CHAPTER 3 Applications of Differentiation

CHAPTER Applications of Differentiation Section. Etrema on an Interval.............. 78 Section. Rolle s Theorem and the Mean Value Theorem. 8 Section. Increasing and Decreasing Functions and the First Derivative Test.............. 87 Section. Concavit and the Second Derivative Test.... 9 Section.5 Limits at Infinit................. 0 Section.6 A Summar of Curve Sketching......... 0 Section.7 Optimization Problems.............. 9 Section.8 Newton s Method................. 9 Section.9 Differentials.................... Review Eercises......................... 7 Problem Solving......................... 5

CHAPTER Applications of Differentiation Section. Etrema on an Interval Solutions to Even-Numbered Eercises. f cos f sin. f f f0 0 f 0 f 0 6. Using the limit definition of the derivative, f f 0 lim 0 0 f f 0 lim 0 0 lim 0 lim 0 0 f0 does not eist, since the one-sided derivatives are not equal. 8. Critical number: 0: neither 0. 0. Critical numbers:, 5. : neither 5: absolute maimum g g 8 Critical numbers: 0, ±. f 6. f sec tan, 0 < < f Critical numbers: ± f sec tan sec sec tan sec sec sin cos cos sec sin On 0,, critical numbers: 7 6, 6 78

Section. Etrema on an Interval 79 8. f Left endpoint: 5, 0, 5 0. f No critical numbers 0, 5 Minimum f,, f Left endpoint: (, 5 Minimum Right endpoint:, Maimum Right endpoint: 5, 5 Maimum. f, 0,. f Left endpoint: Critical number:, 6 Minimum Right endpoint: Note: 0, 0, 6 Maimum is not in the interval. g,, g Left endpoint:, Minimum Critical number: 0, 0 Right endpoint:, Maimum 6. t,, 5 8. From the graph, ou see that t is a critical number. 5 ht t,, 5 t ht t Left endpoint:, Maimum Right endpoint: 5, 5 Minimum Left endpoint:, Minimum Right endpoint: 5, Critical number:, Maimum 0. g sec, 6,. g sec tan Left endpoint: 6, Right endpoint:, Maimum Critical number: 0, Minimum 6,.57 cos,, sin Left endpoint:,.50 Right endpoint:, 7.99 Maimum Critical number: 0, Minimum. (a) Minimum:, 6. (a) Minima:, 0 and, 0 Maimum:, Maimum: 0, Maimum:, Minimum:, 0 (c) Minimum:, (c) Maimum: 0, (d) No etrema (d) Maimum:,

80 Chapter Applications of Differentiation 8. f,, < 5 Left endpoint:, Maimum 0. f, 0, Left endpoint: 0, Minimum Right endpoint: 5, Minimum (, ) (, 7) (0, ) 5 (5, ) 5. (a) Maimum:. (, 8 ) 0 0 f, 0, Minimum: 0, 0,, 0 f, 8 f,, f f f Setting f f 0, we have 0, ±. is the maimum value. Critical number: 6 6 f 0 0 f 0 Minimum Minimum f 8 Maimum:, 8 6. f,, 8. Let f. f is continuous on 0, but does not f f 5 0 5 f 5 0 0 6 f 0 See Eercise. is the maimum value. have a maimum. f is also continuous on, 0 but does not have a minimum. This can occur if one of the endpoints is an infinite discontinuit.

Section. Rolle s Theorem and the Mean Value Theorem 8 50. 5. (a) No 5. (a) No 5 f Yes Yes 5 6 56. v sin, 58. C 00,000, 00 d is constant. dt d d d d dt dt b the Chain Rule C 00,00 C00 600 C 00,000 0 v cos 6 In the interval,,, indicate minimums for ddt and indicates a maimum for ddt. This implies that the sprinkler waters longest when and. Thus, the lawn farthest from the spinkler gets the most water. d dt 00,000 50,000 005 87 > 00 outside of interval C is minimized when 00 units. Yes, if 00, then 87 would minimize C. 60. f The derivative of f is undefined at ever integer and is zero at an noninteger real number. All real numbers are critical numbers. 6. True. This is stated in the Etreme Value Theorem. 6. False. Let f. 0 is a critical number of f. g f k k k is a critical number of g. Section. Rolle s Theorem and the Mean Value Theorem. Rolle s Theorem does not appl to f cot over., since f is not continuous at. f -intercepts: 0, 0,, 0 f 0 at. 6. f -intercepts:, 0, 0, 0 f f 0 at.

8 Chapter Applications of Differentiation 8. f 5,, 0. f f 0 f is continuous on,. f is differentiable on,. Rolle s Theorem applies. f 5 5 0 5 f,, f f 0 f is continuous on,. f is differentiable on,. Rolle s Theorem applies. f 6 5. 6. c value: 5 f, 0, 6. f 0 f 6 0 f is continuous on 0, 6. f is not differentiable on 0, 6 since f does not eist. Rolle s Theorem does not appl. f cos, 0, 8. f 0 f f is continuous on 0,. f is differentiable on 0,. Rolle s Theorem applies. f sin c value: c value: 5 f,, f f 0 f is not continuous on, since f 0 does not eist. Rolle s Theorem does not appl. f cos,, f f 6 f f 0 6 Rolle s Theorem does not appl. 0. f f sec,, f f is continuous on,. f is differentiable on,. Rolle s Theorem applies. f sec tan sec tan 0 0 c value: 0. f, 0, f 0 f 0 f is continuous on 0,. f is differentiable on 0,. (Note: f is not differentiable at 0. ) Rolle s Theorem applies. f 0 7 7 9 c value: 9 0.95 0

Section. Rolle s Theorem and the Mean Value Theorem 8. f sin f f 0 0 f is continuous on, 0. f is differentiable on, 0. Rolle s Theorem applies. cos 6 c value: f 6 cos 6 arccos 0.5756 radian 0.5756,, 0 6. 6 0.0 6 0 Value needed in, 0. C 0 (a) C C6 5 C 0 0 6 9 6 9 0 6 ± 08 6 ± 6 ± In the interval, 6: c.098. 0 0.0 8. 0. f, 0, 6 f is not differentiable at. f a b. f is continuous on, and. f is continuous on, and differentiable on,. differentiable on,. f f 0 f c f f f c

8 Chapter Applications of Differentiation 6. f is continuous on 0, and differentiable on 8. f sin sin is continuous on 0, and differentiable 0,. on 0,. f f 0 0 0 f ± In the interval 0, : c. 0. f sin on, f cos cos 0 cos cos 0 cos cos 0 cos In the interval 0, : c. f f 0 0 0 0 0 cos,, 5 (a) tangent secant f tangent Secant line: slope f f (c) f cos cos 0 c ±, f Tangent lines: f. f 8 5, 0, 5, 5, 80 m (a) 80 5 5 0 50 tangent 5 f secant tangent 0 5 0 Secant line: 5 5 0 f 5 f 5 5 c c 6c 5 0 5 5 f 6 0 c c 6c 5 c 0.67 or c.79 (c) First tangent line: f c m c 9.59 5 0.67 0 5 0.6 Second tangent line: f c m c.5 5.79 0 5 7.5

Section. Rolle s Theorem and the Mean Value Theorem 85. St 00 5 9 t (a) S S0 0 St 00 9 50 t 7 t 8 t 7 005 9 005 9 t 7.95 months 50 7 St is equal to the average value in April. 6. f a f b and fc 0 where c is in the interval a, b. (a) g f k g f k (c) g f k ga gb f a k g f gc 0 Interval: a, b Critical number of g: c ga k gb k f a g f k gc k fc 0 Interval: a k, b k g a k g b f a k g kfk g c k kfc 0 Critical number of g: c k Interval: a k, b k Critical number of g: c k 8. Let Tt be the temperature of the object. Then T0 500 and T5 90. The average temperature over the interval 0, 5 is 90 500 5 0 Fhr. B the Mean Value Theorem, there eists a time to, 0 < t 0 < 5, such that Tt 0. 50. f cos, f 6 cos sin (a) 7 cos sin f and f are both continuous on the entire real line. 7 (c) Since f f 0, Rolle s Theorem applies on,. Since f 0 and f, Rolle s Theorem does not appl on,. (d) lim f 0 lim f 0

86 Chapter Applications of Differentiation 5. f is not continuous on 5, 5. Eample: f, 0, 0 0 5. False. f must also be continuous and differentiable on each interval. Let f. ( 5, ) 5, 5 f () = (5, 5 ) 5 56. True 58. Suppose f is not constant on a, b. Then there eists and in a, b such that f f. Then b the Mean Value Theorem, there eists c in a, b such that fc f f 0. This contradicts the fact that f 0 for all in a, b. 60. Suppose f has two fied points and. Then, b the Mean Value Theorem, there eists c such that fc f c f c c c c c c c. This contradicts the fact that f < for all. c c 6. Let f cos. f is continuous and differentiable for all real numbers. B the Mean Value Theorem, for an interval a, b, there eists c in a, b such that f b f a b a cos b cos a b a fc sin c cos b cos a sin cb a cos b cos a sin c b a cos b cos a b a since sin c.

