Calculus & Its Applications Larry J. Goldstein David Lay Nakhle I. Asmar David I. Schneider Thirteenth Edition

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1 Calculus & Its Applications Larr J. Goldstein David La Nakhle I. Asmar David I. Schneider Thirteenth Edition

2 Pearson Education Limited Edinburgh Gate Harlow Esse CM20 2JE England and Associated Companies throughout the world Visit us on the World Wide Web at: Pearson Education Limited 204 All rights reserved. No part of this publication ma be reproduced, stored in a retrieval sstem, or transmitted in an form or b an means, electronic, mechanical, photocoping, recording or otherwise, without either the prior written permission of the publisher or a licence permitting restricted coping in the United Kingdom issued b the Copright Licensing Agenc Ltd, Saffron House, 6 0 Kirb Street, London ECN 8TS. All trademarks used herein are the propert of their respective owners. The use of an trademark in this tet does not vest in the author or publisher an trademark ownership rights in such trademarks, nor does the use of such trademarks impl an affiliation with or endorsement of this book b such owners. ISBN 0: ISBN 3: British Librar Cataloguing-in-Publication Data A catalogue record for this book is available from the British Librar Printed in the United States of America

3 250 2 = f (t) = f(t) = f (t) t (800) t 0 (800) t (a) Figure 8 Population (in millions) of the United States from 800 to 998. (b) (c) 44. U.S. Population The population (in millions) of the United States (ecluding Alaska and Hawaii) t ears after 800 is given b the function f(t) in Fig. 8(a). The graphs of f (t) andf (t) are shown in Figs. 8(b) and 8(c). (a) What was the population in 925? (b) Approimatel when was the population 25 million? (c) How fast was the population growing in 950? (d) When during the last 50 ears was the population growing at the rate of.8 million people per ear? (e) In what ear was the population growing at the greatest rate? 45. Inde-Fund Fees When a mutual fund compan charges a fee of 0.47% on its inde funds, its assets in the fund are $4 billion. And when it charges a fee of 0.8%, its assets in the fund are $300 billion. (Source: The Boston Globe.) (a) Let denote the fee that the compan charges as a percentage of the inde fund and A() its assets in the fund. Epress A() as a linear function of. (b) In September 2004, Fidelit Mutual lowered its fees on various inde funds from an average of 0.3% to 0.0%. Let R() denote the revenue of the compan from fees when its inde-fund fee is %. Compare the revenue of the compan before and after lowering the fees. [Hint: Revenue is % of assets.] (c) Find the fee that maimizes the revenue of the compan and determine the maimum revenue. 46. Inde-Fund Fees Suppose that the cost function in Eercise 45 is C() = 2.5+, where is the inde-fund fee. (The compan has a fied cost of $ billion, and the cost decreases as a function of the inde-fund fee.) Find the value of that maimizes profit. How well did Fidelit Mutual do before and after lowering the inde-fund fees? Technolog Eercises 47. Draw the graph of f() = in the window [ 2, 6] b [ 0, 20]. It has an inflection point at =2, but no relative etreme points. Enlarge the window a few times to convince ourself that there are no relative etreme points anwhere. What does this tell ou about f ()? 48. Draw the graph of f() = in the window [0, 0] b [ 20, 30]. Algebraicall determine the coordinates of the inflection point. Zoom in and zoom out to convince ourself that there are no relative etreme points anwhere. 49. Draw the graph of f() = in the window [0, 6] b [0, 6]. In what was is this graph like the graph of a parabola that opens upward? In what was is it different? 50. Draw the graph of f() = in the window [0, 25] b [0, 50]. Use the trace feature of the calculator or computer to estimate the coordinates of the relative minimum point. Then determine the coordinates algebraicall. Convince ourself both graphicall (with the calculator or computer) and algebraicall that this function has no inflection points. Solutions to Check Your Understanding 3. Answer: (a) and (d). Curve (b) has the shape of a parabola, but it is not the graph of an function, since vertical lines cross it twice. Curve (c) has two relative etreme points, but the derivative of f() is a linear function, which could not be zero for two different values of. 2. Answer:(a),(c),(d).Curve(b)hasthreerelativeetreme points, but the derivative of f() is a quadratic function, which could not be zero for three different values of. 68

