CHAPTER 3 Applications of Differentiation

Transcription

1 CHAPTER Applications of Differentiation Section. Etrema on an Interval Section. Rolle s Theorem and the Mean Value Theorem. 7 Section. Increasing and Decreasing Functions and the First Derivative Test Section. Concavit and the Second Derivative Test Section.5 Limits at Infinit Section. A Summar of Curve Sketching Section.7 Optimization Problems Section.8 Differentials Review Eercises Problem Solving

2 CHAPTER Applications of Differentiation Section. Etrema on an Interval Solutions to Odd-Numbered Eercises. f. f 7 7 f 8 f 7 7 f0 0 f f f f is undefined. 7. Critical numbers: : absolute maimum 9. Critical numbers:,,., : absolute maimum : absolute minimum f f Critical numbers: 0,. gt t t, t < 5. gt t t t t t t 8 t t Critical number is t 8. h sin cos, 0 < < h sin cos sin sin cos On 0,, critical numbers:,, 5 7. f log ln ln f ln ln ln ln ln f,, f No critical numbers Left endpoint:, 8 Maimum Right endpoint:, Minimum Critical number: 0

3 Section. Etrema on an Interval. f, 0,. f,, f Left endpoint: 0, 0 Minimum Critical number: Right endpoint:, 9 Maimum, 0 Minimum f Left endpoint:, 5 Minimum Right endpoint:, Maimum Critical number: 0, 0 Critical number:, 5. f,, 7. f Left endpoint:, 5 Maimum Critical number: 0, 0 Minimum Right endpoint:, gt t,, t gt t t Left endpoint: Maimum Critical number: 0, 0 Minimum Right endpoint:,, Maimum 9. hs, 0, s hs s Left endpoint: Right endpoint: 0, Maimum, Minimum. e sin, 0, e sin e cos e sin cos Left endpoint: 0, 0 Critical number: Minimum, e, 7. Right endpoint:, 0 Minimum Maimum. f cos, 0, 5. f sin Left endpoint: 0, Maimum Right endpoint: Minimum, tan,, 8 8 sec 8 sec On the interval,, this equation has no solutions. Thus, there are no critical numbers. Left endpoint:,,. Maimum Right endpoint:, Minimum 7. (a) Minimum: 0, Maimum:, Minimum: 0, (c) Maimum:, (d) No etrema

4 Chapter Applications of Differentiation 9. f, 0, < Left endpoint: 0, Minimum. f,, Right endpoint:, Minimum Right endpoint:, Maimum. (a) 5 (,.7) 0 (0.98,.0) Maimum:,.7 (endpoint) Minimum: 0.98, f , 0, f ± f ± f.7 Maimum (endpoint) f Minimum: 0.98,.0 5. (a) 7. (a) (, 5.75) (,.) (.08,.97) (0.57, 0.57).08,.97 Minimum 0.57, 0.57 Minimum

5 Section. Etrema on an Interval 5 9. f, 0, f f f Setting f 0, In the interval 0,, choose we have ± ± is the maimum value. f f f e, 0, f e e e e f e e e Left endpoint: Right endpoint: Maimum value of f0 f 0 f is on 0,. 5. f, 0, 55. f tan f f is continuous on 0, but not on 0,. lim tan. f 9 f f f 5 50 f 0 5 is the maimum value (a) Yes. (a) No 5 5 f No Yes. P VI RI I 0.5I, 0 I 5 P 0 when I 0. P 7.5 when I 5. P I 0 Critical number: I amps When I amps, P 7, the maimum output. No, a 0-amp fuse would not increase the power output. P is decreasing for I >.

6 Chapter Applications of Differentiation 5. S hs s cos sin, csc sec ds d s csc cot csc s csc cot csc 0 cot arcsec radians s S hs s S hs Sarcsec hs s S is minimum when arcsec radians. 7. (a) a b c a b A B The coordinates of B are 500, 0, and those of A are 500, 5. From the slopes at A and B, 500 9% % a b a b 0.0. Solving these two equations, ou obtain a 0000 and b 00. From the points 500, 0 and 500, 5, ou obtain c c. 00 In both cases, c Thus, d For 500 0, d a b c For 0 500, d a b c 0.0. (c) The lowest point on the highwa is 00, 8, which is not directl over the point where the two hillsides come together. 9. True. See Eercise True.

7 Section. Rolle s Theorem and the Mean Value Theorem 7 Section. Rolle s Theorem and the Mean Value Theorem. Rolle s Theorem does not appl to over 0, since f is not differentiable at. f. f -intercepts:, 0,, 0 f 0 at. 5. f 7. -intercepts:, 0, 0, 0 f f 0 at 8 f, 0, f 0 f 0 f is continuous on 0,. f is differentiable on 0,. Rolle s Theorem applies. f 0 c value: 9. f,,. f f 0 f is continuous on,. f is differentiable on,. Rolle s Theorem applies. c f f 0 ±, c f, 8, 8 f 8 f 8 f is continuous on 8, 8. f is not differentiable on 8, 8 since f0 does not eist. Rolle s Theorem does not appl.. f,, f f 0 f is continuous on,. (Note: The discontinuit,, is not in the interval.) f is differentiable on (,. Rolle s Theorem applies. f 0 0 c value: 5 ± 5 ± 5

8 8 Chapter Applications of Differentiation 5. f e, 0, f 0 f 0 f is continuous on 0, and differentiable on 0,, so Rolle s Theorem applies. f e e 7. f sin, 0, f 0 f 0 f is continuous on 0,. f is differentiable on 0,. Rolle s Theorem applies. f cos e 0 c value:. c values:, 9. f sin, 0,. f 0 f 0 f is continuous on 0,. f is differentiable on 0,. Rolle s Theorem applies. arcsin c value: 0.89 f 8 sin cos 0 8 sin cos sin sin 0.89 f tan, 0, f 0 f 0 f is not continuous on 0, since f does not eist. Rolle s Theorem does not appl.. f,, f f 0 f is continuous on,. f is not differentiable on, since f0 does not eist. Rolle s Theorem does not appl.

9 Section. Rolle s Theorem and the Mean Value Theorem 9 5. f tan,, 7. f f 0 f is continuous on,. f is differentiable on,. Rolle s Theorem applies. f sec 0 sec sec ± ± arcsec ± arccos f arcsin,, f f f f0 does not eist. Rolle s Theorem does not appl. ±0.5 radian c values: ±0.5 radian f t t 8t. (a) f f v ft must be 0 at some time in,. ft t 8 0 tangent line (c, f(c )) (a, f(a)) f (b, f) (c, f(c )) t seconds a tangent line b secant line. f, 0, 5. f is continuous on, and differentiable on,. f has a discontinuit at. f f f when Therefore,. c.

