CHAPTER 3 Applications of Differentiation

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

CHAPTER Applications of Differentiation Section. Etrema on an Interval. A: neither B: absolute maimum (and relative maimum) C: neither D: neither E: relative maimum F: relative minimum G: neither. f f f0 0. f 7 7 7. f f 7 7 f 7 0 f f is undefined. 9. Critical number:. Critical numbers:,,. : absolute maimum, : absolute maimum : absolute minimum f f Critical numbers: 0,. gt t t, t < 7. gt t t t t t t t t Critical number: t. h sin cos, 0 < < h sin cos sin sin cos On 0,, critical numbers:,, 9. f,,. f, 0, f No critical numbers f Left endpoint:, Maimum Right endpoint:, Minimum Left endpoint: 0, 0 Minimum Critical number: Right endpoint:, 9 Maimum, 0 Minimum 0

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

0 Chapter Applications of Differentiation. f, 0, < Left endpoint: 0, Minimum. f,, Right endpoint:, Minimum Right endpoint:, Maimum. f,, f 0, ± Maimum: f 0., 0.,. Minimum: f ± 7. (a) 9. (,.7) 0 (0.9,.0) f, 0, f f (b) Maimum:,.7 (endpoint) Minimum: 0.9,.0. 0 f 0 0 f.7 f 9.0 Minimum: 0.9,.0 f.., 0, f. ±. ± 9 9 0.9 Maimum (endpoint) f 0 Setting f 0, In the interval 0,, choose we have 0 0. 0 ± 00 0 ± 0 0 0 0.7. is the maimum value. f 0 0.7

Section. Etrema on an Interval 07. f, 0,. f f 9 f 7 7 f 0 f 0 f 0 is the maimum value. f. (a) Yes (b) No 7. (a) No (b) Yes 9. P VI RI I 0.I, 0 I. P 0 when I 0. P 7. when I. 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 >. v sin, d is constant. dt d d d d dt v cos 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 sprinkler gets the most water. dt b the Chain Rule d dt. True. See Eercise 7.. True 7. If f has a maimum value at c, then fc f for all in I. Hence, fc f for all in I. Thus, f has a minimum value at c. 9. (a) a b c a b The coordinates of B are 00, 0, and those of A are 00,. From the slopes at A and B, 000a b 0.09 000a b 0.0. Solving these two equations, ou obtain a 0,000and b 00. From the points 00, 0 and 00,, ou obtain 0 0,000 00 00 00 c 0,000 00 00 c. 00 7 In both cases, c.7 Thus,. CONTINUED 0,000 7 00. A 00 9% % B 00

0 Chapter Applications of Differentiation 9. CONTINUED (b) 00 00 00 00 00 000 00 00 00 00 00 d 0 0.7.7.7.7 0.7 0 For 00 0, d a b c 0.09. For 0 00, d a b c 0.0. (c) The lowest point on the highwa is 00,, which is not directl over the point where the two hillsides come together. Section. Rolle s Theorem and the Mean Value Theorem. Rolle s Theorem does not appl to f over 0, since f is not differentiable at.. f. f f. But, f is not continuous on,. f -intercepts:, 0,, 0 f 0 at. 7. f 9. f -intercepts:, 0, 0, 0 f f 0 f f 0 at f 0 for c and f 0. f, 0,. f 0 f 0 f is continuous on 0,. f is differentiable on 0,. Rolle s Theorem applies. f 0 c-value: f,, f f 0 f is continuous on,. f is differentiable on,. Rolle s Theorem applies. c 0 ±, c f f. f,, f f f is continuous on,. f is not differentiable on, since f0 does not eist. Rolle s Theorem does not appl.

Section. Rolle s Theorem and the Mean Value Theorem 09 7. 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 c-value: 0 ± ± 9. f sin, 0, f 0 f 0 f is continuous on 0,. f is differentiable on 0,. Rolle s Theorem applies. f cos. f sin, 0, f 0 f 0 f is continuous on 0,. f is differentiable on 0,. Rolle s Theorem applies. c-values:, arcsin c-value: 0.9 f sin cos 0 sin cos sin sin 0.9. 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.