4 4 Curve Sketching (Conclusion) In Section 3, we discussed the main techniques for curve sketching. Here, we add a few finishing touches and eamine some slightl more complicated curves. The more points we plot on a graph, the more accurate the graph becomes. This statement is true even for the simple quadratic and cubic curves in Section 3. Of course, the most important points on a curve are the relative etreme points and the inflection points. In addition, the - and -intercepts often have some intrinsic interest in an applied problem. The -intercept is (0,f(0)). To find the -intercepts on the graph of f(), we must find those values of for which f() = 0. Since this can be a difficult (or impossible) problem, we shall find -intercepts onl when the are eas to find or when a problem specificall requires us to find them. When f() is a quadratic function, as in Eample, we can easil compute the -intercepts (if the eist) either b factoring the epression for f() or b using the quadratic formula. The Quadratic Formula = b ± b 2 4ac 2a The sign ± tells us to form two epressions, one with + and one with. The equation has two distinct roots if b 2 4ac > 0; one double root if b 2 4ac = 0; and no (real) roots if b 2 4ac < 0. EXAMPLE Appling the Second-Derivative Test Sketch the graph of = SOLUTION Let f() = Then f () = 4 f () =. Since f () = 0 onl when = 4, and since f (4) is positive, f() must have a relative minimum point at = 4 (the second derivative test). The relative minimum point is (4,f(4)) = (4, ). The -intercept is (0,f(0)) = (0, 7). To find the -intercepts, we set f() =0and solve for : =0. (0, 7) Figure = (4 2, 0) (4 + 2, 0) (4, ) The epression for f() is not easil factored, so we use the quadratic formula to solve the equation: = ( 4) ± ( 4) 2 4( 2 )(7) 2( 2 ) =4± 2. The -intercepts are (4 2, 0) and (4 + 2, 0). To plot these points, we use the approimation 2.4. (See Fig..) Now Tr Eercise 3 The net eample is interesting. The function that we consider has no critical points. 69

5 EXAMPLE 2 A Function with No Critical Points Sketch the graph of f() = SOLUTION f() = f () = f () = 3 = (3, 7) Searching for critical points, we set f () = 0 and tr to solve for : =0. () If we appl the quadratic formula with a = 2, b = 3, and c = 5, we see that b2 4ac is negative, and so there is no solution to (). In other words, f () is never zero. Thus, the graph cannot have relative etreme points. If we evaluate f () atsome sa, = 0 we see that the first derivative is positive, and so f() is increasing there. Since the graph of f() is a smooth curve with no relative etreme points and no breaks, f() must be increasing for all. (If a function were increasing at = a and decreasing at = b, it would have a relative etreme point between a and b.) Now let us check the concavit. f () = 3 Graph of f() <3 =3 >3 Negative Zero Positive Concave down Concavit reverses Concave up (0, ) Since f () changes sign at = 3, we have an inflection point at (3, 7). The -intercept is (0,f(0)) = (0, ). We omit the -intercept because it is difficult to solve the cubic equation =0. The qualit of our sketch of the curve will be improved if we first sketch the tangent line at the inflection point. To do this, we need to know the slope of the graph at (3, 7): f (3) = 2 (3)2 3(3) + 5 = 2. We draw a line through (3, 7) with slope 2 and then complete the sketch as shown in Figure 2 Fig. 2. Now Tr Eercise 9 EXAMPLE 3 Using the First-Derivative Test Sketch the graph of f() =( 2) 4. SOLUTION f() =( 2) 4 f () =4( 2) 3 f () = 2( 2) 2 Clearl, f () = 0 onl if = 2. So the curve has a horizontal tangent at (2,f(2)) = (2, ). Since f (2) = 0, the second-derivative test is inconclusive. We shall appl the first-derivative test. Note that f () =4( 2) 3 { negative if <2 positive if >2, since the cube of a negative number is negative and the cube of a positive number is positive. Therefore, as goes from left to right in the vicinit of 2, the first derivative changes sign and goes from negative to positive. B the first-derivative test, the point (2, ) is a relative minimum. 70

