x 1.3098 (b) (-oc, -12.1414) U (12.1414, o) (c) x 4.177, x 5503.665 (d) (-3.0281, -3) U (0.5, 0.5991) U (1.9299, 2) 1 e3-1 4. -2 0 is the constant of proportionality which satisfies the equation (instantaneous rate of change of N(t) at time t) = k N(t) It is worth taking some time to compare Equations 6.3 and 6.4. In Equation 6.3, we use P to denote the initial investment; in Equation 6.4, we use No to denote the initial population. In Equation 6.3, r denotes the annual interest rate, and so it shouldn't be too surprising that the k in Equation 6.4 corresponds to a growth rate as well. While Equations 6.3 and 6.4 look entirely different, they both represent the same mathematical concept. EXAMPLE 6.5.2. In order to perform arthrosclerosis research, epithelial cells are harvested from discarded umbilical tissue and grown in the laboratory. A technician observes that a culture of twelve thousand cells grows to five million cells in one week. Assuming that the cells follow The Law of Uninhibited Growth, find a formula for the number of cells, N, in thousands, after t days. SOLUTION. We begin with N(t) = Noekt. Since N is to give the number of cells in thousands, we have No = 12, so N(t) = 12ekt. In order to complete the formula, we need to determine the growth rate k. We know that after one week, the number of cells has grown to five million. Since t 9The average rate of change of a function over an interval was first introduced in Section 2.1. Instanitaneous rates of change are the business of Calculus, as is mentioned on Page 111.  6.5 APPLICATIONS OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS 367 measures days and the units of N are in thousands, this translates mathematically to N(7) = 5000. We get the equation 12e7k = 5000 which gives k In(1250). Hence, N(t) = 12e1n(3) . Of course, in practice, we would approximate k to some desired accuracy, say k 0.8618, which we can interpret as an 86.18% daily growth rate for the cells. Q Whereas Equations 6.3 and 6.4 model the growth of quantities, we can use equations like them to describe the decline of quantities. One example we've seen already is Example 6.1.1 in Section 6.1. There, the value of a car declined from its purchase price of $25,000 to nothing at all. Another real world phenomenon which follows suit is radioactive decay. There are elements which are unstable and emit energy spontaneously. In doing so, the amount of the element itself diminishes. The assumption behind this model is that the rate of decay of an element at a particular time is directly proportional to the amount of the element present at that time. In other words, the more of the element there is, the faster the element decays. This is precisely the same kind of hypothesis which drives The Law of Uninhibited Growth, and as such, the equation governing radioactive decay is hauntingly similar to Equation 6.4 with the exception that the rate constant k is negative. EQUATION 6.5. Radioactive Decay The amount of a radioactive element A at time t is given by the formula A(t) = Aoekt, where A(0) = Ao is the initial amount of the element and k < 0 is the constant of proportionality which satisfies the equation (instantaneous rate of change of A(t) at time t) = k A(t) EXAMPLE 6.5.3. Iodine-131 is a commonly used radioactive isotope used to help detect how well the thyroid is functioning. Suppose the decay of Iodine-131 follows the model given in Equation 6.5, and that the half-life10 of Iodine-131 is approximately 8 days. If 5 grams of Iodine-131 is present initially, find a function which gives the amount of Iodine-131, A, in grams, t days later. SOLUTION. Since we start with 5 grams initially, Equation 6.5 gives A(t) = 5ekt. Since the half-life is 8 days, it takes 8 days for half of the Iodine-131 to decay, leaving half of it behind. Hence, A(8) = 2.5 which means 5e8k = 2.5. Solving, we get k =ln (1) = - 82~ -0.08664, which we can interpret t ln(2) as a loss of material at a rate of 8.664% daily. Hence, A(t) = 5e-- 8 ~ 5e-8664. We now turn our attention to some more mathematically sophisticated models. One such model is Newton's Law of Cooling, which we first encountered in Example 6.1.2 of Section 6.1. In that example we had a cup of coffee cooling from 160°F to room temperature 70°F according to the formula T(t) = 70 + 90e-01lt, where t was measured in minutes. In this situation, we know the physical limit of the temperature of the coffee is room temperature,11 and the differential equation which gives rise to our formula for T(t) takes this into account. Whereas the radioactive decay ioThe time it takes for half of the substance to decay. "1The Second Law of Thermodynamics states that heat can spontaneously flow from a hotter object to a colder one, but not the other way around. Thus, the coffee could not continue to release heat into the air so as to cool below room temperature.  368 EXPONENTIAL AND LOGARITHMIC FUNCTIONS model had a rate of decay at time t directly proportional to the amount of the element which remained at time t, Newton's Law of Cooling states that the rate of cooling of the coffee at a given time t is directly proportional to how much of a temperature gap exists between the coffee at time t and room temperature, not the temperature of the coffee itself. In other words, the coffee cools faster when it is first served, and as its temperature nears room temperature, the coffee cools ever more slowly. Of course, if we take an item from the refrigerator and let it sit out in the kitchen, the object's temperature will rise to room temperature, and since the physics behind warming and cooling is the same, we combine both cases in the equation below. EQUATION 6.6. Newton's Law of Cooling (Warming): The temperature T of an object at time t is given by the formula T(t) = Ta + (To - Ta) e't, where T(O) = To is the initial temperature of the object, Ta is the ambient temperaturea and k > 0 is the constant of proportionality which satisfies the equation (instantaneous rate of change of T(t) at time t) = k (T(t) - Ta) aThat is, the temperature of the surroundings. If we re-examine the situation in Example 6.1.2 with To = 160, Ta = 70, and k = 0.1, we get, according to Equation 6.6, T(t) = 160 + (160 - 70)e-.I which reduces to the original formula given. The rate constant k = 0.1 indicates the coffee is cooling at a rate equal to 10% of the difference between the temperature of the coffee and its surroundings. Note in Equation 6.6 that the constant k is positive for both the cooling and warming scenarios. What determines if the function T(t) is increasing or decreasing is if To (the initial temperature of the object) is greater than Ta (the ambient temperature) or vice-versa, as we see in our next example. EXAMPLE 6.5.4. A 40°F roast is cooked in a 350°F oven. After 2 hours, the temperature of the roast is 125°F. 1. Assuming the temperature of the roast follows Newton's Law of Warming, find a formula for the temperature of the roast T as a function of its time in the oven, t, in hours. 2. The roast is done when the internal temperature reaches 165°F. When will the roast be done? SOLUTION. 1. The initial temperature of the roast is 40°F, so To = 40. The environment in which we are placing the roast is the 350°F oven, so Ta = 350. Newton's Law of Warming tells us T(t) = 350 + (40 - 350)e-k, or T(t) = 350 - 310e-k. To determine k, we use the fact that after 2 hours, the roast is 125°F, which means T(2) = 125. This gives rise to the equation 350 - 310e-2k = 125 which yields k =- j ln ~~ 0.1602. The temperature function is T(t) = 350 - 310e~1n(6) 350 - 310e-o.iso2t  6.5 APPLICATIONS OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS 369 2. To determine when the roast is done, we set T(t) = 165. This gives 350 - 310e-0.1602 165 whose solution is t = - o.16o2 in (6) 3.22. It takes roughly 3 hours and 15 minutes to cook the roast completely. D If we had taken the time to graph y = T(t) in Example 6.5.4, we would have found the horizontal asymptote to be y = 350, which corresponds to the temperature of the oven. We can also arrive at this conclusion by applying a bit of 'number sense'. As t - oc, -0.1602t very big (-) so that e-o.1602t very small (+). The larger the value of t, the smaller e-o.1602t becomes so that T(t) 350 - very small (+), which indicates the graph of y = T(t) is approaching its horizontal asymptote y = 350 from below. Physically, this means the roast will eventually warm up to 350 F.12 The function T is sometimes called a limited growth model, since the function T remains bounded as t - oc. If we apply the principles behind Newton's Law of Cooling to a biological example, it says the growth rate of a population is directly proportional to how much room the population has to grow. In other words, the more room for expansion, the faster the growth rate. The logistic growth model combines The Law of Uninhibited Growth with limited growth and states that the rate of growth of a population varies jointly with the population itself as well as the room the population has to grow. EQUATION 6.7. Logistic Growth: If a population behaves according to the assumptions of logistic growth, the number of organisms N at time t is given by the equation 1 + Ce-kLt' where N(0) = No is the initial population, L is the limiting populationa, C is a measure of how much room there is to grow given by L No and k > 0 is the constant of proportionality which satisfies the equation (instantaneous rate of change of N(t) at time t) - k N(t) (L - N(t)) "That is, as t - oo, N(t) - L The logistic function is used not only to model the growth of organisms, but is also often used to model the spread of disease and rumors.13 EXAMPLE 6.5.5. The number of people N, in hundreds, at a local community college who have heard the rumor 'Carl is afraid of Virginia Woolf' can be modeled using the logistic equation 84 N~) 1 + 2799e-i' where t > 0 is the number of days after April 1, 2009. "at which point it would be more toast than roast. 13Which can be just as damaging as diseases.  370 EXPONENTIAL AND LOGARITHMIC FUNCTIONS 1. Find and interpret N(0). 2. Find and interpret the end behavior of N(t). 3. How long until 4200 people have heard the rumor? 4. Check your answers to 2 and 3 using your calculator. SOLUTION. 1. We find N(0) = 84 = 84 = . Since N(t) measures the number of people who 1+}2799e° 2800 100 heard the rumor in hundreds, N(0) corresponds to 3 people. Since t = 0 corresponds to 1, 2009, we may conclude that on that day, 3 people have heard the rumor.14 have April 2. We could simply note that N(t) is written in the form of Equation 6.7, and identify L = 84. However, to see why the answer is 84, we proceed analytically. Since the domain of N is restricted to t > 0, the only end behavior of significance is t - oc. As we've seen before,15 as t o0, have 1997e-t- 0+ and so N(t) 4 s +) 84. Hence, as t - 00, N(t) - 84. This means that as time goes by, the number of people who will have heard the rumor approaches 8400. 3. To find how long it takes until 4200 people have heard the rumor, we set N(t) 8=4 1+2799e-t = 42 gives t = ln(2799) - 7.937. It takes around 8 days until 4200 heard the rumor. 42. Solving people have 4. We graph y asymptote of y =42 at x= = N(x) using the calculator and see that the line y = 84 is the horizontal the graph, confirming our answer to part 2, and the graph intersects the line ln(2799) 7.937, which confirms our answer to part 3. Ink FS i~n.. .... ILZ |-|=7.9370175 Y=42 Y = f(x) = }2799e_, and Y = f(z) = }2799e_, and y= 84 y= 42 1Or, more likely, three people started the rumor. I'd wager Jeff, Jamie, and Jason started it. So much for telling your best friends something in confidence! 15See, for example, Example 6.1.2.  6.5 APPLICATIONS OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS 371 If we take the time to analyze the graph of y = N(x) above, we can see graphically how logistic growth combines features of uninhibited and limited growth. The curve seems to rise steeply, then at some point, begins to level off. The point at which this happens is called an inflection point or is sometimes called the 'point of diminishing returns'. At this point, even though the function is still increasing, the rate at which it does so begins to decline. It turns out the point of diminishing returns always occurs at half the limiting population. (In our case, when y = 42.) While these concepts are more precisely quantified using Calculus, below are two views of the graph of y = N(x), one on the interval [0, 8], the other on [8, 15]. The former looks strikingly like uninhibited growth; the latter like limited growth. y =f(x) _, for y =f(x) =1+2799e- fo 0 0. Describe the behavior of y as x - 0+ and interpret this physically. 10. The current i measured in amps in a certain electronic circuit with a constant impressed voltage of 120 volts is given by i(t) = 2 - 2e-0 where t > 0 is the number of seconds after the circuit is switched on. Determine the value of i as t - oc. (This is called the steady state current.) 11. If the voltage in the circuit in Exercise 10 above is switched off after 30 seconds, the current is given by the piecewise-defined function 2 - 2e-10t if 0 < t < 30 i t) } (2 - 2e-300) e-10t300 if t > 30 With the help of your calculator, graph y =i(t) and discuss with your classmates the physical significance of the two parts of the graph 0 t < 30 and t;> 30. 22This roast was enjoyed by Jeff and his family on June 10, 2009. This is real data, folks!  378 EXPONENTIAL AND LOGARITHMIC FUNCTIONS 12. In Exercise 7 in Section 2.3, we stated that the cable of a suspension bridge formed a parabola but that a free hanging cable did not. A free hanging cable forms a catenary and its basic eX + e- shape is given by y = 2 Use your calculator to graph this function. What are its domain and range? What is its end behavior? Is it invertible? How do you think it is related to the function given in Exercise 5 in Section 6.3 and the one given in the answer to Exercise 6 in Section 6.4? When flipped upside down, the catenary makes an arch. In fact, the Gateway Arch in St. Louis, Missouri has the shape y = 757.7 - 127.7 (e1 e ) where x and 2 y are measured in feet and -315 < x < 315. What is the highest point on the arch? 13. In Exercise 4a in Section 2.5, we examined the data set given below which showed how two cats and their surviving offspring can produce over 80 million cats in just ten years. It is virtually impossible to see this data plotted on your calculator, so plot x versus ln(x) as was done on page 374. Find a linear model for this new data and comment on its goodness of fit. Find an exponential model for the original data and comment on its goodness of fit. Year x 1 2 3 4 5 6 7 8 9 10 Number of Cats N(x) 12 66 382 2201 12680 73041 420715 2423316 13968290 80399780 14. Each Monday during the registration period before the Fall Semester at LCCC, the Enrollment Planning Council gets a report prepared by the data analysts in Institutional Effectiveness and Planning.23 While the ongoing enrollment data is analyzed in many different ways, we shall focus only on the overall headcount. Below is a chart of the enrollment data for Fall Semester 2008. It starts 21 weeks before "Opening Day" and ends on "Day 15" of the semester, but we have relabeled the top row to be x= 1 through x= 24 so that the math is easier. (Thus, x= 22 is Opening Day.) Week x 1 2 3 4 5 6 7 8 Total Headcount 1194 1564 2001 2475 2802 3141 3527 3790 Week x 9 10 11 12 13 14 15 16 Total Headcount 4065 4371 4611 4945 5300 5657 6056 6478 Week x 17 18 19 20 21 22 23 24 Total Headcount 7161 7772 8505 9256 10201 10743 11102 11181 With the help of your classmates, find a model for this data. Unlike most of the phenomena we have studied in this section, there is no single differential equation which governs the 23The authors thank Dr. Wendy Marley and her staff for this data and Dr. Marcia Ballinger for the permission to use it in this problem.  6.5 APPLICATIONS OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS 379 enrollment growth. Thus there is no scientific reason to rely on a logistic function even though the data plot may lead us to that model. What are some factors which influence enrollment at a community college and how can you take those into account mathematically? 15. As of the writing of this set of exercises, the Enrollment Planning Report for Fall Semester 2009 has only 10 data points for the first 10 weeks of the registration period. Those numbers are given below. Week x 1 2 3 4 5 6 7 8 9 10 Total Headcount 1380 2000 2639 3153 3499 3831 4283 4742 5123 5398 With the help of your classmates, find a model for this data and make a prediction for the Opening Day enrollment as well as the Day 15 enrollment. (WARNING: This year's registration period is one week shorter than it was in 2008 so Opening Day would be x= 21 and Day 15 is x =23.) 16. According to this website, the number of active users of Facebook has grown significantly since its initial launch from a Harvard dorm room in February 2004. The chart below has the approximate number U(x) of active users, in millions, x months after February 2004. For example, the first entry (10, 1) means that there were 1 million active users in December 2004 and the last entry (62, 200) means that there were 200 million active users in April 2009. Month x 10 22 34 38 44 54 59 60 62 Active Users in Millions U(x) 1 5.5 12 20 50 100 150 175 200 With the help of your classmates, find a model for this data.  380 EXPONENTIAL AND LOGARITHMIC FUNCTIONS 6.5.4 ANSWERS 1. (a) A(8) = 2000 (1 + 0.0025)12.8 ~$2040.40 ln(2 (b) t = ~.25 277.29 years 12In (1 + .1225) ( ~2000 (c) P - .25)36 $1985.06 (1+01221238 2. (a) A(8) = 2000 (1 + 0.025)128 ~ $2394.03 (b) t = ~025 30.83 years 12In (1+ 0.1225) ( ~2000 (c) P - ~225) $1869.57 (d) (1 + 0.0225)12 1.0227 so the APY is 2.27% 5. (a) A(t) = Ne n5730t Ne-o00012097t (b) A(20000) ~ 0.088978 - N so about 8.9% remains ln(.42) (c) t n ~. 20 7171 years old -0.00012097 6. A(t) = 2.3eo.0138629t 7. V (t) = 25el"(5 )t ~ 25e- 022314355t 8. (a) T (t) = 75 + 105e-0005005t (b) The roast would have cooled to 140°F in about 95 minutes. 9. From the graph, it appears that as x - 0+, y - oc. This is due to the presence of the ln(x) term in the function. This means Fritzy will never catch Chewbacca, which makes sense since Chewbacca has a head start and Fritzy only runs as fast as he does. y(x) = 4x2 _ lln(x) _  6.5 APPLICATIONS OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS 381 10. The steady state current is 2 amps. 13. The linear regression on the data below is y = 1.74899x + 0.70739 with r2 ~-0.999995. This is an excellent fit. x 1 2 3 4 5 6 7 8 9 10 ln(N(x)) 2.4849 4.1897 5.9454 7.6967 9.4478 11.1988 12.9497 14.7006 16.4523 18.2025 N(x) = 2.02869(5.74879)x = 2.02869ei.74899x with r2 0.999995. This is also an excellent fit and corresponds to our linearized model because ln(2.02869) 0.70739.  382 EXPONENTIAL AND LOGARITHMIC FUNCTIONS  CHAPTER 7 HOOKED ON CONICS 7.1 INTRODUCTION TO CONICS In this chapter, we study the Conic Sections - literally 'sections of a cone.' Imagine a double- napped cone as seen below being 'sliced' by a plane. If we slice the with a horizontal plane the resulting curve is a circle.  384 HOOKED ON CONICS Tilting the plane ever so slightly produces an ellipse. If the plane cuts parallel to the cone, we get a parabola. If we slice the cone with a vertical plane, we get a hyperbola. For a wonderful animation describing the conics as intersections of planes and cones, see Dr. Louis Talman's Mathematics Animated Website.  7.1 INTRODUCTION TO CONICS 385 If the slicing plane contains the vertex of the cone, we get the so-called 'degenerate' conics: a point, a line, or two intersecting lines. We will focus the discussion on the non-degenerate cases: circles, parabolas, ellipses, and hyperbo- las, in that order. To determine equations which describe these curves, we will make use of their definitions in terms of distances.  386 HOOKED ON CONICS 7.2 CIRCLES Recall from geometry that a circle can be determined by fixing a point (called the center) and a positive number (called the radius) as follows. DEFINITION 7.1. A circle with center (h, k) and radius r > 0 is the set of all points (x, y) in the plane whose distance to (h, k) is r. x,y) From the picture, we see that a point (x, y) is on the circle if and only if its distance to (h, k) is r. We express this relationship algebraically using the Distance Formula, Equation 1.1, as r = V(zx- h)2 +(y - k)2 By squaring both sides of this equation, we get an equivalent equation (since r > 0) which gives us the standard equation of a circle. EQUATION 7.1. The Standard Equation of a Circle: The equation of a circle with center (h, k) and radius r > 0 is (x- h)2 + (y - k)2 = r2. EXAMPLE 7.2.1. Write the standard equation of the circle with center (-2, 3) and radius 5. SOLUTION. Here, (h, k) = (-2, 3) and r = 5, so we get - (-2))2 + (y (x + 2)2 + (y 3)2 3)2 (5)2 25 El EXAMPLE 7.2.2. Graph (x + 2)2 + (y - 1)2 = 4. Find the center and radius. SOLUTION. From the standard form of a circle, Equation 7.1, we have that x + 2 is x - h, so h and y - 1 is y - k so k = 1. This tells us that our center is (-2, 1). Furthermore, r2 = 4, so r Thus we have a circle centered at (-2, 1) with a radius of 2. Graphing gives us -2 2.  7.2 CIRCLES 387 4- 3- x 1- -4 3 -2-1 -1- y 1 If we were to expand the equation in the previous example and gather up like terms, instead of the easily recognizable (x + 2)2 + (y - 1)2 = 4, we'd be contending with 2 + 4x + y2 - 2y + 1 = 0. If we're given such an equation, we can complete the square in each of the variables to see if it fits the form given in Equation 7.1 by following the steps given below. To Put a Circle into Standard Form 1. Group the same variables together on one side of the equation and put the constant on the other side. 2. Complete the square on both variables as needed. 3. Divide both sides by the coefficient of the squares. (For circles, they will be the same.) EXAMPLE 7.2.3. Complete the square to find the center and radius of 3x2 - 6x + 3y2 + 4y - 4 = 0. SOLUTION. 3x2 - 6x + 3y2 + 4y - 4 3x2 - 6x + 3y2 + 4y 3 (2 - 2x) + 3y2 + 4y) 3(2-2x+1) +3y2+-y +9-I 3(x - 1)2 + 3 y+ 2j) 3 2 2 (X - 1)2 + y + _ 3) 0 4 4 4 4+ 3(1) + 3 - 25 3 25 9 add 4 to both sides factor out leading coefficients complete the square in x, y factor divide both sides by 3 From Equation 7.1, we identify 3:- 1 as 3:- h, so h = 1, and y + 2 as y - k, so k center is (h, k)= (i,-). Furthermore, we see that r2 =92 so the radius is r - 2. Hence, the te It is possible to obtain equations like (3 - 3)2 + (y + 1)2 = 0 or (3 - 3)2 + (y + 1)2 = -1, neither of which describes a circle. (Do you see why not?) The reader is encouraged to think about what, if  388 HOOKED ON CONICS any, points lie on the graphs of these two equations. The next example uses the Midpoint Formula, Equation 1.2, in conjunction with the ideas presented so far in this section. EXAMPLE 7.2.4. Write the standard equation of the circle which has (-1, 3) and (2, 4) as the endpoints of a diameter. SOLUTION. We recall that a diameter of a circle is a line segment containing the center and two points on the circle. Plotting the given data yields ' / / 4- b 3- y \ -(h, k) / / 1 -2 -1 1 2 3 Since the given points are endpoints of a diameter, we know their the circle. Equation 1.2 gives us midpoint (h, k) is the center of (h, k) (X1 + 2 Y1 + Y2 2 2 -1+2 3+4) 2 A 2 1 7 2' 2 The diameter of the circle is the distance between the given points, so we know that half of the distance is the radius. Thus, r = (Yz2 - i)2 +(y2 - i)2 1 (2-(1)2 + (4 -3)2 -- 32 + 12 2 10 2 2 Finally, since 10 2 10 F y our answer becomes i 2 2 4 F-  7.2 CIRCLES 389 We close this section with the most important1 circle in all of mathematics: the Unit Circle. DEFINITION 7.2. The Unit Circle is the circle centered at (0, 0) with a radius of 1. The standard equation of the Unit Circle is 2 + y2 -=1. EXAMPLE 7.2.5. Find the points on the unit circle with y-coordinate . 2 SOLUTION. We replace y with - in the equation 2 + y2 = 1 to get 2 x 2 + 2 4+2 2 3: 1 4 x = i- 1 2 Our final answers are 2)and (-2 ). D 'While this may seem like an opinion, it is indeed a fact.  390 HOOKED ON CONICS 7.2.1 EXERCISES 1. Find the standard equation of the circle given the center and radius and sketch its graph. (a) Center (-1, -5), radius 10 (c) Center (r, e2), radius 3'91 (b) Center (-3, 1), radius j (d) Center (5, -9), radius ln(8) 2. Complete the square in order to put the equation into standard form. Identify the center and the radius or explain why the equation does not represent a circle. (a) x2 - 4x + y2 +10y =-25 (c) 4x2 + 4y2 - 24y + 36 = 0 (b) -2x2 - 36x - 2y2 - 56 = 0 (d) X2 + x+ y2 -52J 3. Find the standard equation of the circle with center (3, 5) which passes through (-1, -2). 4. Verify that the following 16 points all lie on the Unit Circle: (i1, 0), (0,1il), (i2, i±, (+,+) and (t,+t2) 5. The Giant Wheel at Cedar Point is a circle with diameter 128 feet which sits on an 8 foot tall platform making its overall height is 136 feet.2 Find an equation for the wheel assuming that its center lies on the y-axis. 6. Find an equation for the function represented graphically by the top half of the Unit Circle. Explain how the transformations is Section 1.8 can be used to produce a function whose graph is either the top or bottom of an arbitrary circle. 7. Circles and semi-circles have been used many times throughout the book to illustrate specific concepts. With the help of your classmates, find the Examples or Exercises in which a circle or semi-circle was used to discuss the following. (a) Symmetry (or lack thereof) (b) A graph which failed the Vertical Line Test (c) Local and absolute extrema (d) Transformations (e) Unusual steepness 8. Find a one-to-one function whose graph is half of a circle. (Hint: Think piecewise.) 2Source: Cedar Point's webpage.  7.2 CIRCLES39 391 7.2.2 ANSWERS 1. (a) (x+1)2+(y+5)2 =:100 y (c7(-)2 +( e2)2 2 913 (d x-5)2 + (y + 9)2 - (1n(8))2 yx 5-t I(8 5 5 +ln(8) ()(x+3)2+( 7)l2 1 y -9 + 1n(8)+ -9. x 5 2 x -9- ln(8)+ 2. (a) (x -2) 2+ (y +5)2-=4 Center (2, -5), radius rT:(+ )2 y 2 Center (-9, 0), radius rT:x-3) y - )2 =6 2 5 (c) x2 + (y - 3)2 =0 This is not a circle. Center (-i, ) 3)' radius r = 101 9.x + (y - 72)2 - 4096  392 HOOKED ON CONICS 7.3 PARABOLAS We have already learned that the graph of a quadratic function f(x) =ax2 + bx + c (a # 0) is called a parabola. To our surprise and delight, we may also define parabolas in terms of distance. DEFINITION 7.3. Let F be a point in the plane and D be a line not containing F. A parabola is the set of all points equidistant from F and D. The point F is called the focus of the parabola and the line D is called the directrix of the parabola. Schematically, we have the following. ' ' ' ' V ' ' ' D Each dashed line from the point F to a point on the curve has the same length as the dashed line from the point on the curve to the line D. The point suggestively labeled V is, as you should expect, the vertex. The vertex is the point on the parabola closest to the focus. We want to use only the distance definition of parabola to derive the equation of a parabola and, if all is right with the universe, we should get an expression much like those studied in Section 2.3. Let p denote the directed' distance from the vertex to the focus, which by definition is the same as the distance from the vertex to the directrix. For simplicity, assume that the vertex is (0, 0) and that the parabola opens upwards. Hence, the focus is (0, p) and the directrix is the line y = -p. Our picture becomes y (x,y) p) (0,0) y = -p (x,-p) From the definition of parabola, we know the distance from (0, p) to (x, y) is the same as the distance from (x, -p) to (x, y). Using the Distance Formula, Equation 1.1, we get 'We'll talk more about what 'directed' means later.  7.3 PARABOLAS 393 (x -0)2 + (y-p)2 x2 + (y - p)2 xz2 +(y -p)2 32 + y2 - 2py + p2 x2 V/1(X - - )2 + (y -( p) (y+p)2 (y+p)2 y2 + 2py + p2 4py square both sides expand quantities gather like terms Solving for y yields y =42, which is a quadratic function of the form found in Equation 2.4 with a = 4 and vertex (0,0). We know from previous experience that if the coefficient of x2 is negative, the parabola opens downwards. In the equation y = 4 this happens when p < 0. In our formulation, we say that p is a 'directed distance' from the vertex to the focus: if p > 0, the focus is above the vertex; if p < 0, the focus is below the vertex. The focal length of a parabola is |pl. What if we choose to place the vertex at an arbitrary point (h, k)? We can either use transformations (vertical and horizontal shifts from Section 1.8) or re-derive the equation from Definition 7.3 to arrive at the following. EQUATION 7.2. The Standard Equation of a Verticala Parabola: The equation of a (vertical) parabola with vertex (h, k) and focal length |pl is (x - h)2 = 4p(y - k) If p > 0, the parabola opens upwards; if p < 0, it opens downwards. aThat is, a parabola which opens either upwards or downwards. Notice that in the standard equation of the parabola above, only one of the variables, 3, is squared. This is a quick way to distinguish an equation of a parabola from that of a circle because in the equation of a circle, both variables are squared. EXAMPLE 7.3.1. Graph (x + 1)2 -8(y - 3). Find the vertex, focus, and directrix. SOLUTION. We recognize this as the form given in Equation 7.2. Here, 3 - h is 3 + 1 so h = -1, and y - k is y - 3 so k = 3. Hence, the vertex is (-1, 3). We also see that 4p = -8 so p = -2. Since p < 0, the focus will be below the vertex and the parabola will open downwards. The distance from the vertex to the focus is |pl = 2, which means the focus is 2 units below the vertex. If we start at (-1, 3) and move down 2 units, we arrive at the focus (-1, 1). The directrix, then, is 2 units above the vertex and if we move 2 units up from (-1, 3), we'd be on the horizontal line y = 5.  394 HOOKED ON CONICS y Of all of the information requested in the previous example, only the vertex is part of the graph of the parabola. So in order to get a sense of the actual shape of the graph, we need some more information. While we could plot a few points randomly, a more useful measure of how wide a parabola opens is the length of the parabola's latus rectum.2 The latus rectum of a parabola is the line segment parallel to the directrix which contains the focus. The endpoints of the latus rectum are, then, two points on 'opposite' sides of the parabola. Graphically, we have the following. the latus rectum F D It turns out3 that the length of the latus rectum is |4p|, which, in light of Equation 7.2, is easy to find. In our last example, for instance, when graphing (x + 1)2 = -8(y - 3), we can use the fact that the length of the latus rectum is | - 8| = 8, which means the parabola is 8 units wide at the focus, to help generate a more accurate graph by plotting points 4 units to the left and right of the focus. EXAMPLE 7.3.2. Find the standard form of the parabola with focus (2, 1) and directrix y = -4. SOLUTION. Sketching the data yields, 2No, I'm not making this up. 3Consider this an exercise to show what follows.  7.3 PARABOLAS 395 Y 1 * I x -1 1 3 -1 The vertex lies on this vertical line " -2 I midway between the focus and the directrix -3-I From the diagram, we see the parabola opens upwards. (Take a moment to think about it if you don't see that immediately.) Hence, the vertex lies below the focus and has an x-coordinate of 2. To find the y-coordinate, we note that the distance from the focus to the directrix is 1 - (-4) = 5, which means the vertex lies 5/2 units (halfway) below the focus. Starting at (2, 1) and moving down 5/2 units leaves us at (2, -3/2), which is our vertex. Since the parabola opens upwards, we know p is positive. Thus p = 5/2. Plugging all of this data into Equation 7.2 give us (x-2)2 - 4_5)Y(_3 (x-2)2 = 10 y+ ) 3D If we interchange the roles of x and y, we can produce 'horizontal' parabolas: parabolas which open to the left or to the right. The directrices4 of such animals would be vertical lines and the focus would either lie to the left or to the right of the vertex. A typical 'horizontal' parabola is sketched below. D4 V{ *F 4plural of 'directrix'  396 HOOKED ON CONICS EQUATION 7.3. The Standard Equation of a Horizontal Parabola: The equation of a (horizontal) parabola with vertex (h, k) and focal length |pl is (y - k)2 = 4p(x - h) If p > 0, the parabola opens to the right; if p < 0, it opens to the left. EXAMPLE 7.3.3. Graph (y - 2)2 = 12(x + 1). Find the vertex, focus, and directrix. SOLUTION. We recognize this as the form given in Equation 7.3. Here, x - h is x + 1 so h = -1, and y - k is y - 2 so k = 2. Hence, the vertex is (-1, 2). We also see that 4p = 12 so p = 3. Since p > 0, the focus will be the right of the vertex and the parabola will open to the right. The distance from the vertex to the focus is |plI= 3, which means the focus is 3 units to the right. If we start at (-1, 2) and move right 3 units, we arrive at the focus (2, 2). The directrix, then, is 3 units to the left of the vertex and if we move left 3 units from (-1, 2), we'd be on the vertical line x = -4. Since the length of the latus rectum is |4pl = 12, the parabola is 12 units wide at the focus, and thus there are points 6 units above and below the focus on the parabola. y 8. 