10 0 Chapter Applications of Differentiation 7. f is continuous on 0, and differentiable on 0,. f f 0 0 f 8 7 c f is continuous on 7, and differentiable on 7,. f f f 9 c. f sin is continuous on 0, and differentiable on 0,. f f f cos 0 c. f log ln ln f is continuous on, and differentiable on,. f f 0 f ln ln ln ln e ln ln e c e 5. f. (a) on, tangent f (c) f 0.5 Secant line: slope secant 0 f f 5 In the interval,, c. f c Tangent line: ± 5 0 ±

11 Section. Rolle s Theorem and the Mean Value Theorem 7. f,, 9,, 9, m 9 (a) f 9 Secant line: tangent secant 0 (c) f 9 f 9 f c c c c, f c, m f Tangent line: 0 9. f e cos, 0 f0, f 0 m 0 0 (a) tangent secant f Secant line: 0 f e cos e cos e sin sin (c) Tangent line: fc c.0, fc.8 5. st.9t No. Let f on,. (a) V avg s s msec st is continuous on 0, and differentiable on 0,. Therefore, the Mean Value Theorem applies. f f0 0 and zero is in the interval (, but f f. vt st 9.8t.7 msec t seconds

12 ) Chapter Applications of Differentiation 55. Let Tt be the temperature of the object. Then T0 500 and T5 90. The average temperature over the interval 0, 5 is Fhr. B the Mean Value Theorem, there eists a time t 0, 0 < t 0 < 5, such that Tt f is continuous on 5, 5 and does not satisf 59. False. f has a discontinuit at 0. the conditions of the Mean Value Theorem. f is not differentiable on 5, 5. Eample: f 5, 5) 8 f ) ) ) 5, 5). True. A polnomial is continuous and differentiable everwhere.. Suppose that p n a b has two real roots and. Then b Rolle s Theorem, since p p 0, there eists c in such that pc 0. But p n n, a 0, since n > 0, a > 0. Therefore, p cannot have two real roots. 5. If p A B C, then p A B f b f a b a Ab Bb C Aa Ba C b a Ab a Bb a b a b aab a B b a Ab a B. Thus, A Ab a and b a which is the midpoint of a, b. 7. f cos differentiable on,. f sin f f < for all real numbers. Thus, from Eercise 5, f has, at most, one fied point Section. Increasing and Decreasing Functions and the First Derivative Test. f 8. Increasing on:, Decreasing on:, Increasing on:,,, Decreasing on:,

13 Section. Increasing and Decreasing Functions and the First Derivative Test 5. f 7. f Discontinuit: 0 Test intervals: < < 0 0 < < Sign of f: Conclusion: Increasing Decreasing Increasing on, 0 Decreasing on 0, f > 0 f < 0 g 8 g Critical number: Test intervals: Sign of g: < < g Conclusion: Decreasing Increasing Increasing on:, Decreasing on:, < 0 < < g > 0 9. Domain:, 8 Critical numbers: ± Test intervals: < < < < < < Sign of : < 0 > 0 < 0 Conclusion: Decreasing Increasing Decreasing Increasing on, Decreasing on,,,. f. f 0 f f 0 Critical number: Critical number: Test intervals: < < < < Sign of f: f f < 0 > 0 Conclusion: Decreasing Increasing Increasing on:, Decreasing on:, Relative minimum:, 9 Test intervals: < < < < Sign of f: f f > 0 < 0 Conclusion: Increasing Decreasing Increasing on:, Decreasing on:, Relative maimum:, 5

14 Chapter Applications of Differentiation 5. f f 0 Critical numbers:, Test intervals: < < < < < < Sign of f: f > 0 f f < 0 > 0 Conclusion: Increasing Decreasing Increasing Increasing on:,,, Decreasing on:, Relative maimum:, 0 Relative minimum:, 7 7. f f Critical numbers: 0, Test intervals: < < 0 0 < < < < Sign of f: f < 0 f > 0 f < 0 Conclusion: Decreasing Increasing Decreasing Increasing on: 0, Decreasing on:, 0,, Relative maimum:, Relative minimum: 0, 0 9. f f Critical numbers:, Test intervals: < < < < < < Sign of f: Conclusion: Increasing Decreasing Increasing Increasing on:,,, Decreasing on:, Relative maimum:, 5 Relative minimum:, 5 f > 0 f f < 0 > 0

15 Section. Increasing and Decreasing Functions and the First Derivative Test 5. f. f Critical number: 0 Test intervals: < < 0 0 < < Sign of f: Conclusion: Increasing Increasing Increasing on: No relative etrema, f > 0 f > 0 f f Critical number: Test intervals: < < < < Sign of f: Conclusion: Decreasing Increasing Increasing on:, Decreasing on:, Relative minimum:, 0 f < 0 f > 0 5. f 5 5 f 5 5, Critical number: 5, < 5 > 5 Test intervals: < < 5 5 < < Sign of f: f f > 0 < 0 Conclusion: Increasing Decreasing Increasing on: Decreasing on:, 5 5, Relative maimum: 5, 5 7. f f Critical numbers:, Discontinuit: 0 Test intervals: < < < < 0 0 < < < < Sign of f: f f > 0 < 0 f f < 0 > 0 Conclusion: Increasing Decreasing Decreasing Increasing Increasing on:,,, Decreasing on:, 0, 0, Relative maimum:, Relative minimum:,

16 Chapter Applications of Differentiation 9. f 9 f Critical number: Discontinuities: 0, Test intervals: < < < < 0 0 < < < < Sign of f: f > 0 f f > 0 < 0 f < 0 Conclusion: Increasing Increasing Decreasing Decreasing Increasing on:,,, 0 Decreasing on: 0,,, Relative maimum: 0, 0. f f Critical numbers:, Discontinuit: Test intervals: < < < < < < < < Sign of f: f f > 0 < 0 f f < 0 > 0 Conclusion: Increasing Decreasing Decreasing Increasing Increasing on:,,, Decreasing on:,,, Relative maimum:, 8 Relative minimum:, 0. f e 0.5 f e e e Critical number: Test intervals: < < < < Sign of f: f > 0 Conclusion: Increasing Decreasing Increasing on: Decreasing on:,, Relative maimum:, e f < 0

17 Section. Increasing and Decreasing Functions and the First Derivative Test 7 5. f arcsin, f Critical number: Decreasing on:, No relative etrema 0 Test intervals: < 0 0 < Sign of f: < 0 < 0 Conclusion: Decreasing Decreasing f (Absolute maimum at, absolute minimum at ) f (0, 0) g g ln Critical number: Increasing on: Decreasing on: 0.90 ln Test intervals: < < ln ln < < Sign of f: Conclusion: Increasing Decreasing, ln ln, f > 0 f < 0 Relative minimum: ln, 0.90, 0.9 e ln 9. f log ln ln f ln 0 ln ln ln Critical number: ln 5 (0.7, ) 5 Test intervals: 0 < < ln ln < < Sign of f: Conclusion: Decreasing Increasing Increasing on: Decreasing on: ln, 0, ln f < 0 f > 0 Relative minimum: ln, ln log ln ln lnln, ln 0.7,

18 8 Chapter Applications of Differentiation. f cos, 0 < < f sin 0 Critical numbers:, 5 Increasing on: 0, Test intervals: Sign of f: 0 < < < < 5 < < Conclusion: Increasing Decreasing Increasing Decreasing on:, 5 f > 0, 5, f < 0 5 f > 0 Relative maimum: Relative minimum: 5, 5,. f sin sin, 0 < < Critical numbers:, 7,, f sin cos cos cos sin 0 Test intervals: 0 < < Sign of f: f > 0 < < 7 7 < < f f < 0 > 0 Conclusion: Increasing Decreasing Increasing Decreasing Increasing < < f f < 0 > 0 < < Increasing on: Decreasing on: 0, Relative minima: Relative maima:, 7, 7 7,,,,,,,,,,, 0