0 Chapter Applications of Differentiation 7. f tan,, 9. f t t t f f 0 f is continuous on,. f is differentiable on,. Rolle s Theorem applies. f sec 0 sec sec ± ± arcsec ±0. radian ± arccos (a) (b) f f v ft must be 0 at some time in,. ft t 0 t seconds c-values: ±0. radian 0. 0. 0. 0.. Tangent line (c, f(c )). f is not continuous on the interval 0,. f is not continuous at.. f, 0, f has a discontinuit at. (a, f(a)) Secant line (b, f(b)) (c, f(c )) a b Tangent line f 7. f (a) slope (b) f c (c) Tangent line: secant line: f, f (d) f 7 Secant Tangent 9. f is continuous on, and differentiable. f is continuous on 0, and differentiable on,. on 0,. f f f when Therefore,. c. f f 0 0 f 7 c 7

Section. Rolle s Theorem and the Mean Value Theorem. f is continuous on 7, and. f sin is continuous on 0, and differentiable differentiable on 7,. on 0,. f f 7 7 0 9 f f f 0 0 0 0 0 c f cos 0 9 c 7. f (a),, 9. Tangent 0. Secant f f,, 9,, 9, m 9 (a) Tangent (b) Secant line: slope f f f 9 (b) Secant line: Secant 0 0 (c) f (c) f 9 f 9 f ± ± c In the interval,, c. f c Tangent line: c c c, f c, m f Tangent line: 0 0

Chapter Applications of Differentiation. st.9t 00 (a) v avg s s0 0 (b) st is continuous on 0, and differentiable on 0,. Therefore, the Mean Value Theorem applies. vt st 9.t.7 msec t.7 9..9 00. seconds.7 msec. No. Let f on,.. f f0 0 and zero is in the interval (, but f f. 7. Let St be the position function of the plane. If t 0 corresponds to P.M., S0 0, S. 00 and the Mean Value Theorem sas that there eists a time t 0, 0 < t 0 <., such that St 0 vt 0 00 0. 0.. Appling the Intermediate Value Theorem to the velocit function on the intervals 0, t 0 and t 0,., ou see that there are at least two times during the flight when the speed was 00 miles per hour. 0 < 00 <. 9. Let St be the difference in the positions of the bicclists, St S t S t. Since S0 S. 0, there must eist a time t 0 0,. such that St 0 vt 0 0. At this time, v t 0 v t 0. f 0,, 0 0 < No, this does not contradict Rolle s Theorem. f is not continuous on 0,.. (a) f is continuous on 0, and changes sign, f > 0, f < 0. B the Intermediate Value Theorem, there eists at least one value of in 0, satisfing f 0. (b) There eist real numbers a and b such that 0 < a < b < and f a f b. Therefore, b Rolle s Theorem there eists at least one number c in 0, such that fc 0. This is called a critical number. (c) (d) (e) No, f did not have to be continuous on 0,.. f is continuous on, and does not satisf the conditions of the Mean Value Theorem. f is not differentiable on,. Eample: f ), ) f ) ) ), )

Section. Rolle s Theorem and the Mean Value Theorem. f f is differentiable for all. f and f0, so the Intermediate Value Theorem implies that f has at least one zero c in, 0, fc 0. Suppose f had zeros, fc fc 0. Then Rolle s Theorem would guarantee the eistence of a number a such that fa fc fc 0. But, f > 0 for all. Hence, f has eactl one real zero. 7. f continuous at 0: f continuous at : b a c 9. f differentiable at : a Hence, c. f,,,,, 0 0 < < 0 < f 0 f c f Hence, f. 7. f 7. False. f has a discontinuit at 0. f c f 0 0 c c Hence, f. 7. True. A polnomial is continuous and differentiable everwhere. 77. 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. 79. 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 Ab a B. Thus, A Ab a and b a which is the midpoint of a, b. b aab a B b a. 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

Chapter Applications of Differentiation. Let f cos. f is continuous and differentiable for all. Let 0 < a < b. f satisfies the hpotheses of the real numbers. B the Mean Value Theorem, for an interval Mean Value Theorem on a, b. Hence, there eists c in a, b a, b, there eists c in a, b such that 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. fc fb fa c b a b a. b a Thus b a b a c < b a a. Section. Increasing and Decreasing Functions and the First Derivative Test. (a) Increasing: 0, and, 9. Largest: 0,. (b) Decreasing:, and 9, 0. Largest:, f Increasing on:, Decreasing on:,. Increasing on:,,, Decreasing on:, 7. f sin, 0 < < f cos Critical numbers:, Sign of f: Conclusion: Increasing Decreasing Increasing Increasing on: 0, f Decreasing on:, > 0 Test intervals: 0 < <,, < < f < 0 < < f > 0 9. f f Discontinuit: 0 Test intervals: < < 0 0 < < Sign of f: Conclusion: Increasing Decreasing f > 0 f < 0. g g Critical number: Test intervals: < < < < Sign of g: g < 0 Conclusion: Decreasing Increasing g > 0 Increasing on, 0 Decreasing on 0, Increasing on:, Decreasing on:,