6- 4- 3- 2- 1- * I I I I 1 I I I $3 a> -j - - - -2- -3. -4. 1 2 3 As with circles, not all parabolas will come to us in the forms in Equations 7.2 or 7.3. If we encounter an equation with two variables in which exactly one variable is squared, we can attempt to put the equation into a standard form using the following steps. To Put a Parabola into Standard Form 1. Group the variable which is squared on one side of the equation and put the non-squared variable and the constant on the other side. 2. Complete the square if necessary and divide by the coefficient of the perfect square. 3. Factor out the coefficient of the non-squared variable from it and the constant.  7.3 PARABOLAS 397 EXAMPLE 7.3.4. Consider the equation y2 + 4y + 8x = 4. Put this equation into standard form and graph the parabola. Find the vertex, focus, and directrix. SOLUTION. We need to get a perfect square (in this case, using y) on the left-hand side of the equation and factor out the coefficient of the non-squared variable (in this case, the x) on the other. y2 + 4y + 8x y2 + 4y y2 + 4y + 4 (y + 2)2 (y + 2)2 4 -8x +4 -8x + 4 + 4 complete the square in y only -8x + 8 factor -8(x - 1) Now that the equation is in the form given in Equation 7.3, we see that x - h is x -i1 so h = 1, and y - k is y + 2 so k = -2. Hence, the vertex is (1, -2). We also see that 4p = -8 so that p = -2. Since p < 0, the focus will be the left of the vertex and the parabola will open to the left. The distance from the vertex to the focus is |plI= 2, which means the focus is 2 units to the left of 1, so if we start at (1, -2) and move left 2 units, we arrive at the focus (-1, -2). The directrix, then, is 2 units to the right of the vertex, so if we move right 2 units from (1, -2), we'd be on the vertical line x= 3. Since the length of the latus rectum is |4pl is 8, the parabola is 8 units wide at the focus, so there are points 4 units above and below the focus on the parabola. -2 -1 -1- *-2. -3- -4- -5 -6- y 12 -*3: In studying quadratic functions, we have seen parabolas used to model physical phenomena such as the trajectories of projectiles. Other applications of the parabola concern its 'reflective property' which necessitates knowing about the focus of a parabola. For example, many satellite dishes are formed in the shape of a paraboloid of revolution as depicted below.  398 HOOKED ON CONICS Every cross section through the vertex of the paraboloid is a parabola with the same focus. To see why this is important, imagine the dashed lines below as electromagnetic waves heading towards a parabolic dish. It turns out that the waves reflect off the parabola and concentrate at the focus which then becomes the optimal place for the receiver. If, on the other hand, we imagine the dashed lines as emanating from the focus, we see that the waves are reflected off the parabola in a coherent fashion as in the case in a flashlight. Here, the bulb is placed at the focus and the light rays are reflected off a parabolic mirror to give directional light. x F / / / EXAMPLE 7.3.5. A satellite dish is to be constructed in the shape of a paraboloid of revolution. If the receiver placed at the focus is located 2 ft above the vertex of the dish, and the dish is to be 12 feet wide, how deep will the dish be? SOLUTION. One way to approach this problem is to determine the equation of the parabola sug- gested to us by this data. For simplicity, we'll assume the vertex is (0, 0) and the parabola opens upwards. Our standard form for such a parabola is x2 = 4py. Since the focus is 2 units above the vertex, we know p = 2, so we have x2 = 8y. Visually,  7.3 PARABOLAS 399 y t 12 units wide (6,y) Since the parabola is 12 feet wide, we know the edge is 6 feet from the vertex. To find the depth, we are looking for the y value when x= 6. Substituting x = 6 into the equation of the parabola yields 62 = 8y or y = 36/8 = 9/2 = 4.5. Hence, the dish will be 9/2 or 4.5 feet deep. Q  400 HOOKED ON CONICS 7.3.1 EXERCISES 1. Sketch the graph of the given parabola. Find the vertex, focus and directrix. Include the endpoints of the latus rectum in your sketch. (a) (y - 2)2 = -12(x+ 3) (b) (y+4)2=4x (c) (z-3)2: (d) (x+ )2 -16y 2 (y +) 2. Put the equation into standard form and identify the vertex, focus and directrix. (a) y2 - 10y - 27x + 133 = 0 (b) 25x2 + 20x + 5y - 1 = 0 (c) 2 +2x- 8y+49 = 0 (d) 2y2 + 4y + x - 8 = 0 3. Find an equation for the parabola which fits the given criteria. (a) Vertex (7, 0), focus (0, 0) (b) Vertex (-8, -9), Both (0, 0), (-16,0) are points on the curve (c) Focus (10, 1), directrix x = (d) The endpoints of latus (-2, -7) and (-4, -7) 5 rectum are 4. The graphs of vertical parabolas represent y as a function of x, but those of horizontal parabolas do not because a parabola which open to the left or right fails the Vertical Line Test, Theorem 1.1. However, the top half or bottom half of a horizontal parabola does form the graph of a function which we have studied thoroughly. To what functions am I referring and where in the text have they appeared? Summarize what you should know about these functions. How could we use the transformations in Section 1.8 to better understand the graphs of horizontal parabolas? 5. With the help of your classmates, research spinning liquid mirrors. To get you started, check out this website.  7.3 PARABOLAS 7.3.2 ANSWERS 401 1. (a) Vertex (-3, 2), focus (-6, 2), directrix x = 0 Endpoints of latus rectum (-6, 8), (-6, -4) (c) Vertex (3, 0), focus (3, -4), directrix y = 4 Endpoints of latus rectum (-5, -4), (11, -4) y y 3 2 -- -5-4-3-2-1-- 234 7 01 . -2. -3. -4. - - - . * (d) Vertex (-3, -2), focus (- ,-2) directrix y = -3 Endpoints of latus rectum (- - 2, -2), (-ij, -2) (b) Vertex (0, -4), focus (1, -4) directrix x = -1 Endpoints of latus rectum (1, -2), (1, -6) y - -2. -3- -4- -5- -6. -7- - y 1 2 4 2. (a) (y - 5)2 = 27(x- 4) Vertex (4, 5), focus (1, 5) directrix x = -Q4 (b) (19) y ) Vertex (-j1), focus (-j, ) directrix y _1 (c) (x + 1)2 = 8(y - 6) Vertex (-1, 6), focus (-1, 8) directrix y = 4 (d) (y + 1)2 - - j(z - 10) Vertex (10, -1), focus (k,- directrix x =81 1) 3. (a) (b) (c) y2 = -28(x (x + 8)2 = (y - 1)2 = 1 -7) 4(y + 9) 0O (x - 25) (d) (x (I 1)2 1)2 6(y+ 2) or -6(y + )  402 HOOKED ON CONICS 7.4 ELLIPSES In the definition of a circle, Definition 7.1, we fixed a point called the center and considered all of the points which were a fixed distance r from that one point. For our next conic section, the ellipse, we fix two distinct points and a distance d to use in our definition. DEFINITION 7.4. Given two distinct points F1 and F2 in the plane and a fixed distance d, an ellipse is the set of all points (x, y) in the plane such that the sum of the distance from F1 to (x, y) and the distance from F2 to (x, y) is d. The points F1 and F2 are called the foci" of the ellipse. "the plural of 'focus' di + d2 = d for all (x, y) on the ellipse We may imagine taking a length of string and anchoring it to two points on a piece of paper. The curve traced out by taking a pencil and moving it so the string is always taut is an ellipse. The center of the ellipse is the midpoint of the line segment connecting the two foci. The major axis of the ellipse is the line segment connecting two opposite ends of the ellipse which also contains the center and foci. The minor axis of the ellipse is the line segment connecting two opposite ends of the ellipse which contains the center but is perpendicular to the major axis. The vertices of an ellipse are the points of the ellipse which lie on the major axis. Notice that the center is also the midpoint of the major axis, hence it is the midpoint of the vertices. In pictures we have,  7.4 ELLIPSES 403 Major Axis V1.......................................... V2 F1 F2 An ellipse with center C; foci F1, F2; and vertices V1, V2 Note that the major axis is the longer of the two axes through the center, and likewise, the minor axis is the shorter of the two. In order to derive the standard equation of an ellipse, we assume that the ellipse has its center at (0, 0), its major axis along the x-axis, and has foci (c, 0) and (-c, 0) and vertices (-a, 0) and (a, 0). We will label the y-intercepts of the ellipse as (0, b) and (0, -b) (We assume a, b, and c are all positive numbers.) Schematically, y x 0) (-a,0) Note that since (a, 0) is on the ellipse, it must satisfy the conditions of Definition 7.4. That is, the distance from (-c, 0) to (a, 0) plus the distance from (c, 0) to (a, 0) must equal the fixed distance d. Since all of these points lie on the x-axis, we get distance from (-c, 0) to (a, 0) + distance from (c, 0) to (a, 0) (a + c) + (a - c) 2a d d d  404 HOOKED ON CONICS In other words, the fixed distance d mentioned in the definition of the ellipse is none other than the length of the major axis. We now use that fact (0, b) is on the ellipse, along with the fact that d= 2a to get distance from (-c, 0) to (0, b) + distance from (c, 0) to (0, b) (0- (-c))2+(b-0)2+ (0-c)2+(b-0)2 b2 + c2 + b2 + c2 2 b2 + c2 b2 + c2 2a 2a 2a 2a a From this, we get a2 = b2 + c2, or b2 = a2 - c2, whici (x, y) on the ellipse. Applying Definition 7.4, we get h will prove useful later. Now consider a point distance from (-c, 0) to (x, y) + distance from (c, 0) to (x, y) (x-(-c))2+(y-0)2+ (x-c)2+(y-0)2 (xz+ c)2 + y2+ (xz-c)2+y2 2a 2a 2a In order to make sense of this situation, rearranging.I we need to do some rearranging, squaring, and more (x+c)2+ y2 + (x- c)2+y2 ( z(x + c)2 + y2 (x +c)2 +y2) (X + c)2 + y2 4a (x- c)2 + y2 4a (x- c)2 + y2 a (x-c)2+y2 (a (x - c)2 + y2) a2 ((X - c)2 + y2) a2x2 - 2a2cx + a2c2 + a2y2 a2x2 - c2x2 + a2 2 (a2 - c2) x2 + a 2 y2 2a 2a - 2 (x - c)2 + y2 2 (2a - (x - c)2 + y2) 4a2-4a (x-c)2+y2+(x 4a2 + (x - c)2 - (x + c)2 4a2 - 4cx a2 - cx (a2 - cx)2 a4 - 2a cx + c2x2 a4 - 2a cx + c2x2 a4 -C2C2 a - ac2 a2 (a2 - c2) c)2 + y2 We are nearly finished. Recall that b2 = a2 c2 so that (a2 - c2)x2 + a 2 y2 b2x2 + a2 y2 2 2 a2 b2 a2 (a2 - c2) a2b2 1 'In other words, tons and tons of Intermediate Algebra. Stay sharp, this is not for the faint of heart.  7.4 ELLIPSES 405 This equation is for an ellipse centered at the origin. If the ellipse were centered at a point (h, k), we would get the following EQUATION 7.4. The Standard Equation of an Ellipse: For positive unequal numbers a and b, the equation of an ellipse with center (h, k) is (x-h)2 (y - k)2 a2 + = 1 a2 b2 Some remarks about Equation 7.4 are in order. First note that the values a and b determine how far in the x and y directions, respectively, one counts from the center to arrive at points on the ellipse. Also take note that if a > b, then we have an ellipse whose major axis is horizontal, and hence, the foci lie to the left and right of the center. In this case, as we've seen in the derivation, the distance from the center to the focus, c, can be found by c = a2 - b2. If b > a, the roles of the major and minor axes are reversed, and the foci lie above and below the center. In this case, c =Vb2 - a2. In either case, c is the distance from the center to each focus, and c = bigger denominator - smaller denominator. Finally, it is worth mentioning that if we take the standard equation of a circle, Equation 7.1, and divide both sides by r2, we get EQUATION 7.5. The Alternate Standard Equation of a Circle: The equation of a circle with center (h, k) and radius r > 0 is (x-h)2 (y - k)2 Notice the similarity between Equation 7.4 and Equation 7.5. Both equations involve a sum of squares equal to 1; the difference is that with a circle, the denominators are the same, and with an ellipse, they are different. If we take a transformational approach, we can consider both equations as shifts and stretches of the Unit Circle 2 + y2 = 1 in Definition 7.2. Replacing x with (x - h) and y with (y - k) causes the usual horizontal and vertical shifts. Replacing x with b and y with b causes the usual vertical and horizontal stretches. In other words, it is perfectly fine to think of an ellipse as the deformation of a circle in which the circle is stretched farther in one direction than the other. (x + 1)2 (y - 2)2 EXAMPLE 7.4.1. Graph + = 1. Find the center, the lines which contain the major 9 25 and minor axes, the vertices, and the foci. SOLUTION. We see that this equation is in the standard form of Equation 7.4. Here x - h is x + 1 so h =-1, and y - k is y - 2 so k = 2. Hence, our ellipse is centered at (-1, 2). We see that a2 - 9 so a = 3, and b2 = 25 so b = 5. This means that we move 3 units left and right from the center and 5 units up and down from the center to arrive at points on the ellipse. As an aid to sketching, we draw a rectangle matching this description, called a guide rectangle, and sketch the ellipse inside this rectangle as seen below on the left.  406 HOOKED ON CONICS y x x Since we moved farther in the y direction than in the x direction, the major axis will lie along the vertical line x= -1, which means the minor axis lies along the horizontal line, y = 2. The vertices are the points on the ellipse which lie along the major axis so in this case, they are the points (-1, 7) and (-1, -3). (Notice that these are two of the points we plotted when drawing the ellipse.) To find the foci, we find c =v25 - 9 =v16= 4, which means the foci lie 4 units from the center. Since the major axis is vertical, the foci lie 4 units above and below the center, at (-1, -2) and (-1, 6). Plotting all this information gives the graph seen above on the right. Q EXAMPLE 7.4.2. Find the equation of the ellipse with foci (2, 1) and (4, 1) and vertex (0, 1). SOLUTION. Plotting the data given to us, we have y 1{ * * I 1 1 1 1 > 1 2 3 4 5 X From this sketch, we know that the major axis is horizontal, meaning a > b. Since the center is the midpoint of the foci, we know it is (3, 1). Since one vertex is (0, 1) we have that a = 3, so a2 9. All that remains is to find b2. To that end, we use the fact that c = 1 to get  7.4 ELLIPSES 407 c = a2 - b2 1 = 9-b2 12 = ( 9-b2) 1 = 9-b2 b2 = 8 Substituting all of our findings into the equation (xa2 +(yb2 t e(x-3)2+(y 12 _ 1 tobe ( +) (y1.) 9 8 1, we get our final answer D As with circles and parabolas, an equation may be given which is an ellipse, but isn't in the standard form of Equation 7.4. In those cases, as with circles and parabolas before, we will need to massage the given equation into the standard form. To Put an Ellipse into Standard Form 1. Group the same variables together on one side of the equation and put the constant on the other side. 2. Complete the square in both variables as needed. 3. Divide both sides by the constant term so that the constant on the other side of the equation becomes 1. EXAMPLE 7.4.3. Graph 2 + 4y2 - 2x + 24y + 33 = 0. Find the center, the lines which contain the major and minor axes, the vertices, and the foci. SOLUTION. Since we have a sum of squares and the squared terms have unequal coefficients, it's a good bet we have an ellipse on our hands. (A parabola would have only one squared variable and a circle would have the same coefficient on the squared terms.) We need to complete both squares, and then divide, if necessary, to get the right-hand side equal to 1. x2 + 4y2 - 2x + 24y + 33 x2 - 2x + 4y2 + 24y x2 - 2x + 4 (y2 + 6y) (z2-2x+1) +4(y2+6y+9) (I- 1)2 + 4(y + 3)2 (I- 1)2 + 4(y + 3)2 4 (x - 1)2 (x-12+ (y + 3)2 (x-1)2 (y+ 3)2 4 + 1 0 -33 -33 -33 + 1+ 4(9) 4 4 4 1 1  408 HOOKED ON CONICS Now that this equation is in the standard form of Equation 7.4, we see that x - h is x - 1 so h = 1, and y - k is y + 3 so k = -3. Hence, our ellipse is centered at (1, -3). We see that a2 = 4 so a = 2, and b2 = 1 so b = 1. This means we move 2 units left and right from the center and 1 unit up and down from the center to arrive at points on the ellipse. Since we moved farther in the x direction than in the y direction, the major axis will lie along the horizontal line y = -3, which means the minor axis lies along the vertical line x = 1. The vertices are the points on the ellipse which lie along the major axis so in this case, they are the points (-1, -3) and (3, -3). To find the foci, we find c = v4 - 1 =v/3, which means the foci lie 3 units from the center. Since the major axis is horizontal, the foci lie 3 units to the left and right of the center, at (1 - v/3, -3) and (1+ 3/, -3). Plotting all of this information gives Y 1 2 3 4 X -1- -2 -4- As you come across ellipses in the homework exercises and in the wild, you'll notice they come in all shapes in sizes. Compare the two ellipses below. Certainly, one ellipse is 'rounder' than the other. This notion of roundness is quantified below. DEFINITION 7.5. The eccentricity of an ellipse, denoted e, is the following ratio: distance from the center to a focus edistance from the center to a vertex In an ellipse, the foci are closer to the center than the vertices, so 0 < e < 1. For the ellipse above on the left, e ~0.98; for the ellipse on the right, e ~0.66. In general, the closer the eccentricity is to 0, the more 'circular' the ellipse; the closer the eccentricity is to 1, the more 'eccentric' the ellipse.  7.4 ELLIPSES 409 EXAMPLE 7.4.4. Find the equation of the ellipse whose vertices are (0, i5) with eccentricity e =4. SOLUTION. As before, we plot the data given to us y x From this sketch, we know that the major axis is vertical, meaning b > a. With the vertices located at (0, ±5), we get b= 5 so b- 25. We also know that the center is (0, 0) because the center is the midpoint of the vertices. All that remains is to find a2. To that end, we use the fact that the eccentricity e 1 which means distance from the center to a focus c c 1 e distance from the center to a vertex b 5 4 from which we get c = . To get a2, we use the fact that c=V2 - a2 c - 5 - 25 -a2 i(4) ( 25~a2)2 25 16- 25- a2 a2 =25 --- 2 7516 a 1- ( 22 Substituting all of our findings into the equation v a2l, m b> yies ourtin late eccentricityelde =uwhichl means 16(x-3c)2 (Y=b1)2 +5 375525 As it paablas elipeshav arefeciv proprty -fw mgn h dse ie o ason waes he tewaeseantigfrm n fcu efet ffte opo te lipe n hadtwad theothr ocu. uchgeoety i eplote in th cosrcinoa2-ald'hsern alre' Ifa eso hipesatoe ocs aprsn tndn a heote fcs il ea hefis prona if tey wre tandng ightnex to hem  410 HOOKED ON CONICS -/ */ Fl EXAMPLE 7.4.5. Jamie and Jason want to exchange secrets (terrible secrets) from across a crowded whispering gallery. Recall that a whispering gallery is a room which, in cross section, is half of an ellipse. If the room is 40 feet high at the center and 100 feet wide at the floor, how far from the outer wall should each of them stand so that they will be positioned at the foci of the ellipse? SOLUTION. Graphing the data yields 100 units wide It's most convenient to imagine this ellipse centered at (0,0). Since the ellipse is 100 units wide and 40 units tall, we get a = 50 and b = 40. Hence, our ellipse has the equation 502 + 402 = 1. We're looking for the foci, and we get c = V502 - 402 = V90-0= 30, so that the foci are 30 units from the center. That means they are 50 - 30 = 20 units from the vertices. Hence, Jason and Jamie should stand 20 feet from opposite ends of the gallery. D  7.4 ELLIPSES 411 7.4.1 EXERCISES 1. Graph the ellipse. Find the center, the lines which contain the major and minor axes, the vertices, the foci and the eccentricity. z2 Y 2 (x + 5)2 (y _ 4)2 (a) + =y 1 (c) (1+ + = 169 25 16 1 (b) (x-2)2 (y + 3)2 (d) - 1)2 (y - 3)2 4 9 10 11 2. Put the equation in standard form. Find the center, the lines which contain the major and minor axes, the vertices, the foci and the eccentricity. (a) 12x2 +3y2 - 30y + 39 = 0 (c) x2 - 2x + 2y2 - 12y + 3 = 0 (b) 5x2 + 18y2 - 30x + 72y + 27 = 0 (d) 9x2 +4y2 - 4y - 8 = 0 3. Find the standard form of the equation of the ellipse which has the given properties. (a) Center (3, 7), Vertex (3, 2), Focus (3, 3) (b) All points on the ellipse are in Quadrant IV except (0, -9) and (8, 0)2 (c) Foci (0, i4), Point on curve (2, j) 1 (d) Vertex (-10, 5), Focus (-2, 5), Eccentricity - 2 4. The Earth's orbit around the sun is an ellipse with the sun at one focus and eccentricity e ~ 0.0167. The length of the semimajor axis (that is, half of the major axis) is defined to be 1 astronomical unit (AU). The vertices of the elliptical orbit are given special names: 'aphelion' is the vertex farther from the sun, and 'perihelion' is the vertex closest to the sun. Find the distance in AU between the sun and aphelion and the distance in AU between the sun and perihelion. 5. The graph of an ellipse clearly fails the Vertical Line Test, Theorem 1.1, so the equation of an ellipse does not define y as a function of x. However, much like with circles and horizontal parabolas, we can split an ellipse into a top half and a bottom half, each of which would indeed represent y as a function of x. With the help of your classmates, use your calculator to graph the ellipses given in Exercise 1 above. What difficulties arise when you plot them on the calculator? 6. With the help of your classmates, research whispering galleries and other ways ellipses have been used in architecture and design. 7. With the help of your classmates, research "extracorporeal shock-wave lithotripsy". It uses the reflective property of the ellipsoid to dissolve kidney stones. 2One might also say that the ellipse is "tangent to the axes" at those two points.  412 HOOKED ON CONICS 7.4.2 ANSWERS 92 2 1. (a)1 2 169 25 Center (0,0) Major axis along y = 0 Minor axis along x= 0 Vertices (13, 0), (-13, 0) Foci (12, 0), (-12, 0) 12 1e y -1 -7 -2 -3. -4. -3 1= (b) (x-2)2+(y + 3)2 4 9 Center (2, -3) Major axis along x= Minor axis along y Vertices (2, 0), (2,-6) Foci (2, -3 + 5), (2, 5 y 1 2 -3 -3 x 5) (x +5)2 (y -4)2 (c) 16 1 Center (-5, 4) Major axis along y = 4 Minor axis along x= -5 Vertices (-9, 4), (-1, 4) Foci (-5 + 15, 4), (-5 - 15 x 4. 3. 2 1. I 15, 4) i i i i i i i i i i 9 -8 -7 -6 -5 -4 -3 -2 -1 x e 4 (Xz-1) 2 (y - 3)2 (d) 10 11 Center (1, 3) Major axis along x 1 Minor axis along y = 3 Vertices (1,3+ /II), (1,3 Foci (1, 2), (1, 4) 1 e 11 y 11)  7.4 ELLIPSES 413 2 (y - 5)2 2. (a) -+ 12 3 12 Center (0, 5) Major axis along x= 0 Minor axis along y = 5 Vertices (0, 5 - 12), (0, 5 + 12) Foci (0,2),(0,8) e = 12 (b) (Xz- 3)2+(y + 2)2 1 () 18 5 Center (3, -2) Major axis along y -2 Minor axis along x= 3 Vertices (3 - 18, -2), (3 + 18, -2) Foci (3 - 13, -2), (3 + 13, -2) 13 e 18 ((X-1)2 (y - 3)2 16 8 Center (1, 3) Major Axis along y = Minor Axis along x= Vertices (5, 3), (-3, 3) Foci(1+2 2/,3), (1- e- - 1 3 1 2 , 3) 2 4 (y1)2 (d) T+ 9 Center (0, j) Major Axis along x= 0 (the y-axis) Minor Axis along y = 2 Vertices (0, 2), (0,-1) Foci (0, 1+2 , ( 2 e-3 3. (a) (x - 3)2 (y _72 9 25 ()(x - 8)2 (y + 9)2 (b) 64 81 (c) + =1 9 25 (Xz- 6)2 (y - 5)2 (d) 256 192 4. Distance from the sun to aphelion 1.0167 AU. Distance from the sun to perihelion 0.9833 AU.  414 HOOKED ON CONICS 7.5 HYPERBOLAS In the definition of an ellipse, Definition 7.4, we fixed two points called foci and looked at points whose distances to the foci always added to a constant distance d. Those prone to syntactical tinkering may wonder what, if any, curve we'd generate if we replaced added with subtracted. The answer is a hyperbola. DEFINITION 7.6. Given two distinct points F1 and F2 in the plane and a fixed distance d, a hyperbola is the set of all points (x, y) in the plane such that the absolute value of the difference of the distances between the foci and (x, y) is d. The points F1 and F2 are called the foci of the hyperbola. '- 1F2 ('2, Y2) In the figure above: the distance from F1 to (x1, yi) - the distance from F2 to (xi, Yi) d and the distance from F2 to (x2, Y2) - the distance from F1 to (x2, y2) d Note that the hyperbola has two parts, called branches. The center of the hyperbola is the midpoint of the line segment connecting the two foci. The transverse axis of the hyperbola is the line segment connecting two opposite ends of the hyperbola which also contains the center and foci. The vertices of a hyperbola are the points of the hyperbola which lie on the transverse axis. In addition, we will show momentarily that there are lines called asymptotes which the branches of the hyperbola approach for large x and y values. They serve as guides to the graph. In pictures,  7.5 HYPERBOLAS 415 r -r \ / \ / /\ V F / \ / \ / \ \ / Transverse Axis A hyperbola with center C; foci F1, F2; and vertices V1, V2 and asymptotes (dashed) Before we derive the standard equation of the hyperbola, we need to discuss one further parameter, the conjugate axis of the hyperbola. The conjugate axis of a hyperbola is the line segment through the center which is perpendicular to the transverse axis and has the same length as the line segment through a vertex which connects the asymptotes. In pictures we have \ / Note that in the diagram, we can construct a rectangle using line segments with lengths equal to thep lengtfhsof fthe tra~nsverse a~nd conjugate axes- whose center ipe cnter of fth hyperbola aqnd whoe dagnal ar cntanedintheasypttes Ths uid retagle whchis erysiila t theon wecrate i te Scton .4tohel u grphellpsswil ad u i gaphnghyprbla whenthe ime omes Supos w wshtoderveth euaio ofa yprbla Fr smpicty w sallasum tatth ceneris(0 0, hevetiesar (, ) nd(-, ) ndth fciar (, ) nd(-, ).Welaelth  416 HOOKED ON CONICS endpoints of the conjugate axis (0, b) and (0, -b). (Although b does not enter into our derivation, we will have to justify this choice as you shall see later.) As before, we assume a, b, and c are all positive numbers. Schematically we have Since (a, 0) is on the hyperbola, it must satisfy the conditions of Definition 7.6. That is, the distance from (-c, 0) to (a, 0) minus the distance from (c, 0) to (a, 0) must equal the fixed distance d. Since all these points lie on the x-axis, we get distance from (-c, 0) to (a, 0) - distance from (c, 0) to (a, 0) (a + c) - (c - a) d d 2a = d In other words, the fixed distance d from the definition of the hyperbola is actually the length of the transverse axis! (Where have we seen that type of coincidence before?) Now consider a point (x, y) on the hyperbola. Applying Definition 7.6, we get distance from (-c, 0) to (x, y) - distance from (c, 0) to (x, y) (x-(-c))2+(y-0)2- (x-c)2+(y-0)2 (x+c)2+y2- (x-c)2+y2 2a 2a 2a Using the same arsenal of Intermediate Algebra weaponry we used in deriving the standard formula of an ellipse, Equation 7.4, we arrive at the following.' 'It is a good exercise to actually work this out.  7.5 HYPERBOLAS 417 (az - cz) zz + azyz2 az (az - cz) What remains is to determine the relationship between a, b and c. To that end, we note that since a and c are both positive numbers with a < c, we get a2 < c2 so that a2 - c2 is a negative number. Hence, c2 - a2 is a positive number. For reasons which will become clear soon, we re-write the equation by solving for y2/x2 to get (a2 - c2) 92 + a2y2 (c2 - a2) 2 + a2y2 a2,2 y2 X2 a2 (a2 - c2) -a2 (c2 - a2) (c2 - a2) x2 - (c2- a2) ( a2 a2 (c2 - a2) c2 - a2) x2 (c2 - a2) Y2 (c2 - a2) As x and y attain very large values, the quantity 2 - 0 so that 2 2 . By setting y 2 b b b2 = c2 - a2 we get x2 a 2. This shows that y - i-b as |x| grows large. Thus y i-x are the x 2a a asymptotes to the graph as predicted and our choice of labels for the endpoints of the conjugate axis is justified. In our equation of the hyperbola we can substitute a2 - c2 = -b2 which yields (a2 - c2) x2 + a2y2 -b22 + a2 y2 a2 b2 a2:by a2 (a2 - c2) -a2b2 1 The equation above is for a hyperbola whose center is the origin and which opens to the left and right. If the hyperbola were centered at a point (h, k), we would get the following. EQUATION 7.6. The Standard Equation of a Horizontal' Hyperbola For positive numbers a and b, the equation of a horizontal hyperbola with center (h, k) is (x-h)2 _-(y - k)2 a2 b2 aThat is, a hyperbola whose branches open to the left and right If the roles of x and y were interchanged, then the hyperbola's branches would open upwards and downwards and we would get a 'vertical' hyperbola. EQUATION 7.7. The Standard Equation of a Vertical Hyperbola For positive numbers a and b, the equation of a vertical hyperbola with center (h, k) is: (y-k)2_ (x-h)2 2 - a = 1 b2 a2 The values of a and b determine how far in the x and y directions, respectively, one counts from the center to determine the rectangle through which the asymptotes pass. In both cases, the distance  418 HOOKED ON CONICS from the center to the foci, c, as seen in the derivation, can be found by the formula c = a2 + b2. Lastly, note that we can quickly distinguish the equation of a hyperbola from that of a circle or ellipse because the hyperbola formula involves a difference of squares where the circle and ellipse formulas both involve the sum of squares. EXAMPLE 7.5.1. Graph the equation (x=2)2- 1. Find the center, the lines which contain 4 25 the transverse and conjugate axes, the vertices, the foci and the equations of the asymptotes. SOLUTION. We first see that this equation is given to us in the standard form of Equation 7.6. Here x - h is x - 2 so h = 2, and y - k is y so k = 0. Hence, our hyperbola is centered at (2,0). We see that a2 = 4 so a = 2, and b2 = 25 so b = 5. This means we move 2 units to the left and right of the center and 5 units up and down from the center to arrive at points on the guide rectangle. The asymptotes pass through the center of the hyperbola as well as the corners of the rectangle. This yields the following set up. y 5 ......... .. 4- - 3-- 2-- 1 \ / -H-H- -H-*\H-+-H/h- -2\ / -2-1 1/2, 3 45 6 X -1- -2- / -3- -4/\ - --- Since the y2 term is being subtracted from the 2 term, we know that the branches of the hyperbola open to the left and right. This means that the transverse axis lies along the x-axis. Hence, the conjugate axis lies along the vertical line x= 2. Since the vertices of the hyperbola are where the hyperbola intersects the transverse axis, we get that the vertices are 2 units to the left and right of (2, 0) at (0, 0) and (4, 0). To find the foci, we need c = a2 + b2 = 4 + 25 =29. Since the foci lie on the transverse axis, we move 29 units to the left and right of (2, 0) to arrive at (2 - 29, 0) (approximately (-3.39, 0)) and (2 + 29, 0) (approximately (7.39, 0)). To determine the equations of the asymptotes, recall that the asymptotes go through the center of the hyperbola, (2, 0), as well as the corners of guide rectangle, so they have slopes of i b = i5.Uigtepitsoeeuto of a line, Equation 2.2, yields  7.5 HYPERBOLAS 419 5 y = ( - 2)+0, 2 so we get y = 2z - 5 and y =_-2x + 5. Putting it all together, we get y 5 ...............1 \/ 1- -3 -2 -1 1 3 5 6 7 X -1. -2 - F7 EXAMPLE 7.5.2. Find the equation of the hyperbola with asymptotes y +2x and vertices (+5,0). SOLUTION. Plotting the data given to us, we have y - \ 5 - / \-/ \ ../ \ / \ / -5 /. 5 x / ..\ / \ / \ / . /-5- This graph not only tells us that the branches of the hyperbola open to the left and to the right, it also tells us that the center is (0, 0). Hence, our standard form is a2 b2-  420 HOOKED ON CONICS Since the vertices are (+5, 0), we have a = 5 so a2 = 25. In order to determine b2, we recall that the slopes of the asymptotes are t-j. Since a = 5 and the slope of the line y = 2x is 2, we have that j = 2, so b = 10. Hence, b2 = 100 and our final answer is x2 2 - =1 25 100 As with the other conic sections, an equation whose graph is a hyperbola may be given in a form other than the standard forms in Equations 7.6 or 7.7. In those cases, as with conic sections which have come before, we will need to massage the given equation into one of the forms in Equations 7.6 or 7.7. To Put a Hyperbola into Standard Form 1. Group the same variables together on one side of the equation and put the constant on the other side 2. Complete the square in both variables as needed 3. Divide both sides by the constant term so that the constant on the other side of the equation becomes 1 EXAMPLE 7.5.3. Consider the equation 9y2 - 2 - 6x = 10. Put this equation in to standard form and graph. Find the center, the lines which contain the transverse and conjugate axes, the vertices, the foci, and the equations of the asymptotes. SOLUTION. We need only complete the square on the x, and then divide, if necessary, to get the right-hand side equal to 1. 