6 The -intercept is (0,f(0)) = (0, 5). To find the -intercepts, we set f() =0 and solve for : ( 2) 4 =0 ( 2) 4 = 2= or 2= =3 or =. (See Fig. 3.) Now Tr Eercise 2 A Graph with Asmptotes Graphs similar to the one in the net eample will arise in several applications later in this chapter. EXAMPLE 4 Asmptotes Sketch the graph of f() = +(/), for >0. SOLUTION f() = + f () = 2 (0, 5) We set f () = 0 and solve for : f () = =0 = ( 2) 4 = 2 (, 0) (3, 0) (2, ) Figure 3 = + = ( > 0) 2 = =. (We eclude the case = because we are considering onl positive values of.) The graph has a horizontal tangent at (,f()) = (, 2). Since f () = 2 > 0, the graph is concave up at = and, b the second-derivative test, (, 2) is a relative minimum point. In fact, f () =(2/ 3 ) > 0 for all positive, and therefore the graph is concave up at all points. Before sketching the graph, notice that as approaches zero [a point at which f() is not defined], the term / in the formula for f() is dominant. That is, this term becomes arbitraril large, whereas the term contributes a diminishing proportion to the function value as approaches 0. Thus, f() has the -ais as an asmptote. For large values of, theterm is dominant. The value of f() is onl slightl larger than since the term / has decreasing significance as becomes arbitraril large; that is, the graph of f() is slightl above the graph of =. As increases, the graph of f() has the line = as an asmptote. (See Fig. 4.) Now Tr Eercise 27 (, 2) Figure 4 Summar of Curve-Sketching Techniques. Compute f () andf (). 2. Find all relative etreme points. (a) Find the critical values and critical points: Set f () = 0 and solve for. Suppose that = a is a solution (a critical value). Substitute = a into f() to find f(a), plot the critical point (a, f(a)), and draw a small horizontal tangent line through the point. Compute f (a). The Second-Derivative Test (i) If f (a) > 0, draw a small concave-up arc with (a, f(a)) as its lowest point. The curve has a relative minimum at = a. 7

7 (ii) If f (a) < 0, draw a small concave-down arc with (a, f(a)) as its peak. The curve has a relative maimum at = a. The First-Derivative Test (iii) If f (a) = 0, eamine f () to the left and right of = a to determine if the function changes from increasing to decreasing, or vice versa. If a relative etreme point is indicated, draw an appropriate arc as in parts (i) and (ii). (b) Repeat the preceding steps for each solution to f () =0. 3. Find all the inflection points of f(). (a) Set f () = 0 and solve for. Suppose that = b is a solution. Compute f(b) and plot the point (b, f(b)). (b) Test the concavit of f() to the right and left of b. If the concavit changes at = b, then (b, f(b)) is an inflection point. 4. Consider other properties of the function and complete the sketch. (a) If f() is defined at =0,the-intercept is (0,f(0)). (b) Does the partial sketch suggest that there are -intercepts? If so, the are found b setting f() = 0 and solving for. (Solve onl in eas cases or when a problem essentiall requires ou to calculate the -intercepts.) (c) Observe where f() is defined. Sometimes, the function is given onl for restricted values of. Sometimes, the formula for f() is meaningless for certain values of. (d) Look for possible asmptotes. (i) Eamine the formula for f(). If some terms become insignificant as gets large and if the rest of the formula gives the equation of a straight line, then that straight line is an asmptote. (ii) Suppose that there is some point a such that f() is defined for near a, but not at a (for eample, / at = 0). If f() gets arbitraril large (in the positive or negative sense) as approaches a, the vertical line = a is an asmptote for the graph. (e) Complete the sketch. Check Your Understanding 4 Determine whether each of the following functions has an asmptote as gets large. If so, give the equation of the straight line that is the asmptote.. f() = f() = + 3. f() = 2 EXERCISES 4 Find the -intercepts of the given function.. = = = = = = Show that the function f() = has no relative etreme points. 8. Show that the function f() = is alwas decreasing. Sketch the graphs of the following functions. 9. f() = f() = 3. f() = f() = f() = f() = f() = f() = f() = f() = f() = f() = f() =( 3) f() =( +2) 4 72