Section. Increasing and Decreasing Functions and the First Derivative Test 87 Section. Increasing and Decreasing Functions and the First Derivative Test.. f Increasing on:, Increasing on:, 0,, Decreasing on:, Decreasing on:,, 0, 6. Critical numbers: 0, Discontinuit: Test intervals: < < < < < < 0 0 < < Sign of f: > 0 < 0 < 0 > 0 Conclusion: Increasing Decreasing Decreasing Increasing Increasing on,, 0, Decreasing on,,, 0 8. h 7 h 7 h 0 Critical numbers: ± Test intervals: < < < < < < Sign of h: h < 0 h > 0 h < 0 Conclusion: Decreasing Increasing Decreasing Increasing on, Decreasing on,,, 0. Critical numbers: ± Discontinuit: 0 Test intervals: < < < < 0 0 < < < < Sign of : > 0 < 0 < 0 > 0 Conclusion: Increasing Decreasing Decreasing Increasing Increasing:,,, Decreasing:, 0, 0,

88 Chapter Applications of Differentiation. f 8 0. f 8 0 Critical number: Test intervals: < < < < Sign of f: f f < 0 > 0 Conclusion: Decreasing Increasing Increasing on:, Decreasing on:, Relative minimum:, 6 f 8 f 8 0 Critical number: Test intervals: Sign of f: < < < < f > 0 Conclusion: Increasing Decreasing Increasing on:, Decreasing on:, Relative maimum:, f < 0 6. f 6 5 f Critical numbers: 0, Test intervals: < < 0 0 < < < < Sign of f: f > 0 f < 0 f > 0 Conclusion: Increasing Decreasing Increasing Increasing on, 0,, Decreasing on 0, Relative maimum: 0, 5 Relative minimum:, 7 8. f f Critical numbers:, 0 Test intervals: < < < < 0 0 < < Sign of f: f > 0 f < 0 f > 0 Conclusion: Increasing Decreasing Increasing Increasing on:,, 0, Decreasing on:, 0 Relative maimum:, 0 Relative minimum: 0,

Section. Increasing and Decreasing Functions and the First Derivative Test 89 0. f. f f 8 Critical number: Test intervals: < < < < Sign of f: f f < 0 > 0 Conclusion: Decreasing Increasing f Critical number: Test intervals: Sign of f: 0 < < 0 f < 0 0 < < f > 0 Increasing on:, Decreasing on:, Relative minimum:, Conclusion: Decreasing Increasing Increasing on: 0, Decreasing on:, 0 Relative minimum: 0,. f 6. f Critical number: Test intervals: < < < < Sign of f: Conclusion: Increasing Increasing Increasing on:, No relative etrema f > 0 f > 0 f f, >, < Critical number: Test intervals: < < < < Sign of f: f f < 0 > 0 Conclusion: Decreasing Increasing Increasing on:, Decreasing on:, Relative minimum:, 8. f f Discontinuit: Test intervals: < < < < Sign of f: f f > 0 > 0 Conclusion: Increasing Increasing Increasing on:,,, No relative etrema

90 Chapter Applications of Differentiation 0. f f 6 6 Critical number: 6 Discontinuit: 0 Test intervals: < < 6 6 < < 0 0 < < Sign of f: Conclusion: Decreasing Increasing Decreasing Increasing on: 6, 0 f < 0 Decreasing on:, 6, 0, Relative minimum: 6, f > 0 f < 0. f f 0 Discontinuit: Test intervals: < < < < Sign of f: f f > 0 > 0 Conclusion: Increasing Increasing Increasing on:,,, No relative etrema. f sin cos sin, 0 < < f cos 0 Critical numbers:,, 5, 7 Test intervals: Sign of f: 0 < < < < 7 < < 5 < < < < Conclusion: Increasing Decreasing Increasing Decreasing Increasing Increasing on: Decreasing on: 0, Relative maima:,,,, 5 f > 0, 5, 5, 7,, 7, Relative minima:,, 7, f < 0 f > 0 5 f < 0 7 f > 0

Section. Increasing and Decreasing Functions and the First Derivative Test 9 6. f sin cos, 0 < < f cos sin cos 0 Critical numbers:, Sign of f: f > 0 Test intervals: 0 < < Conclusion: Increasing Decreasing Increasing < < f < 0 < < f > 0 Increasing on: 0, Relative maimum: Decreasing on: Relative minimum:,,,,, 8. f 05 6, 0, 5 5 (a) f 6 5 f 6 f (c) 5 6 0 Critical number: (d) Intervals: 0, f > 0, 5 f < 0 Increasing Decreasing f is increasing when when is negative. f f is positive and decreasing 0. f cos, 0, (a) f sin 8 6 f f π π π π (c) sin 0 (d) Intervals: sin 0, f > 0 Critical number: Increasing f is increasing when, f > 0 Increasing f is positive.

9 Chapter Applications of Differentiation. f t cos t sin t sin t gt, < t <. f is a line of slope f. ft sin t cos t sin t f smmetric with respect to -ais zeros of f: ± Relative maimum: 0, Relative minimum:,,, 6 6 6 6. f is a th degree polnomial f is a cubic polnomial. 8. f has positive slope 6 f 6 6 f In Eercises 50 5, f > 0 on,, f < 0 on, 6 and f > 0 on 6,. 50. g f 5. g f 5. g f 0 g f g f g f 0 g5 f5 > 0 g0 f0 > 0 g8 f < 0 56. Critical number: 5 58. f.5 f is decreasing at. f6 f is increasing at 6. 5, f 5 is a relative minimum. st.9sin t (a) vt 9.8sin t speed If, vt 9.8t. the speed is maimum, 9.8 sin t 60. C (a) t 7 t, t 0 t 0 0.5.5.5 Ct 0 0.055 0.07 0.8 0.7 0.76 0.67 The concentration seems greater near t.5 hours. 0.5 (c) C 7 t tt 7 t 7 t 7 t 0 0 The concentration is greatest when t.8 hours. C 0 when t.8 hours. B the First Derivative Test, this is a maimum.