19 Section. Increasing and Decreasing Functions and the First Derivative Test 9 5. f 9,, (a) f 9 9 (c) f 0 f 8 Critical numbers: ± (d) Intervals:, ±,, 8 0 f < 0 f > 0 f < 0 Decreasing Increasing Decreasing f is increasing when when is negative. f f is positive and decreasing 7. f t t sin t, 0, (a) ft t cos t t sin t (c) tt cos t sin t 0 tt cos t sin t t 0 or t tan t π f f π t t cot t t.889, (graphing utilit) Critical numbers: t.889, t (d) Intervals: 0, , , ft > 0 ft < 0 ft > 0 Increasing Decreasing Increasing f is increasing when is negative. f f is positive and decreasing when 9. (a) f ln, 0, f f f (c) f 0 (d) f > 0 on, ; f < 0 on 0,

20 50 Chapter Applications of Differentiation 5. f 5 f g for all ±. f smmetric about origin zeros of f: 0, 0, ±, 0 f, ±, ± f 0 No relative etrema Holes at, and, (, ) 5 5 (, ) 5 5. f is quadratic f is a line. 55. f has positive, but decreasing slope f f In Eercises 57, f > 0 on,, f < 0 on, and f > 0 on,. 57. g f g f. g f g f g0 f0 < 0 g f < 0 g f 0 g f 0 g0 f0 > 0. f > 0, undefined, < 0, < f is increasing on,. > f is decreasing on,. Two possibilities for f are given below. (a) The critical numbers are in intervals 0.50, 0.5 and 0.5, 0.50 since the sign of changes in these intervals. f is decreasing on approimatel, 0.0, 0.8,, and increasing on 0.0, 0.8. Relative minimum when 0.0. Relative maimum when 0.8. f

21 Section. Increasing and Decreasing Functions and the First Derivative Test 5 7. f, g sin, 0 < < (a) f g f seems greater than g on 0,. 5 (c) Let h f g sin h cos > 0 on 0,. 0 > sin on 0, Therefore, h is increasing on 0,. Since h0 0, h > 0 on 0,. Thus, sin > 0 > sin f > g on 0,. 9. v kr rr krr r 7. v krr r krr r 0 r 0 or R Maimum when r R. P vr R R R, v and R are constant dp R R vr vr R R R dr R R vr R R R R 0 R R Maimum when R R. 7. (a) B 0.98t.879t t.0t B 0 (c) for t.78, or 98, (. thousand bankruptcies) Actual minimum: 98 (. thousand bankruptcies) 75. (a) Use a cubic polnomial f a a a a 0. (c) The solution is a 0 a 0, a, a : f a a a. f. 0, 0: 0 a 0 f a f0 0 (d) (, ), : 8a a f 0 a a f 0 (0, 0)

22 5 Chapter Applications of Differentiation 77. (a) Use a fourth degree polnomial f a a a a a 0. f a a a a (0, 0:, 0:, : 0 a 0 f a f a a a f 0 0 5a 8a 8a f 0 a 8a a f 0 a a a f 0 (c) The solution is a a 0 a 0, a, a,. f (d) 5 (, ) (0, 0) (, 0) True Let h f g where f and g are increasing. Then h f g > 0 since f > 0 and g > False Let f, then f and f onl has one critical number. Or, let f, then f has no critical numbers. 8. False. For eample, f does not have a relative etrema at the critical number Assume that f < 0 for all in the interval a, b and let < be an two points in the interval. B the Mean Value Theorem, we know there eists a number c such that < c <, and fc f f. Since fc < 0 and > 0, then f f < 0, which implies that f < f. Thus, f is decreasing on the interval. 87. Let f n n. Then f n n n n n > 0 since > 0 and n >. Thus, f is increasing on 0,. Since f 0 0 f > 0 on 0, n n > 0 n > n.

23 Section. Concavit and the Second Derivative Test 5 Section. Concavit and the Second Derivative Test.,. Concave upward:, f, Concave upward:,,, Concave downward:, 5. f, 7. Concave upward:,,, Concave downward:, f f f 0 when. The concavit changes at., 8 is a point of inflection. Concave upward:, Concave downward:, 9. f f f f 0 when ±. Test interval: < < < < < < Sign of f: f > 0 f < 0 f > 0 Conclusion: Concave upward Concave downward Concave upward Points of inflection: ±, 0 9. f f f f 0 when,. Test interval: < < < < < < Sign of f: f > 0 f < 0 f > 0 Conclusion: Concave upward Concave downward Concave upward Points of inflection:,,, 0

24 5 Chapter Applications of Differentiation. f, Domain:, f f f > 0 on the entire domain of f (ecept for, for which f is undefined). There are no points of inflection. Concave upward on, 5. f f f when 0, ± 0 Test intervals: < < < < 0 0 < < < < Sign of f: f < 0 f > 0 f < 0 f > 0 Conclusion: Concave downward Concave upward Concave downward Concave upward Points of inflection:,, 0, 0,, 7. f sin, 0 f cos f sin f 0 when 0,,. Point of inflection:, 0 Test interval: 0 < < Sign of f: < < f f < 0 > 0 Conclusion: Concave downward Concave upward 9. f sec f sec, 0 < < f sec tan sec tan Concave upward: 0,,, Concave downward:,,, No points of inflection 0 for an in the domain of f.

25 Section. Concavit and the Second Derivative Test 55. f sin sin, 0 f cos cos f sin sin sin cos f 0 when 0,.8,,.0. Test interval: 0 < <.8.8 < < Sign of f: f f < 0 > 0 < <.0 f f < 0 > 0.0 < < Conclusion: Concave downward Concave upward Concave downward Concave upward Points of inflection:.8,.5,, 0,.,.5. f ln Domain: > 0 5. f f f f f Concave upward 0, Critical numbers: 0, However, f0 0, so we must use the First Derivative Test. f < 0 on the intervals, 0 and 0, ; hence, 0, is not an etremum. f > 0 so, 5 is a relative minimum. 7. f 5 9. f 5 f f f f Critical number: 5 Critical numbers: 0, f5 > 0 Therefore, 5, 0 is a relative minimum. f0 < 0 Therefore, 0, is a relative maimum. f > 0 Therefore,, is a relative minimum.. g. f g 5 g 5 8 Critical numbers: g0 > 0 0, 5, Therefore, 0, 0 is a relative minimum. g < 0 Therefore, 5, 8.7 is a relative minimum. f f 9 Critical number: 0 However, f0 is undefined, so we must use the First Derivative Test. Since f < 0 on, 0 and f > 0 on 0,, 0, is a relative minimum. g 0 Test fails b the First Derivative Test,, 0 is not an etremum.