Section. Increasing and Decreasing Functions and the First Derivative Test. Domain:, Critical numbers: ± Test intervals: < < < < < < Sign of : < 0 > 0 < 0 Conclusion: Decreasing Increasing Decreasing Increasing on, Decreasing on,,,. cos, sin 0 < < 0: sin Critical numbers: 7, Test intervals: 0 < < 7 < < < < Sign of Conclusion: > 0 Increasing < 0 Decreasing > 0 Increasing Increasing on: 7 0,,, Decreasing on: 7, 7 7. f 9. 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:,

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

Section. Increasing and Decreasing Functions and the First Derivative Test 7 7. f 9. f f f Critical number: 0 Critical number: Test intervals: < < 0 0 < < Sign of f: Conclusion: Increasing Increasing f > 0 f > 0 Test intervals: < < < < Sign of f: Conclusion: Decreasing Increasing f < 0 f > 0 Increasing on:, No relative etrema Increasing on: Decreasing on:,, Relative minimum:, 0. f f,, Critical number: < > Test intervals: < < < < Sign of f: f f > 0 < 0 Conclusion: Increasing Decreasing Increasing on: Decreasing on:,, Relative maimum:,. f f Critical numbers: Discontinuit: 0, Test intervals: < < < < 0 0 < < < < Sign of f: Conclusion: Increasing Decreasing Decreasing Increasing f > 0 Increasing on:,,, Decreasing on:, 0, 0, Relative maimum:, Relative minimum:, f < 0 f < 0 f > 0

Chapter Applications of Differentiation. f 9 f 9 9 9 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: Decreasing on:,,, 0 0,,, Relative maimum: 0, 0 7. 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:, Relative minimum:, 0 9. (a) f cos, 0 < < (b) Relative maimum:, Relative minimum:, f sin 0 Critical numbers:, Test intervals: 0 < < Sign of f: f f f > 0 < 0 > 0 Conclusion: Increasing Decreasing Increasing < < < < Increasing on: 0, Decreasing on:,,,

Section. Increasing and Decreasing Functions and the First Derivative Test 9. (a) f sin cos, 0 < < f cos sin 0 sin cos Critical numbers:, Sign of f Conclusion: Increasing Decreasing Increasing Increasing on: Decreasing on: 0, (b) Relative maimum:,,, f > 0, Test intervals: 0 < < Relative minimum:, 0 < < f < 0 (c) < < f > 0. (a) f cos, 0 < < f cos sin 0 cos 0 or Critical numbers:, 7,,,,, sin 0 Sign of f f < 0 Test intervals: 0 < < < < f > 0 Conclusion: Decreasing Increasing Decreasing Increasing < < f < 0 < < f > 0 Increasing on: Decreasing on: (b) Relative maima:, 0, Relative minima: Test intervals: < < < < 7 < < < < Sign of f Conclusion: < 0 Decreasing > 0 Increasing < 0 Decreasing > 0 Increasing,,,,,,,, f,,,,, f, 7,,, 7, 0,, 0,, 0, 7 f (c) 7, 0 0 f

0 Chapter Applications of Differentiation. (a) f sin sin, 0 < < f sin cos cos cos sin 0 Critical numbers:, 7,, Test intervals: Sign of f: 0 < < < < < < < < 7 < < Conclusion: Increasing Decreasing Increasing Decreasing Increasing Increasing on: Decreasing on: 0, (b) Relative minima: Relative maima:, 7, 7 7,,, f > 0,,,,,,,, 0 f < 0 7 f > 0 f < 0 f > 0 7. f 9,, 9. f t t sin t, 0, (a) (b) f 9 9 f 0 f 0 (a) (b) ft t cos t t sin t 0 0 0 0 0 0 tt cos t sin t π f f π t (c) 9 9 0 Critical numbers: (d) Intervals:, f < 0 Decreasing Increasing Decreasing f is increasing when when is negative. f ± f, f > 0 ±, f < 0 is positive and decreasing (c) tt cos t sin t 0 t 0 or t tan t t cot t t.9,.070 (graphing utilit) Critical numbers: (d) Intervals: 0,.9 ft > 0 Increasing Decreasing Increasing f is increasing when when is negative. f t.9, t.070.9,.070.070, ft < 0 ft > 0 f is positive and decreasing