9y2 - z2 - 6x = 10 9y2-1(x2+6x) = 10 9y2 - (x2 + 6x + 9) = 10-1(9) 9y2 - ( + 3)2 1 y2 (x+3)2 1 - 1 = 1 1 9 Now that this equation is in the standard form of Equation 7.7, we see that x - h is x + 3 so h = -3, and y - k is y so k = 0. Hence, our hyperbola is centered at (-3,0). We find that a2 - 1 so a = 1, and b2 = 0so b =1j. This means that we move 1 unit to the left and right of the center and 1/3 units up and down from the center to arrive at points on the guide rectangle. Since the 2 term is being subtracted from the y2 term, we know the branches of the hyperbola open upwards and downwards. This means the transverse axis lies along the vertical line z = -3 and the conjugate axis lies along the c-axis. Since the vertices of the hyperbola are where the hyperbola intersects the transverse axis, we get that the vertices are 1 of a unit above and below (-3, 0) at  7.5 HYPERBOLAS 421 (-3, j) and (-3, -)). To find the foci, we need c =va2 + b2 =v5 + 1 = 10 Since the foci lie on the transverse axis, we move10 units above and below (-3, 0) to arrive at (-3, 3) and (-3, - ). To determine the asymptotes, recall that the asymptotes go through the center of the hyperbola, (-3, 0), as well as the corners of guide rectangle, so they have slopes of b= j. Using the point-slope equation of a line, Equation 2.2, we get y =jx+1 and y -}x -1. Putting it all together, we get Y "c * _1 I I * > -6 ... -1 4 * -1 Hyperbolas can be used in positioning problems, as the next example illustrates. EXAMPLE 7.5.4. Jeff is stationed 10 miles due west of Carl in an otherwise empty forest in an attempt to locate an elusive Sasquatch. At the stroke of midnight, Jeff records a Sasquatch call 9 seconds earlier than Carl. If the speed of sound that night is 760 miles per hour, determine a hyperbolic path along which Sasquatch must be located. SOLUTION. Since Jeff hears Sasquatch sooner, it is closer to Jeff than it is to Carl. Since the speed of sound is 760 miles per hour, we can determine how much closer Sasquatch is to Jeff by multiplying miles 1 hour 760 x x 9 seconds = 1.9 miles hour 3600 seconds This means that Sasquatch is 1.9 miles closer to Jeff than it is to Carl. In other words, Sasquatch must lie on a path where (the distance to Carl) - (the distance to Jeff) = 1.9 This is exactly the situation in the definition of a hyperbola, Definition 7.6. In this case, Jeff and Carl are located at the foci, and our fixed distance d is 1.9. For simplicity, we assume the hyperbola is centered at (0, 0) with its foci at (-5, 0) and (5, 0). Schematically, we have  422 HOOKED ON CONICS Jeff Carl -6-5-4-3-2- 2 3 4 5 6 X -1- -2- -3- -4- -5- -6- We are seeking a curve of the form X _ - = 1 in which the distance from the center to each focus a2 b2 is c = 5. As we saw in the derivation of the standard equation of the hyperbola, Equation 7.6, d = 2a, so that 2a = 1.9, or a = 0.95 and a2 = 0.9025. All that remains is to find b2. To that end, we recall that a2 + b2 = c2 so b2 = c2 - a2 = 25 - 0.9025 = 24.0975. Since Sasquatch is closer to 2 2 Jeff than it is to Carl, it must be on the western (left hand) branch of 0.9025 - 24. 975 1.- In our previous example, we did not have enough information to pin down the exact location of Sasquatch. To accomplish this, we would need a third observer. EXAMPLE 7.5.5. By a stroke of luck, Kai was also camping in the woods during the events of the previous example. He was located 6 miles due north of Jeff and heard the Sasquatch call 18 seconds after Jeff did. Use this added information to locate Sasquatch. SOLUTION. Kai and Jeff are now the foci of a second hyperbola where the fixed distance d can be determined as before miles 1 hour 760 x x 18 seconds = 3.8 miles hour 3600 seconds Since Jeff was positioned at (-5, 0), we place Kai at (-5, 6). This puts the center of the new hyperbola at (-5, 3). Plotting Kai's position and the new center gives us  7.5 HYPERBOLAS 423 Kai * 6- 5- 4- x 3 1 -9 87-65-4-e2f y -& Carl, i :1, - i i i i -.+ L 014 i ;10 .1- -2- -3- -4- -5 2 3 4 5 3 X The second hyperbola is vertical, so it must be of the form (3 - 5 1. As before, the distance d is the length of the major axis, which in this case is 2b. We get 2b = 3.8 so that b = 1.9 and b2 = 3.61. With Kai 6 miles due North of Jeff, we have that the distance from the center to the focus is c = 3. Since a2 + b2 = c2, we get a2 = c2 - b2 = 9 - 3.61 = 5.39. Kai heard the Sasquatch call after Jeff, so Kai is farther from Sasquatch than Jeff. Thus Sasquatch must lie on the southern branch of the hyperbola (y3)2 - (x5)2 =1. Looking at the western branch of the 3.61 5.39 hyperbola determined by Jeff and Carl along with the southern branch of the hyperbola determined by Kai and Jeff, we see that there is exactly one point in common, and this is where Sasquatch must have been when it called. y Kai * 6-- 5-- 4-- x 3-- 2-- 1-- Jeff Carl -9 8 -7 -6 -5 -4-3-2 1 2 3 4 5 6 x Sasquatch1 -2- -3- -4- -5- -6- To determine the coordinates of this point of intersection exactly, we would need techniques for solving systems of non-linear equations. We will see those later in Section 8.7. A calculator can  424 HOOKED ON CONICS be of use in approximating these coordinates using the Intersect command. In order to use this command, however, we first need to solve each of our hyperbolas for y, choose the correct equation to enter into the calculator, and proceed from there. We leave this as an exercise. Q The procedure outlined in the two previous examples is the basis for LOng Range Aid to Navigation (LORAN for short.) While it appears to be losing its popularity due to Global Positioning Satellites (GPS), it remains one of most important applications of hyperbolas.  7.5 HYPERBOLAS 425 7.5.1 EXERCISES 1. Graph the hyperbola. Find the center, the lines which contain the transverse and conjugate axes, the vertices, the foci and the equations of the asymptotes. y2 z2 (y - 3)2 (x - 1)2 (a) = 1 (c) = 1 9 16 11 10 (x-2)2 _ (y+3)2 (x+5)2_ (y-4)2 (b)- = 1 (d)- = 1 4 9 16 1 2. Put the equation in standard form. Find the center, the lines which contain the transverse and conjugate axes, the vertices, the foci and the equations of the asymptotes. (a) 12x2 -3y2+ 30y - 111 = 0 (b) 18y2 - 5x2+72y + 30x - 63 = 0 3. Find the standard form of the equation of the hyperbola which has the given properties. (a) Center (3, 7), Vertex (3, 3), Focus (3, 2) (b) Vertex (0, 1), Vertex (8, 1), Focus (-3, 1) (c) Foci (0,1i5), Point on curve 2, 2 (d) Vertex (-10, 5), Asymptotes y = t (x - 6) + 5 4. The graph of a vertical or horizontal hyperbola clearly fails the Vertical Line Test, Theorem 1.1, so the equation of a vertical of horizontal hyperbola does not define y as a function of x.2 However, much like with circles, horizontal parabolas and ellipses, we can split a hyperbola into pieces, each of which would indeed represent y as a function of x. With the help of your classmates, use your calculator to graph the hyperbolas given in Exercise 1 above. How many pieces do you need for a vertical hyperbola? How many for a horizontal hyperbola? 5. Use your calculator to find the approximate location of Sasquatch in Example 7.5.5. 6. The location of an earthquake's epicenter - the point on the surface of the Earth directly above where the earthquake actually occurred - can be determined by a process similar to how we located Sasquatch in Example 7.5.5. (As we said back in Exercise 4a in Section 6.1, earthquakes are complicated events and it is not our intent to provide a complete discussion of the science involved in them. Instead, we refer the interested reader to a course in Geology or the U.S. Geological Survey's Earthquake Hazards Program found here.) Our technique works only for relatively small distances because we need to assume the Earth is flat in order to use hyperbolas in the plane.) The P-waves ("P" stands for Primary) of an earthquake 2We will see later in the text that the graphs of certain rotated hyperbolas pass the Vertical Line Test. 3Back in the Exercises in Section 1.1 you were asked to research people who believe the world is flat. What did you discover?  426 HOOKED ON CONICS in Sasquatchia travel at 6 kilometers per second.4 Station A records the waves first. Then Station B, which is 100 kilometers due north of Station A, records the waves 2 seconds later. Station C, which is 150 kilometers due west of Station A records the waves 3 seconds after that (a total of 5 seconds after Station A). Where is the epicenter? 7. With the help of your classmates, research the shape of cooling towers for nuclear power plants and other ways hyperbolas have been used in architecture and design. 8. With the help of your classmates, research the Cassegrain Telescope. It uses the reflective property of the hyperbola as well as that of the parabola to make an ingenious telescope. 4Depending on the composition of the crust at a specific location, P-waves can travel between 5 kps and 8 kps.  7.5 HYPERBOLAS 7.5.2 ANSWERS 427 1. (a) y- 1 Center (0,0) Transverse axis on x 0 Conjugate axis on y 0 Vertices (0, 3), (0, -3) Foci (0, 5), (0,-5) Asymptotes y 4=1tx 6- T 5, N 4- ............. 2 p 1 y .. .. . I rn -6 -5 -4 -3- -5a -6- 2 .456 6> - ,A b (-2)2 (y+3)2 1 49 Center (2, -3) Transverse axis on y = Conjugate axis on x= Vertices (0, -3), (4, -3) Foci (2 + 13, -3), (2 - Asymptotes y = t 2(- y -3 2 13, - 2) -3 -3) (C 11 10 Center (1, 3) Transverse axis on x 1 Conjugate axis on y = 3 Vertices (1, 3 + 10), (1, 3 - 10) Foci (1,3+ v2),(1,3- 21) Asymptotes y = t ± (x- 1) + 3 9- N 8. N 7. 3. -1. y - x' -/ * * * N.. / i i i ;14 j - - i - - j j v 1 i i : p - -5 -4 -3,1-2 --1- A -2. ,73 1 4 5 6 7 * N (d) (x+)2 (y_4)2 1 Center (-4, 4) Transverse axis on y = 4 Conjugate axis on x= -4 Vertices (-8, 4), (0, 4) Foci (-4 + 17, 4), (-4 - 17, 4) Asymptotes y =t(( - 2) - 3 . ......................... .5 . -x ~~ . -- ----~1- 1- y - I I I I I I I I Im -11-10-9 -8 -7 -6 -5 -4 -3 -2 -1 1 23  428 HOOKED ON CONICS z2 (y -5 )2 2. (a) = 1) 3 12 Center (0, 5) Transverse axis on y = 5 Conjugate axis on x= 0 Vertices (v/3, 5), (-V3, 5) Foci (/15,5),(-/15,5) Asymptotes y =+t2x + 5 (y + 2)2_ (x-3)2 (b) = 18 5 18 Center (3, -2) Transverse axis on x= 3 Conjugate axis on y = -2 Vertices (3, -2 + v/5), (3, - Foci (3, -2 + v/5), (3, -2 - Asymptotes y = t1 (x - -2- v/5) v5) 3) - 2 (y -7)2 3. (a) 16 16 (b) (x4)2 16 (x - 3)2 9 (y-1)2 33 1 1 ( 2 2 (c) - 9 16 (d) (X 6)2 256 1 (y- 5)2 64 1 5. Sasquatch is located at the point (-0.9629, -0.8113). 6. By placing Station A at (0, -50) and Station B at (0, 50), the two second time difference 2 2 yields the hyperbola =-4- -1 with foci A and B and center (0, 0). Placing Station C 36 2464 at (-150, -50) and using foci A and C gives us a center of (-75, -50) and the hyperbola (x + 75)2_ (y + 50)2 (75) (50)2= 1. The point of intersection of these two hyperbolas which is closer 225 5400 to A than B and closer to A than C is (-57.8444, -9.21336) so that is the epicenter.  CHAPTER 8 SYSTEMS OF EQUATIONS AND MATRICES 8.1 SYSTEMS OF LINEAR EQUATIONS: GAUSSIAN ELIMINATION Up until now, when we concerned ourselves with solving different types of equations there was only one equation to solve at a time. Given an equation f(x) = g(x), we could check our solutions geometrically by finding where the graphs of y = f(x) and y = g(x) intersect. The x-coordinates of these intersection points correspond to the solutions to the equation f(x) = g(x), and the y- coordinates were largely ignored. If we modify the problem and ask for the intersection points of the graphs of y = f(x) and y = g(x), where both the solution to x and y are of interest, we have what is known as a system of equations, usually written as Sy = f(x) y = g(x) The 'curly bracket' notation means we are to find all pairs of points (x, y) which satisfy both equations. We begin our study of systems of equations by reviewing some basic notions from Intermediate Algebra. DEFINITION 8.1. A linear equation in two variables is an equation of the form a1x + a2y = c where ai, a2 and c are real numbers and at least one of a1 and a2 is nonzero. For reasons which will become clear later in the section, we are using subscripts in Definition 8.1 to indicate different, but fixed, real numbers and those subscripts have no mathematical meaning beyond that. For example, 3x - 2 = 0.1 is a linear equation in two variables with ai1= 3, a2 = -2 and c = 0.1. We can also consider x= 5 to be a linear equation in two variables by identifying ai = 1, a2 = 0, and c = 5.1 If a1 and a2 are both 0, then depending on c, we get either an equation which is always true, called an identity, or an equation which is never true, called a contradiction. (If c = 0, then we get 0 = 0, which is always true. If c # 0, then we'd have 0 # 0, which is never true.) Even though identities and contradictions have a large role to play 'Critics may argue that x= 5 is clearly an equation in one variable. It can also be considered an equation in 117 variables with the coefficients of 116 variables set to 0. As with many conventions in Mathematics, the context will clarify the situation.  430 SYSTEMS OF EQUATIONS AND MATRICES in the upcoming sections, we do not consider them linear equations. The key to identifying linear equations is to note that the variables involved are to the first power and that the coefficients of the variables are numbers. Some examples of equations which are non-linear are x2+y = 1, xy = 5 and e2X + ln(y) = 1. We leave it to the reader to explain why these do not satisfy Definition 8.1. From what we know from Sections 1.2 and 2.1, the graphs of linear equations are lines. If we couple two or more linear equations together, in effect to find the points of intersection of two or more lines, we obtain a system of linear equations in two variables. Our first example reviews some of the basic techniques first learned in Intermediate Algebra. EXAMPLE 8.1.1. Solve the following systems of equations. Check your answer algebraically and graphically. 2x-y = 1 3 3 5 . 6x+3y = 9 y= 3 - 2 - _i. 4x 2y =12 9 3 2 x-y = 0 2{ 3x+4y= -2 4 2x-4y = 6 6. x+y = 2 A-3-y = 5 3x-6y = 9 -2x+y =_-2 SOLUTION. 1. Our first system is nearly solved for us. The second equation tells us that y = 3. To find the corresponding value of x, we substitute this value for y into the the first equation to obtain 2x - 3 = 1, so that x= 2. Our solution to the system is (2, 3). To check this algebraically, we substitute x= 2 and y = 3 into each equation and see that they are satisfied. We see 2(2) - 3 = 1, and 3 = 3, as required. To check our answer graphically, we graph the lines 2x - y = 1 and y = 3 and verify that they intersect at (2, 3). 2. To solve the second system, we use the addition method to eliminate the variable x. We take the two equations as given and 'add equals to equals' to obtain 3x+4y -2 + (-3x - y = 5) 3y = 3 This gives us y = 1. We now substitute y = 1 into either of the two equations, say -3x-y = 5, to get -3x - 1 = 5 so that x= -2. Our solution is (-2, 1). Substituting x = -2 and y 1 into the first equation gives 3(-2) + 4(1) =-2, which is true, and, likewise, when we check (-2, 1) in the second equation, we get -3(-2) - 1 =5, which is also true. Geometrically, the lines 3x + 4y =-2 and -3x - y =5 intersect at (-2, 1).  8.1 SYSTEMS OF LINEAR EQUATIONS: GAUSSIAN ELIMINATION 431 y 4 2 (2 ,3 ) 1 -- 2 --(-2, 1) 1 -4 -3 -2 -1 x -H--- -1 1 2 3 4 x -2 3x+4y=-2 2x-y=1 -3x-y= 5 y=3 3. The equations in the third system are more approachable if we clear denominators. We multiply both sides of the first equation by 15 and both sides of the second equation by 18 to obtain the kinder, gentler system {5x-12y = 21 4x+6y = 9 Adding these two equations directly fails to eliminate either of the variables, but we note that if we multiply the first equation by 4 and the second by -5, we will be in a position to eliminate the x term 20x - 48y = 84 + (-20x - 30y =-45) -78y = 39 From this we get y =-. We can temporarily avoid too much unpleasantness by choosing to substitute y = -j into one of the equivalent equations we found by clearing denominators, say into 5x - 12y = 21. We get 5x + 6 = 21 which gives x= 3. Our answer is (3, -1). At this point, we have no choice - in order to check an answer algebraically, we must see if the answer satisfies both of the original equations, so we substitute x = 3 and y - -j into both X - 45= 5 and 9 + Y = 1. We leave it to the reader to verify that the solution is correct. Graphing both of the lines involved with considerable care yields an intersection point of (3, -ij). 4. An eerie calm settles over us as we cautiously approach our fourth system. Do its friendly integer coefficients belie something more sinister? We note that if we multiply both sides of the first equation by 3 and the both sides of the second equation by -2, we are ready to eliminate the x  432 SYSTEMS OF EQUATIONS AND MATRICES 6x - 12y = 18 + (-6x + 12y = -18) 0 = 0 We eliminated not only the x, but the y as well and we are left with the identity 0 = 0. This means that these two different linear equations are, in fact, equivalent. In other words, if an ordered pair (x, y) satisfies the equation 2x - 4y = 6, it automatically satisfies the equation 3x -6y = 9. One way to describe the solution set to this system is to use the roster method2 and write {(x, y) : 2x - 4y = 6}. While this is correct (and corresponds exactly to what's happening graphically, as we shall see shortly), we take this opportunity to introduce the notion of a parametric solution. Our first step is to solve 2x - 4y = 6 for one of the variables, say y = jz - 2. For each value of x, the formula y = 2z - 2 determines the corresponding y-value of a solution. Since we have no restriction on x, it is called a free variable. We let x= t, a so-called 'parameter', and get y =2It - 2. Our set of solutions can then be described as {(t, It - 2) : -oo < t < oo}.3 For specific values of t, we can generate solutions. For example, t = 0 gives us the solution (0, -4); t = 117 gives us (117, 57), and while we can readily check each of these particular solutions satisfy both equations, the question is how do we check our general answer algebraically? Same as always. We claim that for any real number t, the pair (t, jt - 2) satisfies both equations. Substituting x = t and y =2t - 2 into 2x - 4y = 6 gives 2t - 4 (It -2-) = 6. Simplifying, we get 2t -2t+6 = 6, which is always true. Similarly, when we make these substitutions in the equation 3x - 6y = 9, we get 3t - 6 (it -2-) = 9 which reduces to 3t - 3t +9 = 9, so it checks out, too. Geometrically, 2x - 4y = 6 and 3x - 6y = 9 are the same line, which means that they intersect at every point on their graphs. The reader is encouraged to think about how our parametric solution says exactly that. y 2 1 1 -1 1 2 4 5 6 7 1 3 4 x (3,-}) -1 -3 -4 X 7x - 6y = 9+3 23x-6y=9 (Same line.) 2See Section 1.2 for a review of this. 3Note that we could have just as easily chosen to solve 2x - 4y =6 for x to obtain x 2y + 3. Letting y be the parameter t, we have that for any value of t, x 2t + 3, which gives {(2t + 3, t) :-oo < t < oo}. There is no one correct way to parameterize the solution set, which is why it is always best to check your answer.  8.1 SYSTEMS OF LINEAR EQUATIONS: GAUSSIAN ELIMINATION 433 5. Multiplying both sides of the first equation by 2 and the both sides of the second equation by -3, we set the stage to eliminate x 12x + 6y + (-12x - 6y 0 18 -36) -18 As in the previous example, both x and y dropped out of the equation, but we are left with an irrevocable contradiction, 0 = -18. This tells us that it is impossible to find a pair (x, y) which satisfies both equations; in other words, the system has no solution. Graphically, we see that the lines 6x + 3y = 9 and 4x + 2y = 12 are distinct and parallel, and as such do not intersect. 6. We can begin to solve our last system by adding the first two equations x-y + (x+y 0 2) 2x = 2 which gives x= 1. Substituting this into the first equation gives 1 - y = 0 so that y = 1. We seem to have determined a solution to our system, (1, 1). While this checks in the first two equations, when we substitute x= 1 and y = 1 into the third equation, we get -2(1)+(1) = -2 which simplifies to the contradiction -1 = -2. Graphing the lines x-y = 0, x + y = 2, and -2x + y =--2, we see that the first two lines do, in fact, intersect at (1, 1), however, all three lines never intersect at the same point simultaneously, which is what is required if a solution to the system is to be found. y x .x 6x+3y 9 4x±2y= 12 y-x=0 y+x 2 -2x+y =-2 A few remarks about Example 8.1.1 are in order. It is clear that some systems of equations have solutions, and some do not. Those which have solutions are called consistent, those with no  434 SYSTEMS OF EQUATIONS AND MATRICES solution are called inconsistent. We also distinguish the two different types of behavior among consistent systems. Those which admit free variables are called dependent; those with no free variables are called independent.4 Using this new vocabulary, we classify numbers 1, 2 and 3 in Example 8.1.1 as consistent independent systems, number 4 is consistent dependent, and numbers 5 and 6 are inconsistent. The system in 6 above is called overdetermined, since we have more equations than variables.6 Not surprisingly, a system with more variables than equations is called underdetermined. While the system in number 6 above is overdetermined and inconsistent, there exist overdetermined consistent systems (both dependent and independent) and we leave it to the reader to think about what is happening algebraically and geometrically in these cases. Likewise, there are both consistent and inconsistent underdetermined systems,' but a consistent underdetermined system of linear equations is necessarily dependent.8 In order to move this section beyond a review of Intermediate Algebra, we now define what is meant by a linear equation in n variables. DEFINITION 8.2. A linear equation in n variables, x1, x2, ..., xn is an equation of the form a1x1 + a2x2 +... + anxn = c where a1, a2, ... an and c are real numbers and at least one of a1, a2, ... , an is nonzero. Instead of using more familiar variables like x, y, and even z and/or w in Definition 8.2, we use subscripts to distinguish the different variables. We have no idea how many variables may be involved, so we use numbers to distinguish them instead of letters. (There is an endless supply of distinct numbers.) As an example, the linear equation 3x1-z2 = 4 represents the same relationship between the variables x1 and x2 as the equation 3x - y = 4 does between the variables x and y. In addition, just as we cannot combine the terms in the expression 3x - y, we cannot combine the terms in the expression 3x1 - x2. Coupling more than one linear equation in n variables results in a system of linear equations in n variables. When solving these systems, it becomes increasingly important to keep track of what operations are performed to which equations and to develop a strategy based on the kind of manipulations we've already employed. To this end, we first remind ourselves of the maneuvers which can be applied to a system of linear equations that result in an equivalent system.9 4In the case of systems of linear equations, regardless of the number of equations or variables, consistent inde- pendent systems have exactly one solution. The reader is encouraged to think about why this is the case for linear equations in two variables. Hint: think geometrically. 5The adjectives 'dependent' and 'independent' apply only to consistent systems - they describe the type of solu- tions. 6If we think if each variable being an unknown quantity, then ostensibly, to recover two unknown quantities, we need two pieces of information - i.e., two equations. Having more than two equations suggests we have more information than necessary to determine the values of the unknowns. While this is not necessarily the case, it does explain the choice of terminology 'overdetermined'. 7We need more than two variables to give an example of the latter. 8Again, experience with systems with more variables helps to see this here, as does a solid course in Linear Algebra. 9That is, a system with the same solution set.  8.1 SYSTEMS OF LINEAR EQUATIONS: GAUSSIAN ELIMINATION 435 THEOREM 8.1. Given a system of equations, the following moves will result in an equivalent system of equations. " Interchange the position of any two equations. " Replace an equation with a nonzero multiple of itself.a " Replace an equation with itself plus a nonzero multiple of another equation. "That is, an equation which results from multiplying both sides of the equation by the same nonzero number. We have seen plenty of instances of the second and third moves in Theorem 8.1 when we solved the systems Example 8.1.1. The first move, while it obviously admits an equivalent system, seems silly. Our perception will change as we consider more equations and more variables in this, and later sections. Consider the system of equations Ix- jy + z = 1 y{-'1 z = 4 z = -1 Clearly z = -1, and we substitute this into the second equation y - j(-1) = 4 to obtain y =2. Finally, we substitute y = 7 and z = -1 into the first equation to get x - 3 (7) + 2(-1) 1, so that x = . The reader can verify that these values of x, y and z satisfy all three original equations. It is tempting for us to write the solution to this system by extending the usual (x, y) notation to (x, y, z) and list our solution as (g, 7, -1). The question quickly becomes what does an 'ordered triple' like (g, 7, -1) represent? Just as ordered pairs are used to locate points on the two-dimensional plane, ordered triples can be used to locate points in space.10 Moreover, just as equations involving the variables x and y describe graphs of one-dimensional lines and curves in the two-dimensional plane, equations involving variables x, y, and z describe objects called surfaces in three-dimensional space. Each of the equations in the above system can be visualized as a plane situated in three-space. Geometrically, the system is trying to find the intersection, or common point, of all three planes. If you imagine three sheets of notebook paper each representing a portion of these planes, you will start to see the complexities involved in how three such planes can intersect. Below is a sketch of the three planes. It turns out that any two of these planes intersect in a line,"" so our intersection point is where all three of these lines meet. ioYou were asked to think about this in Exercise 13 in Section 1. "I1n fact, these lines are described by the parametric solutions to the systems formed by taking any two of these equations by themselves.  436 SYSTEMS OF EQUATIONS AND MATRICES Since the geometry for equations involving more than two variables is complicated, we will focus our efforts on the algebra. Returning to the system x - 3y + 2z = 1 y -2j = 4 z = -1 we note the reason it was so easy to solve is that the third equation is solved for z, the second equation involves only y and z, and since the coefficient of y is 1, it makes it easy to solve for y using our known value for z. Lastly, the coefficient of x in the first equation is 1 making it easy to substitute the known values of y and z and then solve for x. We formalize this pattern below for the most general systems of linear equations. Again, we use subscripted variables to describe the general case. The variable with the smallest subscript in a given equation is typically called the leading variable of that equation. DEFINITION 8.3. A system of linear equations with variables x1, x2, ... x is said to be in trian- gular form provided all of the following conditions hold: 1. The subscripts of the variables in each equation are always increasing from left to right. 2. The leading variable in each equation has coefficient 1. 3. The subscript on the leading variable in a given equation is greater than the subscript on the leading variable in the equation above it. 4. Any equation without variables' cannot be placed above an equation with variables. anecessarily an identity or contradiction In our previous system, if make the obvious choices x = xi, y = x2, and z = x3, we see that the  8.1 SYSTEMS OF LINEAR EQUATIONS: GAUSSIAN ELIMINATION 437 system is in triangular form.12 An example of a more complicated system in triangular form is x1 - 4x3 + x4 - x6 = 6 x2 + 2x3 = 1 x4 + 3x5 -x6 = 8 x5 + 9x6 = 10 Our goal henceforth will be to transform a given system of linear equations into triangular form using the moves in Theorem 8.1. EXAMPLE 8.1.2. Use Theorem 8.1 to put the following systems into triangular form and then solve the system if possible. Classify each system as consistent independent, consistent dependent, or inconsistent. 3x- y+z = 3 2x+3y-z = 1 3x1+x2+x4 = 6 1. 2x - 4y + 3z = 16 2. 10x - z = 2 3. 2x1 + x2 - x3 = 4 I x - y + z = 5 4x - 9y + 2z = 5 x2 - 3x3- 2x4 = 0 SOLUTION. 1. For definitiveness, we label the topmost equation in the system E1, the equation beneath that E2, and so forth. We now attempt to put the system in triangular form using an algorithm known as Gaussian Elimination. What this means is that, starting with x, we transform the system so that conditions 2 and 3 in Definition 8.3 are satisfied. Then we move on to the next variable, in this case y, and repeat. Since the variables in all of the equations have a consistent ordering from left to right, our first move is to get an x in El's spot with a coefficient of 1. While there are many ways to do this, the easiest is to apply the first move listed in Theorem 8.1 and interchange El and E3. (E1) 3x -y+z = 3 (E1) x-y+z = 5 (E2) 2x - 4y + 3z = 16 Switch El and E3> (E2) 2x - 4y + 3z = 16 (E3) x - y + z = 5 (E3) 3x - y + z = 3 To satisfy Definition 8.3, we need to eliminate the x's from E2 and E3. We accomplish this by replacing each of them with a sum of themselves and a multiple of E1. To eliminate the x from E2, we need to multiply El by -2 then add; to eliminate the x from E3, we need to multiply El by -3 then add. Applying the third move listed in Theorem 8.1 twice, we get {(£1) cr- y +z - 5 (£1) cr- y +z - 5 (£2) 2cr -4y + 3z 16 Replace E2 with -2E1 + E2 (2 2 (£3) 3cr- y + z = 3 Replace E3 with -3E1 + E3 I(£3) 2y - 2z =-12 "I1f letters are used instead of subscripted variables, Definition 8.3 can be suitably modified using alphabetical order of the variables instead of numerical order on the subscripts of the variables.  438 SYSTEMS OF EQUATIONS AND MATRICES Now we enforce the conditions stated in Definition 8.3 for the variable y. To that end we need to get the coefficient of y in E2 equal to 1. We apply the second move listed in Theorem 8.1 and replace E2 with itself times - 1. (E1) x - y + z (E2) -2y + z (E3) 2y - 2z 5 6 .12 Replace E2 with - E2 2 : { (E1) (E2) (E3) x-y+z y- 2z 2y - 2z 5 -3 12 To eliminate the y in E3, we add -2E2 to it. { (E1) (E2) (E3) x-y+z 1 y - z 2y - 2z 5 -3 -12 Replace E3 with -2E2 + E3 (E1) (E2) (E3) x-y+z y - z -z 5 -3 -6 Finally, we apply the second move from Theorem 8.1 one last time and multiply E3 by to satisfy the conditions of Definition 8.3 for the variable z. -1 { (E1) (E2) (E3) x-y+z y -12z -z 5 -3 -6 {(£1) Replace E3 with -1E3 (£2) (E2) (E3) x-y+z y -2 z 5 -3 6 Now we proceed to substitute. Plugging in z = 6 into E2 gives y - 3 = -3 so that y = 0. With y = 0 and z = 6, E1 becomes x - 0 + 6 = 5, or x= -1. Our solution is (-1,0,6). We leave it to the reader to check that substituting the respective values for x, y, and z into the original system results in three identities. Since we have found a solution, the system is consistent; since there are no free variables, it is independent. 2. Proceeding as we did in 1, our first step is to get an equation with x in the El position with 1 as its coefficient. Since there is no easy fix, we multiply El by 2. { (E1) (E2) (E3) 2x + 3y - z lox - z 4x - 9y + 2z 1 2 5 Replace El with jE1 2 { (E1) (E2) (E3) X + 2y - 1z lox - z 4x - 9y + 2z 1 2 5 Now it's time to take care of the x's in E2 and E3. (E1) (E2) (E3) lox - z 4x - 9y + 2z 1 2 5 Replace E2 with -1OE1 + E2 Replace E3 with -4E1 + E3 (E1) (E2) (E3) x +2y - z -15y+4z -15y+4z 1 -3 3  8.1 SYSTEMS OF LINEAR EQUATIONS: GAUSSIAN ELIMINATION 439 Our next step is to get the coefficient of y in E2 equal to 1. To that end, we have (E1) xz+ 2y - 2z (E2) -15y + 4z (E3) -15y + 4z Finally, we rid E3 of y. (E1) x + 2y -2z (E2) y - z (E3) -15y + 4z 1 -3 Replace E2 with -i5 E2 15> (E1) (E2) (E3) Xz+ 2y - 2z y - z -15y+4z 1 1 3 1 1 3 Replace E3 with 15E2 + E3 (E1) x-y+z (E2) y - 2z (E3) 0 5 -3 6 The last equation, 0 = 6, is a contradiction so the system has no solution. According to Theorem 8.1, since this system has no solutions, neither does the original, thus we have an inconsistent system. 3. For our last system, we begin by multiplying El by j to get a coefficient of 1 on xi. (E1) 3x31+3:2 + x4 (E2) 2x1 + x2 - z3 (E3) z2 - 3x3 - 2x4 6 4 0 Replace El with jE1 3 (E1) X1 + 2 + j3z4 (E2) 2x1 + z2 - z3: (E3) z2 - 3x3 - 2x4 2 4 0 Next we eliminate z1 from E2 (E1) z1 + z2 + z4 (E2) 2x1 + x2 - z3: (E3) z2-3x3-2x4 2 4 0 Replace E2 with -2E1 + E2 { (E1) (E2) (E3) Xi + jz2 1 z2- 33 + 143 2 3z4 2 0 0 12 - 313 - 2x4 We switch E2 and E3 to get a coefficient of 1 for x2. (E1) X1 + 2 + j3z4 (E2) j2 - X3 - 2z4 (£3) 323X324 2 0 0 Switch E2 and E3 (E1) (E2) (E3) Xi + 1z2 + 1314 2 - 33 - 2x4 z2 - 3 - 3z4 2 0 0 Finally, we eliminate x2 in E3.  