26 5 Chapter Applications of Differentiation 5. f f f 8 7. f cos, 0 f sin 0 Therefore, f is non-increasing and there are no relative etrema. Critical numbers: ± f < 0 Therefore,, is a relative maimum. f > 0 Therefore,, is a relative minimum. 9. ln Domain: > 0 0 when. (, ) 0 0 > 0 Relative minimum:,. ln Domain: 0 < <, > ln ln 0 when e. ln ln ln ln 0 when e. Relative minimum: e, e Point of inflection: e, e (, e e) ( e, e/) 0 9. f e e f e e 0 when 0. f e e > 0 Relative minimum: 0, (0, ) 0

27 Section. Concavit and the Second Derivative Test f e f e e e 0 when 0,. f e e e 0 when ±. Relative minimum: 0, 0 Relative maimum:, e ± ± e± Points of inflection:., 0.8, 0.58, 0.9 (0, 0) (, e ) 5 0 ( ±, ( ± )e ( ± ) ) 7. f 8 f 8 ln f 8 ln ln f 0 ln ln f ln < 0 Relative maimum: relative maimum f 0 ln ln Point of inflection: ln, 8e ln ln, e ln (, e ln ln (, 8e ln ln ( ( 9. f arcsec f 0 when. 0 when Relative maimum:.7, 0.0 Relative minimum:.7, or ± ±.7. (.7,.77) (.7, 0.0) 5. f 0.,, (a) f 0.5 (c) f f f f0 < 0 0, 0 is a relative maimum. f 5 > 0.,.79 is a relative minimum. f Points of inflection:, 0, 0.5, 0.709,.98, f is increasing when and decreasing when f is concave upward when and concave downward when f < 0. f < 0. f > 0 f > 0

28 58 Chapter Applications of Differentiation 5. f sin sin 5 sin 5, 0, (a) f cos cos cos 5 (c) f 0 when, 5,. f sin sin 5 sin 5 5 f 0 when,,.7, f π π f f π is a relative maimum. < 0,.5 Points of inflection:, 0.7,.7, 0.98, f.985, 0.97, 5, 0.7 Note: 0, 0 and, 0 are not points of inflection since the are endpoints. The graph of f is increasing when and decreasing when f is concave upward when and concave downward when f < 0. f < 0. f > 0 f > 0 f < 0 f > (a) means f decreasing. means f increasing. f increasing means concave upward. f increasing means concave upward. 57. Let f. 59. f f f f f0 0, but 0, 0 is not a point of inflection. 5.. (, 0) (, 0) (, 0) (, 0) 5

29 Section. Concavit and the Second Derivative Test is linear. f f is quadratic. f 8 8 f f is cubic. f concave upwards on,, downward on,. 7. (a) n : n : n : n : f f f f f f 0 f f f f f f No inflection points No inflection points Inflection point:, 0 No inflection points: Relative minimum:, 0 Relative minimum:, Point of inflection Conclusion: If n and n is odd, then, 0 is an inflection point. If n and n is even, then, 0 is a relative minimum. Let f n, f n n, f nn n. For n and odd, n is also odd and the concavit changes at. For n and even, n is also even and the concavit does not change at. Thus, is an inflection point if and onl if n is odd. 9. f a b c d Relative maimum:, Relative minimum: 5, Point of inflection:, f a b c, f a b f 7a 9b c d 98a b c 9a 8b c f 5 5a 5b 5c d f 7a b c 0, f a b 0 9a 8b c a b 0 7a b c 0 a b a b a a, b, c 5, d f 5

30 0 Chapter Applications of Differentiation 7. f a b c d Maimum:, Minimum: 0, 0 (a) f a b c, f a b f 0 0 d 0 f a b c f 0 8a 8b c 0 f0 0 c 0 Solving this sstem ields f a and b a. The plane would be descending at the greatest rate at the point of inflection. f a b 0. 8 Two miles from touchdown. 7. D 5L L 75. D 8 5L L 8 5L L 0 0 or 5L ± L B the Second Derivative Test, the deflection is maimum when 5 L 0.578L. 5 ± L C C C C average cost per unit dc when 00 d B the First Derivative Test, C is minimized when 00 units. 77. S 5000t 8 t St 80,000t 8 t St 80,0008 t 8 t St 0 for t 8.. Sales are increasing at the greatest rate at t. ears. 79. f sin cos, f cos sin, f sin cos, P 0 f f f 0 f P P P 0 P 0 P P The values of f, P, P, and their first derivatives are equal at. The values of the second derivatives of f and P are equal at. The approimations worsen as ou move awa from.

31 Section. Concavit and the Second Derivative Test 8. f arctan, a f P f P f f f P P f f f 8. f sin f cos sin cos sin (, 0) π f sin cos cos sin 0 Point of inflection:, 0 When >, f < 0, so the graph is concave downward. 85. Assume the zeros of f are all real. Then epress the function as f a r r where r, r, r and r are the distinct zeros of f. From the Product Rule for a function involving three factors, we have f a r r r r r r f a r r r r r r a r r r. Consequentl, f 0 if r r r r r r Average of r, r, and r. a b True. Let a b c d, a 0. Then when ba, and the concavit changes at this point. 89. False. f sin cos f cos sin cos sin 0 Critical number: cos sin tan tan 9. False. Concavit is determined b f. ftan.0555 is the maimum value of.

32 Chapter Applications of Differentiation Section.5 Limits at Infinit. f. f 5. f sin No vertical asmptotes No vertical asmptotes No vertical asmptotes Horizontal asmptote: Horizontal asmptote: 0 Horizontal asmptotes: 0 Matches (f) Matches (d) Matches 7. (a) (c) h f lim h (Limit does not eist) h f lim h 5 h f lim h 0 9. (a) (c) lim 0 lim lim (Limit does not eist). (a) 5 lim 0 0. lim lim 0 5 lim (c) 5 lim (Limit does not eist) 5. lim lim lim lim Limit does not eist. 9. lim lim lim, for < 0 we have. lim for < 0, lim lim (

33 Section.5 Limits at Infinit. Since sin for all 0, we 5. lim sin 0 have b the Squeeze Theorem, lim lim sin sin 0 lim 0. sin Therefore, lim 0. lim 7. lim 5e 9. lim e. lim log lim 5 t t arctan t 0 5. (a) f lim lim = = Therefore, and are both horizontal asmptotes. 7. lim sin Let t. sin t lim t 0 t 9. lim. lim lim lim lim lim lim 0. f f lim f 0 Analticall, lim lim

34 Chapter Applications of Differentiation 5. f f lim f Analticall, lim lim f f lim f 5 Analticall, lim f lim lim lim lim f Let t. lim sin sint lim t 0 t lim sint t 0 t

35 Section.5 Limits at Infinit 5 5. Intercepts:, 0, 0, Smmetr: none 5 Horizontal asmptote: since lim lim. 5 Discontinuit: (Vertical asmptote) Intercept: 0, 0 Intercept: 0, 0 Smmetr: origin Smmetr: -ais Horizontal asmptote: 0 Horizontal asmptote: since Vertical asmptote: ± lim lim Relative minimum: 0, Intercept: 0, 0 Smmetr: -ais Horizontal asmptote: Vertical asmptote: ± 8. Domain: > 0 Intercepts: none Smmetr: -ais Horizontal asmptote: 0 since Discontinuit: 0 (Vertical asmptote) lim 5 0 lim.