Section. Increasing and Decreasing Functions and the First Derivative Test. (a) (b) f sin, 0, (c) Critical numbers: f cos f' f (d) Intervals: 0, f < 0 Decreasing Increasing Decreasing f is increasing when is negative., 9 f, 9 f > 0 9, f < 0 is positive and decreasing when f. f f g for all ±. f, ± f smmetric about origin zeros of f: 0, 0, ±, 0, ± f 0 (, ) No relative etrema Holes at, and, (, ). f c is constant f 0. 7. f is quadratic f is a line. 9. f has positive, but decreasing slope f f f. (a) f increasing on, because > 0 on, f decreasing on, because < 0 on, (b) f has a relative minimum at. f f. (a) f increasing on, 0 and, because > 0 there f decreasing on 0, because < 0 there f (b) f has a relative maimum at 0, and a relative minimum at. f In Eercises 70, f > 0 on,, f < 0 on, and f > 0 on,.. g f 7. g f 9. g f g f g0 f0 < 0 g f < 0 g f 0 g f 0 g0 f0 > 0

Chapter Applications of Differentiation 7. f > 0, undefined, < 0, Two possibilities for f are given below. (a) < f is increasing on,. > f is decreasing on,. (b) 7. The critical numbers are in intervals 0.0, 0. and 0., 0.0 since the sign of changes in these intervals. f is decreasing on approimatel, 0.0, 0.,, and increasing on 0.0, 0.. Relative minimum when 0.0. Relative maimum when 0.. f f 7. st.9sin t (a) (b) st.9sin t 9.sin t speed st 9.sin t st 0 0.9 t.9 t 9.t.9 t.9 t 0 The speed is maimum for. 77. f, g sin, 0 < < (a) f g 0... 0... 0.79 0. 0.997 0.909 0.9 0. f seems greater than g on 0,. (b) (c) Let h f g sin f g 0 > sin on 0, h cos > 0 on 0,. Therefore, h is increasing on 0,. Since h0 0, h > 0 on 0,. Thus, sin > 0 > sin f > g on 0,. 79. v kr rr krr r. 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.

Section. Increasing and Decreasing Functions and the First Derivative Test. (a) Using a graphing utilit, t represents 99 M.7t 7.9t.9. (b) 0 (c) Mt 0.t 7.9 0 t. B the First Derivative Test, t. is a minimum. M.. The minimum of the actual data is. at t 7. 0 0. (a) st t t, t 0 vt t (b) vt 0 when t. Moving in positive direction for 0 < t < because vt > 0 on 0 < t <. (c) Moving in negative direction when t >. (d) The particle changes direction at t. 7. (a) st t t t, t 0 vt t 0t 0 ± 00 (b) vt 0 for t ± Particle is moving in a positive direction on 0, and, because 0, 0.., v > 0 on these intervals. (c) Particle is moving in a negative direction on, 0.,. (d) The particle changes direction at t ±. 9. Answers will var. 9. (a) Use a cubic polnomial f a a a a 0. 9. (a) Use a fourth degree polnomial f a a a a a 0. (b) f a a a. 0, 0:, : 0 a 0 f 0 0 0 a f0 0 a a f 0 a a f 0 (c) The solution is a 0 a 0, a, a : (b) f a a a a (0, 0:, 0:, : 0 a 0 f 0 0 0 a f0 0 0 a a a f 0 0 a a a f 0 a a a f 0 a a a f 0 (d) f. (, ) (c) The solution is a a 0 a 0, a, a,. f (0, 0) (d) (, ) (0, 0) (, 0)

Chapter Applications of Differentiation 9. True. Let h f g where f and g are increasing. Then h f g > 0 since f > 0 and g > 0. 97. False. Let f, then f and f onl has one critical number. Or, let f, then f has no critical numbers. 99. False. For eample, f does not have a relative etrema at the critical number 0. 0. 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. 0. 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. 0. Let and be two positive real numbers, 0 < <. Then > f > f Thus, f is decreasing on 0,. Section. Concavit and the Second Derivative Test.,. Concave upward:, f, Concave upward:,,, Concave downward:,. f, 7. Concave upward:,,, Concave downward:, g g g Concave upward:, Concave downward:,