Section. Increasing and Decreasing Functions and the First Derivative Test 9 6. P. 5000, 0 5,000 Test intervals: 0 < <,00,00 < < 5,000 0,000 Sign of P: P < 0 P.,00 0,000 0 Increasing when 0 < <,00 hamburgers. Decreasing when,00 < < 5,000 hamburgers. P > 0 6. R 0.00T T 00 (a) R 0.00T 0.00T T 00 0 5 T 0, R 8.666 00 00 5 The minimum resistance is approimatel R 8.7 at T 0. 66. f sin cos 6 6 The maimum value is approimatel.7. You could use calculus b finding f and then observing that the maimum value of f occurs at a point where f 0. For instance, f0.5 0, and f 0.5.7. 68. (a) Use a cubic polnomial f a a a a 0. f a a a 0, 0:, 000: (c) The solution is 0 a 0 f 0 0 0 a f0 0 000 6a 6a f 000 0 8a 8a f 0 a 0 a 0, a 75, a 5 f 5 75. (d) 00 (, 000) 8 (0, 0) 00

9 Chapter Applications of Differentiation 70. (a) Use a fourth degree polnomial f a a a a a 0. f a a a a (c) The solution is (d), :, :, : f 8 8. (, ) 6 a a a a a 0 f 0 a a a a f 0 a a a a a 0 f 0 a a a a f 0 8a 7a 9a a a 0 f 0 08a 7a 6a a f 0 a 0 8, a, a, a, a 8 (, ) (, ) 6 7. False Let h f g where f g. Then h is decreasing on, 0. 7. True If f is an nth-degree polnomial, then the degree of f is n. 76. False. The function might not be continuous. 78. Suppose f changes from positive to negative at c. Then there eists a and b in I such that f > 0 for all in a, c and f < 0 for all in c, b. B Theorem.5, f is increasing on a, c and decreasing on c, b. Therefore, f c is a maimum of f on a, b and thus, a relative maimum of f. Section. Concavit and the Second Derivative Test., 6 6. Concave upward:, Concave downward:, f, 6 Concave upward: Concave downward:,, 6. 70 5 0 5, 8. 9 Concave upward:,, 0, Concave downward:, 0,, h 5 5 h 5 5 h 0 Concave upward: 0, Concave downward:, 0

Section. Concavit and the Second Derivative Test 95 0. csc,,. csc cot csc csc cot csc cot csc csc cot Concave upward: 0, Concave downward:, 0 f 5 f 6 6 f 6 f 6 0 when. Test interval < < < < Sign of f f < 0 f > 0 Conclusion Concave downward Concave upward Point of inflection:,. f 8 f 8 8 f 0 when 0. However, 0, is not a point of inflection since f 0 for all. Concave upward on, 6. f f f 8 0 f 0 when 0,. Test interval < < 0 0 < < < < Sign of f f > 0 f < 0 f > 0 Conclusion Concave upward Concave downward Concave upward Points of inflection: 0, 0,, 6 8. f, Domain:, f f 6 f > 0 on the entire domain of f (ecept for, for which f is undefined). There are no points of inflection. Concave upward on, 0. f Domain: > 0 f f 5 Point of inflection:,, Test intervals 0 < < < < Sign of f f > 0 Conclusion Concave upward Concave downward f < 0

96 Chapter Applications of Differentiation. f csc, 0 < < f csc cot f 9 csc Concave upward: Concave downward: No points of inflection csc cot 0 for an in the domain of f. 0,,,,. f sin cos, 0 f cos sin f sin cos f 0 when, 7. Test interval: 0 < < < < 7 < < Sign of f: f < 0 f > 0 f < 0 Conclusion: Concave downward Concave upward Concave downward Points of inflection:, 0, 7, 0 7 6. f cos, 0, f sin f cos f 0 when, Test intervals: 0 < < < < < < Sign of f: f < 0 f > 0 f < 0 Conclusion: Concave downward Concave upward Concave downward Points of inflection:,,, 8. f 8 0. f f Critical number: f > 0 Therefore,, is a relative minimum. f 5 f 5 f Critical number: 5 f5 < 0 Therefore, 5, 0 is a relative maimum.

Section. Concavit and the Second Derivative Test 97. f 9 7. f 8 7 f 6 Critical number: However, f 0, so we must use the First Derivative Test. f 0 for all and, therefore, there are no relative etrema. g 8 g g Critical numbers: g 9 < 0, 0 is a relative maimum. g 9 > 0, 0.5 is a relative minimum. g 9 < 0,,, 0 is a relative maimum. 6. f 8. f Critical number: 0 f f0 > 0 Therefore, 0, is a relative minimum. f f There are no critical numbers and is not in the domain. There are no relative etrema. 0. f sin cos, 0 f cos sin cos sin cos cos sin 0 when 6,, 5 6,. f sin cos f f 6 < 0 > 0 f 5 6 < 0 f > 0 Relative maima: Relative minima: 6,, 5,, 6,,

98 Chapter Applications of Differentiation. f 6, 6, 6 (a) f 6 f 0 when 0, ±. (c) 6 f f 6 9 6 f 0 when ± 9. f'' 6 f' f0 > 0 0, 0 is a relative minimum. f± < 0 ±, are relative maima. Points of inflection: ±.758,.05 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 sin, 0, (a) f cos sin Critical numbers:.8,.8 f sin cos cos cos sin sin (c) f' π f f'' cos sin Relative maimum:.8,.85 f is increasing when and decreasing when f is concave upward when and concave downward when f < 0. f < 0. f > 0 f > 0 Relative minimum:.8,.09 Points of inflection: 0.75, 0.8,., 0.7 f < 0 6. (a) means f means f decreasing increasing f decreasing means concave downward f > 0 f decreasing means concave downward 8. (a) The rate of change of sales is increasing. S > 0 The rate of change of sales is decreasing. S > 0, S < 0 50. f'' f f' (c) The rate of change of sales is constant. S C, S 0 (d) Sales are stead. S C, S 0, S 0 (e) Sales are declining, but at a lower rate. S < 0, S > 0 (f) Sales have bottomed out and have started to rise. S > 0

Section. Concavit and the Second Derivative Test 99 5. 5. f f'' (0, 0) (, 0) f' 56. 58. (a) d (0, 0) (, 0) 0 t Since the depth d is alwas increasing, there are no relative etrema. f > 0 (c) The rate of change of d is decreasing until ou reach the widest point of the jug, then the rate increases until ou reach the narrowest part of the jug s neck, then the rate decreases until ou reach the top of the jug. 60. (a) f f f 9 5 Inflection point: 0, 0 6 (0, 0) 6 f does not eist at 0. 6. f a b c d Relative maimum:, Relative minimum:, Point of inflection:, f a b c, f 6a b f 8a b c d 56a b c 8a 6b c f 6a 6b c d f a b c 0, f 8a 8b c 0, f 8a b 0 8a 6b c 8a b 0 a b c 0 6a b 6a b a a, b 9, c, d 6 f 9 6

00 Chapter Applications of Differentiation 6. (a) line OA: 0.06 slope: 0.06 line CB: 0.0 50 slope: 0.0 f a b c d f a b c 000, 60: 60 000 a 000 b 000c d 0.06 000 a 000b c 000, 90: 90 000 a 000 b 000c d 0.0 000 a 000b c ( 000, 60) A 50 00 C (0, 50) (000, 90) B 000 O 000 The solution to this sstem of equations is a.5 0 8, b 0.00005, c 0.075, and d 50..5 0 8 0.00005 0.075 50 (c) 0. 00 00 00 00 00 0. 0 (d) The steepest part of the road is 6% at the point A. 66. S 5.755T 0 8 8.5T 0 6 0.65T 0 0.99987, 0 < T < 5 (a) The maimum occurs when T and S 0.999999. (c) S0 0.998.00 S.000 0.999 0.998 0.997 0.996 5 0 5 0 5 0 T 68. C 00,000 C 00,000 0 when 005 87 70. S (a) 00t 65 t, t > 0 00 B the First Derivative Test, C is minimized when 87 units. 0 0 5 St,000t 65 t St,00065 t 65 t 0 t.65 S is concave upwards on 0,.65, concave downwards on.65, 0. (c) St > 0 for t > 0. As t increases, the speed increases, but at a slower rate.

Section. Concavit and the Second Derivative Test 0 7. f sin cos, f cos sin, f sin cos, P 0 P f 0 f0 f0 6 f P P 6 P 0 0 P P The values of f, P, P, and their first derivatives are equal at 0. The values of the second derivatives of f and P are equal at 0. The approimations worsen as ou move awa from 0. 7. f, f, f f 5 f 6, P P P 6 P P 6 6 f f 8 6 5 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. 76. f 6 6 f 6 6 0 f 6 6 0 Relative etrema:, and 6, 0 Point of inflection, 6 is midwa between the relative etrema of f.