36 Chapter Applications of Differentiation. Intercept: 0, 0 5. Intercepts: ±, 0 Smmetr: none Smmetr: -ais Horizontal asmptote: since Horizontal asmptote: since lim Discontinuit: (Vertical asmptote) lim. lim lim. Discontinuit: 0 (Vertical asmptote) Intercept: 0, 0 Smmetr: none Horizontal asmptote: Vertical asmptote: f 5 5 Domain:,,, Domain:, 0, 0, Intercepts: none Smmetr: origin f No relative etrema Horizontal asmptote: none f No points of inflection Vertical asmptotes: ± (discontinuities) Vertical asmptote: Horizontal asmptote: 7 = 5 =

37 Section.5 Limits at Infinit 7 7. f f 5 0 for an in the domain of f. f 0 when 0. = = Since f > 0 on, 0 and f < 0 on 0,, then 0, 0 is a point of inflection. Vertical asmptotes: ± Horizontal asmptote: f f = 0 f when. Since f > 0 on, and f < 0 on,, then, 0 is a point of inflection. = = Vertical asmptote:, Horizontal asmptote: f f No relative etrema = = f 0 when 0. 5 Point of inflection: 0, 0 Horizontal asmptotes: ± No vertical asmptotes 79. g sin, < < g cos. π (, π ) = sin() Horizontal asmptote: Relative maimum: No vertical asmptotes

38 8 Chapter Applications of Differentiation 8. f e f e Critical numbers:, Relative minimum: Relative maimum:, e,., e,.987 Horizontal asmptote: = (, e) 8. (a) f lim f lim f 0 (c) Since lim f, the graph approaches that of a horizontal line, lim f Yes. For eample, let f f, g (a) 8 (c) 70 f = g f g The graph appears as the slant asmptote. 89. C C C C lim

39 Section.5 Limits at Infinit 9 9. line: 5 m 0 = m + (, ) 9. (a) lim n 0.8 e0.n 0.8 8% P 0.e0.n P 0.08 P e 0.n (a) d A B C A B m m 7 m m (c) lim dm lim dm m m The line approaches the vertical line 0. Hence, the distance approaches. 95. (a) T t 0.00t 0.77t Answers will var. T (c) 90 T T 5 8t 58 t (d) T 0. T (e) lim T 8 8 t (f) The limiting temperature is 8. T has no horizontal asmptote. 99. False. Let f.

40 70 Chapter Applications of Differentiation Section. A Summar of Curve Sketching. f has constant negative slope. Matches (D). The slope is periodic, and zero at 0. Matches (A) 5. (a) f 0(c) for and is increasing on 0,. f > 0 f is negative for < < (decreasing function). f f is positive for > and < (increasing function). f 0 at 0 (Inflection point). f f is positive for > 0 (Concave upwards). is negative for < 0 (Concave downward). (d) f is minimum at 0. The rate of change of f at 0 is less than the rate of change of f for all other values of when when ±., (0, 0) ) =, Horizontal asmptote: < < < < < < < < Conclusion Decreasing, concave down 0 Point of inflection Decreasing, concave up 0 0 Relative minimum Increasing, concave up 0 Point of inflection Increasing, concave down 9. < 0 when. No relative etrema, no points of inflection Intercepts: 7, 0, 0,7 7, 0 Vertical asmptote: Horizontal asmptote: 0, 7

41 ) Section. A Summar of Curve Sketching 7. < 0 if ±. 0 if 0. 0 (0, 0) Inflection point: 0, 0 Intercept: 0, 0 Vertical asmptote: ± Horizontal asmptote: 0 Smmetr with respect to the origin. g g g when 0.9, when ± 0,),.. 788, 0) ) 0. 9, 0. )., 577 ) 0. 85,. 7) ) g0.9 < 0, therefore, 0.9,.0 is relative maimum. g.085 > 0, therefore,.085,.7 is a relative minimum. Points of inflection:,.,,.577 Intercepts: 0,,.788, 0 Slant asmptote: 5. f f 0 when ±. (, ) = f 0 (, ) Relative maimum:, = 0 Relative minimum:, Vertical asmptote: 0 Slant asmptote:

42 7 Chapter Applications of Differentiation 7. 0 when,. 8 8 (0, ) 8 (, ) (, ) 0 < 0 when. Therefore,, is a relative maimum. > 0 when. Therefore,, is a relative minimum. Vertical asmptote: Slant asmptote: 9., Domain:, 8 0 when 8 and undefined when. 0 when and undefined when. ( 8, (0, 0) (, 0) ( Note: is not in the domain. Conclusion < < 8 Increasing, concave down 8 8 < < 0 Relative maimum Decreasing, concave down 0 Undefined Undefined Endpoint. h 9 Domain: h when ± ± h when 0. Relative maimum:, 9 Relative minimum:, 9 Intercepts: 0, 0, ±, 0 Smmetric with respect to the origin Point of inflection: 0, 0 5 ( ), 9. (, 0) (0, 0) (, 0) ( ), 9

43 Section. A Summar of Curve Sketching when and undefined when 0. < 0 when 0. (, 0 0) (, ) 7 ( 8, 0 ) 5 < < < < < < Conclusion Decreasing, concave down 0 Undefined Undefined Relative minimum Increasing, concave down 0 Relative maimum Decreasing, concave down 5. 0 when 0,. 0 when. < < 0 Conclusion Increasing, concave down 0 0 Relative maimum 0 < < Decreasing, concave down 0 Point of inflection < < Decreasing, concave up 0 Relative minimum < < Increasing, concave up ( 0.879, 0) (0, ) (, ) (.5, 0) (, (.7, 0) ) 7. 5 No critical numbers 0 when 0. < < < < Conclusion Decreasing, concave up 0 Point of inflection Decreasing, concave down (0, ) (, 0)

44 ( 7 Chapter Applications of Differentiation 9. f 9 (.785, 0) f when ±.) (, 7 f 8 0 when 0 8 < < < < < < < < f f f Conclusion Increasing, concave down 7 0 Relative maimum Decreasing, concave down 0 Point of inflection Decreasing, concave up 5 0 Relative minimum Increasing, concave up (, 0 ) (.7, 0) (, 5) (0., 0). 0 when 0,. 0 when 0,. < < Conclusion Decreasing, concave up 0 Relative minimum (, 0 ) (0, 0) (, ) (, 7 < < Increasing, concave up 7 0 Point of inflection < < < < Increasing, concave down Point of inflection Increasing, concave up. f f 0 when,. f 0 when 0,. f f f Conclusion < < Decreasing, concave up 0 Relative minimum < < 0 Increasing, concave up Point of inflection 0 < < Increasing, concave down 0 0 Point of inflection < < Increasing, concave up (, ) (0, 0) (, ) (.79, 0)

45 Section. A Summar of Curve Sketching when ±. 0 0 when 0. < < Conclusion Increasing, concave down 0 Relative maimum < < 0 Decreasing, concave down Point of inflection 0 < < Decreasing, concave up 0 Relative minimum < < Increasing, concave up ( 5, 0) (, ) (0, 0) (, ) ( ) 5, undefined at. < < Conclusion Decreasing (0, ), 0 0 Undefined Relative minimum < < Increasing 9. f e f e e e 5 Critical point: 5, 9.7 f e 0 when. Relative maimum: 5, 9.7 Inflection point:, (.7, 9.7) (.,.0) 5. gt 0 e t = 0 gt 0et e t > 0 for all t. gt 0et e t e t 0 at.8. 8 (.8, 5) (0, ) lim gt 0 t 8 0 lim gt 0 t Inflection point:.8, 5

46 7 Chapter Applications of Differentiation. ln, Domain: > ln Critical number: ln e e Relative minimum:.79, 0.79 (, 0) (.8, 0.8) 5. g arcsin Domain: 0, g 8 g (, 0) 5 Relative minimum:, 0 7. f f ln f ln ln Relative maimum: Inflection point: (0, 0) ( 9., e ln ln (, 8e ln ln ( ln,.0.7,.0 ln,.50 ( g log ln, Domain: 0 < < ln g g ln Relative maimum: 0.5 ln = (0.5, ) =, 5. sin sin, 0 8 cos sin Relative maimum: Relative minimum: Inflection points: cos 0 when,. sin 0 when 0,, 5,, 7,., 8 9, 9 8, 9, 5, 9,, 0, 7, 9,, 9 π π π