440 SYSTEMS OF EQUATIONS AND MATRICES (E1) l31+x2 + 3x4 = 2 (E1) x1+ 3x2+ 3x4 = 2 (E2) z2 - 3x3 - 2x4 = 0 Replace E3 j (E2) x2-3x3-2x4 = 0 (E3) j2t3 --E24+=E3 (E3) 0 = 0 Equation E3 reduces to 0 = 0,which is always true. Since we have no equations with x3 or x4 as leading variables, they are both free, which means we have a consistent dependent system. We parametrize the solution set by letting x3= s and x4 = t and obtain from E2 that x2 = 3s + 2t. Substituting this and x4 = t into El, we have x1 + (3s + 2t) + At = 2 which givesx 1 = 2 - s - t. Our solution is the set {(2 - s - t, 2s+3t, s, t) : -o < s, t < 003}.3 We leave it to the reader to verify that the substitutions xi = 2 - s - t, x2 = 3s + 2t, x3 = s and x4 = t satisfy the equations in the original system. D Like all algorithms, Gaussian Elimination has the advantage of always producing what we need, but it can also be inefficient at times. For example, when solving 2 above, it is clear after we eliminated the x's in the second step to get the system (El) x+3y -z 1 (E2) -15y + 4z = -3 (E3) -15y + 4z = 3 that equations E2 and E3 when taken together form a contradiction since we have identical left hand sides and different right hand sides. The algorithm takes two more steps to reach this contradiction. We also note that substitution in Gaussian Elimination is delayed until all the elimination is done, thus it gets called back-substitution. This may also be inefficient in many cases. Rest assured, the technique of substitution as you may have learned it in Intermediate Algebra will once again take center stage in Section 8.7. Lastly, we note that the system in 3 above is underdetermined, and as it is consistent, we have free variables in our answer. We close this section with a standard 'mixture' type application of systems of linear equations. EXAMPLE 8.1.3. Lucas needs to create a 500 milliliters (mL) of a 40% acid solution. He has stock solutions of 30% and 90% acid as well as all of the distilled water he wants. Set-up and solve a system of linear equations which determines all of the possible combinations of the stock solutions and water which would produce the required solution. SOLUTION. We are after three unknowns, the amount (in mL) of the 30% stock solution (which we'll call x), the amount (in mL) of the 90% stock solution (which we'll call y) and the amount (in mL) of water (which we'll call w). We now need to determine some relationships between these variables. Our goal is to produce 500 milliliters of a 40% acid solution. This product has two defining characteristics. First, it must be 500 mL; second, it must be 40% acid. We take each 13Here, any choice of s and t will determine a solution which is a point in 4-dimensional space. Yeah, we have trouble visualizing that, too.  8.1 SYSTEMS OF LINEAR EQUATIONS: GAUSSIAN ELIMINATION 441 of these qualities in turn. First, the total volume of 500 mL must be the sum of the contributed volumes of the two stock solutions and the water. That is amount of 30% stock solution + amount of 90% stock solution + amount of water = 500 mL Using our defined variables, this reduces to x+ y + w = 500. Next, we need to make sure the final solution is 40% acid. Since water contains no acid, the acid will come from the stock solutions only. We find 40% of 500 mL to be 200 mL which means the final solution must contain 200 mL of acid. We have amount of acid in 30% stock solution + amount of acid 90% stock solution = 200 mL The amount of acid in x mL of 30% stock is 0.30x and the amount of acid in y mL of 90% solution is 0.90y. We have 0.30x + 0.90y = 200. Converting to fractions,14 our system of equations becomes { x+y+w =500 ox+ Ny = 200 We first eliminate the x from the second equation { (E1) x + y + w = 500 Replace E2 with - El +E2 (E1) x + y + w = 500 (E2) 0x+1Ny = 200 (E2) jy- ow = 50 Next, we get a coefficient of 1 on the leading variable in E2 { (E1) x + y + w = 500 Replace E2 with jE2 (E1) x +y w = 500 (E2) jy -ow = 50 ' (E2) y 2-5w = 0 Notice that we have no equation to determine w, and as such, w is free. We set w = t and from E2 get y =2t + 2 3. Substituting into El gives x+ (2t+ 230) + t = 500 so that x= -it+ 13. This system is consistent, dependent and its solution set is {(-jt + 1 3, It + 2, t) : -oo < t < oo}. While this answer checks algebraically, we have neglected to take into account that x, y and w, being amounts of acid and water, need to be nonnegative. That is, x > 0, y > 0 and w > 0. The constraint x > 0 gives us -It+ 1 0>0, or t <2500. From y 3 0, we get 2t+ 2 0 0 or t > -_ . The condition z > 0 yields t > 0, and we see that when we take the set theoretic intersection of these intervals, we get 0 < t < 25900. Our final answer is {(-2t + 12 It + 250t) : 0 < t 2500}. Of what practical use is our answer? Suppose there is only 100 mL of the 90% solution remaining and it is due to expire. Can we use all of it to make our required solution? We would have y = 100 so that jt + 20= 100, and we get t =100 This means the amount of 30% solution required is X = -1t + 00 - (3) + = 0 mL, and for the water, w = t = 100 mL. The reader is invited to check that mixing these three amounts of our constituent solutions produces the required 40% acid mix. "4We do this oniy because we believe students can use all of the practice with fractions they can get!  442 SYSTEMS OF EQUATIONS AND MATRICES 8.1.1 EXERCISES 1. Put the following systems of linear equations into triangular form and then solve the sys- tem if possible. Classify each system as consistent independent, consistent dependent, or inconsistent. (a){ 5+ y = 17 2x1+2-12x3-x4 = 16 (a+ y = 5-Xi +z212 - 4x4 = -5 x+y+z 3 3x1+2x2-16x3-3x4 25 (b) 2x - y + z = 0 x1+ 2x2-54 = 11 -3x+5y +7z = 7 ( x1 - 3 = -2 4x-y+z = 5 (g) 2x2-x4 = 0 (c) 2y + 6z=301 -2x2+3= 0 x+z = 5 -x3+x4 = 1 4x - y + z = 5 1 - x2 - 5x3+3x4 = -1 x+z = 6 (h) x2+5x3-3x4 = 1 {) x+ y + z = -17 x1 - 2x2 - 10x3+6x4 = -1 y-3z = 0 2. Find two other forms of the parametric solution to Exercise ic above by reorganizing the equations so that x or y can be the free variable. 3. At The Old Home Fill'er Up and Keep on a-Truckin' Cafe, Mavis mixes two different types of coffee beans to produce a house blend. The first type costs $3 per pound and the second costs $8 per pound. How much of each type does Mavis use to make 50 pounds of a blend which costs $6 per pound? 4. At The Crispy Critter's Head Shop and Patchouli Emporium along with their dried up weeds, sunflower seeds and astrological postcards they sell an herbal tea blend. By weight, Type I herbal tea is 30% peppermint, 40% rose hips and 30% chamomile, Type II has percents 40%, 20% and 40%, respectively, and Type III has percents 35%, 30% and 35%, respectively. How much of each Type of tea is needed to make 2 pounds of a new blend of tea that is equal parts peppermint, rose hips and chamomile? 5. Discuss with your classmates how you would approach Exercise 4 above if they needed to use up a pound of Type I tea to make room on the shelf for a new canister. 6. Discuss with your classmates why it is impossible to mix a 20% acid solution with a 40% acid solution to produce a 60% acid solution. If you were to try to make 100 mL of a 60% acid solution using stock solutions at 20% and 40%, respectively, what would the triangular form of the resulting system look like?  8.1 SYSTEMS OF LINEAR EQUATIONS: GAUSSIAN ELIMINATION 443 8.1.2 ANSWERS 1. Because triangular form is not unique, we give only one possible answer to that part of the question. Yours may be different and still be correct. (a) z + y = 5 Consistent independent l y = 7 Solution (-2, 7) Sx-3y-3z =- Consistent independent (b) y +4z = 2 Solution (1,2,0) z = 0 X- y + 4 z = 4Consistent dependent (c) y + 3z = 15 Solution (-t+5,-3t+15,t) 0 = 0 for all real numbers t x - 4 iy + z = Inconsistent (d) y + 3z = 15 No solution 0= 1 x + y + z = -17 Consistent dependent y - 3z = 0 Solution (-4t - 17, 3t, t) for all real numbers t (f) { X1 + 3x2 -j3 - z4 = 39 2 + 4x3 - 34 = 2 0 = 0 0 = 0 Consistent dependent Solution (8s -t+ 7,-4s+3t+2,s,t) for all real numbers s and t X2 - jz4 (g) 1 13 - 2 14 14 -2 Consistent independent Solution (1, 2, 3, 4) - 0 = 1 = 4 (h) { x2 - 5x3 + 3x4 = -1 2 + 53 - 34 1 0 = 1 0 = 0 Inconsistent No solution 2. If x is the free variable then the solution is (it, t, -At + 5). solution is (t, 3t, -t + 5) and if y is the free variable then the 3. Mavis needs 20 pounds of $3 per pound coffee and 30 pounds of $8 per pound coffee. 4. 5 -l t pounds of Type I, 4 - jt pounds of Type II and t pounds of Type III where 0 < t <3.  444 SYSTEMS OF EQUATIONS AND MATRICES 8.2 SYSTEMS OF LINEAR EQUATIONS: AUGMENTED MATRICES In Section 8.1 we introduced Gaussian Elimination as a means of transforming a system of linear equations into triangular form with the ultimate goal of producing an equivalent system of linear equations which is easier to solve. If take a step back and study the process, we see that all of our moves are determined entirely by the coefficients of the variables involved, and not the variables themselves. Much the same thing happened when we studied long division in Section 3.2. Just as we developed synthetic division to streamline that process, in this section, we introduce a similar bookkeeping device to help us solve systems of linear equations. To that end, we define a matrix as a rectangular array of real numbers. We typically enclose matrices with square brackets, '[' and ']', and we size matrices by the number of rows and columns they have. For example, the size (sometimes called the dimension) of 3 0 -1 2 -5 10 is 2 x 3 because it has 2 rows and 3 columns. The individual numbers in a matrix are called its entries and are usually labeled with double subscripts: the first tells which row the element is in and the second tells which column it is in. The rows are numbered from top to bottom and the columns are numbered from left to right. Matrices themselves are usually denoted by uppercase letters (A, B, C, etc.) while their entries are usually denoted by the corresponding letter. So, for instance, if we have _ 3 0 -1 2 -5 10 then an1= 3, a12 =0, a13 = -1, a21= 2, a22 = -5, and a23 = 10. We shall explore matrices as mathematical objects with their own algebra in Section 8.3 and introduce them here solely as a bookkeeping device. Consider the system of linear equations from number 2 in Example 8.1.2 (E1) 2x+3y-z = 1 (E2) 10x - z = 2 (E3) 4x - 9y + 2z = 5 We encode this system into a matrix by assigning each equation to a corresponding row. Within that row, each variable and the constant gets its own column, and to separate the variables on the left hand side of the equation from the constants on the right hand side, we use a vertical bar, I. Note that in E2, since y is not present, we record its coefficient as 0. The matrix associated with this system is 3: y z c (E1)- [ 2 3 -1 1] (E2) - 10 0 -12 (E3)- [ 4 -9 2 5]_  8.2 SYSTEMS OF LINEAR EQUATIONS: AUGMENTED MATRICES 445 This matrix is called an augmented matrix because the column containing the constants is appended to the matrix containing the coefficients.1 To solve this system, we can use the same kind operations on the rows of the matrix that we performed on the equations of the system. More specifically, we have the following analog of Theorem 8.1 below. THEOREM 8.2. Row Operations: Given an augmented matrix for a system of linear equations, the following row operations produce an augmented matrix which corresponds to an equivalent system of linear equations. " Interchange any two rows. " Replace a row with a nonzero multiple of itself. a " Replace a row with itself plus a nonzero multiple of another row.b aThat is, the row obtained by multiplying each entry in the row by the same nonzero number. bWhere we add entries in corresponding columns. As a demonstration of the moves in Theorem 8.2, we revisit some of the steps that were used in solving the systems of linear equations in Example 8.1.2 of Section 8.1. The reader is encouraged to perform the indicated operations on the rows of the augmented matrix to see that the machinations are identical to what is done to the coefficients of the variables in the equations. We first see a demonstration of switching two rows using the first step of part 1 in Example 8.1.2. (E1) 3x-y+z (E2) 2x - 4y + 3z (E3) x-y+z 3 16 5 Switch El and E3 (El) (E2) (E3) x-y+z 2x - 4y + 3z 3x - y + z 1 5 313 1 3] 5 16 3 3 2 1 -1 -4 -1 1 3 1 3 16 5 I Switch RI and R3 1 S2 3 -1 -4 -1 Next, we have a demonstration of replacing a row with a nonzero multiple of itself using the first step of part 3 in Example 8.1.2. { (E1) (E2) (E3) 3x1 2x1 x2 - + x2 + x2 3x3 - + 14 - X3 - 214 6 4 0 Replace El with 1El 3 { (E1) (E2) (E3) 1 1 3 2 1 0 1 x1 + 32 + 34 2x1 + x2 - x3 2 - 33 - 2x4 0 3 2 -1 0 4 -3 -2 0] 2 4 0 3 1 0 1 6 Replace RI with jRl 2 1 -1 0 4 > 0 1 -3 -2 0 _ Finally, we have an example of replacing a row with itself plus second step from part 2 in Example 8.1.2. a multiple of another row using the 'We shall study the coefficient and constant matrices separately in Section 8.3.  446 SYSTEMS OF EQUATIONS AND MATRICES 3(E1) 1+ - z = 2 (E1) 3 1+y - 1z (E2) 10x - z = 2 Replace E2 with -10E1 + E2 (E2) -15y + 4z = -3 (E3) 4x - 9y + 2z = 5 Replace E3 with -4E1 + E3 (E3) -15y + 4z = 3 1 3 -1 1 -- 1 3 1 1 - 10 0 -1 Replace R2 with -10R1 + R2: 0 -1 4 -31 20 0 120 21 2 4 21 4-9 2 5 Replace R3 with -4R1 +R3 0 -15 4 3 The matrix equivalent of 'triangular form' is row echelon form. The reader is encouraged to refer to Definition 8.3 for comparison. Note that the analog of 'leading variable' of an equation is 'leading entry' of a row. Specifically, the first nonzero entry (if it exists) in a row is called the leading entry of that row. DEFINITION 8.4. A matrix is said to be in row echelon form provided all of the following conditions hold: 1. The first nonzero entry in each row is 1. 2. The leading 1 of a given row must be to the right of the leading 1 of the row above it. 3. Any row of all zeros cannot be placed above a row with nonzero entries. To solve a system of a linear equations using an augmented matrix, we encode the system into an augmented matrix and apply Gaussian Elimination to the rows to get the matrix into row-echelon form. We then decode the matrix and back substitute. The next example illustrates this nicely. EXAMPLE 8.2.1. Use an augmented matrix to transform the following system of linear equations into triangular form. Solve the system. 3x -y+z = 8 x+2y-z = 4 2x+3y-4z = 10 SOLUTION. We first encode the system into an augmented matrix. 3x-y+z = 8 3 -1 1 8 I x+2y- z = 4 Encode into the matrix 1 2 -1 4 2x+3y-4z = 10 2 3 -4 10 Thinking back to Gaussian Elimination at an equations level, our first order of business is to get x in E1 with a coefficient of 1. At the matrix level, this means getting a leading 1 in R1. This is in accordance with the first criteria in Definition 8.4. To that end, we interchange R1 and R2. 3 -1 1 8 1 2 -1 4 1 2 -1 4 witch R1andR2  8.2 SYSTEMS OF LINEAR EQUATIONS: AUGMENTED MATRICES 447 Our next step is to eliminate the x's from E2 and E3. From a matrix standpoint, this means we need 0's below the leading 1 in R1. This guarantees the leading 1 in R2 will be to the right of the leading 1 in R1 in accordance with the second requirement of Definition 8.4. [ 1 3 2 2 -1 3 1 4 1 8 4 10 I Replace R2 with -3R1 + R2 Replace R3 with -2R1 + R3 1 0 0 2 -7 -1 -1 4 4 -4 -2 2 Now we repeat the above process for the variable y which means we need to get the leading entry in R2 to be 1. [1 2 -1 4] Replace R2 with - R2 [1 0 -7 4 -4 7 > 0 0 -1 -2 2 _0 To guarantee the leading 1 in R3 is to the right of the leading 1 column of R3. 2 -1 4 1 -4 4 _7 7i -1 -2 2] in R2, we get a 0 in the second 1 0 0 2 1 -1 -1 4 7 4 4 2 I Replace R3 with R2 + R3 1 0 0 2 1 0 -1 4 7 18 7 4 4 7 18 7 ] Finally, we get the leading entry in R3 to be 1. [ 1 2 -1 4] 1 4 4 Replace R3 with -18R3 0 1 -7 7 > 0 0 0 18 18 '0 7 7 _ 2 1 0 -1 4 1 4 4 7 -1 I Decoding from the matrix gives a system in triangular form 1 0 0 2 1 0 -1 4 1 4 4 7 -1 Decode from the matrix x + 2y- -z = 4 7 7 z = -1 We get z =-1, y =7z + = (-1) + = 0 and x =-2y + z +4 final answer of (3, 0, -1). We leave it to the reader to check. -2(0) + (-1) + 4 3 for a As part of Gaussian Elimination, we used row operations to obtain 0's beneath each leading 1 to put the matrix into row echelon form. If we also require that 0's are the only numbers above a leading 1, we have what is known as the reduced row echelon form of the matrix. DEFINITION 8.5. A matrix is said to be in reduced row echelon form provided both of the following conditions hold: 1. The matrix is in row echelon form. 2. The leading is are the only nonzero entry in their respective columns.  448 SYSTEMS OF EQUATIONS AND MATRICES Of what significance is the reduced row echelon form of a matrix? To illustrate, let's take the row echelon form from Example 8.2.1 and perform the necessary steps to put into reduced row echelon form. We start by using the leading 1 in R3 to zero out the numbers in the rows above it. 1 2 -1 4 1 2 0 3 0 1 - 4 ] Replace R1 with R3+R1 [ 0 1 0 0 070 1 7- Replace R2 with 7 R3 + R2 001- Finally, we take care of the 2 in R1 above the leading 1 in R2. 1 2 0 3 1 0 0 3 0 1 0 0 Replace R1 with -2R2i+ R1 0 1 0 0 [0 0 1 -1 Rp R h 0 0 1 -1 To our surprise and delight, when we decode this matrix, we obtain the solution instantly without having to deal with any back-substitution at all. 1 0 0 3 x = 3 0 1 0 0 Decode from the matrix { 0 0 0 1 -1_ z = -1 Note that in the previous discussion, we could have started with R2 and used it to get a zero above its leading 1 and then done the same for the leading 1 in R3. By starting with R3, however, we get more zeros first, and the more zeros there are, the faster the remaining calculations will be.2 It is also worth noting that while a matrix has several3 row echelon forms, it has only one reduced row echelon form. The process by which we have put a matrix into reduced row echelon form is called Gauss-Jordan Elimination. EXAMPLE 8.2.2. Solve the following system using an augmented matrix. Use Gauss-Jordan Elimi- nation to put the augmented matrix into reduced row echelon form. 2 - 3x1+ x4 = 2 2x1 + 4x3 = 5 4x2 - x4 = 3 SOLUTION. We first encode the system into a matrix. (Pay attention to the subscripts!) z2-3x1+z4 2 -3 1 0 1 2 2x1 + 4x3 _=5 Encode into the matrix -2 0 4 0 5 4x2-x - 4 = 3 0 4 0 -1 3 Next, we get a leading 1 in the first column of R1. -3 1 0 1 2 Replace R1 with -R1 [1 0 - j ] 2 0 4 05 3> 2 0 4 0 5 2Carl also finds starting with R3 to be more symmetric, in a purely poetic way. sinfinite, in fact  8.2 SYSTEMS OF LINEAR EQUATIONS: AUGMENTED MATRICES Now we eliminate the nonzero entry below our leading 1. 449 [ 1 2 0 1 0 4 0 4 0 1 0 -1 21 3 Replace R2 with -2R1 + R2 3] [. 1 0 0 1 2 4 0 4 0 1 2 -1 2 3 19 3 ] We proceed to get a leading 1 in R2. -1 1 0 1 2 - 3 3 3 Replace R2 with !R2 0 2 4 2 19 2 33 3 0 4 0 -1 3 We now zero out the entry below the leading 1 in R2. 1- 1 1 0 1 2 - 0 1 6 1 1 Replace R3 with -4R2 + R3 0 4 0 -1 3 . Next, it's time for a leading 1 in R3. -[ 1 0 0 1 3 1 4 0 6 0 1 3 1 -1 2 3 19 2 3 ] -[ 1 0 0 1 1 0 0 6 -24 1 1 -5 2 19 -35 ] 1 0 0 1 1 0 0 6 -24 1 21 3 3 1 1 -5 -35] Replace R3 with -1R3 24 > -1 1 0 1 2 - 3 3 3 0 1 6 1 L9 0 0 1 5 3_ 0 f 24 24] row echelon form, we start with the The matrix is now in row echelon form. To get the reduced last leading 1 we produced and work to get 0's above it. 1 0 0 1 3 1 0 0 6 1 1 2 - 3 19 1 1 5 35 24 24 _ Replace R2 with -6R3 + R2 1 0 0 1 3 1 0 0 0 1 Lastly, we get a 0 above the leading 1 of R2. 1 2 3 3 1 3 4 4 5 35 24 24 5 5 12 12 1 3 - 4 5 35 24 24 ] [1 1 0 1 2 3 3 3 0 1 0 -1 3 0 0 1 5 35 24 24 ] Replace RI with }R2 + RI -[ 1 0 0 0 1 0 0 0 1 ] At last, we decode to get 5 5 - _1r_ - 5 1 0 0 1- 5-2 xl12xn4 = 1- 0 1 0 - 123 Decode from the matrix> 2 1 -4 5 3515 35 0 0 1 2 2z3 + 24x4 =- 2 We have that x4 is free and we assign it the parameter t. We obtain x3= -t + 2, x2 = jt + j, and x1 = 2t - 2. Our solution is {(1t - , 4t + 4, -4t + 2,t) : -oo [ X2 1 _ x4 _ Since the row operations for both processes are the same, all of the arithmetic on the left hand side of the vertical bar is identical in both problems. The only difference between the two processes is what happens to the constants to the right of the vertical bar. As long as we keep these separated into columns, we can combine our efforts into one 'super-sized' augmented matrix and describe the above process as A 1 0 Gauss Jordan Elimination 1 X2 0 1 2 x3z4 We have the identity matrix 12 appearing as the right hand side of the first super-sized augmented matrix and the left hand side of the second super-sized augmented matrix. To our surprise and delight, the elements on the right hand side of the second super-sized augmented matrix are none other than those which comprise A-1. Hence, we have [ A 1 2 ] Gauss Jordan Elimination [ 12 IA-1 In other words, the process of finding A-1 for a matrix A can be viewed as performing a series of row operations which transform A into the identity matrix of the same dimension. We can view this process as follows. In trying to find A--1, we are trying to 'undo' multiplication by the matrix A. The identity matrix in the super-sized augmented matrix [A|I] keeps a running memory of all of the moves required to 'undo' A. This results in exactly what we want, A-1. We are now ready  8.4 SYSTEMS OF LINEAR EQUATIONS: MATRIX INVERSES 475 to formalize and generalize the foregoing discussion. We begin with the formal definition of an invertible matrix. DEFINITION 8.11. An n x n matrix A is said to be invertible if there exists a matrix A-1, read 'A inverse', such that A-1A = AA-1 = In. Note that, according to our definition, invertible matrices are square, and as such, the conditions in Definition 8.11 force the matrix A-1 to be same dimensions as A, that is, n x n. Since not all matrices are square, not all matrices are invertible. However, just because a matrix is square doesn't guarantee it is invertible. (See the exercises.) Our first result summarizes some of the important characteristics of invertible matrices and their inverses. THEOREM 8.6. Suppose A is an n x n matrix. 1. If A is invertible then A-1 is unique. 2. A is invertible if and only if AX = B has a unique solution for every n x r matrix B. The proofs of the properties in Theorem 8.6 rely on a healthy mix of definition and matrix arith- metic. To establish the first property, we assume that A is invertible and suppose the matrices B and C act as inverses for A. That is, BA = AB = In and CA = AC = I. We need to show that B and C are, in fact, the same matrix. To see this, we note that B = InB = (CA)B = C(AB) = CIn = C. Hence, any two matrices that act like A-1 are, in fact, the same matrix.4 To prove the second property of Theorem 8.6, we note that if A is invertible then the discussion on page 472 shows the solution to AX = B to be X = A-1B, and since A-1 is unique, so is A-1B. Conversely, if AX = B has a unique solution for every n x r matrix B, then, in particular, there is a unique solution Xo to the equation AX = In. The solution matrix Xo is our candidate for A-1. We have AXo = In by definition, but we need to also show XoA = In. To that end, we note that A (XoA) = (AXo) A = InA = A. In other words, the matrix XoA is a solution to the equation AX = A. Clearly, X = In is also a solution to the equation AX = A, and since we are assuming ev- ery such equation as a unique solution, we must have XoA = In. Hence, we have XoA = AXo = In, so that Xo = A-1 and A is invertible. The foregoing discussion justifies our quest to find A-1 using our super-sized augmented matrix approach [ A 1 I Gauss Jordan Elimination> [ i A We are, in essence, trying to find the unique solution to the equation AX = In using row operations. What does all of this mean for a system of linear equations? Theorem 8.6 tells us that if we write the system in the form AX = B, then if the coefficient matrix A is invertible, there is only one solution to the system - that is, if A is invertible, the system is consistent and independent.5 We 41f this proof sounds familiar, it should. See the discussion following Theorem 5.2 on page 281. 51t can be shown that a matrix is invertible if and only if when it serves as a coefficient matrix for a system of equations, the system is always consistent independent. It amounts to the second property in Theorem 8.6 where the matrices B are restricted to being ni x 1 matrices. We note for the interested reader that, owing to how matrix multiplication is defined, being able to find unique solutions to AX =B for ni x 1 matrices B gives you the same statement about solving such equations for ni x r matrices - since we can find a unique solution to them one column at a time.  476 SYSTEMS OF EQUATIONS AND MATRICES also know that the process by which we find A-1 is determined completely by A, and not by the constants in B. This answers the question as to why we would bother doing row operations on a super-sized augmented matrix to find A-1 instead of an ordinary augmented matrix to solve a system; by finding A-1 we have done all of the row operations we ever need to do, once and for all, since we can quickly solve any equation AX = B using one multiplication, A-1B. We illustrate this in our next example. 3 1 2 EXAMPLE 8.4.1. Let A = 0 -1 5 2 1 4 1. Use row operations to find A-1. Check your answer by finding A--1A and AA-1. 2. Use A-1 to solve the following systems of equations 3x+y+2z = 26 3x+y+2z = 4 3x+y+2z = 1 (a) -y+5z = 39 (b) -y+5z = 2 (c) -y+5z = 0 2x+y+4z = 117 2x+y+4z = 5 2x+y+4z = 0 SOLUTION. 1. We begin with a super-sized augmented matrix and proceed with Gauss-Jordan elimination. 3 1 2 1 0 0 1 1 21 0 0 0 -1 5 0 1 0 Replace Ri 0 -1 5 0 1 0 L 2 1 4 0 0 1 withR1 2 1 4 0 0 1 1 1 2 1 1 2 1 0 0 0 -1 5 0 10 0 -1 5 0 1 0 2 1 4 0 0 1 -2R1+R3 0- 0 1_ 1 81 0 012 0 0 0 -1 5 0 1 0 Replace 0 1 -5 0 -1 0 0 1 8] 2 0 with(-1)R2 0 1 8 2 0 1 3 3 3 3 3 3- 1 1 2 1 0 0 1 1 2 1 0 0 3 3 3RpaeR Fwt 0 1 -5 0 -1 0 ReplaceR3 0 1 -5 0 -1 0 0 1 8 -2 0 1 -n2+R3 L 0 0 1 2 1 1 3 3 3 3 39 3 9 1 1 2 1 0 0 1 1 2 1 0 0 3 3 3 13pl13 133 3- 0 1 -5 0 -1 03± 0> 0 1 0 - 1 8 0 0 1 2 placeR2with 0 0 1 -2 5R3 + R2 '1~  8.4 SYSTEMS OF LINEAR EQUATIONS: MATRIX INVERSES 477 1 13 0 1 0 0 0 0 1 17 39 10 13 2 13 2 39 8 13 1 13 2 13 15 13 3 13 ] Replace RI with -iR2 + R1 . 1 0 0 0 1 0 0 0 1 9 13 10 13 2 13 2 13 8 13 1 13 7- 13 15 13 3 13. We find A-1 [ A-1A= 2 13 8 13 B 9 10 13 2 13 71 B13 15 T cekiu 13 *Tchcou 131 answer, we compute 2 B3 8 B3 1 13 7 15 3 13 ] [ 3 0 2 1 -1 1 21 5 4] [ 1 0 0 0 1 0 0 0 1 ] I3/ and AA-1= 3 0 2 1 1 1 21 5 4] S9 - 2 B3 8 B3 1 13 7 B3 15 B3 3 13 ] [. 1 0 0 0 1 0 0 0 1 ] I3 / 2. Each of the systems in this part has A as its coefficient matrix. The only difference between the systems is the constants which is the matrix B in the associated matrix equation AX = B. We solve each of them using the formula X = A-1B. (a) X = A-1B [= (b) X = A--1B = (c) X = A--1B = 9 13 10 13 2 13 9 13 10 13 2 13 9 10 2 13 2 13 8 13 1 13 2 13 8 13 1 13 2 8 13 1 7 13 15 13 B 3 13 7 15 13 3 13 7 15 3 13 ] ] ] [ [ [ 26 39 117 2 5 1- 0 0 ] [ [ -39 91 [ 26] 13 19 .We 13 . 1- 13 2 13- Our solution is (-39, 91, 26). get (-,3) ,13 ) . Vefind (9,-130' El In Example 8.4.1, we see that finding one inverse matrix can enable us to solve an entire family of systems of linear equations. There are many examples of where this comes in handy 'in the wild', and we chose our example for this section from the field of electronics. We also take this opportunity to introduce the student to how we can compute inverse matrices using the calculator. 6Note that the solution is the first column of the A-1. The reader is encouraged to meditate on this 'coincidence'.  478 SYSTEMS OF EQUATIONS AND MATRICES EXAMPLE 8.4.2. Consider the circuit diagram below.7 We have two batteries with source voltages VB1 and VB2, measured in volts V, along with six resistors with resistances R1 through R6, measured in kiloohms, kQ. Using Ohm's Law and Kirchhoff's Voltage Law, we can relate the voltage supplied to the circuit by the two batteries to the voltage drops across the six resistors in order to find the four 'mesh' currents: i1, i2, i3 and i4, measured in milliamps, mA. If we think of electrons flowing through the circuit, we can think of the voltage sources as providing the 'push' which makes the electrons move, the resistors as obstacles for the electrons to overcome, and the mesh current as a net rate of flow of electrons around the indicated loops. R5 14 R1 R2 R6 V1 - ii R3 2 R4 2 __-VB2 The system of linear equations associated with this circuit is (R1 + R3)i1-R3i2-R1i =V1 -Ri1 + (R2 + R3 + R4)i2- R4i3-R2i4= 0 -R82+ (R4+ R6)is-8R6i4 = - /2 -R1i1 -R2i2-R6i3 + (R1 + R2 + R5 + R6) i4= 0 1. Assuming the resistances are all 1kg, find the mesh currents if the battery voltages are (a) VB1= 10V and VB2 = 5V (b) VB1= 10V and VB2 =0V (c) VB1 = OV and VB2 = 1OV (d) VB1= 10V and VB2 =10V 2. Assuming V4B1 10V and 1482 =5V, find the possible combinations of resistances which would yield the mesh currents you found in 1(a). 7The authors wish to thank Don Anthan of Lakeland Community College for the design of this example.  8.4 SYSTEMS OF LINEAR EQUATIONS: MATRIX INVERSES SOLUTION. 1. Substituting the resistance values into our system of equations, we get 479 { 221 -22 -24 - V 11 S+ 312 -13 i4 = 0 12+ 223 - 24 - - VB2 i1 - i2 - is + 4i4 0 This corresponds to the matrix equation AX = B where A =[ 2 -1 0 -1] 1 3 -1 -1 o -1 2 -1 -1 -1 -1 4] X -[ _L 1 22 i3 14 1 j B= _L VB1 0 VB2 0 1 When we input the matrix A into the calculator, we find [A]-1 [[1.625 1.25 1.... [1.25 1.5 1.... [1.125 1.25 1.... [1 1 1 .. [A]-1 ... 1.25 1.125 1] ... 1.5 1.25 1] ... 1.25 1.625 1] 1 1 1]] I from which we have A-1 F L 1.625 1.25 1.125 1 1.25 1.5 1.25 1 1.125 1.25 1.625 1 1 1 1 1_ To solve the four systems given to us, we find X by the given values of VB1 and VB2 A-1B where the value of B is determined 1 (a) B = L (a) For V1l = 10] 0 -5 0j 1 (b) B ~F 10 0 0 0 1 j 1(c) F Br1 L 0 0 10 0 1 1 (d) B =F 10 0 10 0 1 10V and VB2 5V, the calculator gives i1 10.625 mA, i2 6.25 mA, 13 = 3.125 mA, and i4 = 5 mA.  480 SYSTEMS OF EQUATIONS AND MATRICES (b) By keeping VB1 = 10V and setting VB2 = OV, we are essentially removing the effect of the second battery. We get i1 = 16.25 mA, i2 = 12.5 mA, i3 = 11.25 mA, and i4=10 mA. [A] -1[B] [[10.625] [6.25 ] [3.125 ] [5 ]] Solution to 1(a) [A]-1[B] [[16.25] [12.5 ] [11.25] [10 ]] Solution to 1(b) (c) Part (c) is a symmetric situation to part (b) in so much as we are zeroing out VB1 and making VB2 = 10. We find i1 = -11.25 mA, i2 = -12.5 mA, i3 = -16.25 mA, and i4 =-10 mA, where the negatives indicate that the current is flowing in the opposite direction as is indicated on the diagram. The reader is encouraged to study the symmetry here, and if need be, hold up a mirror to the diagram to literally 'see' what is happening. (d) For VB1 = 1OV and VB2 = 1OV, we get i1 = 5 mA, i2 = 0 mA, i3 = -5 mA, and i4 = 0 mA. The mesh currents i2 and i4 being zero is a consequence of both batteries 'pushing' in equal but opposite directions, causing the net flow of electrons in these two regions to cancel out. [A]-1[B] [[-11.25] [-12.5 ] [-16.25] [-10 ]] [A]-1[B] [[5 ] [0 ] [-5] [0 ]] Solution to 1(d) = 5V, ii = 10.625 mA, i2 = 6.25 mA, the resistance values. Rewriting our Solution to 1(c) 2. We now turn the tables and are given VB1 = 10V, VB2 is = 3.125 mA and i4 = 5 mA and our unknowns are system of equations, we get { 5.625R1 1.25R2 - 4.375R3 -3.125R4 -5.625R1 - 1.25R2 + 5R5 + 4.375R3 + 3.125R4 - 1.875R6 + 1.875R6 10 0 -5 0 The coefficient matrix for this system is 4 x 6 (4 equations with 6 unknowns) and is therefore not invertible. We do know, however, this system is consistent, since setting all the resis- tance values equal to 1 corresponds to our situation in problem la. This means we have an  8.4 SYSTEMS OF LINEAR EQUATIONS: MATRIX INVERSES 481 underdetermined consistent system which is necessarily dependent. To solve this system, we encode it into an augmented matrix [ 5.25 0 4.375 0 1.25 0 0 -5.625 -1.25 -4.375 0 0 0 0 0 3.125 0 0 -3.125 0 -1.875 0 5 1.875 10 0 -5 0 ] and use the calculator to write in reduced row echelon form 1 0 0 0 0 1 0 0 0.7 -3.5 0 0 0 0 1 0 0 0 0 1 0 -1.5 0.6 0 1.7 -4 1.6 1 Decoding this system from the matrix, we get { R1 + 0.7R3 3.5R3 - 1.5R6 R4 + 0.6R6 R5 1.7 -4 1.6 1 We have can solve for R1, R2, R4 and R5 leaving R3 and R6 as free variables. Labeling R3 = s and R6 = t, we have R1 = -0.7s + 1.7, R2 = 3.5s + 1.5t - 4, R4 = -0.6t + 1.6 and R5 = 1. Since resistance values are always positive, we need to restrict our values of s and t. We know R3 = s > 0 and when we combine that with R1 = -0.7s + 1.7 > 0, we get 0 < s < 7. Similarly, R6 = t > 0Oand with R4 = -0.6t + 1.6 > 0, we find 0 < t < 3.In order visualize the inequality R2 = 3.5s + 1.5t - 4 > 0, we graph the line 3.5s + 1.5t - 4 = 0 on the st-plane and shade accordingly.8 Imposing the additional conditions 0 < s < 7 and 0 < t < g, we find our values of s and t restricted to the region depicted on the right. Using the roster method, the values of s and t are pulled from the region {(s, t) : 0 < s < 7, 0 < t < g, 3.5s + 1.5t - 4 > 0}. The reader is encouraged to check that the solution presented in 1(a), namely all resistance values equal to 1, corresponds to a pair (s, t) in the region. 