0 Chapter Applications of Differentiation 78. p a b c d p a b c p 6a b 6a b 0 b a The sign of p changes at ba. Therefore, ba, pba is a point of inflection. p a b a b 7a b b 9a c a b When p, a, b, c 0, and d. 0 0 0 0 0 7 b bc d 7a a d The point of inflection of p is 0, 0, 0. 80. False. f has a discontinuit at 0. 8. True sinb Slope: b cosb b b Assume b > 0 8. False. For eample, let f. Section.5 Limits at Infinit. f. f 6. f 5 No vertical asmptotes No vertical asmptotes No vertical asmptotes Horizontal asmptotes: ± Horizontal asmptote: Horizontal asmptote: Matches (c) Matches (a) Matches (e) 8. f 0 0 0 0 0 0 0 0 5 0 6 f 8.8 98.0 998.0 9,998 99,998,999,998 0 0 lim f (Limit does not eist.)

Section.5 Limits at Infinit 0 0. 8 f 0 0 0 0 0 0 5 0 6 0 7 f 8. 8.00 8 8 8 8 8 0 0 5 lim f 8. f 0 0 0 0 0 0 0 0 5 0 6 f 5.0.000.0.0 0 0 5 lim f. (a) (c) h f 5 7 5 7 lim h (Limit does not eist) h f 5 7 5 7 lim h 5 h f 5 7 5 7 lim h 0 6. (a) (c) lim 0 lim lim (Limit does not eist) 8. (a) lim 0 lim 5 5 5 0. lim 9 7 9 (c) 5 lim (Limit does not eist). lim 0. lim (Limit does not eist) 6. lim for < 0, lim lim (

0 Chapter Applications of Differentiation 8. lim lim lim, for < 0 we have 0. cos cos lim lim. lim cos cos 0 Note: 0 cos lim 0 b the Squeeze Theorem since cos.. f 6 6. lim tan tan t lim t 0 t lim t 0 sin t t cos t lim f 9 lim f 6 9 Let t. Therefore, and are both horizontal asmptotes. 8. lim lim lim 0. lim 9 lim 9 9 9 lim lim lim 9 9 9 6 for < 0 we have 0. 0 0 0 0 0 0 0 5 0 6 0 f.0 5. 50. 500. 5000. 50,000. 500,000. lim lim 0 50 0 Limit does not eist.

Section.5 Limits at Infinit 05. f 0 0 0 0 0 0 0 5 0 6.000 0.8 0.0 0.0 0.00 0.00 0.00 lim 0 0 5 6. is a critical number. f < 0 for <. f > 0 for >. lim f lim f 6 6 For eample, let f 6. 0. 8. (a) The function is even: The function is odd: lim f 5 f 5 lim 8 6 50. Intercepts: Smmetr: none Horizontal asmptote: since lim lim. Discontinuit: (Vertical asmptote) 5, 0, 0, 5 6 5. 9 Intercept: 0, 0 Smmetr: origin Horizontal asmptote: 0 Vertical asmptote: ± 5 6 5 6 5 6 5 5. 9 Intercept: 0, 0 Smmetr: -ais Horizontal asmptote: since lim lim 9 9. 5 5 5 5 Discontinuities: ± (Vertical asmptotes) Relative maimum: 0, 0

06 Chapter Applications of Differentiation 56. Intercept: 0, 0 Smmetr: -ais Horizontal asmptote: Relative minimum: 0, 0 5 58. Intercepts: none Smmetr: -ais Horizontal asmptote: 0 since lim 0 lim. Discontinuit: 0 (Vertical asmptote) 5 5 5 5 5 60. 6. Intercept: 0, 0 Intercept:, 0 Smmetr: origin Horizontal asmptote: 0 since Smmetr: none Horizontal asmptote: since lim 0 lim. Discontinuities: ± (Vertical asmptotes) lim lim. Discontinuit: 0 (Vertical asmptote) 5 5 5 5 5 5 6. 66. Intercepts: ±, 0 Smmetr: -ais Horizontal asmptote: Vertical asmptote: 0 5 5 5 Domain:,,, Intercepts: none Smmetr: origin Horizontal asmptotes: ± since lim, lim Vertical asmptotes: ± (discontinuities) 5. 5 5 5

Section.5 Limits at Infinit 07 68. f = = f 0 when 0. (0, 0) f Since f0 < 0, then 0, 0 is a relative maimum. Since f 0, nor is it undefined in the domain of f, there are no points of inflection. = Vertical asmptotes: ± Horizontal asmptote: 70. f f 0 when. (, 9 f 6 f < 0,, 9 = = Since then is a relative maimum. Since f 0, nor is it undefined in the domain of f, there are no points of inflection. = 0 ( Vertical asmptotes:, Horizontal asmptote: 0 7. f f 0 when 0,. (0, ) f (, 0 when 0.5, 0.657,.879. f0 < 0 Therefore, 0, is a relative maimum. ( 0.657, 0.9) (0.5, 0.80) (.879, 0.9) ( f > 0 Therefore,, is a relative minimum. Points of inflection: 0.5, 0.80, 0.657, 0.9 and.879, 0.9 Horizontal asmptote: 0

08 Chapter Applications of Differentiation 7. g = g 6 6 g 8 5 = No relative etrema. Point of inflection: 0, 0. Horizontal asmptotes: ± No vertical asmptotes sin 76. f Hole at 0, cos sin f There are an infinite number of relative etrema. In the interval,, ou obtain the following. Relative minima: ±.5, 0.869, ±5.5, 0.65 Relative maima: ±.87, 0.5 Horizontal asmptote: 0 No vertical asmptotes 5 6 78. f (a) f = g, g (c) 70 6 6 80 80 f g 70 The graph appears as the slant asmptote. 80. lim 00 v v v v c 00 0 00% 8..5t.6t 5.70 t (a) 5 0 0 00 Yes. lim t.5

Section.5 Limits at Infinit 09 8. S 00t 65 t, t > 0 (a) 0 5 0 0 Yes. lim S 00 00 t 86. p a lim lim n n... a a 0 q b m m... b b 0 Divide p and q b m. Case : If p n < m: lim lim q p Case : If m n: lim lim q p Case : If n > m: lim q a n mn... a m a 0 m b m... b m b 0 m a n... a m a 0 m b m... b m b 0 m a n nm... a a 0 lim m m b m... b m b 0 m 0... 0 0 b m... 0 0 0 0 b m a n... 0 0 b m... 0 0 a n b m. ±... 0 b m... 0 ±. 88. False. Let, then 0. Thus, and 0. Finall, and 0. Let p a b, then p0. Thus, and p0 Finall, and p0 a b. 8. Therefore, p a b p a f 8,, < 0 0 and f 0, f,, f,, < 0 > 0 and f0, and < 0 > 0 and f0. f < 0 for all real, but f increases without bound.

0 Chapter Applications of Differentiation Section.6 A Summar of Curve Sketching. The slope of f approaches as 0, and approaches as 0. Matches (C). The slope is positive up to approimatel.5. Matches (B) 6. (a) 0,,, (c) (d) (e), 8. 0 when ±. Horizontal asmptote: 0 0 when 0, ±. ( (, ), ) (, ) (, (0, 0) ) Conclusion < < Decreasing, concave down 0 Point of inflection < < Decreasing, concave up 0 Relative minimum < < 0 Increasing, concave up 0 0 0 Point of inflection 0 < < Increasing, concave down < < 0 Relative maimum Decreasing, concave down 0 Point of inflection < < Decreasing, concave up

Section.6 A Summar of Curve Sketching 0. 9 0 60 Therefore, Intercept: 9 0 when 0 when 0 9 < 0 Vertical asmptotes: Horizontal asmptote: Smmetric about -ais 5 0, 9 0, 9 = is a relative maimum. ±. f f < 0 when 0. f 0 Intercept: Vertical asmptote: 0 Horizontal asmptote: = (, 0) 5, 0 = 0 5 0, 5 = 5 = ( 9 ). f 6. f 6 6 0 when. f 9 > 0 if 0. Therefore,, 6 is a relative minimum. Intercept;, 0 Vertical asmptote: 0 Slant asmptote: (, 0) 8 6 (, 6) = 8 6 6 8 = 0 6 f f 0 f 8 0 when 0 Intercept: 0, 0 Relative maimum:, Relative minimum:, Inflection point: 0, 0 Vertical asmptotes: Slant asmptote: 8 6 8 6 6 8 0 ± when 0, ±