Section. Concavit and the Second Derivative Test 9. tan,,. sec sec tan Concave upward:, 0 Concave downward: 0, f f f 0 when. The concavit changes at., is a point of inflection. Concave upward:, Concave downward:,. 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 7. 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,

Chapter Applications of Differentiation 9. 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,,. f sin, 0 f cos f sin f 0 when 0,,. Test interval: 0 < < Sign of f: < < f f < 0 > 0 Conclusion: Concave downward Concave upward Point of inflection:, 0. f sec, 0 < < f sec f sec tan sec tan Concave upward: 0,,, Concave downward:,,, No points of inflection 0 for an in the domain of f.. f sin sin, 0 f cos cos f sin sin sin cos f 0 when 0,.,,.0. Test interval: 0 < <.. < < Sign of f: f f < 0 > 0 < <.0.0 < < f f < 0 > 0 Conclusion: Concave downward Concave upward Concave downward Concave upward Points of inflection:.,.,, 0,.,.

Section. Concavit and the Second Derivative Test 7 7. f 9. f f 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, is a relative minimum. f f f Critical number: f > 0 Therefore,, 0 is a relative minimum.. f. f f Critical numbers: 0, f0 < 0 Therefore, 0, is a relative maimum. f > 0 Therefore,, is a relative minimum. g g g Critical numbers: g0 > 0 0,, Therefore, 0, 0 is a relative minimum. g. < 0 Therefore,,.7 is a relative maimum. g 0 Test fails. B the First Derivative Test,, 0 is not an etremum.. 9. f 7. 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. f cos, 0 f sin 0 Therefore, f is non-increasing and there are no relative etrema. f f f Critical numbers: ± f < 0 Therefore,, is a relative maimum. f > 0 Therefore,, is a relative minimum.. f 0.,, (a) f 0. f 9.. 0. 0 9 (b) f0 < 0 0, 0 is a relative maimum. f > 0.,.79 is a relative minimum. Points of inflection:, 0, 0., 0.709,.9, 0.909 (c) f f f f is increasing when f is concave upward when when f < 0. f > 0 f > 0 f < 0. and decreasing when and concave downward

Chapter Applications of Differentiation. f sin sin sin, 0, (a) f cos cos cos f 0 when,,. f sin sin sin f 0 when,,.7,.9 (c) f (b) f < 0 Points of inflection: is a relative maimum.,., 0.7,.7, 0.9,.9, 0.97,, 0.7 Note: 0, 0 and, 0 are not points of inflection since the are endpoints. π π f π f f > 0 f > 0 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. (a) means f decreasing 7. Let f. f increasing means concave upward f f0 0, but 0, 0 is not a point of inflection. (b) f > 0 means f increasing f increasing means concave upward 9. f f f. f f f. (, 0) (, 0). (, 0) (, 0) f 7. is linear. f is quadratic. f is cubic. f concave upwards on,, downward on,. f f

Section. Concavit and the Second Derivative Test 9 9. (a) n : n : n : n : f f f f f f 0 f f f f f f No inflection points No inflection points Relative minimum:, 0 Inflection point:, 0 No inflection points: Relative minimum:, 0 9 9 9 9 9 9 Point of inflection 9 9 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. (b) 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.. f a b c d Relative maimum:, Relative minimum:, Point of inflection:, f a b c, f a b f 7a 9b c d 9a b c 9a b c f a b c d f 7a b c 0, f a b 0 9a b c a b 0 7a b c 0 a b a b a a, b, c, d f. 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 a b c 0 f0 0 c 0 Solving this sstem ields f a and b a. (b) The plane would be descending at the greatest rate at the point of inflection. f a b 0. Two miles from touchdown.

0 Chapter Applications of Differentiation. D L L 7. D L L L L 0 0 or L ± L B the Second Derivative Test, the deflection is maimum when L 0.7L. ± L C 0. 000 C C 0. 000 C average cost per unit dc 000 0. 0 when 00 d B the First Derivative Test, C is minimized when 00 units. 9. S 000t 0 t t, (a) t 0... S.. 097..7 9.0 7. (b) Increasing at greatest rate when t.. 000 0 0 Increasing at greatest rate when t.. (c) S 000t t St 0,000t t St 0,000 t t for t ± St 0. Hence, t. rs. 7. 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.