8See Section 2.4 for a review of this procedure.  482 SYSTEMS OF EQUATIONS AND MATRICES S S 3! 16 27 -2 -1 1 2 t -2 -1 I 1 2 4 t -1 - -I -1 i f 3 The region where 3.5s + 1.5t - 4 > 0 The region for our parameters s and t.  8.4 SYSTEMS OF LINEAR EQUATIONS: MATRIX INVERSES 483 8.4.1 EXERCISES 1. Find the inverse of the matrix or state that the matrix is not invertible. (a) A [1 2] ()A= 3 4 _ (b)B[12 -7 (b) = 5 3 (c) C [ 6 15 (c)C = 14 35 (d) D[2 -1 (e) D = 16 -9 3 0 (e) E = 2 -1 -3 2 4 6 -3 (f) F 3 4 -3 L1 2 6_ 1 2 3 (g) G = 2 3 11 3 4 19 1 0 -3 (h H 2 -2 8 -5 0 16 1 0 4 0 7 0 1 ] 4] 3 -5_ 2. Use a matrix inverse to solve the following systems of linear equations. (a) { J+ 7y 5x + 12y 26 39 (b) 3x + 7y 5x + 12y 0 ( 3x + 7y -1 { 5x + 12y -7 5 3. Use the inverse of E from Exercise I above to solve the following systems of linear equations. (a) { 3x+4z 2x -y+3z -3x + 2y - 5z 1 0 (b) 0 3x+4z 2x - y + 3z -3x + 2y - 5z 0 3x+4z 1 (c) 2x - y + 3z 0 -3x + 2y - 5z 0 0 1 4. Using the matrices A, B and D from Exercise 1, show AB = D and D-1 = B-1A-1. That is, show that (AB)-1 = B-1A-1. 5. Let M and N be invertible n x n matrices. Show that (MN)-1 = N-1M-1 and compare your work to Exercise 8 in Section 5.2. 6. This exercise is a continuation of Example 8.3.3 in Section 8.3 and gives another application of matrix inverses. Recall that given the position matrix P for a point in the plane, the matrix RP corresponds to a point rotated 450 counterclockwise from P where 2 f/~ (a) Find R-1  484 SYSTEMS OF EQUATIONS AND MATRICES (b) If RP rotates a point counterclockwise 450, what should R-1P do? Check your answer by finding R-1P for various points on the coordinate axes and the lines y = x and y =_-x. (c) Find R-1P where P corresponds to a generic point P(x, y). Verify that this takes points on the curve y =-j to points on the curve 2 _Y2 -4. 7. Matrices can be used in cryptography. Check out this website for a nice example of how this is done. With the help of your classmates, see if you can create a 3 x 3 matrix M with integer entries such that M-1 also has integer entries and then use your matrix to encode the message.  8.4 SYSTEMS OF LINEAR EQUATIONS: MATRIX INVERSES 485 8.4.2 ANSWERS -2 1 1. (a) A--' 3 i.1 2 2 _ (b) B-1- 3 7 1 L5 12 J (c) C is not invertible 9 1 (d) D-1 = 2 2 8 -1 8 -1 8 (e) E-1 = 1 -3 1 -6 5 7 (f) F4 4- (g) G is not invertible 21 4 1 16 -90 (h)H = 5 _-36 0 3 0] 1 35 7 I j - I 0 1 0 0 -7 1] 4 -13 -3 2. The coefficient matrix is B-1 from Exercise 1 above so the inverse we need is (B-1)-1 = B. The solutions to the systems are (a) [ (b) [ (c) [ 3. (a) [ (b) [ (c) [ 12 -5 12 -5 12 -5 -1 1 1 -1 1 1 -1 1 1 -7 3 -7 3 -7 3 I I I [ [ [ 26 391 39 J -I Sox = 097 -1=[-35 Sox= 39 and y 7andy= -13. -3. -119 and y = 50. =-1,y=1andz=1. 8 -3 -6 8 -3 -6 8 -3 -6 4 -1 -3 4 -1 -3 4 -1 -3 I I I [ [ [ 11 0 0] 0 1 0 0 0 1 [ [ [ -1 1 1 8 -3 -6 4] So x So x .3 and z =-6. So x 4, y -1 and z =-3.  486 SYSTEMS OF EQUATIONS AND MATRICES 8.5 DETERMINANTS AND CRAMER'S RULE 8.5.1 DEFINITION AND PROPERTIES OF THE DETERMINANT In this section we assign to each square matrix A a real number, called the determinant of A, which will eventually lead us to yet another technique for solving consistent independent systems of linear equations. The determinant is defined recursively, that is, we define it for 1 x 1 matrices and give a rule by which we can reduce determinants of n x n matrices to a sum of determinants of (n - 1) x (n - 1) matrices.1 This means we will be able to evaluate the determinant of a 2 x 2 matrix as a sum of the determinants of 1 x 1 matrices; the determinant of a 3 x 3 matrix as a sum of the determinants of 2 x 2 matrices, and so forth. To explain how we will take an n x n matrix and distill from it an (n - 1) x (n - 1), we use the following notation. DEFINITION 8.12. Given an n x n matrix A where n > 1, the matrix Aij is the (n - 1) x (n - 1) matrix formed by deleting the ith row of A and the jth column of A. For example, using the matrix A below, we find the matrix A23 by deleting the second row and third column of A. A Delete R2 and C3 23 3 1 We are now in the position to define the determinant of a matrix. DEFINITION 8.13. Given an n x n matrix A the determinant of A, denoted det(A), is defined as follows " If n = 1, then A = [an] and det(A) = det ([anl]) = an. " If n > 1, then A = [aij]nxn and det(A) = det ([aij]nn) = an det (A11) - a12 det (A12) + - ... + (-1)1+nain det (Ain) There are two commonly used notations for the determinant of a matrix A: 'det(A)' and 'lA' We have chosen to use the notation det(A) as opposed to A because we find that the latter is often confused with absolute value, especially in the context of a 1 x 1 matrix. In the expansion anl det (A11) - a12 det (A12) +- ... + (-1)1+nain det (Ain), the notation '+-. .. + (-1)1+nain' means that the signs alternate and the final sign is dictated by the sign of the quantity (-1)1+n. Since the entries anl, a12 and so forth up through ain comprise the first row of A, we say we are finding the determinant of A by 'expanding along the first row'. Later in the section, we will develop a formula for det(A) which allows us to find it by expanding along any row. Applying Definition 8.13 to the matrix A = 2 1 we get 'We will talk more about the term 'recursively' in Section 9.1.  8.5 DETERMINANTS AND CRAMER'S RULE 487 det (A) = detI1L24- = 4 det (All) - (-3) det (A12) = 4det([1]) + 3det([2]) = 4(1)+3(2) = 10 For a generic 2 x 2 matrix AL = c dwe get det(A) = det([a b1) = adet (All) - bdet (A12) = a det ([d]) - b det ([c]) = ad - bc This formula is worth remembering EQUATION 8.1. For a 2 x 2 matrix, det([a )ad-bc 3 1 2 Applying Definition 8.13 to the 3 x 3 matrix A =[0 -1 5 we obtain 2 1 4_ 3 1 2 det(A) = det(0 -1 5 2 1 4 = 3 det (All) - 1 det (A12) + 2 det (A13) = 3 det([-i4]) det([ 4])+2det([ 1J) = 3((-1)(4) - (5)(1)) - ((0)(4) - (5)(2)) + 2((0)(1) - (-1)(2)) = 3(-9) - (-10) + 2(2) = -13 To evaluate the determinant of a 4 x 4 matrix, we would have to evaluate the determinants of four 3 x 3 matrices, each of which involves the finding the determinants of three 2 x 2 matrices. As you can see, our method of evaluating determinants quickly gets out of hand and many of you may be reaching for the calculator. There is some mathematical machinery which can assist us in calculating determinants and we present that here. Before we state the theorem, we need some more terminology.  488 SYSTEMS OF EQUATIONS AND MATRICES I ---I DEFINITION 8.14. Let A be minor of A, denoted Mig is defined by Ci = (-1)i+JM 2 an n x n matrix and Aig be defined as in Definition 8.12. The ij defined by Mig = det (Ag3). The ij cofactor of A, denoted Ci3 is = (-1)i+J det (Agg). We note that in Definition 8.13, the sum anl det (All) - ale det (A12) + - ... + (-1)1+na1 det (Ain) can be rewritten as a11(-1)1+1 det (A11) + a12(-1)1+2 det (A12) + ... + a1n(-1)1+n det (Ain) which, in the language of cofactors is a11C11 + a12C12 + ... + a1nC1n We are now ready to state our main theorem concerning determinants. THEOREM 8.7. Properties of the Determinant: Let A = [aj].. " We may find the determinant by expanding along any row. That is, for any 1 < k < n, det(A) = ak1Ck1 + ak2Ck2 + ... + aknCkn " If A' is the matrix obtained from A by: - interchanging any two rows, then det (A') - det (A). - replacing a row with a nonzero multiple (say c) of itself, then det(A') = c det(A) - replacing a row with itself plus a multiple of another row, then det(A') = det(A) " If A has two identical rows, or a row consisting of all 0's, then det(A) = 0. " If A is upper or lower triangulara then det(A) is the product of the entries on the main diagonal. 1) " If B is an n x n matrix, then det(AB) = det(A) det(B). " det (A") = det(A)n for all natural numbers n. 1 " A is invertible if and only if det(A) $ 0. In this case, det (A-1) = . det (A) aSee Exercise 5 in 8.3. bSee page 461 in Section 8.3. Unfortunately, while we can easily demonstrate the results in Theorem 8.7, the proofs of most of these properties are beyond the scope of this text. We could prove these properties for generic 2 x 2 or even 3 x 3 matrices by brute force computation, but this manner of proof belies the elegance and  8.5 DETERMINANTS AND CRAMER'S RULE 489 symmetry of the determinant. We will prove what few properties we can after we have developed some more tools such as the Principle of Mathematical Induction in Section 9.3.2 For the moment, let us demonstrate some of the properties listed in Theorem 8.7 on the matrix A below. (Others will be discussed in the Exercises.) 3 1 2 A = 0 -1 5 2 1 4 We found det(A) = -13 by expanding along the first row. Theorem 8.7 guarantees that we will get the same result if we expand along the second row. (Doing so would take advantage of the 0 there.) 3 1 2 det( 0 -1 5 = 0j) + (-1)C2 + 50, 2 1 4 (-1)(-1)2+2 det (A22) + 5(-1)2+3 det (A23) - det(-I5])5det([:21) ((3)(4)- (2)(2)) - 5((3)(1) - (2)(1)) -8-5 = -13 / In general, the sign of (-1)i+j in front of the minor in the expansion of the determinant follows an alternating pattern. Below is the pattern for 2 x 2, 3 x 3 and 4 x 4 matrices, and it extends naturally to higher dimensions. The reader is cautioned, however, against reading too much into these sign patterns. In the example above, we expanded the 3 x 3 matrix A by its second row and the term which corresponds to the second entry ended up being negative even though the sign attached to the minor is (+). These signs represent only the signs of the (-1)i±J in the formula; the sign of the corresponding entry as well as the minor itself determine the ultimate sign of the term in the expansion of the determinant. To illustrate some of the other properties in Theorem 8.7, we use row operations to transform our 3 x 3 matrix A into an upper triangular matrix, keeping track of the row operations, and labeling 2For a very elegant treatment, take a course in Linear Algebra. There, you will most likely see the treatment of determinants logically reversed than what is presented here. Specifically, the determinant is defined as a function which takes a square matrix to a real number and satisfies some of the properties in Theorem 8.7. From that function, a formula for the determinant is developed.  490 SYSTEMS OF EQUATIONS AND MATRICES each successive matrix.3 3 1 2 3 1 2 3 1 2 0 -1 5 Replace R3 0 -1 5 Replace R3 with 0 -1 5 2 1 4 with -3R1+R3 0 1 l8a2+R3 0 0 135 A B C Theorem 8.7 guarantees us that det(A) = det(B) = det(C) since we are replacing a row with itself plus a multiple of another row moving from one matrix to the next. Furthermore, since C is upper triangular, det(C) is the product of the entries on the main diagonal, in this case det(C) = (3)(-1) ( ) -13. This demonstrates the utility of using row operations to assist in calculating determinants. This also sheds some light on the connection between a determinant and invertibility. Recall from Section 8.4 that in order to find A-1, we attempt to transform A to In using row operations [ A In ] Gauss Jordan Elimination> [ _iA-1 As we apply our allowable row operations on A to put it into reduced row echelon form, the determinant of the intermediate matrices can vary from the determinant of A by at most a nonzero multiple. This means that if det(A) $ 0, then the determinant of A's reduced row echelon form must also be nonzero, which, according to Definition 8.4 means that all the main diagonal entries on A's reduced row echelon form must be 1. That is, A's reduced row echelon form is In, and A is invertible. Conversely, if A is invertible, then A can be transformed into In using row operations. Since det (In) = 1 # 0, our same logic implies det(A) $ 0. Basically, we have established that the determinant determines whether or not the matrix A is invertible.4 It is worth noting that when we first introduced the notion of a matrix inverse, it was in the context of solving a linear matrix equation. In effect, we were trying to 'divide' both sides of the matrix equation AX = B by the matrix A. Just like we cannot divide a real number by 0, Theorem 8.7 tells us we cannot 'divide' by a matrix whose determinant is 0. We also know that if the coefficient matrix of a system of linear equations is invertible, then system is consistent and independent. It follows, then, that if the determinant of said coefficient is not zero, the system is consistent and independent. 8.5.2 CRAMER'S RULE AND MATRIX ADJOINTS In this section, we introduce a theorem which enables us to solve a system of linear equations by means of determinants only. As usual, the theorem is stated in full generality, using numbered unknowns zi, x2, etc., instead of the more familiar letters x, y, z, etc. The proof of the general case is best left to a course in Linear Algebra. 3Essentially, we follow the Gauss Jordan algorithm but we don't care about getting leading l's. 4As we will see in Section 8.5.2, determinants (specifically cofactors) are deeply connected with the inverse of a matrix.  8.5 DETERMINANTS AND CRAMER'S RULE 491 THEOREM 8.8. Cramer's Rule: Suppose AX = B is the matrix form of a system of n linear equations in n unknowns where A is the coefficient matrix, X is the unknowns matrix, and B is the constant matrix. If det(A) $ 0, then the corresponding system is consistent and independent and the solution for unknowns x1, x2, ... x is given by: - det(Aj) ' det(A) where A3 is the matrix A whose jth column has been replaced by the constants in B. In words, Cramer's Rule tells us we can solve for each unknown, one at a time, by finding the ratio of the determinant of A3 to that of the determinant of the coefficient matrix. The matrix Ai is found by replacing the column in the coefficient matrix which holds the coefficients of xz with the constants of the system. The following example fleshes out this method. EXAMPLE 8.5.1. Use Cramer's Rule to solve for the indicated unknowns. 1. Solve ( 2x1 - 3x2 = 51i + 2 = 2x - 3y + z 2. Solve x - y + z 3x - 4z 4 for x1 and x2 -2 -1 1 for z. 0 SOLUTION. 1. Writing this system in matrix form, we find A=[2 -3] 51 X= Z To find the matrix A1, we remove the column of the coefficient matrix A which holds the coefficients of xi and replace it with the corresponding entries in B. Likewise, we replace the column of A which corresponds to the coefficients of x2 with the constants to form the matrix A2. This yields Al -[ - 1] A 2 -4 1 Computing determinants, we get det(A) = 17, det (A1) -2 and det (A2) -24, so that det (A1) det(A) 2 17 det (A2) det(A) 24 17 The reader can check that the solution to the system is (- -).  492 SYSTEMS OF EQUATIONS AND MATRICES 2. To use Cramer's Rule to find z, we identify x3 as z. We have 2 -3 1x -1 2 -3 -1 A= 1 -1 1 X= y B= 1 A3=Az 1 -1 1 3 0 -4 z 0 3 0 0 Expanding both det(A) and det (Az) along the third rows (to take advantage of the 0's) gives det(Az) -12 6 det(A) -10 5 The reader is encouraged to solve this system for x and y similarly and check the answer. D Our last application of determinants is to develop an alternative method for finding the inverse of a matrix. Let us consider the 3 x 3 matrix A which we so extensively studied in Section 8.5.:1 3 1 2 A = 0 -1 5 2 1 4 We found through a variety of methods that det(A) = -13. To our surprise and delight, its inverse below has a remarkable number of 13's in the denominators of its entries. This is no coincidence. 9 2 7_ A-1= 10 8 15 13 13 13] 2 1 3 Recall that to find A-1, we are essentially solving the matrix equation AX =13, where X = [x]33 is a 3 x 3 matrix. Because of how matrix multiplication is defined, the first column of I3 is the product of A with the first column of X, the second column of I3 is the product of A with the second column of X and the third column of I3 is the product of A with the third column of X.6 In other words, we are solving three equations z11 1 x12 0 xi3 0 A x21 = 0 A z = 1 A z23 = 0 x31_ 0 _x32_ 0 _z3 _ 1 We can solve each of these systems using Cramer's Rule. Focusing on the first system, we have 1 1 2 3 1 2 3 1 1 A1= 0 -1 5 A2= 0 0 5 A3= 0 -1 0 0 1 4_ 2 0 4_ 2 1 0_ 5We are developing a method in the forthcoming discussion. As with the discussion in Section 8.4 when we developed the first algorithm to find matrix inverses, we ask that you indulge us. 6The reader is encouraged to stop and think this through.  8.5 DETERMINANTS AND CRAMER'S RULE 493 If we expand det (A1) along the first row, we get det (A1) = det - det I + 2 det(L00 - = det[1 5]) Amazingly, this is none other than the C11 cofactor of A. The reader is invited to check this, as well as the claims that det (A2) = C12 and det (A3) = C13.' (To see this, though it seems unnatural to do so, expand along the first row.) Cramer's Rule tells us det (A1) z1 = det(A) C11 det(A)' X21 det (A2) det(A) C12 det(A)' X31 det (A3) det(A) C13 det(A) So the first column of the inverse matrix X is: [ 11 X21 X31 I Cn det(A) C12 det(A) C13 det(A) 1 det(A) C1] C12 C13 Notice the reversal of the subscripts going from the unknown to the corresponding cofactor of A. This trend continues and we get [ X12 X22 z32 j 1 det(A) C21 C22 C23] X13 X23 1:33 j 1 det(A) C31 C32 C33_ Putting all of these together, we have obtained a new and surprising formula for A-1, namely 1 det(A) C11 C12 C13 C21 C22 C23 C31 C32 C33] To see that this does indeed yield A-1, we find all of the cofactors of A C11 C12 C13 -9, C21 10, C22 2, C23 -2, C31 8, C32 -1, C33 7 -15 -3 And, as promised, 7In a solid Linear Algebra course you will learn that the properties in Theorem 8.7 hold equally well if the word 'row' is replaced by the word 'column'. We're not going to get into column operations in this text, but they do make some of what we're trying to say easier to follow.  494 SYSTEMS OF EQUATIONS AND MATRICES 9 2 7 1 C1C21 C3 1 -9 -2 7 A-1- =C12 C22 C32 = 10 8 -15 = 10 3 13 det(A)C13 C23 C 2 -1 -3 -z2 1 3 ---- . 13 13 13 . To generalize this to invertible n x n matrices, we need another definition and a theorem. Our definition gives a special name to the cofactor matrix, and the theorem tells us how to use it along with det(A) to find the inverse of a matrix. DEFINITION 8.15. Let A be an n x n matrix, and Ci3 denote the ij cofactor of A. The adjoint of A, denoted adj(A) is the matrix whose ij-entry is the ji cofactor of A, C. That is Cn1C21 ... Cn1 Cn Cn... Cn2I a d j ( A ) =I.2. " . C1f C2n ... Cn_ This new notation greatly shortens the statement of the formula for the inverse of a matrix. THEOREM 8.9. Let A be an invertible n x n matrix. Then 1 A--' = 1 adj(A) det(A) For 2 x 2 matrices, Theorem 8.9 reduces to a fairly simple formula. EQUATION 8.2. For an invertible 2 x 2 matrix, a b - 1 d -b c d] ad-bc[-c a _ The proof of Theorem 8.9 is, like so many of the results in this section, best left to a course in Linear Algebra. In such a course, not only do you gain some more sophisticated proof techniques, you also gain a larger perspective. The authors assure you that persistence pays off. If you stick around a few semesters and take a course in Linear Algebra, you'll see just how pretty all things matrix really are - in spite of the tedious notation and sea of subscripts. Within the scope of this text, we will prove a few results involving determinants in Section 9.3 once we have the Principle of Mathematical Induction well in hand. Until then, make sure you have a handle on the mechanics of matrices and the theory will come eventually.  8.5 DETERMINANTS AND CRAMER'S RULE 495 8.5.3 EXERCISES 1. Compute the determinant of the following matrices. Exercise 1 in Section 8.4.) (Some of these matrices appeared in (a) B= (b) C= (c) Q [ (d) L =[ (e) F [ 12 -7 -5 3 6 151 14 35 _ 1 2x __ 1 ln(x) x3 3 _ 1-31n(x) X4 -4 4 6 -3 3 4 -3 1 2 6] (f) G= [ (g) V= (h) H [ 1 2 3 - 2 3 3 11 4 19 ij 1 0 9 -4 1 0 2 -2 5 0 1 0 k 5 -2 -3 8 16 4 I 0 7 0 1 ] 2. Use Cramer's Rule to solve the system of linear equations. { 3+7y () 5x + 12y 26 39 (b) { x+y+z 2x - y + z -3x+5y+7z 3 0 7 3. Use Cramer's Rule to solve for x4 in the following system of linear equations. { I - 3 2x2 -4 1- 2x2 + 13 -z3 + 14 -2 0 0 1 4. Find the inverse of the following matrices using their determinants and adjoints. 5. 1 (a) B 12 -7 1 (a)B= -5 3 Let R= 7 3 11 2 _' (a) Show that det(RS) (bc) Show that det(U) 4 6 -3 (b) F = 3 4 -3 1 2 6_ T-[ 11 2 - 3] and U= 36 15 1[ -5 S 6 9 _ = det(R) det(S) -det(R) -3 det(S)  496 SYSTEMS OF EQUATIONS AND MATRICES 6. For M and N below, show that det(M) = 0 and det(N) = 0. 1 2 3 1 2 3 M 1 2 3 , N=44 5 6 4 5 6 0 0 0 7. Let A be an arbitrary invertible 3 x 3 matrix. (a) Show that det(Is) = 1.8 (b) Using the facts that AA-1- =13 and det(AA-1) = det(A) det(A-1), show that 1 det(A-1) = det (A) 8. The purpose of this exercise is to introduce you to the eigenvalues and eigenvectors of a matrix.9 We begin with an example using a 2 x 2 matrix and then guide you through some exercises using a 3 x 3 matrix. Consider the matrix 6 15 C =] C 14 35 from Exercise 1 above. We know that det(C) = 0 which means that CX = 02x2 does not have a unique solution. So there is a nonzero matrix Y such that CY = 02x2. In fact, every matrix of the form -5t Y = 2 t is a solution to CX = 02x2, so there are infinitely many matrices such that CX = 02x2. But consider the matrix 3 Xu= X41[7 It is NOT a solution to CX = 02x2, but rather, CX4 = 6 15 3 123 41 3 * 14 35 7 287 7 In fact, if Z is of the form Z = 7 t 8If you think about it for just a moment, you'll see that det(I,) = 1 for any natural number n. The formal proof of this fact requires the Principle of Mathematical Induction (Section 9.3) so we'll stick with nr= 3 for the time being. 9This material is usually given its own chapter in a Linear Algebra book so clearly we're not able to tell you everything you need to know about eigenvalues and eigenvectors. They are a nice application of determinants, though, so we're going to give you enough background so that you can start playing around with them.  8.5 DETERMINANTS AND CRAMER'S RULE 497 then 6 15 7t 123t a3t CZ -14 -5 =-41 7 =41Z 1 35_ t 41t t for all t. The big question is "How did we know to use 41?" We need a number A such that CX = AX has nonzero solutions. We have demonstrated that A = 0 and A = 41 both worked. Are there others? If we look at the matrix equation more closely, what we really wanted was a nonzero solution to (C - A12)X =02x2 which we know exists if and only if the determinant of C - AI2 is zero.10 So we computed det(C-A12) =det 14 1 (6-A)(35-A)-14.15=A2-41A ( 2) ~14 35 - A ( This is called the characteristic polynomial of the matrix C and it has two zeros: A = 0 and A = 41. That's how we knew to use 41 in our work above. The fact that A = 0 showed up as one of the zeros of the characteristic polynomial just means that C itself had determinant zero which we already knew. Those two numbers are called the eigenvalues of C. The corresponding matrix solutions to CX = AX are called the eigenvectors of C and the 'vector' portion of the name will make more sense after you've studied vectors. Okay, you should be mostly ready to start on your own. In the following exercises, you'll be using the matrix 1 2 3 G = 2 3 11 3 4 19 from Exercise 1 above. (a) Show that the characteristic polynomial of G is p(A) =_-A(A - 1)(A - 22). That is, compute the determinant of G - A13. (b) Let Go = G. Find the parametric description of the solution to the system of linear equations given by GX = 03X3. (c) Let G1 = G - 13. Find the parametric description of the solution to the system of linear equations given by G1X = 03x3. Show that any solution to G1X = 03x3 also has the property that GX = IX. (d) Let G22 = G - 2213. Find the parametric description of the solution to the system of linear equations given by G22X = 03X3. Show that any solution to G22X = 03X3 also has the property that GX = 22X. ioThink about this.  498 498 SYSTEMS OF EQUATIONS AND MATRICES 8.5.4 ANSWERS (a) (b) (c) (d) det (B) det(C) tQ det(L) 9 ~1 -x7 (e) det(F) (f) det(G) (g) det(V) (h) det(H) -12 0 20i + 43j + 4k -2 2.(a) x =3 9,y 13 (b) x=1, y=2, z=0 3. x 44 4. (a) B 1 (b F 1 3 5 J7 5 7 1 7 9 1 11 2i  8.6 PARTIAL FRACTION DECOMPOSITION 499 8.6 PARTIAL FRACTION DECOMPOSITION This section uses systems of linear equations to rewrite rational functions in a form more palatable to Calculus students. In College Algebra, the function 92 - x- 6 (1) is written in the best form possible to construct a sign diagram and to find zeros and asymptotes, but certain applications in Calculus require us to rewrite f(x) as z+7 1 6(2 f (z)= 2 2 (2) X2+1 xx2 If we are given the form of f(x) in (2), it is a matter of Intermediate Algebra to determine a common denominator to obtain the form of f (x) given in (1). The focus of this section is to develop a method by which we start with f(x) in the form of (1) and 'resolve it into partial fractions' to obtain the form in (2). Essentially, we need to reverse the least common denominator process. Starting with the form of f(x) in (1), we begin by factoring the denominator 2 -x-6 x2-x-6 x4 + x2 x2 (x2 + 1) We now think about which individual denominators could contribute to obtain 2 (2 + 1) as the least common denominator. Certainly 2 and 2 + 1, but are there any other factors? Since 2 + 1 is an irreducible quadratic1 there are no factors of it that have real coefficients which can contribute to the denominator. The factor 2, however, is not irreducible, since we can think of it as 2 = xx= (x - 0)(x - 0), a so-called 'repeated' linear factor.2 This means it's possible that a term with a denominator of just x contributed to the expression as well. What about something like x (2 + 1)? This, too, could contribute, but we would then wish to break down that denominator into x and (2 + 1), so we leave out a term of that form. At this stage, we have guessed 2 -x-6 2-x-6 ? ? +2 x2(x2 +1) x X2+ x2+1 Our next task is to determine what form the unknown numerators take. It stands to reason that since the expression 4-26 is 'proper' in the sense that the degree of the numerator is less than the degree of the denominator, we are safe to make the ansatz that all of the partial fraction resolvents are also. This means that the numerator of the fraction with x as its denominator is just a constant and the numerators on the terms involving the denominators 2 and 2+ 1 are at most linear polynomials. That is, we guess that there are real numbers A, B, C, D and £ so that z2-z-6 z2-z-6 A Bz+C Dz+E x4+x2 x2(x2+1) xz x2 + 2+1 1Recall this means it has no real zeros; see Section 3.4. 2Recall this means x 0 is a zero of multiplicity 2.  500 SYSTEMS OF EQUATIONS AND MATRICES However, if we look more closely at the term z2C, we see that Bx+C _C= _+ 2 = $ + 2. The term J has the same form as the term A which means it contributes nothing new to our expansion. Hence, we drop it and, after re-labeling, we find ourselves with our new guess: x2-x-6 2x6 A B Cx+D 4+x2 x2(x2+1) - 2 + +2 x2+1 Our next task is to determine the values of our unknowns. Clearing denominators gives 2 -x-6=Ax(2+1) +B(2+1) +(Cx+D)z2 Gathering the like powers of x we have x2 -x-6 = (A+C)x3+ (B+D)x2+Ax+B In order for this to hold for all values of x in the domain of f, we equate the coefficients of corresponding powers of x on each side of the equation3 and obtain the system of linear equations (E1) A + C = 0 From equating coefficients of x3 (E2) B + D = 1 From equating coefficients of 2 (E3) A = -1 From equating coefficients of x (E4) B = -6 From equating the constant terms To solve this system of equations, we could use any of the methods presented in Sections 8.1 through 8. 5, but none of these methods are as efficient as the good old-fashioned substitution you learned in Intermediate Algebra. From E3, we have A = -1 and we substitute this into El to get C = 1. Similarly, since E4 gives us B = -6, we have from E2 that D = 7. Our final answer is 2 -x-6 2-x-6 1 6 x +7 x4 + x2 -x2 (2 + 1) - 72+ 2 + 1 which matches the formula given in (2). As we have seen in this opening example, resolving a rational function into partial fractions takes two steps: first, we need to determine the form of the decomposition, and then we need to determine the unknown coefficients which appear in said form. Theorem 3.16 guarantees that any polynomial with real coefficients can be factored over the real numbers as a product of linear factors and irreducible quadratic factors. Once we have this factorization of the denominator of a rational function, the next theorem tells us the form the decomposition takes. The reader is encouraged to review the Factor Theorem (Theorem 3.6) and its connection to the role of multiplicity to fully appreciate the statement of the following theorem. 3We will justify this shortly.  8.6 PARTIAL FRACTION DECOMPOSITION 501 THEOREM 8.10. Suppose R(x) =N(x) is a rational function where the degree of N(x) less than D(x) the degree of D(x) a and N(x) and D(x) have no common factors. " If c is a real zero of D of multiplicity m which corresponds to the linear factor ax + b, the partial fraction decomposition includes A1 A2 Am + + + a~~ ax+b (ax+b)2 (ax+b) for real numbers A1, A2, ... Am. " If c is a non-real zero of D of multiplicity m which corresponds to the irreducible quadratic ax2 + bx + c, the partial fraction decomposition includes B1x+C1 B2x+C2 + Bmx+Cm +...(ax2e+bx+c)m ax2 +bx+c (ax2+bx+c)2 - ( for real numbers B1, B2, ... Bm and C1, C2, ... Cm. aIn other words, R(x) is a proper rational function. The proof of Theorem 8.10 is best left to a course in Abstract Algebra. Notice that the theorem provides for the general case, so we need to use subscripts, A1, A2, etc., to denote different unknown coefficients as opposed to the usual convention of A, B, etc.. The stress on multiplicities is to help us correctly group factors in the denominator. For example, consider the rational function 3x - 1 (x2 - 1) (2 - x - x2) Factoring the denominator to find the zeros, we get (x + 1)(x - 1)(1 - x)(2 + x). We find x= -1 and x =-2 are zeros of multiplicity one but that x = 1 is a zero of multiplicity two due to the two different factors (x - 1) and (1 - x). One way to handle this is to note that (1 - x) = -(X - 1) so 3x- 1 (x+1)(x-1)(1-x)(2+x) 3x-1 -(x - 1)2(x + 1)(x + 2) 1-3x (z -1)2( + 1)( + 2) from which we proceed with the partial fraction decomposition 1-3x (z -1)2( + 1)( + 2) A B C D - 1 + (z-1)2+x+1+x+2 Turning our attention to non-real zeros, we note that the tool of choice to determine the irreducibil- ity of a quadratic ax2 + bx+ c is the discriminant, b2 - 4ac. If b2 - 4ac < 0, the quadratic admits a pair of non-real complex conjugate zeros. Even though one irreducible quadratic gives two distinct non-real zeros, we list the terms with denominators involving a given irreducible quadratic only once to avoid duplication in the form of the decomposition. The trick, of course, is factoring the denominator or otherwise finding the zeros and their multiplicities in order to apply Theorem 8.10.  502 SYSTEMS OF EQUATIONS AND MATRICES We recommend that the reader review the techniques set forth in Sections 3.3 and 3.4. Next, we state a theorem that if two polynomials are equal, the corresponding coefficients of the like powers of x are equal. This is the principal by which we shall determine the unknown coefficients in our partial fraction decomposition. THEOREM 8.11. Suppose anzx + an_1xn- + ... a2x2 + a1x + ao = bmzm + mmixm-1 + ... b2x2 + b1x + be for all xin an open interval I. Then n = m and a2 = b2 for all i = 1 ... n. Believe it or not, the proof of Theorem 8.11 is a consequence of Theorem 3.14. Define p(x) to be the difference of the left hand side of the equation in Theorem 8.11 and the right hand side. Then p(x) = 0 for all x in the open interval I. If p(x) were a nonzero polynomial of degree k, then, by Theorem 3.14, p could have at most k zeros in I, and k is a finite number. Since p(x) = 0 for all the x in I, p has infinitely many zeros, and hence, p is the zero polynomial. This means there can be no nonzero terms in p(x) and the theorem follows. Arguably, the best way to make sense of either of the two preceding theorems is to work some examples. EXAMPLE 8.6.1. Resolve the following rational functions into partial fractions. z+5 1. R(x) = 2 x - -1 2x2 --1 3. R(x) 3 - -z 2+x 5. R(x) x =3 +652+9 x4+6x2+9 6. R(x) X +2 $4 +16 2. R(x) 3 z3-2x2 + 4x3 4. R(x) = 2 SOLUTION. 