( Chapter Applications of Differentiation 8. 5 5 Relative maimum: Relative minimum: Intercept: 8 5 6 0 0, 5 6,.8990 6, 7.8990 0 when ± 6. ( + 6,.899 8 ( ( + 6, 7.899 = 8 8 8 = Vertical asmptote: Slant asmptote: 0. g 9 Domain: 9. g g 6 0 when 6 9 9 < 0 when 6 Relative maimum: Intercepts: 0, 0, 9, 0 Concave downward on, 9 ( 6, 6 ) 0 8 6 (0, 0) (9, 0) 8 6 6 8 0 6, 6 6 Domain: 8 0 when ± 6 6 0 when 0 Relative maimum:, 8 Relative minimum:, 8 Intercepts: 0, 0, ±, 0 Smmetric with respect to the origin Point of inflection: 0, 0 (, 8) 0 8 6 (, 0) (0, 0) (, 0) 5 5 (, 8) 6 8 0. 0 when 0, 5 (, 0) (, ) undefined for < 0 for all Concave downward on, and, 5 5 5 (, 0) Relative maimum: 0,,, Relative minimum:, 0 Intercepts: 0,,, 0,.80, 0,.80, 0

Section.6 A Summar of Curve Sketching 6. 0 when ± 0 when 0 < < Conclusion Decreasing, concave up (, 0) (, ( ( 0, (, 0) ( 0 Relative minimum < < 0 Increasing, concave up 0 0 Point of inflection 0 < < Increasing, concave down 0 0 Relative maimum < < Decreasing, concave down 8. f f 0 when. f 0 when. f f f Conclusion < < Increasing, concave down 0 0 Point of inflection < < Increasing, concave up (, ) (0, 5 ) ( 0.87, 0) 0. f 5 f 5 5(, 6 ) 0 when ±. f 6 0 when. (, 0) 9 6 6 9 < < < < < < < < f f f Conclusion Increasing, concave down 6 0 Relative maimum Decreasing, concave down 0 0 Point of inflection Decreasing, concave up 6 0 Relative minimum Increasing, concave up ( +, 6 ) Intercepts: 0, 0,, 0,, 0, 5, 0

Chapter Applications of Differentiation. 6 5 0 when 0, ±. 6 0 when ± < <. Conclusion Decreasing, concave up 7 6 5 (, 0 ( 5,, 0 ( ( ( 0, 5 ) 5 ( ) (, ) 0 Relative minimum < < Increasing, concave up 0 0 Point of inflection < < 0 Increasing, concave down 0 5 0 Relative maimum 0 < < Decreasing, concave down 0 0 Point of inflection < < Decreasing, concave up 0 Relative minimum < < Increasing, concave up. f 8 8 6 5 f 6 6 (, 0) (5, 0) 0 when,. f 8 6 0 when,. < < < < < < < < f f f Conclusion Decreasing, concave up 0 0 0 Point of inflection Decreasing, concave down 6 0 Point of inflection Decreasing, concave up 7 0 Relative minimum Increasing, concave up (0, 5) 0 8 6 7 (, 6) (, 7)

Section.6 A Summar of Curve Sketching 5 6. 5 5 0 when. 0 0 when. < < Conclusion Increasing, concave down 0 0 0 Point of inflection < < Increasing, concave up (, 0) 8. 6 5 6 5 6 5 5 5 6 5 (, ) 0 when and undefined when, 5. 6 5 6 5 5 undefined when, 5. 5 (, 0) (5, 0) 5 6 < < < < < < 5 5 5 < < Conclusion Decreasing, concave up 0 Undefined Undefined Relative minimum, point of inflection Increasing, concave down 0 Relative maimum Decreasing, concave down 0 Undefined Undefined Relative minimum, point of inflection Increasing, concave up 0. cos cos, 0 sin sin sin cos 0 when 0,, cos cos cos cos cos cos 0 when cos ± 8 Therefore, 0.5678 or 5.75,.057 or.0775. Relative maima: Relative minimum:,, 5,,, 5. 0.8, 0.59. Inflection points: 0.5678, 0.6,.057, 0.9, 5.75, 0.6,.0775, 0.9 π π

6 Chapter Applications of Differentiation. cot, 0 < < csc 0 when csc cot 0 when Relative maimum:,, 5 5 5 π Relative minimum:, Point of inflection:, Vertical asmptotes: 0,. sec 8 tan 8, < < sec 8 tan 8 8 sec 8 8 Relative minimum:, 0 5 5 5 5 (, ) 6. g cot, < < sin cos g sin g0 does not eist. But lim cot lim. 0 0 tan Vertical asmptotes: ±, ± Intercepts:, 0,, 0,, 0,, 0 Smmetric with respect to -ais. Decreasing on 0, and, π π π π 8. f 5 50. 9 9 6 f 5 8 8 6 f 5. is constant. f is linear. f is quadratic. f 6 6 f '', vertical asmptote 0 horizontal asmptote ± horizontal asmptotes 0, 0 point of inflection f '

Section.6 A Summar of Curve Sketching 7 5. 0 00 80 60 f 0 8 6 f 6 6 9 5 6 8 0 (an vertical translate of f will do) 56. (an vertical translate of the segments of f will do) 58. If f in 5, 5, then f and f 7 is the least possible value of f. If f in 5, 5, then f and f is the greatest possible value of f. 60. g 5 6. g Vertical asmptote: none Horizontal asmptote: 7, Undefined, if if The rational function is not reduced to lowest terms. 6 6 8 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, 6. g 8 5 5 5 5 8 0 0 The graph appears to approach the slant asmptote.

8 Chapter Applications of Differentiation 66. f tansin (a) (c) Periodic with period (e) On 0,, the graph of f is concave downward. f tansin tansin tansin f Smmetr with repect to the origin (d) On,, there is a relative maimum at and a relative minimum at, tan., tan 68. Vertical asmptote: 70. Vertical asmptote: 0 Horizontal asmptote: none Slant asmptote: 7. f a a aa, a 0 f a a aa 0 when a. f a > 0 for all. (a) Intercepts: Relative minimum: 0, 0, a, 0 a, Points of inflection: none a = a = 5 a = a = 7. Tangent line at P: 0 f 0 0 (a) Let 0: 0 f 0 0 (c) (e) -intercept: Let 0: -intercept: f 0 0 f 0 0 0 0 f 0 0 f 0 f 0 0 f 0 f 0, 0 Normal line: 0 f 0 0 0 f 0 0 0 f 0 0 0 0 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 Let 0: 0 f 0 0 (d) (f) (h) -intercept: -intercept: 0 0 f 0 f 0 0 f 0 0, f 0 0 f 0 Let 0: 0 f 0 0 0 0 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

Section.7 Optimization Problems 9 Section.7 Optimization Problems. Let and be two positive numbers such that S.. Let and be two positive numbers such that 9. P S S S 9 dp d S 0 when S. d P d < 0 when S. P is a maimum when S. ds d 9 0 when 8. d S 8 d > 0 when 8. S is minimum when 8 and. 6. Let and be two positive numbers such that 00. P 00 00 dp 00 0 when 5. d d P d < 0 when 5. P is a maimum when 50 and 5. 8. Let be the length and the width of the rectangle. P P P A P P da d P 0 when P. d A d < 0 when P. A is maimum when P units. (A square!) 0. Let be the length and the width of the rectangle. A dp d A d P A > 0 when A. d P is minimum when A centimeters. (A square!) P A A A 0 when A.