Section. Concavit and the Second Derivative Test 7. f, f 0 f, f f0 f, f0 P P P 0 P 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 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. 77. 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 0 79. True. Let a b c d, a 0. Then when ba, and the concavit changes at this point.. False. Concavit is determined b f. For eample, let f and c. fc f > 0, but f is not concave upward at c. f g g > 0.. f and g are concave upward on a, b implies that and are increasing on a, b, and hence > 0 and Thus, f g > 0 f g is concave upward on a, b b Theorem.7. f

Chapter Applications of Differentiation Section. Limits at Infinit. lim f means that f. approaches as becomes large. f No vertical asmptotes. f No vertical asmptotes Horizontal asmptote: Horizontal asmptote: 0 Matches (f) f < Matches (d) 7. f sin No vertical asmptotes Horizontal asmptote: 0 f > Matches (b) 9. f 0 0 0 0 0 0 0 0 0 0 0 f 7..0.00.000 lim f 0. f 0 0 0 0 0 0 0 0 0 0 0 f.9.999 0. lim f f 0 0 0 0 0 0 0 0 f..99.9999.999999 0 lim f. (a) (b) (c) h f 0 0 lim h (Limit does not eist) h f 0 0 lim h h f 0 0 lim h 0 7. (a) (b) (c) lim 0 lim lim (Limit does not eist)

Section. Limits at Infinit 9. (a) lim (b) lim (c) lim (Limit does not eist) 0. lim lim 0 0. lim lim 0 0. lim lim Limit does not eist 7. lim lim lim, for < 0 we have 9. lim for < 0, lim lim. Since sin for all 0, we. lim sin 0 have b the Squeeze Theorem, lim lim sin sin 0 lim 0. sin Therefore, lim 0. lim. f lim lim Therefore, and are both horizontal asmptotes. = = 7. f lim lim f 9 f Therefore, and are both horizontal asmptotes. 9 9. lim sin Let t. sin t lim t 0 t

Chapter Applications of Differentiation. lim. lim lim lim lim lim lim 0. lim lim lim lim 7. 0 0 0 0 0 0 0 0 f 0. 0.0 0.00 0.00 0.00 0.00 lim lim lim lim 9. 0 0 0 0 0 0 0 0 f 0.79 0.00 0.00 0.00 0.00 0.00 0.00. (a) Let t. lim sin sint lim t 0 t f f lim sint t 0 t (b) lim f lim f 0 (c) Since lim f, the graph approaches that of a horizontal line, lim f 0.. Yes. For eample, let f.

Section. Limits at Infinit. 7. Intercepts:, 0, 0, Intercept: 0, 0 Smmetr: none Smmetr: origin Horizontal asmptote: since Horizontal asmptote: 0 lim lim. Discontinuit: (Vertical asmptote) Vertical asmptote: ± 9. 9 Intercept: 0, 0 Smmetr: -ais Horizontal asmptote: since lim lim 9 9. Relative minimum: 0, 0. Intercept: 0, 0 Smmetr: -ais Horizontal asmptote: Vertical asmptotes: ± Relative maimum: 0, 0. Domain: > 0 Intercepts: none Smmetr: -ais Horizontal asmptote: 0 since lim 0 lim. Discontinuit: 0 (Vertical asmptote)

Chapter Applications of Differentiation. Intercept: 0, 0 7. Intercepts: ±, 0 Smmetr: none Smmetr: -ais Horizontal asmptote: since Horizontal asmptote: since lim Discontinuit: (Vertical asmptote) lim. lim lim. Discontinuit: 0 (Vertical asmptote) 9. Intercept: 0 ;, 0 Smmetr: none Horizontal asmptote: 7 Vertical asmptote: 0 7. 7. f 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: 0 0 Horizontal asmptote: 7 = = 0 0

Section. Limits at Infinit 7 7. f f = 0 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: 0 77. f f 0 = 0 f = = 7 0 when. Since f > 0 on, and f < 0 on,, then, 0 is a point of inflection. Vertical asmptotes:, Horizontal asmptote: 0 79. f f No relative etrema = = f 0 when 0. Point of inflection: 0, 0 Horizontal asmptotes: ± No vertical asmptotes. g sin, < < g cos. π (, π ) = sin() Horizontal asmptote: sin Relative maimum: No vertical asmptotes.09 0

Chapter Applications of Differentiation. f, g (a) (b) f (c) 70 f = g g 0 70 The graph appears as the slant asmptote. 0. C 0. 00 7. C C lim Nt t Et c lim t C 0. 00 00 lim 0. 0. 9..t.t.70 t (a) (b) Yes. lim t. 0 0 00 9. (a) T t 0.00t 0.77t. 9. line: m 0 (b) 90 T = m + 0 0 0 (, ) (c) 90 0 T T 0 0 t t (a) (b) d A B C A B m m 7 m m (d) T 0. T 0.0 (e) lim T t (f) No. The limiting temperature is. T has no horizontal asmptote. (c) lim dm lim dm m m The line approaches the vertical line 0. Hence, the distance approaches.