1. We begin by factoring the denominator to find 2x2 - x - 1 = (2x + 1)(x and x = 1 are both zeros of multiplicity one and thus we know 1). We get x 1 x+5 2x2 - x - 1 x+5 (2x + 1)(x - 1) A B 2x+x1 -1 Clearing denominators, we get x+5 = A(x-1)+B(2x+1) so that x:+5 Equating coefficients, we get the system (A+2B)x+B-A. { A+2B -A+B 1 5 This system is readily handled using the Addition Method from Section 8.1, and after adding both equations, we get 3B = 6 so B = 2. Using back substitution, we find A = -3. Our answer is easily checked by getting a common denominator and adding the fractions. x+5 2x2 - 3 - 1 2 3 x - 1 2x + 1  8.6 PARTIAL FRACTION DECOMPOSITION 2. Factoring the denominator gives x3 - as a zero of multiplicity one and x = 503 2x2+ = x (2 - 2x + 1) = z(z-1)2 which gives z 1 as a zero of multiplicity two. We have 0 3 3 A B C 3 - 2x2 +x (z- 1)2 x x-1 (cv-1)2 Clearing denominators, we get 3 = A(z - 1)2 + Bz(z - 1) + Cz, which, after gathering up the like terms becomes 3 = (A + B)z2 + (-2A - B + C)z + A. Our system is { A+B -2A-B+C 0 0 A = 3 Substituting A= 3 into A + B= 0 gives B= -2A - B + C = 0 gives C = 3. Our final answe 3 _3 X3 - 2x2 + xv --3, and substituting both for A and B in r is 3 3 - 1 (c-1)2 3. The denominator factors as x (2 - x + 1). We see immediately that xc= 0 is a zero of multiplicity one, but the zeros of 2 - x + 1 aren't as easy to discern. The quadratic doesn't factor easily, so we check the discriminant and find it to be (-1)2 - 4(1)(1) = -3 < 0. We find its zeros are not real so it is an irreducible quadratic. The form of the partial fraction decomposition is then 3 3 A Bz+C c3-c2+cv c(c2-cv+1) x+xc2-xv+l Proceeding as usual, we clear denominators and get 3 3 = (A + B)92 + (-A + C)z + A. We get A (2 - x + 1) + (B + C)c or A+B -A+C A 0 0 From A=3 and A + B =0, we get B -3. From -A + C =0, we get C=A=3. We get 3 c3_ 2 + 3 3-3x c x2-x+l 4. Since 4x3 isn't proper, we use long division and we get a quotient of 4x with a remainder of 8x. That is, x_ = 4x + 8x so we now work on resolving 82 into partial fractions. The quadratic c2 -2, though it doesn't factor nicely, is, nevertheless, reducible. Solving c2 -2 = 0  504 SYSTEMS OF EQUATIONS AND MATRICES gives us x = i2, and each of these zeros must be of multiplicity one since Theorem 3.14 enables us to now factor 2 - 2 = (x - 2) (x + 2). Hence, 8x X-2 8x (x- v2)(x+2 A B + Clearing fractions, we get 8x = A (x+ 2) + B (x-- 2) or 8x = (A + B)x+ (A - B)v2. We get the system { A+B (A - B) 2 8 0 From (A - B) 2= 0, we get A = B, which, when substituted into A + B = 8 gives B = 4. Hence, A = B = 4 and we get 4x3 X2-2 8x x+ 2 2 4 4 4x+ + x + 2 x - 2 5. At first glance, the denominator D(x) = 4+6x2+9 appears irreducible. However, D(x) has three terms, and the exponent on the first term is exactly twice that of the second. Rewriting D(x) (2)2 + 6x2 + 9, we see it is a quadratic in disguise and factor D(x) -(2 +3)2. Since 2 + 3 clearly has no real zeros, it is irreducible and the form of the decomposition is x3+5:-1 x4 + 6x2 + 9 x3+5x-1 (12+3)2 Ax+B Cx+D X2+3 +(32+3)2 When we clear denominators, we find x3 + 5x - 1= (Ax+ B) (x2 + 3) + Cx + D which yields x3 + 5x - 1 = Ax3 + Bx2 + (3A + C)x + 3B + D. Our system is A B 3A + C 3B+D 1 0 5 -1 We have A = 1 and B = 0 from which we get C = 2 and D -1. Our final answer is x3+5x-1 x4 + 6x2 + 9 x 2x-1 2+3 (32:2+3)2 6. Once again, the difficulty in our last example is factoring the denominator. In an attempt to get a quadratic in disguise, we write x4 + 16 = (x2)2 +42 = (x2)2 + 8x2 + 42 - 8x2 = (x2 + 4)2  8.6 PARTIAL FRACTION DECOMPOSITION and obtain a difference of two squares: (2 + 4)2 and 8x2 = (2xv/)2. Hence, 505 x4 +16 = (x2 +4 - 2x 2)(x2+4+2x 2) (x2 - 2x +4)(x2+ 2x2+4) The discrimant of both of these quadratics works out to be -8 < 0, which means they are irreducible. We leave it to the reader to verify that, despite having the same discriminant, these quadratics have different zeros. The partial fraction decomposition takes the form 8x2 8x2 Ax + B Cx + D x4+ 16 (x2 - 2xv/ 2+ 4)(x2 + 2xv 2+ 4) x2 - 2xv/2 + 4 z2 + 2xv5 + 4 We get 8x2 = (Ax+ B) (x2 + 2xv2+ 4) + (Cxz+ D) (x2 - 2xv/2+ 4) or 8x2 = (A+C)x3 +(2A2+ B - 2C2+ D)x2 + (4A+ 2B2+4C - 2D 2)x +4B+4D which gives the system { A+C 2Av/5+ B - 2Cv/2+ D 4A + 2Bv/2+ 4C - 2Dv/ 2 4B + 4D 0 8 0 0 We choose substitution as the weapon of choice to solve this system. From A + C = 0, we get A = -C; from 4B + 4D = 0, we get B = -D. Substituting these into the remaining two equations, we get { -2C 2 - D - 2Cv/2+ D -4C - 2Dv/2+ 4C - 2Dv 2 8 0 or { -4Cv/25 -4Dv 2 8 0 We get C answer is 2 so that A -C = v2/5and D = 0 which means B -D = 0. Our final 8x2 x4 + 16 x2 - 2xv/+ 4 xv2 x2+2xv+ 2+4 E  506 SYSTEMS OF EQUATIONS AND MATRICES As we stated at the beginning of this section, the technique of resolving a rational function into partial fractions is a skill needed for Calculus. However, we hope to have shown you that it is worth doing if, for no other reason, it reinforces a hefty amount of algebra. In the end, if you take only one thing away from this section, it's that 8 8 8 x2 -9 x2 -9 For if the above were true, this section wouldn't be here.  8.6 PARTIAL FRACTION DECOMPOSITION 507 8.6.1 EXERCISES 1. Find only the form needed to begin the process of partial fraction decomposition. Do not create the system of linear equations or attempt to find the actual decomposition. 7 (a) ( 3)(x + 5) 5x + 4 (b) x(x-2)(2 -x) m ()(7x - 6)(x2 + 9) az:2+ bz+ c (d) z3(5z + 9)(3x2 + 7x + 9) A polynomial of degree < 9 (x+ 4)5(x2 + 1)2 (f) A polynomial of degree < 7 x(4x - 1)2(X2 + 5)(9x2 + 16) 2. Find the partial fraction decomposition of the following rational expressions. (a) _2x (a 2 _ 1 (b) 7x+43 3x2 + 19x - 14 (c) 11x2 - 5: - 10 5x3 - 5x2 (d) -2x2 + 20x - 68 33 +4x2+4x +16 -x2 + 15 (e) 4x4 + 40x2 + 36 f -21x2 +3: - 16 3x3 + 4x2 - 3x + 2 (g) 54 - 34x3 + 70x2 - (z -3)2 - 33x 19 3:6 + 5x5+ 16x4 + 80x3 - 2x2 + 6x - 43 (3 + 5 2 + 16x + 80 -732 - 76x - 208 3:3 + 18x2 + 108x + 216 -10x4+x3 --19x2 + x- 10 4x3 - 932 + 123 + 12 (k) 4 - 4x3 + 8x2 - 16x + 16 2x2 + 3x + 14 (2+2x +9)(z2+3z+5)  508 508 SYSTEMS OF EQUATIONS AND MATRICES 8.6.2 ANSWERS 1. (a) xA+xB Ab-+B +c x x -2 (x -2)2 A Bx+C (7x -6 2 + 9 A B C D Ex+F (x +X2 +X3 5x + 9 3x2+ 7x+ 9 eA B C D £ Fx+G Hx+I (x+4+ (x+4)2 (x +4)3 +Qx+ 4)4 + x+ 4)5 + 2 +1 + (2+1)2 A +B +C +Dx+E +Fx+G x41 (+x1+2x2 +5+9x2+16 2x _ 1 1 2. (a)x2 1 +1+ - (b) -7x+43 _ 5 4 (c 1lx2-5x-10 +3-2 _ _4 () 5x3 - 5x2 x x2 5(x - 1) ()-292 + 20x - 68 x3 +4x2+4x+16 9 7x-8 x+4 +x2+4 -x2+15 1 3 (e) 4x4 +40x2 +36- 2(x2 +1) -4(x2 +9) (f) -21x2 X-16 6 3x3+x2-x+2 x+2 5x4 -34x3 +70x2 -33x -19 (g) (x -3)2 3x:+ 5 3x2 - 2x + 1 9 1 5x2-x+1x-3 (x-3)2 +6x-43 3+x+l 3 +216 x5 (h) x6 + 5x5 +16x + 80x - 2x213+5291x+8 (i ~792 - 76x - 208 S 3+ 18x2 +108x +216 J -10x4 + x3 - 19x2 + x 10 7 8 4 x+6 + (x+6)2&(v+6)3 10 1x x+x2+1+(x21)2 4x3 -39x2 +12x + 12 _ 1 4 3x+ 1 (k X443+ 8x2 16 + 16 x -2 +(x -2)2 +X2+4 9 2-1-~ x114 1 I (1) (3:2 ( I _______+ 1 J(+2x+9)(2+x+5) 3x2+2x:+9 32+x:+5  8.7 SYSTEMS OF NON-LINEAR EQUATIONS AND INEQUALITIES 509 8.7 SYSTEMS OF NON-LINEAR EQUATIONS AND INEQUALITIES In this section, we study systems of non-linear equations and inequalities. Unlike the systems of linear equations for which we have developed several algorithmic solution techniques, there is no general algorithm to solve systems of non-linear equations. Moreover, all of the usual hazards of non-linear equations like extraneous solutions and unusual function domains are once again present. Along with the tried and true techniques of substitution and elimination, we shall often need equal parts tenacity and ingenuity to see a problem through to the end. You may find it necessary to review topics throughout the text which pertain to solving equations involving the various functions we have studied thus far. To get the section rolling we begin with a fairly routine example. EXAMPLE 8.7.1. Solve the following systems of equations. Verify your answers algebraically and graphically. ( 2+y2 = 4 3 2+2 =4 1. 4x2 + 9y2 = 36 3. y - 2x = 0 2. 2+y2 = 4 r2+y2=4 2 4x2 _ 9 2 = 36 4' y -X2 = 0 SOLUTION: 1. Since both equations contain 2 and y2 only, we can eliminate one of the variables as we did in Section 8.1. (E1) x2 + y2 = 4 Replace E2 with (E1) z2 + y2 - 4 (E2) 4x2- 9y2 = 36 -4E1+E2 (E2) 5y2 = 20 From 5y2 = 20, we get y2 = 4 or y - +2. To find the associated x values, we substitute each value of y into one of the equations to find the resulting value of x. Choosing x2 + y2 4, we find that for both y = -2 and y = 2, we get x= 0. Our solution is thus {(0, 2), (0, -2)}. To check this algebraically, we need to show that both points satisfy both of the original equations. We leave it to the reader to verify this. To check our answer graphically, we sketch both equations and look for their points of intersection. The graph of x2 + y2 = 4 is a circle centered at (0, 0) with a radius of 2, whereas the graph of 4x2+9y2 = 36, when written in the standard form 2+ = 1 is easily recognized as an ellipse centered at (0, 0) with a major axis along the x-axis of length 6 and a minor axis along the y-axis of length 4. We see from the graph that the two curves intersect at their y-intercepts only, (0, +2). 2. We proceed as before to eliminate one of the variables { (E) 92+ 2 = 4 Replace E2 with f(E) 92+ y2 4 (E2) 4x2+ 9y2 =36 -4E1+±E2 ' (E2) -l3y2 =20  510 SYSTEMS OF EQUATIONS AND MATRICES Since the equation -13y2 = 20 admits no real solution, the system is inconsistent. To verify this graphically, we note that x2 + y2 = 4 is the same circle as before, but when writing the second equation in standard form, - y = 1, we find a hyperbola centered at (0, 0) opening to the left and right with a transverse axis of length 6 and a conjugate axis of length 4. We see that the circle and the hyperbola have no points in common. 1 --1 - -1..-1. f 92+2 = 4 x92+y4 Graphs for 49 2_36 Graphs for 49 22 2 36 3. Since there are no like terms among the two equations, elimination won't do us any good. We turn to substitution and from the equation y - 2x = 0, we get y = 2x. Substituting this into 2 + y2 = 4 gives 2 + (2x)2 = 4. Solving, we find 52 = 4 or x = + 2. Returning to the equation we used for the substitution, y = 2x, we find y = 2Vwhenz= ,s one solution is 2v/-,-4v/5). Similarly, we find the other solution to be (- 2, - 4 ). We leave it to the reader that both points satisfy both equations, so that our final answer is { ( -<4)-( -< 4v--)}. The graph of 2 + y2 = 4 is our circle from before and the graph of y - 2x = 0 is a line through the origin with slope 2. Though we cannot verify the numerical values of the points of intersection from our sketch, we do see that we have two solutions: one in Quadrant I and one in Quadrant III as required. 4. While it may be tempting to solve y - x2 = 0 as y =cx2 and substitute, we note that this system is set up for elimination.' { (E1) c2 + y2 = 4 Replace E2 with (E1) 2 + y2 -4 (E2) y - x2 = 0 E1+E2 (E2) y2 + y = 4 From y2 + y = 4 we get y2 + y - 4 = 0 which gives y _ - . Due to the complicated 2 nature of these answers, it is worth our time to make a quick sketch of both equations to head off any extraneous solutions we may encounter. We see that the circle x2 + y2 = 4 intersects the parabola y = ,2 exactly twice, and both of these points have a positive y value. Of the two olutons or , ony y -1 7is positive, so to get our solution, we substitute this 1We encourage the reader to solve the system using substitution to see that you get the same solution.  8.7 SYSTEMS OF NON-LINEAR EQUATIONS AND INEQUALITIES 511 into y -9x2 = 0 and solve for x. We get x = -1+17 _-2+217 Our solution is 2 2 -2+217 -1+ 7 ( -2+217 -1+ which we leave to the reader to verify. r32 -1 12 3 - 2 - f 2 2 4 (x2 + y2 = 4 x2 +y2= 4 Graphs for x+ 2x 0Graphs for +Y2 ' y-2x =0 y-z 36 A couple of remarks about Example 8.7.1 are in order. First note that, unlike systems of linear equations, it is possible for a system of non-linear equations to have more than one solution without having infinitely many solutions. In fact, while we characterize systems of nonlinear equations as being 'consistent' or 'inconsistent,' we generally don't use the labels 'dependent' or 'independent'. Secondly, as we saw with number 4, sometimes making a quick sketch of the problem situation can save a lot of time and effort. While in general the curves in a system of non-linear equations may not be easily visualized, it sometimes pays to take advantage when they are. Our next example provides some considerable review of many of the topics introduced in this text. EXAMPLE 8.7.2. Solve the following systems of equations. Verify your answers algebraically and graphically, as appropriate. 1 2 + 2xy - 16 = 0 y + 4e2x = 1 z(x- 2) = x y2 + 2xy- 16 = 0 2 y2 + 2ex = 1 3. yz y (x-2)2+y2 - 1 SOLUTION. 1. At first glance, it doesn't appear as though elimination will do us any good since it's clear that we cannot completely eliminate one of the variables. The alternative, solving one of the equations for one variable and substituting it into the other, is full of unpleasantness. Returning to elimination, we note that it is possible to eliminate the troublesome xy term, and the constant term as well, by elimination and doing so we get a more tractable relationship between x and y {(E) 2 +2xy -16 =0 Replace E2 wt (E1) 2 +2xy -16 =0 S(£2) y2 +2xy -16 =0 -E1+ E2 { (E£2) y2 -2 =0  512 SYSTEMS OF EQUATIONS AND MATRICES We get y2 - 2 = 0 ory +x. Substituting y = x into Elwe get2 + 2x2 - 16 = 0 so that 2 = L or x = . On the other hand, when we substitute y = -x into El, we get 2 - 2x2 - 16 = 0 or 2 = -16 which gives no real solutions. Substituting each of xz=+i3 3 into the substitution equation y = x yields the solution {(4, 4 ,-4, - )}. We leave it to the reader to show that both points satisfy both equations and now turn to verifying our solution graphically. We begin by solving 2+2xy- 16= 0 for y to obtain y = 162x 2. This function is easily graphed using the techniques of Section 4.2. Solving the second equation, y2 + 2xy - 16 = 0, for y, however, is more complicated. We use the quadratic formula to obtain y - t vx2 + 16 which would require the use of Calculus or a calculator to graph. Believe it or not, we don't need either because the equation y2 + 2xy -16 = 0 can be obtained from the equation 2 + 2xy - 16 = 0 by interchanging y and x. Thinking back to Section 5.2, this means we can obtain the graph of y2 + 2xy - 16 = 0 by reflecting the graph of 2 + 2xy - 16 = 0 across the line y = x. Doing so confirms that the two graphs intersect twice: once in Quadrant I, and once in Quadrant III as required. y 4 3- 2-- 1- -4 -3 -2 -1 1 2 3 4x -2- -3- -4 The graphs of 2 + 2xy - 16 = 0 and y2 + 2xy - 16 = 0 2. Unlike the previous problem, there seems to be no avoiding substitution and a bit of algebraic unpleasantness. Solving y + 4e2x - 1 for y, we get y 1 - 4e2x which, when substituted into the second equation, yields (1 - 4e2x)2 + 2ex 1. After expanding and gathering like terms, we get 16e4x -8e2x +2ex = 0. Factoring gives us 2ex (8e3x - 4ex +1) = 0, and since 2ex # 0 for any real x, we are left with solving 8e3x - 4ex + 1 = 0. We have three terms, and even though this is not a 'quadratic in disguise', we can benefit from the substitution ua= ex. The equation becomes 8u3 -4u+1 = 0. Using the techniques set forth in Section 3.3, we find u = 1 is a zero and use synthetic division to factor the left hand side as (u - 2) (8u2 + 4u - 2). We use the quadratic formula to solve 8a2 + 4a - 2 =0 and find a = 14 Since a eX, we now must solve eX =2 n er From er =1,w e z=I j - ln(2). As for eX =-j , we first note that - < 0, 50 eX - has no real solutions. We are  8.7 SYSTEMS OF NON-LINEAR EQUATIONS AND INEQUALITIES 513 left with e' -14 , so that x in (-i j). We now return to y = 1 - 4e2x to find the accompanying y values for each of our solutions for x. For x ln(2), we get y = 1 - 4e2x 1 - 4e-21n(2) =1 - 4en(1/4) = 1-4(l) =0 For x = ln ( ,1+ we have y 1 - 4e2x 1~421-) 1-4e 4 1-4 e 4 1-4-+ 1-4(38 2 We get two solutions, {(0, - ln(2)), (ln (-1. , -1J)}. It is a good review of the prop- erties of logarithms to verify both solutions, so we leave that to the reader. We are able to sketch y = 1 - 4e2x using transformations, but the second equation is more difficult and we resort to the calculator. We note that to graph y2 + 2ex = 1, we need to graph both the positive and negative roots, y = +'1 - 2e. After some careful zooming,2 we confirm our solutions. ZEro i\ K=--.6931472 Y=0 \ Intekrs-ection N H= -1.i74359 -.51D0339 - 1 - 4e2x and y = i:/1 - 2ex. The graphs of y 3. Our last system involves three variables and gives some insight on how to keep such systems organized. Labeling the equations as before, we have 2The calculator has trouble confirming the solution (- ln(2), 0) due to its issues in graphing square root functions. If we mentally connect the two branches of the thicker curve, we see the intersection.  514 SYSTEMS OF EQUATIONS AND MATRICES El z(x -2) = x E2 yz = y E3 (x - 2)2 + y2 1 The easiest equation to start with appears to be E2. While it may be tempting to divide both sides of E2 by y, we caution against this practice because it presupposes y + 0. Instead, we take E2 and rewrite it as yz - y = 0 so y(z - 1) = 0. From this, we get two cases: y = 0 or z = 1. We take each case in turn. CASE 1: y = 0. Substituting y = 0 into El and E3, we get { El z(x - 2) = x E3 (x - 2)2= Solving E3 for x gives x = 1 or x = 3. Substituting these values into El gives z = -1 when x = 1 and z = 3 when x = 3. We obtain two solutions, (1, 0, -1) and (3, 0, 3). CASE 2: z = 1. Substituting z = 1 into El and E3 gives us { El (1)(x - 2) = x E3 (1 - 2)2 + y2 1 Equation El gives us x - 2 = x or -2 = 0, which is a contradiction. This means we have no solution to the system in this case, even though E3 is solvable and gives y = 0. Hence, our final answer is {(1, 0, -1), (3, 0, 3)}. These points are easy enough to check algebraically in our three original equations, so that is left to the reader. As for verifying these solutions graphically, they require plotting surfaces in three dimensions and looking for intersection points. While this is beyond the scope of this book, we provide a snapshot of the graphs of our three equations near one of the solution points, (1, 0, -1). Example 8.7.2 showcases some of the ingenuity and tenacity mentioned at the beginning of the section. Sometimes you just have to look at a system the right way to find the most efficient method to solve it. Sometimes you just have to try something.  8.7 SYSTEMS OF NON-LINEAR EQUATIONS AND INEQUALITIES 515 We close this section discussing how non-linear inequalities can be used to describe regions in the plane which we first introduced in Section 2.4. Before we embark on some examples, a little motivation is in order. Suppose we wish to solve x2 <4 _ 2. If we mimic the algorithms for solving nonlinear inequalities in one variable, we would gather all of the terms on one side and leave a 0 on the other to obtain x2 + y2 - 4 < 0. Then we would find the zeros of the left hand side, that is, where is x2 + y2 - 4 = 0, or x2 + y2 = 4. Instead of obtaining a few numbers which divide the real number line into intervals, we get an equation of a curve, in this case, a circle, which divides the plane into two regions - the 'inside' and 'outside' of the circle - with the circle itself as the boundary between the two. Just like we used test values to determine whether or not an interval belongs to the solution of the inequality, we use test points in the each of the regions to see which of these belong to our solution set.3 We choose (0, 0) to represent the region inside the circle and (0, 3) to represent the points outside of the circle. When we substitute (0, 0) into x2 + y2 - 4 < 0, we get -4 < 4 which is true. This means (0, 0) and all the other points inside the circle are part of the solution. On the other hand, when we substitute (0, 3) into the same inequality, we get 5 < 0 which is false. This means (0, 3) along with all other points outside the circle are not part of the solution. What about points on the circle itself? Choosing a point on the circle, say (0, 2), we get 0 < 0, which means the circle itself does not satisfy the inequality.4 As a result, we leave the circle dashed in the final diagram. y 2 SI-2 2 X The solution to x2 <4 2 We put this technique to good use in the following example. EXAMPLE 8.7.3. Sketch the solution to the following nonlinear inequalities in the plane. 1. y2-4 4 which we rewrite as x2+ y2 - 4 > 0. The points which satisfy x2+ y2 - 4 = 0 form our friendly circle x2+ y2 4. Using test points (0, 0) and (0, 3) we find that our solution comprises the region outside the circle. As far as the circle itself, the point (0, 2) satisfies the inequality, so the circle itself is part of the solution set. Moving to the inequality 2 - 2x + y2 - 2y < 0, we start with 2 - 2x + y2 - 2y = 0. Completing the squares, we obtain (x - 1)2 + (y - 1)2 = 2, which is a circle centered at (1, 1) with a radius of 2. Choosing (1, 1) to represent the inside of the circle, (1, 3) as a point outside of the circle and (0, 0) as a point on the circle, we find that the solution to the inequality is the inside of the circle, including the circle itself. Our final answer, then, consists of the points on or outside of the circle 92 + y2 - 4 which lie on or  8.7 SYSTEMS OF NON-LINEAR EQUATIONS AND INEQUALITIES 517 inside the circle (x - 1)2 + (y - 1)2 = 2. To produce the most accurate graph, we need to find where these circles intersect. To that end, we solve the system { (E1) (E2) x2 2 + y2 2x + y2 -2y 4 0 We can eliminate both the 2 and y2 by replacing E2 with -El + E2. Doing so produces -2x - 2y = -4. Solving this for y, we get y = 2 - x. Substituting this into El gives 2 + (2 - x)2 = 4 which simplifies to 2 + 4 - 4x + 2 = 4 or 2x2 - 4x = 0. Factoring yields 2x(x - 2) which gives x= 0 or x= 2. Substituting these values into y = 2 - x gives the points (0, 2) and (2, 0). The intermediate graphs and final solution are below. y 1 y y 3- 3- 2- -1 x -3 -2 - -3 -2 -i 2 i i i i i le i ), x -2- -3- -3- 2 + y2 > 4 2 -2x+y2 -2y<0 Solution to the system. El  518 SYSTEMS OF EQUATIONS AND MATRICES 8.7.1 EXERCISES 1. Solve the following systems of nonlinear equations. Use a graph to help you avoid any potential extraneous solutions. (a) x2 + 4y2 (b{ y/x+1 -y (b xz + 4y2 ({ x+2y2 =- (c) x2 + 4y 2 d f (x- 2)2 + y2 d) 2+ 4y2 1 4 0 4 x2 2+y2 = 25 (e) { =y - = 1 {2 + y22 25 () 2 + (y - 3)2 = 10 (g) y = z+ y = 10x - 2 x2 + y2 = 25 (h) 4x2 - 9y = 0 3y2 - 16x = 0 2 4 1 4 2. Consider the system of nonlinear equations below 4 3 3 2 x y If we let u = and v = then the system becomes {4u + 3v = 3u + 2v = 1 -1 1 1 This associated system of linear equations can then be solved using any of the presented earlier in the chapter to find that u = -5 and v = 7. Thus x 1=a Y = 1 _ 1 v-7. techniques - and We say that the original system is linear in form because its equations are not linear but a few basic substitutions reveal a structure that we can treat like a system of linear equations. Each system given below is linear in form. Make the appropriate substitutions to help you solve for x and y. { 4e' + 3e-- (a) 3e' + 2e-- 1 -1 3. Solve the following system { 4x3 + 3/y -= (b) 3x3+2 y 2 + /Y +log2(z) 3x2 - 2 1/9 + 2log2(z) -4x2 + Iy/ - 31og2(z) 1 1 f 4ln(x) + 3y2 S3ln(x) + 2y2 1 _1 { 6 5 11  8.7 SYSTEMS OF NON-LINEAR EQUATIONS AND INEQUALITIES 4. Sketch the solution to each system of nonlinear inequalities in the plane. 519 ( x2-y2 < 1 (a) 2 + 4y2 4 { z2+y2 < 25 (b) x2 + (y- 3)2 > 10 ( { f (x-2)2+y2 < 1 (c)2 + 4y2 < 4 (d) { > x3x2 ( x+2y2 > 2 ( 2 + 4y2 2 4 xz2 + y2 > 25 y -x < 1 5. Systems of nonlinear equations show up in third semester Calculus in the midst of some really cool problems. The system below came from a problem in which we were asked to find the dimensions of a rectangular box with a volume of 1000 cubic inches that has minimal surface area. The variables x, y and z are the dimensions of the box and A is called a Lagrange multiplier. With the help of your classmates, solve the system. 2y + 2z 2x + 2z 2y + 2x xyz Ayz Axz Axy 1000 6. According to Theorem 3.16 in Section 3.4, the polynomial p(x) x= 4 + 4 can be factored into the product linear and irreducible quadratic factors. In this exercise, we present a method for obtaining that factorization. (a) Show that p has no real zeros. (b) Because p has no real zeros, its factorization must be of the form (2+ ax + b)(2+ cx + d) where each factor is an irreducible quadratic. Expand this quantity and gather like terms together. (c) Create and solve the system of nonlinear equations which results from equating the coefficients of the expansion found above with those of x4 + 4. You should get four equations in the four unknowns a, b, c and d. Write p(x) in factored form. 7. Factor q(x) =x4 + 6x2 - 5x + 6. 51f using A bothers you, change it to w when you solve the system.  00 I-- cl LZ) 9 tQ + IQ CL IV A OZ01 00 VIA z O 0J~ C' C I I - c I Cs 0 zz z 9 01 -H -- .:/ /' N N Ca t 01 Q1 C1 H 0 0 H cr  -1 C1 Q 1+ IQ1 II II + o II I II o IQ I + + + IA IV H-l1 IA V 00 00 A V 0 00 t 0 ~~-- I s N. F I : : v w 01  522 SYSTEMS OF EQUATIONS AND MATRICES  CHAPTER 9 SEQUENCES AND THE BINOMIAL THEOREM 9.1 SEQUENCES When we first introduced a function as a special type of relation in Section 1.4, we did not put any restrictions on the domain of the function. All we said was that the set of x-coordinates of the points in the function F is called the domain, and it turns out that any subset of the real numbers, regardless of how weird that subset may be, can be the domain of a function. As our exploration of functions continued beyond Section 1.4, we saw fewer and fewer functions with 'weird' domains. It is worth your time to go back through the text to see that the domains of the polynomial, rational, exponential, logarithmic and algebraic functions discussed thus far have fairly predictable domains which almost always consist of just a collection of intervals on the real line. This may lead some readers to believe that the only important functions in a College Algebra text have domains which consist of intervals and everything else was just introductory nonsense. In this section, we introduce sequences which are an important class of functions whose domains are the set of natural numbers.1 Before we get to far ahead of ourselves, let's look at what the term 'sequence' means mathematically. Informally, we can think of a sequence as an infinite list of numbers. For example, consider the sequence 1 3 9 27 2' 4'8' 16 As usual, the periods of ellipsis, ..., indicate that the proposed pattern continues forever. Each of the numbers in the list is called a term, and we call 1 the 'first term', -j the 'second term', 8 the 'third term' and so forth. In numbering them this way, we are setting up a function, which we'll call a per tradition, between the natural numbers and the terms in the sequence. 'Recall that this is the set {1, 2, 3, ...}.  524 SEQUENCES AND THE BINOMIAL THEOREM n a(n) 1 1 2 2 3 4 3 9 8 4 27 16 In other words, a(n) is the nth term in the sequence. We formalize these ideas in our definition of a sequence and introduce some accompanying notation. DEFINITION 9.1. A sequence is a function a whose domain is the natural numbers. The value a(n) is often written as an and is called the nth term of the sequence. The sequence itself is usually denoted {an}°_1. Applying the notation provided in Definition 9.1 to the sequence given (1), we have a1 = j, a2 -4, a3 = and so forth. Now suppose we wanted to know a117, that is, the 117th term in the sequence. While the pattern of the sequence is apparent, it would benefit us greatly to have an explicit formula for an. Unfortunately, there is no general algorithm that will produce a formula for every sequence, so any formulas we do develop will come from that greatest of teachers, experience. In other words, it is time for an example. EXAMPLE 9.1.1. Write the first four terms of the following sequences. 1.a= t>1 4. 2.1)b =.,b k 2 ,k 0 5. a1= 7, an+ =2-an, n 2 3. {2n -1}°° 6. fo = 1, f= n-f_1, nJ1 SOLUTION. 1. Since we are given n > 1, the first four terms of the sequence are a1, a2, a3 and a4. Since the notation ai means the same thing as a(1), we obtain our first term by replacing every occurrence of n in the formula for an with n = 1 to get a1 = 51 =-}. Proceeding similarly, 52-1 5 3-1 25 54-1 - 125 we get a2 3= 2 1=, a3 33 1= and a4 =34 - 81 2. For this sequence we have k > 0, so the first four terms are b0, b1, b2 and b3. Proceeding as before, replacing in this case the variable k with the appropriate whole number, we get b-(- =1, bi = =(1)1 -, b-(2)+1 and b = -3 - C-4. (This sequence is called an alternating sequence since the signs alternate between + and -. The reader is encouraged to think what component of the formula is producing this effect.)  9.1 SEQUENCES 525 3. From {2n - 1}° 1, we have that an = 2n - 1, n> 1. We get a1= 1, a2 = 3, a3S = 5 and a4 = 7. (The first four terms are the first four odd natural numbers. The reader is encouraged to examine whether or not this pattern continues indefinitely.) 4. Proceeding as in the previous problem, we set a3 =+(- ,jl 2. We find a2 =1, a3 = 0, a4 = jand a5 =0. 5. To obtain the terms of this sequence, we start with a1= 7 and use the equation an+1 = 2-an for n > 1 to generate successive terms. When n = 1, this equation becomes a1+1 = 2 - a1 which simplifies to a2 = 2-a1= 2-7 = -5. When n = 2, the equation becomes a2+ 1 =2-a2 so we geta3 = 2-a2 = 2 - (-5) =7. Finally, when n = 3, we get a3+1 = 2-a3so a4= 2 - a3= 2 -7 =-5. 6. As with the problem above, we are given a place to start with fo = 1 and given a formula to build other terms of the sequence. Substituting n = 1 into the equation f = n - f_1, we get fi = 1 - fo = 1 - 1 = 1. Advancing to n = 2, we get f2 = 2 - fi= 2. 1 =2. Finally, f3=3-f2= 3. =6. D Some remarks about Example 9.1.1 are in order. We first note that since sequences are functions, we can graph them in the same way we graph functions. For example, if we wish to graph the sequence {bk}O from Example 9.1.1, we graph the equation y = b(k) for the values k > 0. That is, we plot the points (k, b(k)) for the values of k in the domain, k = 0, 1, 2, .... The resulting collection of points is the graph of the sequence. Note that we do not connect the dots in a pleasing fashion as we are used to doing, because the domain is just the whole numbers in this case, not a collection of intervals of real numbers. If you feel a sense of nostalgia, you should see Section 1.2. y 3 1 0 -2 1 2 3 X -1 3 Graphing y= bk =()k k>0 2k +1 - Speaking of {bk}%, the astute and mathematically minded reader will correctly note that this technically isn't a sequence, since according to Definition 9.1, sequences are functions whose domains are the natural numbers, not the whole numbers, as is the case with {bk}%%. In other words, to satisfy Definition 9.1, we need to shift the variable k so it starts at k = 1 instead of k = 0. To see how we can do this, it helps to think of the problem graphically. What we want is to shift the graph of y =b(k) to the right one unit, and thinking back to Section 1.8, we can accomplish this by replacing k with k - 1 in the definition of {bk}%%. Specifically, let ck bk_1 where k- 1>a0. W e =(1ii 1 - ,iko1 where now k > 1. We leave to the reader to verify that {ck}b) generates the same list of numbers as does {bk, but the former satisfies Definition  526 SEQUENCES AND THE BINOMIAL THEOREM 9.1, while the latter does not. Like so many things in this text, we acknowledge that this point is pedantic and join the vast majority of authors who adopt a more relaxed view of Definition 9.1 to include any function which generates a list of numbers which can be then be matched up with the natural numbers.2 Finally, we wish to note the sequences in parts 5 and 6 are examples of sequences described recursively. In each instance, an initial value of the sequence is given which is then followed by a recursion equation - a formula which enables us to use known terms of the sequence to determine other terms. The terms of the sequence in part 6 are given a special name: fn is called n-factorial. We will study factorials in greater detail in Section 9.4. The world famous Fibonacci Numbers are defined recursively and are explored in the exercises. While none of the sequences worked out to be the sequence in (1), they do give us some insight into what kinds of patterns to look for. Two patterns in particular are given in the next definition. DEFINITION 9.2. Arithmetic and Geometric Sequences: Suppose {af}k is a sequence" " If there is a number d so that an+1 = an + d for all n > k, then {afh}k is called an arithmetic sequence. The number d is called the common difference. " If there is a number r so that an+1 = ran for all n> k, then {af}ik is called an geometric sequence. The number r is called the common ratio. aNote that we have adjusted for the fact that not all 'sequences' begin at n = 1. Both arithmetic and geometric sequences are defined in terms of recursion equations. In English, an arithmetic sequence is one in which we proceed from one term to the next by always adding the fixed number d. The name 'common difference' comes from a slight rewrite of the recursion equation from an+1 = an + d to an+1 - an = d. Analogously, a geometric sequence is one in which we proceed from one term to the next by always multiplying by the same fixed number r. If r # 0, we can rearrange the recursion equation to get a +1 = r, hence the name 'common ratio.' Some sequences are arithmetic, some are geometric and some are neither as the next example illustrates.3 EXAMPLE 9.1.2. Determine if the following sequences are arithmetic, geometric or neither. If arithmetic, find the common difference d; if geometric, find the common ratio r. 5-1 >3. {2n - 1}0 1. an =3 , n>21n= (-1)k 1 3 9 27 2. bk =2k+1 k 0 4.,.8.1 . SOLUTION. A good rule of thumb to keep in mind when working with sequences is "When in doubt, write it out!" Writing out the first several terms can help you identify the pattern of the sequence should one exist. 