0 Chapter Applications of Differentiation. From the graph, it is clear that 8, 0 is the closest point on the graph of f to, 0. f 8,, 0. 6 8 0 f, 5, d 5 0 5 0 5 8 8 9 Since d is smallest when the epression inside the radical is smallest, ou need to find the critical numbers of g 8 9 g 8 8 9 B the First Derivative Test, ields a minimum. Hence,, is closest to 5,. 6. v F 8. 00 is the perimeter. (see figure) 0.0v df 0.0v dv 0.0v 0 when v 00.66. B the First Derivative Test, the flow rate on the road is maimized when v mph. A 00 8 50 da d 8 50 0 when 5. d A d 6 < 0 when 5. A is a maimum when 5 feet and 00 feet. 0. (a) Height, Length & Width Volume 8 (d) 5 5 800 97 0 5 5 980 6 6 6 6 86 V, 0 < < 00 0 0 The maimum volume seems to be 0. (c) dv d 6 0 when, is not in the domain. d V 6 d d V < 0 when. d When, V 0 is maimum.

Section.7 Optimization Problems. (a) P r 00 00 00 Length, Width, Area, 0 0 0 0 50 00 0 00 0 00 0 00 0 00 50 60 00 60 60 00 60 58 0 00 0 57 0 00 0 09 0 00 0 7 0 00 0 58 50 00 50 59 The maimum area of the rectangle is approimatel 59 m. (c) (e) A 00 00 000 0 00 0 Maimum area is approimatel 59.55 m 50 m.. You can see from the figure that A and A 6. 6 da d 6 0 when. d A < 0 when. d 6. 6 5 = 6 (, ) 5 6 A is a maimum when and.

Chapter Applications of Differentiation 6. (a) A base height 6 h h 6 h h da dh 6 h h h 6 h 6 h h h 6 h h h 8 6 h h h 6 h da when h, which is a maimum b the First Derivative Test. dh 0 Hence, the sides are 6 h, an equilateral triangle. Area sq. units. h 6 h cos h h 8 h 8 6 h tan h Area 6 h h α h 6 h 8 + h h tan 6 cos tan A 6cos sec cos sin tan 0 cos sec cos sin tan cos sin tan sin sin 0 and A. (c) Equilateral triangle 8. A r see figure da d r r 0 when r. ( ( ( r,, r B the First Derivative Test, A is maimum when the rectangle has dimensions r b r. ( r, 0) ( r, 0) (

( Section.7 Optimization Problems 0. 6 6 A 6 6 08 da 08 0 d 08 6, 6 Dimensions: 9 9 9 + +. V r h V 0 cubic units or h V 0 r S r rh r V 0 r ds dr r V 0 r 0 when r V 0 h units. V 0 V 0 V 0 V 0 V 0 r B the First Derivative Test, this will ield the minimum surface area.. V r r 08 08 r see figure V r 08 r 08r r dv dr 6r 6r 6r6 r 0 when r 6 and 6. d V dr 6 r < 0 when r 6. Volume is maimum when 6 inches and r 6.59 inches. r 6. V h r r see figure dv d r r r r 0 when 0 and r 6r. r h (, (, r r B the First Derivative Test, the volume is a maimum when 6r and h r. Thus, the maimum volume is V r r r. (

Chapter Applications of Differentiation 8. No. The volume will change because the shape of the container changes when squeezed. 0. V 000 r r h h 000 r r Let k cost per square foot of the surface area of the sides, then k cost per square foot of the hemispherical ends. C kr krh k 8r r 000 r r k 6 dc dr k r 6000 r 0 when r 5 B the Second Derivative Test, we have d C dr k,000 r > 0 when r 5 Therefore, these dimensions will produce a minimum cost. 5.66 feet and h.55 feet.. r 6000 r. (a) Let be the side of the triangle and the side of the square. A cot 5, 0 0. A 5 0 60 9 cot where 0 When 0, A 5, when 60 9, A 0.87, and when 0, A 9.5. Area is maimum when all 0 feet are used on the square. (c) Let be the side of the pentagon and the side of the heagon. Let be the side of the square and the side of the pentagon. A cot.7077 5, 0 5. A.75768 5 0.6 5 cot When 0, A 7.58, when.6, A.0, and when 5, A 5. Area is maimum when all 0 feet are used on the pentagon. 5 where 5 0 (d) Let be the side of the heagon and r the radius of the circle. A 5 cot 5 cot A 5 cot 5 6 cot 5 5 5 6 6 where 5 6 0 0 5 6, 0. 0 5 6 0.78 A 6 cot 6 r where 6 r 0 0 A 6 0, 0 0. 0.075 When 0, A 8.868, when.075, A.09, and when, A 7.58. Area is maimum when all 0 feet are used on the heagon When 0, A.8, when.78, A 5.8, and when 0, A 8.868. Area is maimum when all 0 feet are used on the circle. In general, using all of the wire for the figure with more sides will enclose the most area.

Section.7 Optimization Problems 5. Let A be the amount of the power line. A h da d 0 when. (0, h) h d A d > 0 for. The amount of power line is minimum when. (, 0) (, 0) 6. f g 6 on 0, (a) 9 (c) f, Tangent line at, is f d f g 6 6 d 0 8 g 0, in 0, The maimum distance is d when.. g, Tangent line at, 0 is 0 8. The tangent lines are parallel and vertical units apart. (d) The tangent lines will be parallel. If d f g, then d 0 f g implies that f g at the point where the distance is maimum. 8. Let F be the illumination at point P which is units from source. F ki ki d df d ki ki d 0 when ki ki d. I I d d I I di I I d I I I d F d 6kI 6kI d > 0 when d I I I. This is the minimum point. 50. (a) T θ + Q dt d 0 CONTINUED T hours

6 Chapter Applications of Differentiation 50. CONTINUED (c) v v dt d v 0 v sin v depends on T v v v v onl. v (d) Cost C C C C From above, sin C C C C 5. T d d a 5. v v dt d v d a v d a 0 Since a and d a sin d sin C k k C k k 0 we have Since sin v sin v 0 sin v sin v. d T d d v d d v d a > 0 this condition ields a minimum time. Or, use Eercise 50(d): Thus,. θ + sin C C 0. 56. V r h r r dv dr r r r r r 88r r r r96 r r 0 when r 0, 6. B the First Derivative Test, V is maimum when r 6 and h. Area of circle: A Lateral surface area of cone: S 66 86 Area of sector: 86 r 7 86 7 6.5 radians or 66

Section.7 Optimization Problems 7 58. Let d be the amount deposited in the bank, i be the interest rate paid b the bank, and P be the profit. P 0.d id d ki since d is proportional to i P 0.ki iki k0.i i dp di k0.i i 0 when i 0. 0.08. d P k0. 6i < 0 when i 0.08 Note: k > 0. di The profit is a maimum when i 8%. 60. (a) P 0 s 6s 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 ds 0 < 0 s 0 ields a maimum. The maimum profit occurs when s 0, which corresponds to $0,000 P $,600,000. 6. 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. S m 5m 6 0m Using a graphing utilit, ou can see that the minimum occurs when m 0.. Line 0. S S 0. 50. 6 00..7 mi. 0 0 0 (0.,.5) m

8 Chapter Applications of Differentiation 6. (a) Label the figure so that r h. Note that sin and cos The area A of the h r r. Then, the area A is 8 times the area of the region cross equals the sum of two large rectangles minus given b OPQR: the common square in the middle. h O A 8 h hh P θ r 8 r r r 8r r R Q h A 8r 8 8 0 r 8 r 8 8r r r A h h 8h h A r cos sin cos 0 cos tan r sin sin sin 8r sin cos r sin cos sin arctan.075 or 6. r r r r r 5 5 r r 0 5r ± 5r 0r 0 Take positive value. Quadratic in. r 5 ± 5. 0 r 5 5 0 0.85065r Critical number (c) Note that r 5 5 0 and A 8r r 8 r r 5 5 0 0 5 5 r 0 8 r 0 0 r 5 5r r 8 5 r 5 r 5 5 r r 5 5 5 5 r 5 Using the angle approach, note that tan, sin and 5 Thus, A r sin sin r 5 r 5 r r 5 5. 0 5 r 5 5 5 r sin cos 5.