47 Section. A Summar of Curve Sketching sec 0 when ± sec tan 0 when 0. Relative maimum: Relative minimum:, Inflection point: 0, 0 Vertical asmptotes: ±, tan, < <. 55. csc sec, 0 < < sec tan csc cot 0 Relative minimum: Vertical asmptotes: 8, 0, π π π 57. g tan, < < 59. f 0 9 g sin cos cos 0 when 0. 0 g cos sin cos 5 5 Vertical asmptotes:,,, Intercepts:, 0, 0, 0,, 0 Smmetric with respect to -ais. 0 0 vertical asmptote 0 horizontal asmptote Minimum:.0, 9.05 Increasing on 0, Points of inflection: and, ±.80, 0 Maimum:.0, 9.05 Points of inflection:.8, 7.8,.8, 7.8 π π π π

48 78 Chapter Applications of Differentiation. 7. ln 0, 0 point of inflection ± horizontal asmptotes Vertical asmptotes:, 0 Slant asmptote: 5. Vertical asmptote: 5 7. Vertical asmptote: 5 Horizontal asmptote: 0 Slant asmptote: f is cubic. f is quadratic. f f f is linear. f 7. f f (an vertical translate of f will do) 7. f 8 8 f 8 (an vertical translate of f will do) 75. Since the slope is negative, the function is decreasing on, 8, and hence f > f 5.

49 Section. A Summar of Curve Sketching f h Vertical asmptote: none Horizontal asmptote: 9, Undefined, if if The rational function is not reduced to lowest terms. 9 The graph crosses the horizontal asmptote. If a function has a vertical asmptote at c, the graph would not cross it since f c is undefined. hole at, 8. f The graph appears to approach the slant asmptote. 8. f cos, 0, (a) On 0, there seem to be 7 critical numbers: 0.5,.0,.5,.0,.5,.0,.5 f cos cos sin 0 Critical numbers, 0.97,,.98, 5,.98, 7. The critical numbers where maima occur appear to be integers in part (a), but approimating them using shows that the are not integers. f 85. f a b (a) The graph has a vertical asmptote at b. If a > 0, the graph approaches as b. If a < 0, the graph approaches as b. The graph approaches its vertical asmptote faster as a 0. As b varies, the position of the vertical asmptote changes: b. Also, the coordinates of the minimum a > 0 or maimum a < 0 are changed. 87. g ln f, f > 0 g f f (a) Yes. If the graph of g is increasing, then g > 0. Since f > 0, ou know that f g f and, thus, f > 0. Therefore, the graph of f is increasing. No. Let f (positive and concave up). g ln is not concave up.

50 80 Chapter Applications of Differentiation 89. f n (a) For n even, f is smmetric about the -ais. For n odd, f is smmetric about the origin. The -ais will be the horizontal asmptote if the degree of the numerator is less than. That is, n 0,,,. (c) n gives as the horizontal asmptote. (d) There is a slant asmptote if n 5: 5. (e) n 0 5 M 0 N 5 9. Tangent line at P: 0 f 0 0 (a) Let 0: 0 f 0 0 (c) (e) -intercept: Normal line: 0 f 0 0 Let 0: -intercept: f 0 0 f f 0 0 f 0 f 0 0 f 0 f 0, 0 0 f f f 0 0 f 0 f 0 0 f 0 f 0, 0 BC 0 f 0 f 0 0 f 0 f 0 (g) AB 0 0 f 0 f 0 f 0 f 0 (d) (f) (h) Let 0: 0 f 0 0 -intercept: Let 0: 0 f 0 0 -intercept: 0 0 f 0 f 0 0 f 0 0, f 0 0 f f 0 0, 0 0 f 0 PC 0 f 0 f 0 f 0 f 0 f 0 f 0 PC f 0 f 0 AP f 0 f 0 0 AP f 0 f 0 f 0 In Eercises 9-95 f a b c d f a b c f a b f 0 when b ± b ac a Furthermore, lim f if and onl if a > 0. lim f if and onl if a < 0. b ± b ac. a 9. Since lim f, a > 0. Also, since there is onl 95. Since lim f, a > 0. Also, since there are two one critical point, the discriminant is zero. critical points, the discriminant is positive. b ac 0 b ac b ac > 0 b > ac

51 Section.7 Optimization Problems 8 Section.7 Optimization Problems. (a) First Number, Second Number Product, P First Number, Second Number Product, P The maimum is attained near 50 and 0. (c) (d) P (55, 05) 0 The solution appears to be 55. (e) dp 0 0 when 55. d d P d < 0 P is a maimum when The two numbers are 55 and 55.. Let and be two positive numbers such that 9. S 9 ds 9 0 when 9. d d S 8 d > 0 when 9. S is a minimum when Let be a positive number. S ds d 0 when. d S d > 0 when. The sum is a minimum when and.

52 8 Chapter Applications of Differentiation 7. Let be the length and the width of the rectangle A 50 da 50 0 when 5. d d A < 0 when 5. d A is maimum when 5 meters. 9. Let be the length and the width of the rectangle. P 8 dp d 8 0 when 8. d P 5 > 0 when 8. d P is minimum when 8 feet.. d 0. 7 Since d is smallest when the epression inside the radical is smallest, ou need onl find the critical numbers of B the First Derivative Test, the point nearest to, 0 is 7, 7. f 7. f d 7 Since d is smallest when the epression inside the radical is smallest, ou need onl find the critical numbers of f 7. f 0 B the First Derivative Test, the point nearest to, is,. (, ) (, ) d d (, (, 0) ( 5. dq d kq 0 kq 0 k 7. 80,000 (see figure) d Q d kq 0 k kq 0 0 when Q 0. d Q d k < 0 when Q 0. dqd is maimum when Q 0. S 0,000 of fence needed. d S 70,000 > 0 when 00. d where S is the length ds 0,000 0 when 00. d S is a minimum when 00 meters and 00 meters.

53 Section.7 Optimization Problems 8 9. (a) A area of side area of Top V lengthwidthheight (a) A 50 square inches (a) V 99 cubic inches A square inches V cubic inches (c) A.5 50 square inches (c) V.5 7 cubic inches (c) S V 75 V 75 0 ±5 B the First Derivative Test, 5 ields the maimum volume. Dimensions: (A cube!). (a) V s, 0 < < s dv s s d s s 0 when s, s s is not in the domain. d V 8s d d V d < 0 when s. V s 7 is maimum when 5. If the length is doubled, V 7s. Volume is increased b a factor of 8. 7s 8. A da 8 d 0 when d A d < 0 when The area is maimum when feet and feet. 8

54 ( 8 Chapter Applications of Differentiation 5. (a) (c) 0 0 L A 8, > Area A ± 0, (select ) Then and A. Vertices: 0, 0,, 0, 0, (.587,.) 0 L is minimum when.587 and L.. 7. A 5 see figure da d (, when.5. B the First Derivative Test, the inscribed rectangle of maimum area has vertices ±5, 0, ±5, 5. Width: 5 Length: 5 ; A 0 da d see figure when 0. B the First Derivative Test, the dimensions b are 0 b 0 (approimatel 7.77 b 7.77). These dimensions ield a minimum area. + +

55 Section.7 Optimization Problems 85. V r h cubic inches or h r (a) Radius, r Height Surface Area Radius, r 0. Height 0. Surface Area (c) S r rh (d) rr h r r r r r (.5,.) (e) 0 The minimum seems to be. for r.5. ds dr r r 0 when r.5 in h.0 in. r Note: Notice that h r r The minimum seems to be about. for r..