Section. Limits at Infinit 9 9. f 97. lim (a) (b) lim f L f (a) For 0., we need M > 0 such that whenever < 0. f L > M. < B graphing <..,. and., b smmetr (c) Let M > 0. For > M: > 0 0 0 > > ou see that the inequalit is satisfied for >.7, so let M. > > f L > (d) Similarl, N. (b) For 0., the inequalit becomes.9 < <.. From the corresponding graph ou obtain M 7. 99. lim Let > 0 be given. We need M > 0 0. 0. lim Let > 0. We need to find N < 0 such 0. such that that f L 0 < > let M >,,. Hence, for > M, we have > whenever > M. > < f L <. f L 0 Let < > <. N. Hence, for > < < N <, < whenever < N. < f L <.

0 Chapter Applications of Differentiation 0. 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 0. False. Let f (See Eercise.). 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 ±. Section. A Summar of Curve Sketching. f has constant negative slope. Matches (d). The slope is periodic, and zero at 0. Matches (a). (a) f 0(c) for and is increasing on 0,. f > 0 f f is negative for < < (decreasing function). is positive for > and < (increasing function). f (b) 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. 7. 0 when 0. 0 when ±., (0, 0) ) =, Horizontal asmptote: < < < < 0 0 0 < < < < 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

) Section. A Summar of Curve Sketching 9.. No relative etrema, no points of inflection Intercepts: < 0 when. 7, 0, 0,7 Vertical asmptote: Horizontal asmptote: Inflection point: 0, 0 Intercept: < 0 if ±. 0 if 0. 0, 0 Vertical asmptotes: ± Horizontal asmptote: 0 Smmetr with respect to the origin 7, 0 0 0, 7 (0, 0). g g g 0 when 0.9,.0 0 when ± 0,),.. 7, 0) ) 0. 9, 0. )., 77 ) 0.,. 7) ) g0.9 < 0, therefore, 0.9,.0 is relative maimum. g.0 > 0, therefore,.0,.7 is a relative minimum. Points of inflection:,.,,.77 Intercepts: 0,,.7, 0 Slant asmptote:. f f 0 when ±. (, ) = f 0 (, ) Relative maimum:, = 0 Relative minimum:, Vertical asmptote: 0 Slant asmptote:

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

Section. A Summar of Curve Sketching. 0 when and undefined when 0. < 0 when 0. (, 0 0) (, ) 7 (, 0 ) < < 0 0 0 < < < < Conclusion Decreasing, concave down 0 Undefined Undefined Relative minimum Increasing, concave down 0 Relative maimum Decreasing, concave down. 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 ( 0.79, 0) (0, ) (, ) (., 0) (, (.7, 0) ) 7. No critical numbers 0 when 0. < < 0 0 0 < < Conclusion Decreasing, concave up 0 Point of inflection Decreasing, concave down (0, ) (, 0)

( Chapter Applications of Differentiation 9. 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 < < 0 0 0 < < Increasing, concave down 0 0 0 Point of inflection Increasing, concave up. 0 when ±. 0 0 when 0. < < Conclusion Increasing, concave down 0 Relative maimum < < 0 Decreasing, concave down 0 0 0 Point of inflection 0 < < Decreasing, concave up 0 Relative minimum < < Increasing, concave up (, 0) (, ) (0, 0) (, ) ( ), 0. 0 undefined at. < < Conclusion Decreasing (0, ), 0 0 Undefined Relative minimum < < Increasing. f 0 0 9 0 vertical asmptote 0 horizontal asmptote Minimum:.0, 9.0 0 Maimum:.0, 9.0 Points of inflection:., 7.,., 7.