Section.8 Newton s Method 9 Section.8 Newton s Method. f f n n f n f n f n f n n f n f n 5 5.5 0.5 5.0 0.05.5. f tan f sec 0. n n f n f n f n f n n f n f n 0.000 0.00.00 0.099 0.0007 0.0007 0.0007.0000 0.0007 0.0000 6. f 5 f 5 Approimation of the zero of f is 0.755. n n f n f n f n f n n f n f n 0.5000 0.688.5 0.57 0.857 0.857 0.96.698 0.086 0.7707 0.7707 0.06.76 0.05 0.755 0.755 0.00.67 0.000 0.759 8. f f Approimation of the zero of f is.88. n n f n f n f n f n n f n f n 5 0.00 0.598 0.707.89.89 0.0005 0.5858.00085.88 0. f f 6 Approimation of the zero of f is 0.797. n n f n f n f n f n n f n f n 6 0.667 0.8 0.8 0.57.66 0.078 0.7955 0.7955 0.0068.7969 0.008 0.797 0.797 0.0000.7798 0.0000 0.797

0 Chapter Applications of Differentiation. f f Approimation of the zero of f is 0.897. n n f n 0.5 0.9 0.08 f n 5.58 f n f n 0. 0.006 n f n f n 0.9 0.897 0.897 0.000.76 0.0000 0.897 Approimation of the zero of f is.070. n n f n f n f n f n n f n f n 0.0769.0769.0769 0.075.975 0.009.070.070 0.000.790 0.0000.070. f cos f sin Approimation of the zero of f is 0.866. n n f n f n f n f n n f n f n 0.9000 0.07. 0.0 0.8666 0.8666 0.00.05 0.00 0.8655 0.8655 0.000.0087 0.0000 0.8655 6. h f g h Point of intersection of the graphs of f and g occurs when.89. n n h n h n h n h n n h n h n.9000 0.006 0.95 0.0067.89.89 0.0000 0.9 0.0000.89 8. h f g cos h sin One point of intersection of the graphs of f and g occurs when 0.8. Since f and g cos are both smmetric with respect to the -ais, the other point of intersection occurs when 0.8. n n h n h n h n h n n h n h n 0.8000 0.0567.7 0.05 0.85 0.85 0.0009.8 0.000 0.8 0. f n a 0 f n n i i i n a n i n n i n in a n i n n n i a n n i

Section.8 Newton s Method. i i 5 i. i i 5 i i i i.0000.500.6.6 i.5000.667.66.66 5.6 5.66 6. f tan f sec Approimation of the zero:. n n f n f n f n f n n f n f n.0000 0.5.00 0.97.97.97 0.009.0000 0.009.6.6 0.0000.0000 0.0000.6 8. f 0. f f 0; therefore, the method fails. n n f n f n f n f n n f n f n 0 f sin cos f cos sin Fails because f 0. n n f n f n 0. Newton s Method could fail if fc 0, or if the initial value is far from c.. Let g f cot g csc. The fied point is approimatel 0.86. n n g n g n g n g n n g n g n.0000 0.579. 0.8 0.856 0.856 0.00.7668 0.0087 0.860 0.860 0.000.70 0.0000 0.860 6. f sin, f cos (a) (d) (.8, 0.97) (, 0.) (6.086, 0) π π (., 0) (c).8 f 6.086 f f. f The -intercepts correspond to the values resulting from the first iteration of Newton s Method. (e) If the initial guess is not close to the desired zero of the function, the -intercept of the tangent line ma approimate another zero of the function.

Chapter Applications of Differentiation 8. (a) n n n n n n i i i 0.000 0.00 0. 0. i 0.000 0.0900 0.0909 0.0909 0. 0.09 0. f sin, 0, f cos sin 0 Letting F f, we can use Newton s Method as follows. F cos sin (.09,.80) n n F n F n F n F n n F n F n π π π.0000 0.0770.6509 0.090.090.090 0.0007.70 0.000.088 Approimation to the critical number:.09. f,, d 7 8 5 d is minimum when D 7 8 5 is minimum. n g D 8 g n g n g n g n g n n g n g n (0.59, 0.80) (, ) 0.5000 0.5000 7.0000 0.09 0.59 0.59 0.005 7.6 0.000 0.59 0.59 0.000 7.59 0.0000 0.59 0.59 Point closest to, is approimatel 0.59, 0.80.. Maimize: C t t 50 t n C t t 00t 50 Let f t t 00t 50 f t 6t 00. 50 t 0 n f n f n f n f n n f n f n.5000.75 95.0000 0.06.86.86 0.0658 90.8 0.000.86 Since f 5 and f 5 575, the solution is in the interval, 5. Approimation: t.86 hours

Section.8 Newton s Method 6. 70 0.808 7.97 7.8 0.8, 5 Let f 0.808 7.97 7.8 59.57 f. 5.98 7.8. From the graph, choose and.5. Appl Newton s Method. n n f n f n f n f n n f n f n.0000 5.0750 7.70 0.5.5.5 0.805.589 0.008.9.9 0.0006.9 0.0000.9 n n f n f n f n f n n f n f n.5000.675.8760 0.878.6878.6878 0.86 8.550 0.06.676.676 0.0009 8.50 0.0000.676 The zeros occur when.9 and.676. These approimatel correspond to engine speeds of revmin and 676 revmin. 8. True 50. True 5. f sin Domain:, and are both zeros. f cos Let. sin n n f n f n f n f n n f n f n.0000 0..796 0.6.6.6 0.0090.698 0.0055.6.6 0.0000.6 0.0000.6 Zeros: ±,.

Chapter Applications of Differentiation Section.9 Differentials. f 6 6.9.99.0. f Tangent line at 8, : f 6 T 9.660.55.5.85.605.65.55.5.85.5 9. f.9.99.0. f Tangent line at, : f T.78.07..77.9.789.07..77.96 f f 6. f csc f csc cot Tangent line at, csc : f f csc csc cot csc cot csc.9.99.0. f csc T csc cot csc.0567.098.0998.09.585.09.097.0998.08.50 8. f, f, 0, d 0. f f f 0. f 0 0. 0 0.0 d f d f00. 00. 0 0. f, f,, d 0.0 f f f.0 f.0 0.0 d f d f0.0 0.0 0.0

Section.9 Differentials 5.. d d d 9 d 9 d 9 d 6. 8. sin d d d d cos sin d 0. sec d sec tan sec d. (a) f.9 f 0. f f0. 0.. f.0 f 0.0 f f0.0 sec tan tan d 0.0 0.96. (a) f.9 f 0. f f0.6. (a) 00. f.0 f 0.0 f f0.0 00.0 g.9 g 0.07 g g0.07 8 0.07 7.79 g. g 0. g g0. 8 0. 8. 8. (a) g.9 g 0.07 g g0.070. 8 50.07 7.65 g. g 0. g g0. 8 50. 8.5 A bh, b 6, h 50 db dh ±0.5 da b dh h db A da 6±0.5 50±0.5 ±0.75 square centimeters. inches. (a) C 56 centimeters d ±0.0 inch C dc ±. centimeters (a) V dv d ±0.0 ±.96 cubic inches S 6 ds d ±0.0 C r r C da C dc A r C C 56±..6 ±. square inches da A.6 0.0857.857% 56 da A C dc C dc C 0.0 dc C 0.0 0.05.5%

6 Chapter Applications of Differentiation 6. P 500 77 000, changes from 5 to 0 dp 500 77d 577 d 577 50 5 60 Approimate percentage change: dp P 60 00 00.7% 57.50 8. V r, r 00 cm, dr 0. cm0. V dv r dr 00 0. 8000 cm E IR R E I dr E I di dr R EI di di EI I dr R di I di I. See Eercise.. h 50 tan A baseheight 9.5cot 9.5 5.5 cot da 5.5 csc d da csc A d cot sin cos 0.5 sin 6.75cos 6.75 d 0.00 sin 0.669cos 0.669 7.5.79 radians dh 50 sec d dh 50 sec.79 0.06 50 tan.79 d 9.96 0.06.9886 d d 0.08 θ 50 ft h 0.009.09% in radians 6. Let f, 7, d f f fd d 6 7 7.960 7 Using a calculator, 6.965 8. Let f,, d 0.0. f f f d d f.99 0.0 7 0.7 6.7 Using a calculator:.99 6.709