56 ( 8 Chapter Applications of Differentiation. Let be the sides of the square ends and the length of the package. P V dv d 8 0 when 8. d V < 0 when 8. d The volume is maimum when 8 inches and 08 8 inches. 5. V h r r see figure dv d r r r r rr 0 r r rr 0 (0, r) rr r r r 9 r r r 9 8r h r 0, r (, r B the First Derivative Test, the volume is a maimum when r and h r r r. Thus, the maimum volume is V 8r 9 r r cubic units No, there is no minimum area. If the sides are and, then 0 0. The area is A 0 0. This can be made arbitraril small b selecting V r r h h r r r rh S r rh r r r r r ds dr 8 r r 0 when r 9. cm. d S dr 8 r r 8 r r r 8 r > 0 when r 9 cm. The surface area is minimum when r 9 cm and h 0. The resulting solid is a sphere of radius r. cm.

57 Section.7 Optimization Problems 87. Let be the length of a side of the square and the length of a side of the triangle A da d d A 9 d 8 > 0. Let S be the strength and k the constant of proportionalit. Given h w, h w, S kwh S kw57 w k57w w ds dw k57 w 0 when w 8, h 8. d S kw < 0 when w 8. dw These values ield a maimum. A is minimum when 0 9 and R v 0 sin g dr d d R d v 0 g cos 0 when v 0 g sin < 0 when B the Second Derivative Test, R is maimum when.,.. 7. sin h s s h, 0 < < sin tan h h tan s tan sec sin di k sin sin cos cos cos d I k sin k sin k s sec sin cos k cos cos sin α h ft s α k cos sin Since 0 when,, or when sin ±. is acute, we have sin h tan feet. Since d Id k sin 9 sin 7 < 0 when sin, this ields a maimum.

58 88 Chapter Applications of Differentiation 9. S, L Time T 0 dt d 0 0 S = + L = + ( ( Q You need to find the roots of this equation in the interval 0,. B using a computer or graphics calculator, ou can determine that this equation has onl one root in this interval. Testing at this value and at the endpoints, ou see that ields the minimum time. Thus, the man should row to a point mile from the nearest point on the coast. 5. T v 0 v dt d v v 0 0 Since sin and 0 sin θ θ Q we have sin v sin v 0 sin v sin v. Since d T d v v 0 > 0 this condition ields a minimum time. 5. C k k C k k 0 Or, use Eercise 50(d): Thus, θ + kilometers sin C C 0. w 8.5 w (a) Using a graphing utilit, ou find that minimum when 7 7, 8 No, the answer would not change. w w c θ θ w 89 8 c w and c c is a

59 Section.7 Optimization Problems pt pt 000 pt e t e t e t ; p 8.5 elkmonth e t e t e t 0 when t. months S m 5m 0m ds m 5m 5 0m 0 8m 8 0 when m dm. Line: S C , C Approimation: 0.5 units, or 05 units 858. mi. (a) P 0 s s 00 dp ds 0 s s ss 0 0 when 0, s 0. 0 d P ds s 5 d P ds 0 > 0 s 0 ields a minimum. d P 0 < 0 s 0 ields a maimum. ds The maimum profit occurs when s 0, which corresponds to $0,000 P $,00,000. d P ds s 0 when s 0. 5 The point of diminishing returns occurs when s 0, which corresonds to $0,000 being spent on advertising.

60 90 Chapter Applications of Differentiation 5. (a) f c f c 0ce c 0c e c c c c e e c (c) A 0 e e e ce c c e c ce c ce c c e A f c 0 e ee 0 e e e (d) The maimum area is.59 for.8 and f.57. c e lim c 0 lim c Section.8 Differentials. f f f Tangent line at, : f f T f f 5 f Tangent line at, : f f 80 T f sin f cos f sin Tangent line at, sin : f f T cos sin sin cos cos sin 7. f, f,, d 0. f f f. f 0.05 d fd f

61 Section.8 Differentials 9 9. f, f,, d 0.0 f f f 0.99 f d f d f d d d d 5. d d d 7. ln ln d d d 9. cot. d cot csc d cot cot d cos d sin d. arcsin 5. (a) f.9 f 0. f f0. d arcsin d f.0 f 0.0 f f (a) f.9 f 0. f f0.9. (a) f.0 f 0.0 f f g.9 g 0.07 g g g. g 0. g g (a) g.9 g 0.07 g g g. g 0. g g A d ± da d A da ± ± 8 square inches 5. A r r r dr ± A da r dr 8± ±7 square inches

62 9 Chapter Applications of Differentiation 7. (a) d ±0.05 centimeters A da d 5±0.05 Percentage error: da A 5 centimeter 9. ±.5 square centimeters da A ± % d d 0.05 d % (a) r inches r dr ±0.0 inches V r dv r dr ±0.0 ±.88 cubic inches S r ds 8r dr 8±0.0 ±0.9 square inches (c) Relative error: Relative error: dv V r dr dr r r % ds S 8r dr dr r r %. P 00e 00, changes from 5 to 0. dp 00 e 00 Approimate percentage change: dp P 00 e00 d 5 e % V r h 0r, r 5 cm, h 0 cm, dr 0. cm V dv 80r dr cm d ±5 ±0.5 (a) h 9.5 csc dh 9.5 csc cot d dh h cot d dh cot h Converting to radians, cot % in radians. θ h b 9.5 dh h cot d 0.0 d d 0.0 cot 0.0 tan 0.0 tan tan % in radians

63 Review Eercises for Chapter 9 7. r v 0 sin 9. Let f, 00, d 0.. v 0 00 ftsec changes from 0 to dr 00 cos d 0 d 0 r dr feet 80 cos 0 9 feet 8080 f f fd d f Using a calculator: Let f, 5, d. f f f d d f Let f,, d 0.0, f. Then f.0 f fd Using a calculator, In general, when 0, d 57. True 59. True approaches. Review Eercises for Chapter. A number c in the domain of f is a critical number if fc 0 or is undefined at c. f f ( c) is undefined. f ( c) = 0. g 5 cos, 0, g 5 sin 8 (.8, 7.57) 0 when sin 5. (.7, 0.88) Critical numbers: 0.,.7 Left endpoint: 0, 5 Critical number: 0., 5. Critical number:.7, 0.88 Minimum Right endpoint:, 7.57 Maimum

64 9 Chapter Applications of Differentiation 5. Yes. f f 0. f is continuous on,, differentiable on,. 7. f (a) f 0 for. c satisfies fc 0. 0 f f 7 0 f is not differentiable at. 9. f b f a b a f, 8. f 8 7 fc c 7 c f b f a b a f sin fc sin c c 0 f cos,. f A B C f A B f f A B A B fc Ac B A B Ac A c Midpoint of, 5. f f 7 Critical numbers: and 7 Interval: < < < < 7 7 < < Sign of f: f > 0 f < 0 f > 0 Conclusion: Increasing Decreasing Increasing 7. h Domain: 0, Interval: Sign of h: 0 < < h < 0 < < h > 0 h Conclusion: Decreasing Increasing Critical number:

65 Review Eercises for Chapter ft t t ft t t ln t t tln ft 0: tln t ln t ln Critical number Interval: Sign of ft: < t < ln ft > 0 ln < t < ft < 0 Conclusion: Increasing Decreasing. ht t 8t ht t 8 0 when t. Relative minimum:, Test Interval: < t < < t < Sign of ht: ht < 0 ht > 0 Conclusion: Decreasing Increasing. cost sint v sint cost (a) When t inch and v inches/second. 8, sint cost 0 when sint Therefore, sint and cost The maimum displacement is 5 5. (c) Period: inch. Frequenc: cost tant. 5. f cos, 0 f sin f cos 0 when,. Points of inflection:,,, Test Interval: 0 < < Sign of f: f < 0 f > 0 f < 0 Conclusion: Concave downward Concave upward Concave downward < < < < 7. g g g Critical numbers: 0, ± (, ) (0, 0) (, ) g0 > 0 g ± 8 < 0 Relative minimum at 0, 0 Relative maimums at ±,

66 9 Chapter Applications of Differentiation 9. 7 (5, f(5)) 5 (, f()) (, 0) (0, 0) 5 7. The first derivative is positive and the second derivative is negative. The graph is increasing and is concave down.. (a) D 0.00t 0.5t.9t 0.8t (c) Maimum at.9, Minimum at., (d) Outlas increasing at greatest rate at the point of inflection 9.8, lim lim cos 7. lim 0, since 5 cos h. f Discontinuit: Discontinuit: 0 lim lim lim Vertical asmptote: Horizontal asmptote: Vertical asmptote: 0 Horizontal asmptote:. 5 f 5. g ln e e Horizontal asmptote: 0 (to the right) Horizontal asmptotes: 0 (to the left) 5 (to the right) No vertical asmptotes 7. f 9. f Relative minimum:, 08 Relative maimum:, 08 Relative minimum: 0.55,.077 Relative maimum:.55, Vertical asmptote: 0. Horizontal asmptote: 0

67 Review Eercises for Chapter f Domain:, ; Range:, 5, ) ) f 0 when. f Therefore,, is a relative maimum. Intercepts: 0, 0,, f, Domain:,, Range: 8, 8 Domain:, ; Range: 8, 8 f 0 when ± and undefined when ±. f, 8 8 (, 0) (, 0) 8 8 (0, 0) 8, 8 f > 0 Therefore,, 8 is a relative minimum. f < 0 Therefore,, 8 is a relative maimum. Point of inflection: 0, 0 Intercepts:, 0, 0, 0,, 0 Smmetr with respect to origin 55. f Domain:, ; Range:, f 5 0 when, 5,. f 5 0 when, ±. 5 f > 0 Therefore,, 0 is a relative minimum. f 5 < 0 Therefore, 5, 5 5 is a relative maimum. Points of inflection:, 0, 5 Intercepts: 0, 9,, 0,, 0, 0.0, 5, 0. (, 0),. 5 (.9, 0.) (, 0) (.7, 0.0) ( (

68 98 Chapter Applications of Differentiation 57. f Domain:, ; Range:, f f 0 when and undefined when, 0. 5 is undefined when 0,. B the First Derivative Test, 0 is a relative maimum and is a relative minimum. 0, 0 is a point of inflection., 5, 0) ) ),.59) 0, 0) ) Intercepts:, 0, 0, f Domain:,,, ; Range:,,, f < 0 if. f Horizontal asmptote: Vertical asmptote: Intercepts:, 0, 0,. f Domain:, ; Range: 0, f 8 0 when 0. f 8 0 when ±. (, ( 5 (0, ), ( ( f0 < 0 Therefore, 0, is a relative maimum. Points of inflection: ±, Intercept: 0, Smmetric to the -ais Horizontal asmptote: 0

69 ) Review Eercises for Chapter 99. f 0 Domain:, 0, 0, ; Range:,,, 5 (, ) f 0 when ±. (, ) 5 0 f f < 0 Therefore,, is a relative maimum. f > 0 Therefore,, is a relative minimum. Vertical asmptote: 0 Smmetric with respect to origin 5. f 9 Domain:, ; Range: 0, f 9 9 f 9 is undefined at ±. 9 f0 < 0 Therefore, 0, 9 is a relative maimum. 0 when 0 and is undefined when ±. 0 5, 0) 0, 9) ) ), 0) Relative minima: ±, 0 Points of inflection: ±, 0 Intercepts: ±, 0, 0, 9 Smmetric to the -ais 7. h e 9. h e h e Horizontal asmptote: 0 (to the left) Critical point: 0, (relative maimum) Inflection point:, e, 0.7 g ln g ln g Domain: > Relative minimum: Alwas concave upward., 0.8 ( ), e (0, ) (, 0)

70 00 Chapter Applications of Differentiation 7. f 0 log f f 0 ln ln 0 ln ln Relative maimum: 0 ln ln Inflection point: e,. e, 5.78,.57 e ln e, 5e ( ln ( e/, 5e / ln ( 8 ( 7. f cos, ) Domain: 0, ; Range:, f sin 0, f is increasing. f cos 0 when,. Points of inflection: Intercept: 0,,,,, ) 0, ), ) arctan Relative maimum: Relative minimum: 0 (a) The graph is an ellipse: Maimum:, Minimum:, Inflection point: 0, 0, arctan, arctan 0 8 d d d d 0 d 8 d d d 8 8 (, + arctan (, arctan ( ( (, ) (, ) The critical numbers are and. These correspond to the points,,,,,, and,. Hence, the maimum is, and the minimum is,.

71 Review Eercises for Chapter Let t 0 at noon. L d 00 t 0t 0,000 00t t dl t 0 when t.9 hr. dt Ship A at 0.98, 0; Ship B at 0, 9.8 d 0,000 00t t 098. when t.9 :55 P.M.. d km (0, 0) B d (0, 0 t) (00 t, 0) A (00, 0) 8. We have points 0,,, 0, and, 8. Thus, Let m 8 0 f L or 8 f when 0, 5 minimum. Vertices of triangle: 0, 0, 5, 0, 0, (0, ) (, 8) (, 0) A Average of basesheight s s s see figure da d s s s s s s s s 0 when s. s s A is a maimum when s. s s s+ s s s s 85. You can form a right triangle with vertices 0, 0,, 0 and 0,. Assume that the hpotenuse of length L passes through,. Let m 0 0 or f L. f 7 0 when 0 or. L.05 feet 0

72 0 Chapter Applications of Differentiation 87. csc csc L L L L csc 9 csc dl csc cot 9 sec tan 0 d L or L csc see figure 9 or L 9 csc L π θ ( θ 9 csc 9 sec L θ ( tan tan sec csc tan sec tan L 9 Compare to Eercise 7 using a 9 and b. ft.07 ft 89. Total cost Cost per hournumber of hours T v v 550 v 0 v dt 550 dv 0 v v,000 0v 0 when v mph. d T 00 so this value ields a minimum. dv v > 0 when v cos cos 9. d sin cos d d sin cos d S r. dr r ±0.05 ds 8r dr 89±0.05 ±.8 square cm ds S 00 8r dr dr r r ± ±0.5% 9 V r dv r dr 9 ±0.05 ±8. cubic cm dv V 00 r dr dr r r ± ±0.8% 9