2Math fans will delight to know we are basically talking about the 'countably infinite' subsets of the real number line when we do this. 3Sequences which are both arithmetic and geometric are discussed in the Exercises.  9.1 SEQUENCES 527 1. From Example 9.1.1, we know that the first four terms of this sequence are 1, 5, 2 and 125 To see if this is an arithmetic sequence, we look at the successive differences of terms. We find that a2 - ai l= -j =3 and a3- a2 = 27- - = 7. Since we get different numbers, there is no 'common difference' and we have established that the sequence is not arithmetic. To investigate whether or not it is geometric, we compute the ratios of successive terms. The first three ratios 5 25 125 a2= =5 a 5 and 4a4 81 a1 13 a2 5 3 a3 suggest that the sequence is geometric. To prove it, we must show that a +1 = r for all n. 5(n+l)-1 an+_ 3n+1 5" 3" 5 an 5n-1 3n+1 5n-1 3 3n This sequence is geometric with common ratio r = 3. 2. Again, we have Example 9.1.1 to thank for providing the first four terms of this sequence: 1, -j, j and -}. We find bi - -3 and b2 - bi = . Hence, the sequence is not arithmetic. To see if it is geometric, we compute 1=-j and = -5. Since there is no 'common ratio,' we conclude the sequence is not geometric, either. 3. As we saw in Example 9.1.1, the sequence {2n - 1}°OL1 generates the odd numbers: 1, 3, 5, 7, .... Computing the first few differences, we find a2 - ai1= 2, a3 - a2 = 2, and a4 - a3= 2. This suggests that the sequence is arithmetic. To verify this, we find an+1- an =(2(n + 1) - 1) - (2n - 1)= 2n + 2 - 1 - 2n + 1 2 This establishes that the sequence is arithmetic with common difference d = 2. To see if it is geometric, we compute ai= 3 and a - =. Since these ratios are different, we conclude the sequence is not geometric. 4. We met our last sequence at the beginning of the section. Given that a2 - ai = -j and as - a2= 1, the sequence is not arithmetic. Computing the first few ratios, however, gives us a= -, 3 a3 - and a4= - -. Since these are the only terms given to us, we assume that the pattern of ratios continue in this fashion and conclude that the sequence is geometric. D We are now one step away from determining an explicit formula for the sequence given in (1). We know that it is a geometric sequence and our next result gives us the explicit formula we require.  528 SEQUENCES AND THE BINOMIAL THEOREM EQUATION 9.1. Formulas for Arithmetic and Geometric Sequences: " An arithmetic sequence with first term a and common difference d is given by an = a+(n -1)d, n>21 " A geometric sequence with first term a and common ratio r # 0 is given by an = ar"-1, n >1 While the formal proofs of the formulas in Equation 9.1 require the techniques set forth in Section 9.3, we attempt to motivate them here. According to Definition 9.2, given an arithmetic sequence with first term a and common difference d, the way we get from one term to the next is by adding d. Hence, the terms of the sequence are: a, a + d, a + 2d, a + 3d, .... We see that to reach the nth term, we add d to a exactly (n - 1) times, which is what the formula says. The derivation of the formula for geometric series follows similarly. Here, we start with a and go from one term to the next by multiplying by r. We get a, ar, ar2, ar3 and so forth. The nth term results from multiplying a by r exactly (n - 1) times. We note here that the reason r = 0 is excluded from Equation 9.1 is to avoid an instance of 00 which is an indeterminant form.4 With Equation 9.1 in place, we finally have the tools required to find an explicit formula for the nth term of the sequence given in (1). We know from Example 9.1.2 that it is geometric with common ratio r = - 2. The first term is a = jso by Equation 9.1 we get an = ar"-1 -=1(-) for n> 1. After a touch of simplifying, we get an -2f-1 for n> 1. Note that we can easily check our answer by substituting in values of n and seeing that the formula generates the sequence given in (1). We leave this to the reader. Our next example gives us more practice finding patterns. EXAMPLE 9.1.3. Find an explicit formula for the nth term of the following sequences. 1. 0.9, 0.09, 0.009, 0.0009, ... 2 2 2 2. - 2,- 5''3 7 2 4 8 3. 1,... '7' 13' 19' SOLUTION. 1. Although this sequence may seem strange, the reader can verify it is actually a geometric sequence with common ratio r 0.1 - 1. With a=0.9 - 9, we get an - ( for n > 0. Simplifying, we get an = y n r> 1. There is more to this sequence than meets the eye and we shall return to this example in the next section. 2. As the reader can verify, this sequence is neither arithmetic nor geometric. In an attempt to find a pattern, we rewrite the second term with a denominator to make all the terms 4See the footnotes on page 171 in Section 3.1 and page 316 of Section 6.1.  9.1 SEQUENCES 529 appear as fractions. We have j, j, - , - 3,. . .. If we associate the negative '-' of the last two terms with the denominators we get 5, j, 2,.O,.... This tells us that we can tentatively sketch out the formula for the sequence as an = - where do is the sequence of denominators. Looking at the denominators 5, 1, -3, -7,..., we find that they go from one term to the next by subtracting 4 which is the same as adding -4. This means we have an arithmetic sequence on our hands. Using Equation 9.1 with a = 5 and d = -4, we get the nth denominator by the formula do = 5+ (n - 1)(-4) = 9 - 4n for n> 1. Our final answer is an = 4n, 1. 3. The sequence as given is neither arithmetic nor geometric, so we proceed as in the last problem to try to get patterns individually for the numerator and denominator. Letting cn and do denote the sequence of numerators and denominators, respectively, we have an = o. After some experimentation,5 we choose to write the first term as a fraction and associate the negatives '-' with the numerators. This yields 1, 7, 4, 9,8.... The numerators form the sequence 1, -2, 4, -8, ... which is geometric with a = 1 and r = -2, so we get cn = (-2)m-1, for n > 1. The denominators 1, 7, 13, 19, ... form an arithmetic sequence with a = 1 and d = 6. Hence, we get do = 1 + 6(n - 1) = 6n - 5, for n > 1. We obtain our formula for a= = )-, for n> 1. We leave it to the reader to show that this checks out. Q While the last problem in Example 9.1.3 was neither geometric nor arithmetic, it did resolve into a combination of these two kinds of sequences. If handed the sequence 2, 5,10,17,..., we would be hard-pressed to find a formula for an if we restrict our attention to these two archetypes. We said before that there is no general algorithm for finding the explicit formula for the nth term of a given sequence, and it is only through experience gained from evaluating sequences from explicit formulas that we learn to begin to recognize number patterns. The pattern 1, 4, 9, 16,... is rather recognizable as the squares, so the formula an = n2, n > 1 may not be too hard to determine. With this in mind, it's possible to see 2,5,10,17, ... as the sequence 1 + 1, 4 + 1, 9 + 1, 16 + 1,..., so that an = n2 + 1, n 1. Of course, since we are given only a small sample of the sequence, we shouldn't be too disappointed to find out this isn't the only formula which generates this sequence. For example, consider the sequence defined by b = -1n4 + n_3 - 2 + 2rn - 5, n> 1. The reader is encouraged to verify that it also produces the terms 2, 5, 10, 17. In fact, it can be shown that given any finite sample of a sequence, there are infinitely many explicit formulas all of which generate those same finite points. This means that there will be infinitely many correct answers to some of the exercises in this section.6 Just because your answer doesn't match ours doesn't mean it's wrong. As always, when in doubt, write your answer out. As long as it produces the same terms in the same order as what the problem wants, your answer is correct. We would be remiss to close this section without mention of the utility of sequences in everyday life. Indeed, sequences play a major role in the Mathematics of Finance, as we have already seen with Equation 6.2 in Section 6.5. Recall that if we invest P dollars at an annual percentage rate r and compound the interest n times per year, the formula for Ak, the amount in the account after k 5Here we take 'experimentation' to mean a frustrating guess-and-check session. 6For more on this, see When Every Answer is Correct: Why Sequience: and Numbler Pa ttern: F>ail the Test by Professor Don White of Kent State University.  530 SEQUENCES AND THE BINOMIAL THEOREM compounding periods, is Ak = P (1+ L)k= [P(1+L) (1+ ,) k>1. We now spot this as a geometric sequence with first term P (1+ n) and common ratio (1+ ). In retirement planning, it is seldom the case that an investor deposits a set amount of money into an account and waits for it to grow. Usually, additional payments of principal are made at regular intervals and the value of the investment grows accordingly. This kind of investment is called an annuity and will be discussed in the next section once we have developed more mathematical machinery.  9.1 SEQUENCES 531 9.1.1 EXERCISES 1. Write out the four terms of the following sequences. (a) an=2n-1,n>0 (g) ai= 3, an+1=an-1, n>1 j0j+1) -d - (b) d = (-1) 2 ,> 21 (h) do=12, dm o, m>21 (c) {5k-2}%-1 (i) b1= 2, bk+1 = 3bk + 1, k > 1 (d) n2+1 (j) co=-2, ci= -±,)m± -1 (e) 1 (e) { =1 (k) ai1= 117, an+1 = n->"2 (f) (1) so = 1, sn+1 = xn+1 + sn, rn 0 (m) F = 1, F1 = 1, F = Fn-1 + Fn-2, n 2 (This is the famous Fibonacci Sequence ) 2. Find an explicit formula for the nth term of the following sequences. Use the formulas in Equation 9.1 as needed. (a) 3, 5, 7, 9, ... (d) 1, 3j, }, , ... (g) 0.9, 0.99, 0.999, 0.9999, ... (b) 1, - 2,4 , -}8, . .. (e) 1, I , 9I, 1 , . . . (h) 27, 64, 125, 216, .. . (c) 1, , , ,.(f) , - , , -XX.7(i) 1,0,1,0,... 3. Find a sequence which is both arithmetic and geometric. (Hint: Start with an = c for all n.) 4. Show that a geometric sequence can be transformed into an arithmetic sequence by taking the natural logarithm of the terms. 5. Thomas Robert Malthus is credited with saying, "The power of population is indefinitely greater than the power in the earth to produce subsistence for man. Population, when unchecked, increases in a geometrical ratio. Subsistence increases only in an arithmetical ratio. A slight acquaintance with numbers will show the immensity of the first power in comparison with the second." (See this webpage for more information.) Discuss this quote with your classmates from a sequences point of view. 6. This classic problem involving sequences shows the power of geometric sequences. Suppose that a wealthy benefactor agrees to give you one penny today and then double the amount she gives you each day for 30 days. So, for example, you get two pennies on the second day and four pennies on the third day. How many pennies do you get on the 30th day? What is the total dollar value of the gift you have received? 7. Research the terms 'arithmetic mean' and 'geometric mean.' With the help of your classmates, show that a given term of a arithmetic sequence ak k 2 is the arithmetic mean of the term immediately preceding, ak_1 it and immediately following it, ak1 State and prove an analogous result for geometric sequences.  532 SEQUENCES AND THE BINOMIAL THEOREM 8. Discuss with your classmates how the results of this section might change if we were to examine sequences of other mathematical things like complex numbers or matrices. Find an explicit formula for the nth term of the sequence i, -1, -i, 1, i, .... List out the first four terms of the matrix sequences we discussed in Exercise 3 in Section 8.3.  9.1 SEQUENCES 9.1.2 ANSWERS 533 1. (a) (b) (c) (d) (e) (f) (g) 2. (al) (b) (c) 0,1,3,7 -1, -1,1,1 3,)8,13,)18 x2 x3 x4 x,2 T3 0, ln(2), ln(3) ,ln (4) 3 3214 (h) 12, 0.12, 0.0012, 0.000012 (i) 2,7,22,67 (k) 117, 117', 117, 117 (1) 1,x+ 1,92+x+1,x3+x2+x+1 (in) 1, 1,2, 3 1+22,nr> 1 (-)2l 1 1> 1 2n-1 n 1'1> 1 (e) an 1 2> 1 (f) (n1)lx2n-1121 (g) an (h) an (i) an (n + 2)3, n12 1 1±(-1) n-i12 > 1 2  534 SEQUENCES AND THE BINOMIAL THEOREM 9.2 SUMMATION NOTATION In the previous section, we introduced sequences and now we shall present notation and theorems concerning the sum of terms of a sequence. We begin with a definition, which, while intimidating, is meant to make our lives easier. DEFINITION 9.3. Summation Notation: Given a sequence {af}"k and numbers m and p satisfying k < m < p, the summation from m to p of the sequence {an} is written p Z an =am +am++... +ap rn~m The variable n is called the index of summation. The number m is called the lower limit of summation while the number p is called the upper limit of summation. In English, Definition 9.3 is simply defining a short-hand notation for adding up the terms of the sequence {af}"k from am through ap. The symbol E is the capital Greek letter sigma and is shorthand for 'sum'. The index of summation tells us which term to start with and which term to end with. For example, using the sequence an = 2n - 1 for n > 1, we can write the sum a3 + a4 + a5 + a6 as 6 E(2n - 1) = (2(3) - 1) + (2(4) - 1) + (2(5) - 1) + (2(6) - 1) n=3 = 5+7+9+11 = 32 The index variable is considered a 'dummy variable' in the sense that it may be changed to any letter without affecting the value of the summation. For instance, 6 6 6 E(2n - _) = (2k -1) = (2j -1) n=3 k=3 j=3 One place you may encounter summation notation is in mathematical definitions. For example, summation notation allows us to define polynomials as functions of the form n f (x)= axk k=o for real numbers ak, k = 0,1, ... n. The reader is invited to compare this with what is given in Definition 3.1. Summation notation is particularly useful when talking about matrix operations. For example, we can write the product of the ith row RZ of a matrix A = [aij]mxm and the jth column C3 of a matrix B =[bi],xr as Ri - Cj =Zaibkg k=1  9.2 SUMMATION NOTATION 535 Again, the reader is encouraged to write out the sum and compare it to Definition 8.9. Our next example gives us practice with this new notation. EXAMPLE 9.2.1. 1. Find the following sums. 13 (a) 100S0 k=1 5 (_ijn+1 n=1 1)n 2. Write the following sums using summation notation. (a) 1+3+5+...+117 1 1 1 1 (b) 12-g+ 34**117 (c) 0.9+0.09+0.009+...0.0-.-09 n - 1 zeros SOLUTION. 1. (a) We substitute k = 1 into the formula lok and add successive terms until we reach k = 4. 4 13 E100k k=1 13 13 13 13 1001 +1002 + ioo + i004 0.13 + 0.0013 + 0.000013 + 0.00000013 0.13131313 (b) We proceed as in part (a), replacing the index n, but not the variable x, with the values 1 through 5 and adding the resulting terms. 1 (_1)1+1 ( _1)2+1 1) = (x 1 2 (_-1)3+1 3 ( 1) +( 4 (+ 1) 1 )5 + 4 (x - 1)4 + 5 (z- 1) ( -1)2 (X _-1)3 2 3 (x-1)4 (x-1)5 4 + 5 2. The key to writing these sums with summation notation is to find the pattern of the terms. To that end, we make good use of the techniques presented in Section 9.1.  536 SEQUENCES AND THE BINOMIAL THEOREM (a) The terms of the sum 1, 3, 5, etc., form an arithmetic sequence with first term a = 1 and common difference d = 2. We get a formula for the nth term of the sequence using Equation 9.1 to get an = 1+(n -1)2 = 2n -1, n > 1. At this stage, we have the formula for the terms, namely 2n - 1, and the lower limit of the summation, n = 1. To finish the problem, we need to determine the upper limit of the summation. In other words, we need to determine which value of n produces the term 117. Setting an = 117, we get 2n - 1 = 117 or n = 59. Our final answer is 59 1 + 3+ 5+ .. .+117 = E(2n -1) n=1 (b) We rewrite all of the terms as fractions, the subtraction as addition, and associate the negatives '-' with the numerators to get 1 -1 1 -1 1 1 2 3 4 117 The numerators, 1, -1, etc. can be described by the geometric sequence1 c = (-1)n-1 for n >1, while the denominators are given by the arithmetic sequence2 do = n for n 1. Hence, we get the formula a (-1) for our terms, and we find the lower and upper limits of summation to be n = 1 and n = 117, respectively. Thus 1 1 1 1 m (_1 )_1 1- + +-...+11 E= 12+3 4 ~117 n n=1 n (c) Thanks to Example 9.1.3, we know that one formula for the nth term is an = $ for n> 1. This gives us a formula for the summation as well as a lower limit of summation. To determine the upper limit of summation, we note that to produce the n - 1 zeros to the right of the decimal point before the 9, we need a denominator of lO. Hence, n is the upper limit of summation. Since n is used in the limits of the summation, we need to choose a different letter for the index of summation.3 We choose k and get S9 0.9+0.09+0.009+...0.0---09 = 1 10 n - 1 zeros k=1 The following theorem presents some general properties of summation notation. While we shall not have much need of these properties in Algebra, they do play a great role in Calculus. Moreover, 1This is indeed a geometric sequence with first term a =1 and common ratio r =-1. 2It is an arithmetic sequence with first term a =1 and common difference d =1. 3To see why, try writing the summation using 'in' as the index.  9.2 SUMMATION NOTATION 537 there is much to be learned by thinking about why the properties hold. We invite the reader to prove these results. To get started, remember, "When in doubt, write it out!" THEOREM 9.1. Properties of Summation Notation: Suppose {an} and {bn} are sequences so that the following sums are defined. P p p e Z (ank bn)=VE ank EZbn n=m n=m n=m p p * S c an = c E an, for any real number c. n=m n=m P i p e an = an + an, for any natural number m < j < j + 1 < p. n=m n=m n=j+1 p p+r *E an = > an-r, for any whole number r. n=m n=m+r We now turn our attention to the sums involving arithmetic and geometric sequences. Given an arithmetic sequence ak = a + (k - 1)d for k > 1, we let S denote the sum of the first n terms. To derive a formula for S, we write it out in two different ways S = a + (a + d) + ... + (a + (n - 2)d) + (a + (n - 1)d) S = (a + (n - 1)d) + (a + (n - 2)d) + ... + (a + d) + a If we add these two equations and combine the terms which are aligned vertically, we get 2S = (2a + (n - 1)d) + (2a + (n - 1)d) + ..". + (2a + (n - 1)d) + (2a + (n - 1)d) The right hand side of this equation contains n terms, all of which are equal to (2a + (n - 1)d) so we get 2S = n(2a + (n - 1)d). Dividing both sides of this equation by 2, we obtain the formula S = - (2a + (n - 1)d) 2 If we rewrite the quantity 2a + (n - 1)d as a + (a + (n - 1)d) = ai + an, we get the formula S=(na1 + anr1 A helpful way to remember this last formula is to recognize that we have expressed the sum as the product of the number of terms n~ and the average of the first and nth terms. To derive the formula for the geometric sum, we start with a geometric sequence a = ar1, k > 1, and let S once again denote the sum of the first n~ terms. Comparing S and rS, we get  538 SEQUENCES AND THE BINOMIAL THEOREM S = a + ar + ar2 + rS = ar + ar2 + ... + arn-2 + ar-1 ... + arn-2 + ar"-1 + arm Subtracting the second equation from the first forces all of the terms except a and ar" to cancel out and we get S - rS = a - ar"m. Factoring, we get S(1 - r) = a (1 - r"). Assuming r - 1, we can divide both sides by the quantity (1 - r) to obtain S = a If we distribute a through the numerator, we get a - ar" = ai - an+1 which yields the formula = ai - an+i 1 -r In the case when r = 1, we get the formula S=a+a +...+ a n times na Our results are summarized below. EQUATION 9.2. Sums of Arithmetic and Geometric Sequences: " The sum S of the first n terms of an arithmetic sequence ak = a + (k - 1)d for k > 1 is n k=1 (a1 + an2) n 2 ) 2(2a + (n - 1)d) " The sum S of the first n terms of a geometric sequence ak = ark-1 for k > 1 is 1. S=Zak k=1 n 2. SE ak k=1 1-asi= a rr 'if r -1. n Sa = na, if r = 1. k=1 While we have made an honest effort to derive the formulas in Equation 9.2, formal proofs require the machinery in Section 9.3. An application of the arithmetic sum formula which proves useful in Calculus results in formula for the sum of the first n natural numbers. The natural numbers themselves are a sequence4 1, 2, 3, ... which is arithmetic with a = d = 1. Applying Equation 9.2, n(n +1) 1+2+3+...+n = 2 4This is the identity function on the natural numbers!  9.2 SUMMATION NOTATION 539 So, for example, the sum of the first 100 natural numbers is -oo20 3 = 5050." An important application of the geometric sum formula is the investment plan called an annuity. Annuities differ from the kind of investments we studied in Section 6.5 in that payments are deposited into the account on an on-going basis, and this complicates the mathematics a little.' Suppose you have an account with annual interest rate r which is compounded n times per year. We let i = n denote the interest rate per period. Suppose we wish to make ongoing deposits of P dollars at the end of each compounding period. Let Ak denote the amount in the account after k compounding periods. Then A1 = P, because we have made our first deposit at the end of the first compounding period and no interest has been earned. During the second compounding period, we earn interest on A1 so that our initial investment has grown to A1(1 + i) = P(1 + i) in accordance with Equation 6.1. When we add our second payment at the end of the second period, we get A2 =A1(1+ i)+ P =P(1+ i) + P=P(1+ i) 1 +l . The reason for factoring out the P(1 + i) will become apparent in short order. During the third compounding period, we earn interest on A2 which then grows to A2(1 + i). We add our third payment at the end of the third compounding period to obtain A3 =A2(1+ i) + P =P(1+ i) 1+ 1 i (1+i) + P =P(1+ i)2 1+ 1 +(1 i)2 During the fourth compounding period, A3 grows to A3(1+i), and when we add the fourth payment, we factor out P(1 + i)3 to get A4=P(1 + i)3 1+ 1 + i + (1+)2 + (1+)3 This pattern continues so that at the end of the kth compounding, we get 1 1 1 Ak =P(1 +i)k- 1 +1+ (+ i)2+ ... + +)- The sum in the parentheses above is the sum of the first k terms of a geometric sequence with a = 1 and r =li. Using Equation 9.2, we get 1 1 + 1 1 (1 + i)k (1+ i) (1 - (1+ i)-k) 1++i (1+i)2 (l+i)k-1 _1 1+ i Hence, we get 5There is an interesting anecdote which says that the famous mathematician Carl Friedrich Gau s: was given this problem in primary school and devised a very clever solution. 6The reader may wish to re-read the discussion on compound interest in Section 6.5 before proceeding.  540 SEQUENCES AND THE BINOMIAL THEOREM Ak = p(l + i)k-I ( (I + i) (I (I + i)-k) i P ((1 + i)k - 1) i If we let t be the number of years this investment strategy is followed, then k = nt, and we get the formula for the future value of an ordinary annuity. EQUATION 9.3. Future Value of an Ordinary Annuity: Suppose an annuity offers an annual interest rate r compounded n times per year. Let i = n be the interest rate per compounding period. If a deposit P is made at the end of each compounding period, the amount A in the account after t years is given by P ((1 + i)"t - 1) A=. i The reader is encouraged to substitute i = n into Equation 9.3 and simplify. Some familiar equations arise which are cause for pause and meditation. One last note: if the deposit P is made a the beginning of the compounding period instead of at the end, the annuity is called an annuity- due. We leave the derivation of the formula for the future value of an annuity-due as an exercise for the reader. EXAMPLE 9.2.2. An ordinary annuity offers a 6% annual interest rate, compounded monthly. 1. If monthly payments of $50 are made, find the value of the annuity in 30 years. 2. How many years will it take for the annuity to grow to $100,000? SOLUTION. 1. We have r = 0.06 and n = 12 so that i = = 006= 0.005. With P = 50 and t = 30, A. 50 ((1 + 0.005)(12)(30) - 0.005 1) ~ 50225.75 Our final answer is $50,225.75. 2. To find how long it will take for the annuity to grow to $100,000, we set A = 100000 and solve for t. We isolate the exponential and take the natural logarithm of both sides of the equation. 10000 1 (1.005)1 ln ((1.005)12t 12t 1n(1.005 0 0 50 ((1 + 0.005)12t - 1) 0.005 (1.005)12t _ 1 2t 11 ) = ln(11) ) = ln(11) ln(11) 40.06 121n(1.005)  9.2 SUMMATION NOTATION 541 This means that it takes just over 40 years for the investment to grow to $100,000. Comparing this with our answer to part 1, we see that in just 10 additional years, the value of the annuity nearly doubles. This is a lesson worth remembering. D We close this section with a peek into Calculus by considering infinite sums, called series. Consider the number 0.9. We can write this number as 0.9 = 0.9999... = 0.9 + 0.09 + 0.009 + 0.0009 + .. . From Example 9.2.1, we know we can write the sum of the first n of these terms as S9 0. 9 - -9 =.9 + 0.09 + 0.009+ . ..0.0 ---0 9 E 0 10k n nines n - 1 zeros k=1 Using Equation 9.2, we have 1 =110k 10 1 1 0n+1 10 It stands to reason that 0.9 is the same value of 1- 0+1 as n -oc. Our knowledge of exponential expressions from Section 6.1 tells us that 10+1 - 0 as n- oc, So 1 - 10n+1 - 1. We have just argued that 0.9 = 1, which may cause some distress for some readers. Any non-terminating decimal can be thought of as an infinite sum whose denominators are the powers of 10, so the phenomenon of adding up infinitely many terms and arriving at a finite number is not as foreign of a concept as it may appear. We end this section with a theorem concerning geometric series. THEOREM 9.2. Geometric Series: Given the sequence ak = ark-1 for k > 1, where Irl < 1, 00 a+ a+ar2+... = ark-1 -1ar 1-r k=1 If Irl > 1, the sum a + ar+ ar2 + ... is not defined. The justification of the result in Theorem 9.2 comes from taking the formula in Equation 9.2 for the sum of the first n terms of a geometric sequence and examining the formula as n - oc. Assuming Irl < 1 means -1 < r < 1, so r -0 asn-n oo. Hence asn -cx0, ar - 1 = 1 - r 1 r As to what goes wrong when Irl > 1, we leave that to Calculus as well, but will explore some cases in the exercises. 7To make this more palatable, it is usually accepted that 0.3= so that 0.9 =3 (0.3) =3 (i) =1. Feel better?  542 SEQUENCES AND THE BINOMIAL THEOREM 9.2.1 EXERCISES 1. Find the following sums. 9 (a) E(5g + 3) g=4 (b) 81 k=3 5 (c) > 2i j=0 2 (d) >(3k - 5)xk k=0 4 100 (f) n(-1" n=1 2. Rewrite the sum using summation notation. (a) 8+11+14+17+20 (b) 1-2+3-4+5-6+7-8 3 5 7 (d) 1+2+4+---+229 (e +5 6 (e) 2 + + + ( + ( (f) - ln(3) + ln(4) - ln(5) + - - - + ln(20) (g) 1 + 9-16 + 25-36 (h) j(x-5) + (x_-5)2+ (zx-5)3+ 1(x-5) 4 3. Find the sum of the first 10 terms of the following sequences. (a) an = 3+ 5n (b) bn = (2)n (c) cn - -2n + Q)' 4. Express the following repeating decimals as a fraction of integers. (a) 0.7 (b) 0.13 (c) 10.159 (d) -5.867 5. If monthly payments of $300 are made to an ordinary annuity with an APR of 2.5% com- pounded monthly what is the value of the annuity after 17 years? 6. Prove the properties listed in Theorem 9.1. 7. Show that the formula for the future value of an annuity due is A = P(1+i) . 8. Discuss with your classmates the problems which arise in trying to find the sum of the following geometric series. When in doubt, write them out! (a) =2-1 k=1 (b) (1.0001)k-1 k=1 o0 k=1  9.2 SUMMATION NOTATION 9.2.2 ANSWERS 543 1. (a) 213 (b) 28 (c) 63 (d) -5 -2x +9x (f) 0 5 2. (a) Z (3k+5) k~1 8 k~1 (e k~l 20 k=3 E k 4 k~1 1)2k-1 29 (d Z2 kl k~1 (h 4~j~ 5)k 3. (a) 305 7 4. (a) - 99 5. $76,163.67 b 1023 (b) 1024 17771050 (c 59049 (C) 3383 333 ()_5809 d)-990  544 SEQUENCES AND THE BINOMIAL THEOREM 9.3 MATHEMATICAL INDUCTION The Chinese philosopher Confucius is credited with the saying, "A journey of a thousand miles begins with a single step." In many ways, this is the central theme of this section. Here we introduce a method of proof, Mathematical Induction, which allows us to prove many of the formulas we have merely motivated in Sections 9.1 and 9.2 by starting with just a single step. A good example is the formula for arithmetic sequences we touted in Equation 9.1. Arithmetic sequences are defined recursively, starting with ai1= a and then an± = an + d for n> 1. This tells us that we start the sequence with a and we go from one term to the next by successively adding d. In symbols, a, a + d, a+ 2d,a+ 3d,a+ 4d + The pattern suggested here is that to reach the nth term, we start with a and add d to it exactly n - 1 times, which lead us to our formula an = a + (n - 1)d for n> 1. But how do we prove this to be the case? We have the following. The Principle of Mathematical Induction (PMI): Suppose P(n) is a sentence involving the natural number n. IF 1. P(1) is true and 2. whenever P(k) is true, it follows that P(k + 1) is also true THEN the sentence P(n) is true for all natural numbers n. The Principle of Mathematical Induction, or PMI for short, is exactly that - a principle.' It is a property of the natural numbers we either choose to accept or reject. In English, it says that if we want to prove that a formula works for all natural numbers n, we start by showing it is true for n = 1 (the 'base step') and then show that if it is true for a generic natural number k, it must be true for the next natural number, k + 1 (the 'inductive step'). The notation P(n) acts just like function notation. For example, if P(n) is the sentence (formula) 'n2 + 1 = 3', then P(1) would be '12 + 1 = 3', which is false. The construction P(k + 1) would be '(k + 1)2 + 1 = 3'. As usual, this new concept is best illustrated with an example. Returning to our quest to prove the formula for an arithmetic sequence, we first identify P(n) as the formula an = a + (n - 1)d. To prove this formula is valid for all natural numbers n, we need to do two things. First, we need to establish that P(1) is true. In other words, is it true that ai1= a + (1 - 1)d? The answer is yes, since this simplifies to ai1= a, which is part of the definition of the arithmetic sequence. The second thing we need to show is that whenever P(k) is true, it follows that P(k + 1) is true. In other words, we assume P(k) is true (this is called the 'induction hypothesis') and deduce that P(k + 1) is also true. Assuming P(k) to be true seems to invite disaster - after all, isn't this essentially what we're trying to prove in the first place? To help explain this step a little better, we show how this works for specific values of n. We've already established P(1) is true, and we now want to show that P(2) 'Another word for this you may have seen is 'axiom.'  9.3 MATHEMATICAL INDUCTION 545 is true. Thus we need to show that a2 = a+ (2 -1)d. Since P(1) is true, we have ai1= a, and by the definition of an arithmetic sequence, a2 = a1+d = a+d = a+(2-1)d. So P(2) is true. We now use the fact that P(2) is true to show that P(3) is true. Using the fact that a2 = a+(2 -1)d, we show a3 = a+(3-1)d. Since a3 = a2+d, we get a3 =(a+(2-1)d)+d = a+2d = a+(3-1)d, so we have shown P(3) is true. Similarly, we can use the fact that P(3) is true to show that P(4) is true, and so forth. In general, if P(k) is true (i.e., ak = a+(k-1)d) we set out to show that P(k+1) is true (i.e., ak+1 = a + ((k + 1) - 1)d). Assuming ak = a + (k - 1)d, we have by the definition of an arithmetic sequence that ak+1 = ak + d so we get ak+1 = (a + (k - 1)d) + d = a + kd = a + ((k + 1) - 1)d. Hence, P(k + 1) is true. In essence, by showing that P(k + 1) must always be true when P(k) is true, we are showing that the formula P(1) can be used to get the formula P(2), which in turn can be used to derive the formula P(3), which in turn can be used to establish the formula P(4), and so on. Thus as long as P(k) is true for some natural number k, P(n) is true for all of the natural numbers n which follow k. Coupling this with the fact P(1) is true, we have established P(k) is true for all natural numbers which follow n = 1, in other words, all natural numbers n. One might liken Mathematical Induction to a repetitive process like climbing stairs.2 If you are sure that (1) you can get on the stairs (the base case) and (2) you can climb from any one step to the next step (the inductive step), then presumably you can climb the entire staircase.3 We get some more practice with induction in the following example. EXAMPLE 9.3.1. Prove the following assertions using the Principle of Mathematical Induction. 