Review Eercises for Chapter 7 50. Let f tan, 0, d 0.05, f sec.5. Propagated error f f, Then relative error and the percent error d 00. d, f 0.05 f 0 f0d tan 0.05 tan 0 sec 00.05 0 0.05. 5. True, d d a 56. False Let f,, and d. Then and f f f f d f d. Thus, d > in this eample. Review Eercises for Chapter. (a) (c) f f 6 6 6 6 At least si critical numbers on 6, 6. f f (d) Yes. Since f f and f f, the Mean Value sas that there eists at least one value c in, such that fc f f (e) No, lim f eists because f is continuous at 0, 0. 0 (f) Yes, f is differentiable at... f, 0, 6. No. f is not differentiable at. f No critical numbers Left endpoint: 0, 0 Minimum Right endpoint:, 5 Maimum 8. No; the function is discontinuous at 0 which is in the interval,. 0. f b f a b a f,. f fc c c f b f a b a f, 0 f fc c c 6 0 0

8 Chapter Applications of Differentiation. f b f a b a f f 0 5 fc c 5 c Midpoint of 0, 6. g g Critical number: Interval < < < < Sign of g g > 0 g > 0 Conclusion Increasing Increasing 8. f sin cos, 0 f cos sin 5 Critical numbers:, Interval 0 < < Sign of f f > 0 f < 0 f > 0 Conclusion Increasing Decreasing Increasing < < 5 5 < < 0. g sin, 0, Test Interval g cos 0 when, Relative maimum:, Relative minimum:, Sign of g 0 < < g > 0 g < 0 g > 0 Conclusion Increasing Decreasing Increasing < < < <. (a) A sinkm t B coskm t Akm coskm t Bkm sinkm t 0 when sinkm t coskm t A B tankm t A B. Therefore, sinkm t coskm t When v 0, A A A A B B A B. A B B B A B A B. Period: km Frequenc: km km

Review Eercises for Chapter 9. f 6 f f 6 0 when 0. Point of inflection: 0, 6 Test Interval < < 0 0 < < Sign of f f < 0 f > 0 Conclusion Concave downward Concave upward 6. ht t t Domain:, 8. ht ht t h 8 > 0 0 t t, 5 is a relative minimum. 7 6 5 5 6 7 0. C Q s r dc d Qs r 0 Qs r Qs r Qs r. (a) S 0.t.655t 0.905t.89 5 0 (c) St 0 when t.7. This is a maimum b the First Derivative Test. (d) No, because the t coefficient term is negative.. lim lim 5 5 0 6. lim lim 8. g 5 0. f 5 lim 5 lim 5 lim lim Horizontal asmptote: 5 lim lim lim lim Horizontal asmptotes: ±

0 Chapter Applications of Differentiation. f. Relative minima: (0, 0,, 0,, 0 Relative maima:.577, 0.8, 0., 0.8 g cos cos Relative minima: k, 0.9 where k is an integer. Relative maima: k, 8.9 where k is an integer. 0 5 5 0 6. f Domain:, ; Range:, 7 f 0 when 0,. f 0 when 0,. f < 0 Therefore,, 7 is a relative maimum. Points of inflection: 0, 0,, 6 Intercepts: 0, 0,, 0 0 5 0 5 0 5 5 8. f Domain:, ; Range: 0, f 0 when 0, ±. f 0 when ±. f0 < 0 Therefore, 0, 6 is a relative maimum. f± > 0 Therefore, ±, 0 are relative minima. Points of inflection: ±, 69 Intercepts:, 0, 0, 6,, 0 Smmetr with respect to -ais 0 (0, 6) (, 0) (, 0) 8 50. f Domain:, ; Range: 6,875 f f 7 f 7 > 0 7 0 when, 7. 6 0 when,. Therefore, 7, 6,875 56 is a relative minimum. Points of inflection:, 0,, 65 6 Intercepts:, 0, 0,,, 0 (, 0) 56, 0 (0, ) 0 60 (, 0) ( ) 7, 6.875 56

Review Eercises for Chapter 5. f Graph of Eercise 9 translated units to the right replaces b., 0 is a relative maimum., is a relative minimum. (, 0) (, 0), 0 is a point of inflection. Intercepts:, 0,, 0 (, / ) 5. f Domain:, ; Range:, f 0 when ±. f 0 when 0, ±. (, ) (, ) f < 0 Therefore,, is a relative maimum. f > 0 Therefore,, is a relative minimum. Points of inflection:,, 0, 0,, Intercept: 0, 0 Smmetric with respect to the origin Horizontal asmptote: 0 56. f Domain:, ; Range: 0, f f± < 0 Therefore, f0 > 0 ±, are relative maima. Therefore, 0, 0 is a relative minimum. ± 6 Points of inflection: Intercept: 0, 0 Smmetric to the -ais Horizontal asmptote: 0 0 when 0, ±. f 0 5, 0.9, ± 8 6, 0.0 0 when ± 6 ±. (, ) (, ) (0, 0)

Chapter Applications of Differentiation 58. f Domain:, 0, 0, ; Range:, f 0 when. f f > 0 Therefore,, Point of inflection:, 0 Intercept:, 0 Vertical asmptote: 0 0 when. is a relative minimum. 60., f,, Domain:, Range:, Intercept: 0, (0, ) < > (, 0) (, ) 6. f sin sin Domain:, ; Range:, f cos cos cos cos 0 Critical Numbers: ±, 0 f sin sin sin cos 0 when 0, ±, ±0.0. B the First Derivative Test: is a relative minimum., is a relative maimum., Points of inflection: 0.0, 0.6, 0.0, 0.6, ±, 0, 0, 0 Intercepts: (, 0, 0, 0,, 0 Smmetric with respect to the origin 6. f n, n is a positive integer. (a) f n n The function has a relative minimum at 0, 0 when n is even. f nn n The function has a point of inflection at 0, 0 when n is odd and n.

( ( Review Eercises for Chapter 66. Ellipse: 6, A 8 (, da d 0 when 7 6. 8 The dimensions of the rectangle are b 7. 68. We have points 0,,, 0, and, 5. Thus, Let m 5 0 f L 5 5 0 5 or. (0, ) f 50 00 0 0 00 0 when 0 or 00. L (, 5) 5 (, 0) L 5 00 5 00 00 5.7 feet 70. Label triangle with vertices 0, 0, a, 0, and b, c. The equations of the sides of the triangle are cb and cb a a. Let, 0 be a verte of the inscribed rectangle. The coordinates of the upper left verte are, cb. The -coordinate of the upper right verte of the rectangle is cb. Solving for the -coordinate of the rectangle s upper right verte, ou get b a b a Finall, the lower right verte is c b c a b a b a b a b a, 0 b. Width of rectangle: Height of rectangle: a a a b. b a a b b c see figure b A WidthHeight a a b b c b a a b c b da d a a b c b c b a ac ac b b b 0 when b. A b a a b b c b b a c ac ac Area of triangle (, c b = c b ( ( b, c) = c ( a) b a (0, 0) (, 0) ( a, 0) a b ( a, 0 b ( a a b, c b b (

Chapter Applications of Differentiation 7. You can form a right triangle with vertices 0,, 0, 0, and, 0. Choosing a point a, b on the hpotenuse (assuming the triangle is in the first quadrant), the slope is Let m b 0 a b 0 b a a. f L b a. a ab a 0 when 0, a ab. Choosing the nonzero value, we have b a b. L a ab b a b a a b a b b a b meters f b a ab a 7. Using Eercise 7 as a guide we have L a csc and L b sec. Then when tan ab, sec a b b, csc a b a and L L L a csc b sec a a b This matches the result of Eercise 7. a dld a csc cot b sec tan 0 b a b b a b. 76. Total cost Cost per hournumber of hours T v 500 7.50 0 v 85 v 50 v dt 85 dv 50 v v,50 50v 0 when v 750 56 6. mph. d T 650 > 0 when v 56 so this value ields a minimum. dv v 78. f From the graph, ou can see that f has one real zero. f f changes sign in, 0. n n f n f n f n f n n f n f n 0.5000 0.50.7500 0.055 0.55 0.55 0.009.697 0.00 0.5 On the interval, 0: 0.5.