1. The sum formula for arithmetic sequences: Z(a + (j - 1)d) - (2a + (n - 1)d). j=1 2. For a complex number z, (2)"j T=h for n> 1. 3. 3" > 100n for n> 5. 4. Let A be an n x n matrix and let A' be the matrix obtained by replacing a row R of A with cR for some real number c. Use the definition of determinant to show det(A') = c det(A). SOLUTION. 1. We set P(n) to be the equation we are asked to prove. For n = 1, we compare both sides of the equation given in P(n) 11 E(a +(j - 1)d) E -(2a+(1-1)d) j=1 a (1 a +(1 -1)d -~ -(2a) 2 a - 2Falling dominoes is the most widely used metaphor in the mainstream College Algebra books. 3This is how Carl climbed the stairs in the Cologne Cathedral. Well, that, and encouragement from Kai.  546 SEQUENCES AND THE BINOMIAL THEOREM This shows the base case P(1) is true. Next we assume P(k) is true, that is, we assume j=1 k 1)d) -(2a + (k-)d) 2 and attempt to use this to show P(k + 1) is true. Namely, we must show k+1 (a+ j=1 1)d) = 2-(2a + (k+1-1)d) To see how we can use P(k) in this case to prove P(k + 1), we note that the sum in P(k + 1) is the sum of the first k + 1 terms of the sequence ak in P(k) is the sum of the first k terms. We compareI k+1 (a +(j - 1)d) j=1 summing the first k + 1 terms k E(a +(j - 1)d) + (a+(k+1-1)d) j=1 summing the first k terms adding the (k + 1)st term k -(2a+ (k -1)d)+(a+ kd) 2 Using P(k) k(2a+(k -1)d)+2(a+kd) 2 2ka+ 2a+ k2d+kd 2 = a + (k - 1)d for k > 1 while the sum both side of the equation in P(k + 1). - 2 (2a + (k + 1 1)d) 2 (2a + kd) , (k+1)(2a+kd) 2 , 2ka+k2d+2a+kd 2 2ka+ 2a+ k2d+ kd 2 Since all of our steps on both sides of the string of equations are reversible, we conclude that the two sides of the equation are equivalent and hence, P(k + 1) is true. By the Principle of Mathematical Induction, we have that P(n) is true for all natural numbers n. 2. We let P(n) be the formula ()Th = 7. The base case P(1) is (z)1 z= , which reduces to z = z which is true. We now assume P(k) is true, that is, we assume (Z)k =9 and attempt to show that P(k+1) is true. Since (z)k+1 _()k z, we can use the induction hypothesis and  9.3 MATHEMATICAL INDUCTION 547 write (z)k 9= .Hence,_(z)k+1 (z)k = zk. We now use the product rule for conjugates4 to write zk =zkz= zk+1. This establishes (z)k+1 = zk+1, so that P(k + 1) is true. Hence, by the Principle of Mathematical Induction, ()Th =7 for all n> 1. 3. The first wrinkle we encounter in this problem is that we are asked to prove this formula for n > 5 instead of n > 1. Since n is a natural number, this means our base step occurs at n = 6. We can still use the PMI in this case, but our conclusion will be that the formula is valid for all n > 6. We let P(n) be the inequality 3" > 100n, and check that P(6) is true. Comparing 36 = 729 and 100(6) = 600, we see 36 > 100(6) as required. Next, we assume that P(k) is true, that is we assume 3k > 100k. We need to show that P(k + 1) is true, that is, we need to show 3k+1 > 100(k + 1). Since 3k+1 = 3 - 3k, the induction hypothesis gives 3k+1 = 3 - 3k > 3(100k) = 300k. We are done if we can show 300k > 100(k + 1) for k > 6. Solving 300k> 100(k + 1) we get k > 1. Since k > 6, we know this is true. Putting all of this together, we have 3k+1 = 3 - "3 > 3(100k) = 300k > 100(k + 1), and hence P(k + 1) is true. By induction, 3" > 100n for all n> 6. 4. To prove this determinant property, we use induction on n, where we take P(n) to be that the property we wish to prove is true for all n x n matrices. For the base case, we note that if A is a 1 x 1 matrix, then A = [a] so A' = [ca]. By definition, det(A) = a and det(A') = ca so we have det(A') = c det(A) as required. Now suppose that the property we wish to prove is true for all k x k matrices. Let A be a (k +1) x (k +1) matrix. We have two cases, depending on whether or not the row R being replaced is the first row of A. CASE 1: The row R being replaced is the first row of A. By definition, n det(A') = ZEa'pC'p p=1 where the lp cofactor of A' is C'p = (-1)(1+p) det (A'p) and A'p is the k x k matrix obtained by deleting the 1st row and pth column of A'.5 Since the first row of A' is c times the first row of A, we have a's,= c aip. In addition, since the remaining rows of A' are identical to those of A, A',= Aip. (To obtain these matrices, the first row of A' is removed.) Hence det (A'p) = det (Aip), so that C' = Cip. As a result, we get n n a det(A') = EZ apCp = Zc aipCip = c aipCip = c det(A), p=1 p=1 p=1 as required. Hence, P(k + 1) is true in this case, which means the result is true in this case for all natural numbers n> 1. (You'll note that we did not use the induction hypothesis at all in this case. It is possible to restructure the proof so that induction is only used where 4See Exercise 6 in Section 3.4. 5See Section 8.5 for a review of this notation.  548 SEQUENCES AND THE BINOMIAL THEOREM it is needed. While mathematically more elegant, it is less intuitive, and we stand by our approach because of its pedagogical value.) CASE 2: The row R being replaced is the not the first row of A. By definition, n det(A') Z= a'pCi, p=1 where in this case, a',= aip, since the first rows of A and A' are the same. The matrices A'pand Aip, on the other hand, are different but in a very predictable way - the row in A'p which corresponds to the row cR in A' is exactly c times the row in A1, which corresponds to the row R in A. In other words, A'p and A1 are k x k matrices which satisfy the induction hypothesis. Hence, we know det (A'p) = c det (Aip) and C', = c Cip. We get n n a det(A')= iapCp = airc C1 = c apC1p = c det(A), p=1 p=1 p=1 which establishes P(k + 1) to be true. Hence by induction, we have shown that the result holds in this case for n> 1 and we are done. Q While we have used the Principle of Mathematical Induction to prove some of the formulas we have merely motivated in the text, our main use of this result comes in Section 9.4 to prove the celebrated Binomial Theorem. The ardent Mathematics student will no doubt see the PMI in many courses yet to come. Sometimes it is explicitly stated and sometimes it remains hidden in the background. If ever you see a property stated as being true 'for all natural numbers n', it's a solid bet that the formal proof requires the Principle of Mathematical Induction.  9.3 MATHEMATICAL INDUCTION 549 9.3.1 EXERCISES 1. Prove the following assertions using the Principle of Mathematical Induction. (a)Sj2 n(n + 1)(2n + 1) 6 j=1 (b) ja n2(n +1)2 j=1 (c) 2n > 500n for n> 12 (d) 3"n > na for n> 4 (e) Use the Product Rule for Absolute Value to show I1 x I for all real numbers x and all natural numbers n> 1 (f) Use the Product Rule for Logarithms to show log (x") = n log(x) for all real numbers x > 0 and all natural numbers n> 1. a 0 n a" 0 (g) [ 0]_ [ ] _ for n > 1. 2. Prove Equations 9.1 and 9.2 for the case of geometric sequences. That is: (a) For the sequence ai1= a, an±= raan, n > 1, prove an = ar"-1, n> 1. (b) ar"-1 = a (' r, if r # 1, ar--1 = na, if r = 1. j=1 j=1 3. Prove that the determinant of a lower triangular matrix is the product of the entries on the main diagonal. (See Exercise 5 in Section 8.3.) Use this result to then show det (In) = 1 where In is the n x n identity matrix. 4. Discuss the classic 'paradox' All Horses are the Same Color problem with your classmates.  550 SEQUENCES AND THE BINOMIAL THEOREM 9.3.2 SELECTED ANSWERS .2 1. (a) Let P(n) be the sentence S2 j=1 n(n +1) (2n +1) 6 . For the base case, n 1, we get 1j2 ? (1)(1 + 1)(2(1) + 1 6 j=1 12 1J We now assume P(k) is true and use it to show P(k + 1) is true. We have k+1l2 j=1 3 + (k + 1)2 j=1 k(k + 1) (2k + 1) +k+12 6 Using P(k) k(k + 1)(2k + 1) 6(k + 1)2 6 6 k(k + 1)(2k + 1) + 6(k + 1)2 6 (k + 1)(k(2k + 1) + 6(k + 1)) 6 (k + 1) (2k2 + 7k + 6) 6 (k+1)(k+2)(2k+3) 6 ? (k + 1)((k + 1) + 1)(2(k + 1) + 1) 6 ? (k + 1)(k + 2)(2k + 3) 6 ? (k + 1)(k + 2)(2k + 3) 6 ? (k + 1)(k+2)(2k + 3) 6 ? (k + 1)(k+2)(2k + 3) 6 ? (k + 1)(k+2)(2k + 3) 6 7 (k + 1)(k+2)(2k + 3) 6 (k + 1)(k + 2)(2k + 3) 7 6 By induction, >3j2 j=1 n(n + 1)(2n + 1) 6 is true for all natural numbers n> 1. (d) Let P(n) be the sentence 3" > n3. Our base case is n = 4 and we check 34 = 81 and 43 = 64 so that 34 > 43 as required. We now assume P(k) is true, that is 3k > k3, and try to show P(k + 1) is true. We note that 3k+1 = 3 - "3 > 3k3 and so we are done if we can show 3k3> (k + 1)3 for k > 4. We can solve the inequality 3x3 > (x + 1)3 using the techniques of Section 5.3, and doing so gives us x> _r~ 2.26. Hence, for k > 4, 3k+1 = 3 3k > 3k3 > (k +1)3 so that 3k+1 > (k +1)3. By induction, 3n > n3 is true for all natural numbers n> 4.  9.3 MATHEMATICAL INDUCTION 551 (f) Let P(n) be the sentence log (x") = n log(x). For the duration of this argument, we assume x > 0. The base case P(1) amounts checking that log (z9) = 1 log(x) which is clearly true. Next we assume P(k) is true, that is log (xk) = k log(x) and try to show P(k + 1) is true. Using the Product Rule for Logarithms along with the induction hypothesis, we get log (k+1)= log (zk .-x = log (zk) + log(x) = k log(x) + log(x) = (k + 1) log(x) Hence, log (xk+1) = (k + 1) log(x). By induction log (x) = n log(x) is true for all x > 0 and all natural numbers n> 1. 3. Let A be an n x n lower triangular matrix. We proceed to prove the det(A) is the product of the entries along the main diagonal by inducting on n. For n = 1, A = [a] and det(A) = a, so the result is (trivially) true. Next suppose the result is true for k x k lower triangular matrices. Let A be a (k + 1) x (k + 1) lower triangular matrix. Expanding det(A) along the first row, we have n det(A) = ZaipC1p p=1 Since alp= 0 for 2 < p < k + 1, this simplifies det(A)= alC11. By definition, we know that C11= (-1)1+1 det (Al) = det (All) where Al1 is k x k matrix obtained by deleting the first row and first column of A. Since A is lower triangular, so is Al1 and, as such, the induction hypothesis applies to All. In other words, det (A11) is the product of the entries along A1n's main diagonal. Now, the entries on the main diagonal of Al1 are the entries a22, a33, ..., a(k+1)(k+1) from the main diagonal of A. Hence, det(A) = a11 det (A) = al1 (a22a33 . .. a(k+1)(k+1)) ana22a33 --. a(k+1)(k+1) We have det(A) is the product of the entries along its main diagonal. This shows P(k + 1) is true, and, hence, by induction, the result holds for all n x n upper triangular matrices. The n x n identity matrix In is a lower triangular matrix whose main diagonal consists of all l's. Hence, det (In) = 1, as required.  552 SEQUENCES AND THE BINOMIAL THEOREM 9.4 THE BINOMIAL THEOREM In this section, we aim to prove the celebrated Binomial Theorem. Simply stated, the Binomial Theorem is a formula for the expansion of quantities (a+ b)" for natural numbers n. In Elementary and Intermediate Algebra, you should have seen specific instances of the formula, namely (a+b)1 = a+b (a+b)2 = a2+2ab+b2 (a+ b)3 = a3+3a2b+3ab2+ b3 If we wanted the expansion for (a+ b)4 we would write (a+ b)4 = (a+ b)(a+ b)3 and use the formula that we have for (a+b)3 to get (a+b)4 = (a+b) (a3 + 3a2b + 3ab2 + b3) = a4+4a3b+6a2b2+4ab3+b4. Generalizing this a bit, we see that if we have a formula for (a + b)k, we can obtain a formula for (a + b)k+1 by rewriting the latter as (a + b)k+1 = (a + b)(a + b)k. Clearly this means Mathematical Induction plays a major role in the proof of the Binomial Theorem.1 Before we can state the theorem we need to revisit the sequence of factorials which were introduced in Section 9.1. DEFINITION 9.4. Factorials: For a whole number n, n factorial, denoted n!, is the term fn of the sequence fo = 1, fn = n - fn_1, n > 1. In other words, 0! = 1 and n! = n(n - 1)! for n > 1. After 0! = 1 the next four terms, written out in detail, are 1! = 1 -"0! = 1 - 1 = 1, 2! = 2 - 1! =2 -1 =2, 3! = 3 -2!= 3 -2 -1 = 6 and 4! = 4.- 3! = 4 -"3 -"2.- 1 = 24. From this, we see a more informal way of computing n!, which is n! = n"- (n - 1) - (n - 2) - - - 2. 1 with 0! = 1 as a special case. Our first example familiarizes us with some of the basic computations involving factorials. EXAMPLE 9.4.1. 1. Simplify the following expressions. 3!2! (b)7! 1000! (k+2)! 0! 5! 998! 2! (k-i)!' - 2. Prove n! > 3n for all n> 7. SOLUTION: 1. We keep in mind the mantra, "When in doubt, write it out!" as we simplify the following. (a) We have been programmed to react with alarm to the presence of a 0 in the denominator, but in this case 0! =1, so the fraction is defined after all. As for the numerator, 3!= 3 -2 -1 =6and 2! =2.-1= 2, so we have 3 = (6)(2) =12. 'It's pretty much the reason Section 9.3 is in the book.  9.4 THE BINOMIAL THEOREM 553 (b) We have 7!= 7 6 5 4. 3 2. 1= 5040 while 5!= 5 4 3 2. 1= 120. Dividing, we get 7! _ 5040 -42. While this is correct, we note that we could have saved ourselves some 5! 120, of time had we proceeded as follows 5! 5-4-3-2-1 -6 --- 7642 In fact, should we want to fully exploit the recursive nature of the factorial, we can write 7! 7 -6 -5! =7 -6 ,-1 -= 42 5! 5! - (c) Keeping in mind the lesson we learned from the previous problem, we have 1000! 1000.-999 - 998! 1000-999 -998i 999000 _____ _ ___________ _ ___________ -499500 998! 2! 998! - 2! 9 98 t-2! 2 (d) This problem continues the theme which we have seen in the previous two problems. We first note that since k + 2 is larger than k - 1, (k + 2)! contains all of the factors of (k - 1)! and as a result we can get the (k - 1)! to cancel from the denominator. To see this, we begin by writing out (k + 2)! starting with (k +2) and multiplying it by the numbers which precede it until we reach (k - 1): (k + 2)! = (k + 2)(k + 1)(k)(k - 1)!. As a result, we have (k + 2)! (k + 2)(k ))(k - 1)! (k+2)(k+1)(k)-(k+-)(k+2) (k - 1)! (k - 1)! --) The stipulation k > 1 is there to ensure that all of the factorials involved are defined. 2. We proceed by induction and let P(n) be the inequality n! > 3n. The base case here is n = 7 and we see that 7! = 5040 is larger than 37 = 2187, so P(7) is true. Next, we assume that P(k) is true, that is, we assume k! > 3k and attempt to show P(k+ 1) follows. Using the properties of the factorial, we have (k + 1)! = (k + 1)k! and since k! > 3k, we have (k + 1)! > (k + 1)3k. Since k > 7, k + 1 > 8, so (k + 1)3k > 8 .3 > 3 - 3k= 3k+1. Putting all of this together, we have (k + 1)! = (k + 1)k! > (k + 1)3k > 3k+1 which shows P(k + 1) is true. By the Principle of Mathematical Induction, we have n! > 3n for all n> 7. Q Of all of the mathematical animals we have discussed in the text, factorials grow most quickly. In problem 2 of Example 9.4.1, we proved that n! overtakes 3' at n = 7. 'Overtakes' may be too polite a word, since n! thoroughly trounces 3" for n> 7, as any reasonable set of data will show. It can be shown that for any real number x > 0, not only does n! eventually overtake z", but the ratio - __0 as n a oo.z Applications of factorials in the wild often involve counting arrangements. For example, if you have fifty songs on your mp3 player and wish arrange these songs in a playlist in which the order of the 2This fact is far more important than you could ever possibly imagine.  554 SEQUENCES AND THE BINOMIAL THEOREM songs matters, it turns out that there are 50! different possible playlists. If you wish to select only ten of the songs to create a playlist, then there are4; such playlists. If, on the other hand, you just want to select ten song files out of the fifty to put on a flash memory card so that now the order no longer matters, there are 40!0! ways to achieve this.3 While some of these ideas are explored in the Exercises, the authors encourage you to take courses such as Finite Mathematics, Discrete Mathematics and Statistics. We introduce these concepts here because this is how the factorials make their way into the Binomial Theorem, as our next definition indicates. DEFINITION 9.5. Binomial Coefficients: Given two whole numbers n and j with n > j, the binomial coefficient . (read, n choose j) is the whole number given by n) n! If j!(n-j)! The name 'binomial coefficient' will be justified shortly. For now, we can physically interpret ( )as the number of ways to select j items from n items where the order of the items selected is unimportant. For example, suppose you won two free tickets to a special screening of the latest Hollywood blockbuster and have five good friends each of whom would love to accompany you to the movies. There are (5) ways to choose who goes with you. Applying Definition 9.5, we get 5 5! 5! 5.-4 =1 2 2!(5 - 2)! 2!3! 2 So there are 10 different ways to distribute those two tickets among five friends. (Some will see it as 10 ways to decide which three friends have to stay home.) The reader is encouraged to verify this by actually taking the time to list all of the possibilities. We now state anf prove a theorem which is crucial to the proof of the Binomial Theorem. THEOREM 9.3. For natural numbers n and j with n > j, n n n +1 The proof of Theorem 9.3 is purely computational and uses the definition of binomial coefficients, the recursive property of factorials and common denominators. 3 For reference, 3Fr efrec50! =30414093201713378043612608166064768844377641568960512000000000000, 50! =37276043023296000, and 50! =10272278170 40!10!  9.4 THE BINOMIAL THEOREM 555 7n n n! n! j -1j (j - )(n - (j -1) j! (n - j)! n!n! (j-1)!(n- j + 1)! j!(n - j)! n! n! (j-1)!(n- j + 1)(n - j)! j(j - 1)!(n - j)! n! j n!(n - j +1) j(j - 1)!(n - j + 1)(n -j)! j(j - 1)!(n - j + 1)(n - j)! n! j +n!(n - j +1) j!(n - j + 1)! j!(n - j +1)! n!j + n!(n - j +1) j!(n - j +1)! n! (j + (n - j + 1)) j!(n - j +1)! (n + 1)n! j!(n + 1 -j))! (n+ 1)! j!((n + 1) - )! We are now in position to state and prove the Binomial Theorem where we see that binomial coefficients are just that - coefficients in the binomial expansion. THEOREM 9.4. Binomial Theorem: For nonzero real numbers a and b and natural numbers n, (a + b)"=Z 0a-ib To get a feel of what this theorem is saying and how it really isn't as hard to remember as it may first appear, let's consider the specific case of n = 4. According to the theorem, we have (a +b)4 = a 4-jb S( 4-0b0 + a4-1b1 + a4-2b2 + ( 4-33 + ( 4-4b4 (ja + (aaba (b)a2 2 a 3 a (+)b4  556 SEQUENCES AND THE BINOMIAL THEOREM We forgo the simplification of the coefficients in order to note the pattern in the expansion. First note that in each term, the total of the exponents is 4 which matched the exponent of the binomial (a+ b)4. The exponent on a begins at 4 and decreases by one as we move from one term to the next while the exponent on b starts at 0 and increases by one each time. Also note that the binomial coefficients themselves have a pattern. The upper number, 4, matches the exponent on the binomial (a + b)4 whereas the lower number changes from term to term and matches the exponent of b in that term. This is no coincidence and corresponds to the kind of counting we discussed earlier. If we think of obtaining (a + b)4 by multiplying (a + b)(a + b)(a + b)(a + b), our answer is the sum of all possible products with exactly four factors - some a, some b. If we wish to count, for instance, the number of ways we obtain 1 factor of b out of a total of 4 possible factors, thereby forcing the remaining 3 factors to be a, the answer is (1). Hence, the term (4)a3b is in the expansion. The other terms which appear cover the remaining cases. While this discussion gives an indication as to why the theorem is true, a formal proof requires Mathematical Induction.' To prove the Binomial Theorem, we let P(n) be the expansion formula given in the statement of the theorem and we note that P(1) is true since 1 (a+b)1 I E (1)a-ib-iy j=0 a+ b E ()a1-0b0+ ({)al-bl a+b = a+b/ Now we assume that P(k) is true. That is, we assume that we can expand (a + b)k using the formula given in Theorem 9.4 and attempt to show that P(k + 1) is true. (a+b)k+l = (a+b)(a+b)k =(a+ b ak-b = aE (Ikak-ib + bE (Ikak-ilb = (j) kak+1-j + (i)ak-ibj+1 Our goal is to combine as many of the terms as possible within the two summations. As the counter j in the first summation runs from 0 through k, we get terms involving ak+1, akb, ak-lb2, ..., abk. In the second summation, we get terms involving ak b, ak-lb2, . . ., abk, bk+1. In other words, apart from the first term in the first summation and the last term in the second summation, we have 4ada fair amount of tenacity and attention to detail.  9.4 THE BINOMIAL THEOREM 557 terms common to both summations. Our next move is to 'kick out' the terms which we cannot combine and rewrite the summations so that we can combine them. To that end, we note and E j=0 k E j=0 (.7) ak+1-jbj ak+Z + j=1 (ka k+1-j gj (k) ( ak -jbj+1 j=0 C.7) ak bj+1 + bk+l so that (a + b)k+= ak+1 + ak+1-j+ (k-1 ak-i j1 + bk+l i=O We now wish to write S (j)ak±1-jt + E (k) ak-ib+1 j=1 Zj=0 as a single summation. The wrinkle is that the first summation starts with j = 1, while the second starts with j = 0. Even though the sums produce terms with the same powers of a and b, they do so for different values of j. To resolve this, we need to shift the index on the second summation so that the index j starts at j = 1 instead of j = 0 and we make use of Theorem 9.1 in the process. k-1 E (i ) i=O k -1+1 j1(~ii1)ak - (j1)bQ+j1)+1 k ak+1-j l j We can now combine our two sums using Theorem 9.1 and simplify using Theorem 9.3 k kk-1 E ( )ak+1j + E()ak-jj+1 ji=1 j= j=1 ( ak+ k j=1 Si ak+1-j k- . . ak 1-b j=1 k + ( 1J 1 ak+1-j bj Using this and the fact that ( k1)1 1 and (i) = 1, we get  558 SEQUENCES AND THE BINOMIAL THEOREM (a + b)k+1 k (k+ 1 ak+1b + (k+ 11 ak+1-i + ( + Da)bk+1 k+1 (k+ 1)a(k+1)-jb which shows that P(k + 1) is true. Hence, by induction, we have established that the Binomial Theorem holds for all natural numbers n. EXAMPLE 9.4.2. Use the Binomial Theorem to find the following. 1. (x - 2)4 2. 2.13 3. The term containing x3 in the expansion (2x + y)5 SOLUTION. 1. Since (x- 2)4 = (x + (-2))4, we identify a = x, b -2 and n = 4 and obtain ( - 2) -4- (= 4-0(-2)o+ ()X4-1(2)'+ )X4-2(-2)2+ )X4-3(_2)3+ =-X4 - 8x3 + 24x2 - 32x + 16 ()4-4 2. At first this problem seem misplaced, but we can write 2.13 = (2 + 0.1)3. Identifying a= b =0.1 1and n= 3, we get 2, 1 3 (2+10 \.)23-j 10O 100 ( 0) 2 3 - 0 ( 1 0 - + 1 J 2 3 - 1 ( 1 - + 1 J 2 3 - 2 ( 2O - 3) 2 3 - 3 1 J 3 12 6 1 8+10 100 1000 8+ 1.2 + 0.06 + 0.001 9.261  9.4 THE BINOMIAL THEOREM 559 3. Identifying a = 2x, b = y and n = 5, the Binomial Theorem gives (2x + y) 5 = (2x)5--yj Since we are concerned with only the term containing x3, there is no need to expand the entire sum. The exponents on each term must add to 5 and if the exponent on x is 3, the exponent on y must be 2. Plucking out the term j = 2, we get (5 )(2x)5-2y2 = 1O(2x)3y2 = 80x3y2 We close this section with Pascal's Triangle, named in honor of the mathematician Blaise Pascal. Pascal's Triangle is obtained by arranging the binomial coefficients in the triangular fashion below. (A) ( ) ( ) Since () =1and ( 0) 1 for all whole numbers , we get that each row of Pascal's Triangle begins and ends with 1. To generate the numbers in the middle of the rows (from the third row onwards), we take advantage of the additive relationship expressed in Theorem 9.3. For instance, (S) + (i) =(i), ( ) + ( ) =( ) and so forth. This relationship is indicated by the arrows in the array above. With these two facts in hand, we can quickly generate Pascal's Triangle. We start with the first two rows, 1 and 1 1. From that point on, each successive row begins and ends with 1 and the middle numbers are generated using Theorem 9.3. Below we attempt to demonstrate this building process to generate the first five rows of Pascal's Triangle.  560 SEQUENCES AND THE BINOMIAL THEOREM 1 1 - 1+1 1 1 1 1 1 1 1 2 1 1 1 1 1 2 1 1 3 3 1 1 1 1 1+2 2 1 2+1 1 1 1 1 1 1+3 2 1 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 3 3 1 3+3 3+1 To see how we can use Pascal's Triangle to expedite the Binomial Theorem, suppose we wish to expand (3x - y)4. The coefficients we need are (4) for j = 0, 1, 2, 3, 4 and are the numbers which form the fifth row of Pascal's Triangle. Since we know that the exponent of 3x in the first term is 4 and then decreases by one as we go from left to right while the exponent of -y starts at 0 in the first term and then increases by one as we move from left to right, we quickly obtain (3x - 1)4 (1)(3x)4 + (4)(3x)3(-y) + (6)(3x)2(-y)2 + 4(3x)(-y)3 + 1(-y)4 81x4 - 108x3y + 54x22 - 12xy3 + Y4 We would like to stress that Pascal's Triangle is a very quick method to expand an entire binomial. If only a term (or two or three) is required, then the Binomial Theorem is definitely the way to go.  9.4 THE BINOMIAL THEOREM 561 9.4.1 EXERCISES 1. Simplify the following expressions. (a) (3!)2 (d) 4!!2! 10! (b) 71!( (n+ 1)! (117) (e) , n > 0. (h) n! 0 7! (c) 233 (f)k -1)! kn 3.('k > 1.(in > 2 (k+2)!' - n 2 2. Use Pascal's Triangle to expand the following. (a) (x + 2)5 (b) (2x - 1)4 (c) (}X + y2)3 (d) (x - z-1)4 3. Use Pascal's Triangle to simplify the following powers of complex numbers. (a) (1+ 2i)4 ( 2 4 (b) (-1+i3)3 (c) 2 2 4. Use the Binomial Theorem to find the indicated term in the following expansions. (a) The term containing x3 in the expansion (2x - y)5 (b) The term containing x117 in the expansion (x + 2)118 (8 (c) The term containing X2 in the expansion (2x - 3)8 (d) The term containing x7 in the expansion (2x - x3 (e) The constant term in the expansion (x + z-1)8 5. Use the Prinicple of Mathematical Induction to prove n! > 2n for n> 4. 6. Prove E ( =j 2" for all natural numbers n. (HINT: Use the Binomial Theorem!) j=0 7. With the help of your classmates, research Patterns and Properties of Pascal's Triangle. 8. You've just won three tickets to see the new film, '8.9.' Five of your friends, Albert, Beth, Chuck, Dan, and Eugene, are interested in seeing it with you. With the help of your class- mates, list all the possible ways to distribute your two extra tickets among your five friends. Now suppose you've come down with the flu. List all the different ways you can distribute the three tickets among these five friends. How does this compare with the first list you made? What does this have to do with the fact that ( ) - (5)?  562 562 SEQUENCES AND THE BINOMIAL THEOREM 9.4.2 ANSWERS 1. (a) 36 (d) 1260 (b) 720 (e) n2+ 1 (c) 105 (f) k(k+1)k 2 2. (a) (x +2)5-=x5 +10x4 +40x3 +80x2+ 80x +32 (b) (2x -i)4 16x4 - 32x3 + 2492 - 8x + 1 (C) ( 3X +X2 - 27x3 + 3x2y2 + xy 4 + y 6 (dl) (x~x1)4~x4_4x2+6_4x_2+x_4 3. (a) -7 -24i (b) 8 4. (a) 80x3y2 (b) 236x117 (c) -24x2I (g) 56 (h) 1 (i) m(2-1) (c) -1 (d) -40x-7 (e) 70  INDEX nth root definition, 297 index, 297 radicand, 297 u-substitution, 201 x-axis, 1 x-coordinate, 2 x-intercept, 23 y-axis, 1_ y-coordinate, 2 y-intercept, 23 abscissa, 2 absolute value, 117 annuity annuity-due, 540 introduction to, 539 ordinary, 540 asymptote horizontal, 222, 223 slant (oblique), 244 vertical, 222 average rate of change, 110 binomial coefficient, 554 Binomial Theorem, 555 bisection method, 205 Cartesian coordinate plane, Cauchy's Bound, 197 circle definition, 386 from slicing a cone, 383 standard equation, 386 standard equation, alternate, 405 common base, 318 complex number conjugate, 210 definition, 209 complex numbers imaginary unit, i, 209 conic sections definition, 383 continuous, 174 coordinates, Cartesian, 2 cost average, 255 fixed, start-up, 108 variable, 109 Cramer's Rule, 491 dependent variable, 41 Descartes' Rule of Signs, 202 difference quotient, 53 discriminant, 134 distance definition, 6 distance formula, 7 domain applied, 46 definition of, 33 implied, 44 ellipse center, 402 definition, 402 eccentricity, 408 I 563  564 INDEX foci, 402 from slicing a cone, 384 guide rectangle, 405 major axis, 402 minor axis, 402 reflective property, 409 standard equation, 405 vertices, 402 ellipsis (. ..),19 equation contradiction, 429 graph, 22 identity, 429 linear of n variables, 434 linear of two variables, 429 factorial, 526, 552 function (absolute) maximum, 66 (absolute, global) minimum, 66 argument, 41 composite, 265 constant, 64, 106 decreasing, 64 definition as a relation, 31 difference, 50 difference quotient, 53 domain, 33 even, 60 exponential, 316 graphing, 58 identity, 115 increasing, 64 inverse, 280 linear, 106 local (relative) maximum, 66 local (relative) minimum, 66 logarithmic, 320 notation, 41 odd, 60 one-to-one, 282 piecewise-defined, 46 polynomial, 169 product, 50 quadratic, 129 quotient, 50 range, 33 sum, 50 zero, 60 functions algebraic, 299 Fundamental Theorem of Algebra, The, 212 graph function, 58 graphing equation, 22 greatest integer function, 48 growth model limited, 369 logistic, 369 uninhibited, 366 Henderson-Hasselbalch Equation, 341 Horizontal Line Test (HLT), 282 hyperbola asymptotes, 414 branch, 414 center, 414 conjugate axis, 415 definition of, 414 foci, 414 from slicing a cone, 384 guide rectangle, 415 standard equation horizontal, 417 vertical, 417 transverse axis, 414 vertices, 414 independent variable, 41 induction base step, 544 induction hypothesis, 544 inductive step, 544 inflection point, 371  INDEX 565 integers, 48 interest compound, 363 compound continuously, 365 simple, 363 line horizontal, 17 parallel, 113 perpendicular, 14 point-slope form, 105 slope-intercept form, 105 vertical, 17 logarithm common, 320 natural, 320 leading entry, 446 main diagonal, 461 matrix multiplication, 460 minor, 488 product of row and column, 460 reduced row echelon form, 447 row echelon form, 446 scalar multiplication, 456 size, 444 square matrix, 462 sum, 454 zero product property for scalar multiplica- tion, 457 midpoint definition, 8 midpoint formula, 8 matrix additive identity, 455 additive inverse, 455 adjoint, 494 associative law of addition, 455 associative property of matrix multiplication, 461 associative property of scalar multiplication, 457 augmented, 445 cofactor, 488 commutative law of addition, 455 definition, 444 determinant, 486 dimension, 444 distributive property of matrix multiplica- tion over matrix addition, 461 distributive property of scalar multiplication over matrix addition, 457 distributive property of scalar multiplication over scalar addition, 457 entry, 444 equality, 454 identity of matrix multiplication, 461 identity of scalar multiplication, 457 invertible, 475 natural base, 318 ordered pair, 2 ordinate, 2 parabola, 131 parabola axis of symmetry, 131 definition, 392 focal length, 393 from slicing a cone, 384 latus rectum, 394 reflective property, 397 standard equation horizontal, 396 vertical, 393 vertex, 131, 392 paraboloid, 397 partial fractions, 499 Pascal's Triangle, 559 piecewise-defined function, 46 polynomial division dividend, 188 divisor, 188 factor, 188 quotient, 188  566 INDEX remainder, 188 synthetic division, 190 polynomial function completely factored over the complex numbers, 213 over the real numbers, 213 constant term, 170 degree, 170 end behavior, 173 leading coefficient, 170 leading term, 170 variations in sign, 201 zero multiplicity, 18 price-demand equation, 109 principal, 363 profit, 134 projection x-axis, 34 quadrants, 4 quadratic formula, 133 quadratic function general form, 132 irreducible quadratic, 213 standard form, 132 range definition of, 33 rational functions, 219 Rational Zeros Theorem, 197' reflecton, 6 regression coefficient of determination, 160 correlation coefficient, 160 least squares line, 159 quadratic, 162 total squared error, 159 relation algebraic description, 16 definition, 14 graphical description, 14 list description, 14 roster description, 14 revenue, 111 secant line, 110 sequence arithmetic common difference, 526 definition, 526 definition, 524 geometric common ratio, 526 definition, 526 sequences alternating, 524 index of summation, 534 lower limit of summation, 534 recursion, 526 summation notation, 534 term, 524 upper limit of summation, 534 series, 541 sign diagram, 149 slope definition, 101 rate of change, 104 smooth, 174 symmetry about the x-axis, 4 about the y-axis, 4 about the origin, 4 system of equations back-substitution, 440 coefficient matrix, 466 consistent, 433 constant matrix, 466 definition, 429 dependent, 434 free variable, 432 Gauss-Jordan Elimination, 448 Gaussian Elimination, 437 inconsistent, 434 independent, 434 leading variable, 436  INDEX 567 overdetermined, 434 parametric solution, 432 triangular form, 436 underdetermined, 434 unknowns matrix, 466 Triangle Inequality, 125 Unit Circle definition, 389 variable dependent, 41 independent, 41 variation constant of proportionality, 258 direct, 258 inverse, 258 joint, 258 Vertical Line Test (VLT), 31 zero of a function, 60