Section. A Summar of Curve Sketching 7. 7 0, 0 point of inflection ± horizontal asmptotes 9. sin sin, 0 cos cos cos cos cos sin sin cos sin cos sin cos cos sin sin cos sin 0: cos 0, impossible sin 0 sin, π π π sin sin 0 sin sin sin 0 0,, cos cos ±,, 7, Relative maimum: Relative minimum: Inflection points:, 0, 7,, 9, 9,, 9,, 9, 9, 9 sin cos cos sin sin cos cos sin sin cos cos sin cos

Chapter Applications of Differentiation. tan, < < sec 0 when ± sec tan 0 when 0. Relative maimum:,. π π π Relative minimum: Inflection point: 0, 0, Vertical asmptotes: ±. csc sec, 0 < < sec tan csc cot 0 Relative minimum: Vertical asmptotes: 0,,. g tan, < < g g sin cos cos cos sin cos 0 when 0. 7. f is cubic. f f is quadratic. is linear. f f Vertical asmptotes:,,, Intercepts:, 0, 0, 0,, 0 Smmetric with respect to -ais. f Increasing on 0, and, Points of inflection: ±.0, π 0 0 π π π

Section. A Summar of Curve Sketching 7 9.. f f f f (an vertical translate of f will do) (an vertical translate of f will do). Since the slope is negative, the function is decreasing on,, and hence f > f.. f Vertical asmptote: none Horizontal asmptote: 7. sin h Vertical asmptote: none Horizontal asmptote: 0 9 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. Yes, it is possible for a graph to cross its horizontal asmptote. It is not possible to cross a vertical asmptote because the function is not continuous there. 9. h. f, Undefined, if if The rational function is not reduced to lowest terms. The graph appears to approach the slant asmptote. hole at,. f The graph appears to approach the slant asmptote.

Chapter Applications of Differentiation. f cos, 0, (a). (b) 0 0. On 0, there seem to be 7 critical numbers: 0.,.0,.,.0,.,.0,. f cos cos sin 0 Critical numbers, 0.97,,.9,,.9, 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 7. Vertical asmptote: 9. Vertical asmptote: Horizontal asmptote: 0 Slant asmptote: 9 7. 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. (b) 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. 7. f n (a) For n even, f is smmetric about the -ais. For n odd, f is smmetric about the origin. (b) 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 :. (e) n 0 M 0 N 7. (a) 70 77. 9 as, and as. 0 (b) When t 0, N bacteria. (c) N is greatest at t 7.. (d) Nt is greatest when t.. (e) (Find the t-value of the point of inflection.),0 lim Nt 9 t 7 bacteria 9 9

Section.7 Optimization Problems 9 Section.7 Optimization Problems. (a) First Number, Second Number Product, P 0 0 0 00 0 000 0 0 0 00 0 00 0 0 0 00 0 00 0 0 0 00 0 00 0 0 0 00 0 000 0 0 0 00 0 000 (b) First Number, 0 0 0 0 0 0 70 0 90 Second Number 0 0 0 0 0 0 0 0 0 0 0 0 0 70 0 0 0 90 Product, P 00 0 000 00 0 00 00 0 00 00 0 00 00 0 000 00 0 000 700 70 00 00 0 00 900 90 00 00 0 00 000 00 000 The maimum is attained near 0 and 0. (c) (d) P 0 0 00 0 0 (, 0) 0 The solution appears to be. (e) dp 0 0 when. d d P d < 0 P is a maimum when 0. The two numbers are and.. 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. d S > 0 when. d S is minimum when and.

0 Chapter Applications of Differentiation 7. Let and be two positive numbers such that 00. P 00 00 dp 00 0 when. d d P < 0 when. d P is a maimum when 0 and. 9. Let be the length and the width of the rectangle. 00 0 A 0 da 0 0 when. d d A < 0 when. d A is maimum when meters.. Let be the length and the width of the rectangle. P dp d 0 when. d P d > 0 when. P is minimum when 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 7 0 7 (, ) d (, 0). d 7. 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,. dq d kq 0 kq 0 k 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. (, ) d (, (

Section.7 Optimization Problems 9. 0,000 (see figure) S 0,000 ds 0,000 0 when 00. d d S 70,000 d > 0 when 00. where S is the length of fence needed. S is a minimum when 00 meters and 00 meters.. (a) A area of side area of Top (b) V lengthwidthheight () A 0 square inches () V 99 cubic inches () A 0 square inches () V cubic inches () A. 0 square inches () V. 7 cubic inches (c) S 0 0 0 V 7 V 7 0 ± B the First Derivative Test, ields the maimum volume. Dimensions:. (A cube!). A da d 0 when d A d. < 0 when. The area is maimum when feet and feet.

( Chapter Applications of Differentiation. (a) (b) 0 0 0 L, > (c) Area A A ± 0, (select ) Then and A. Vertices: 0, 0,, 0, 0, 0 (.7,.) 0 0 0 L is minimum when.7 and L.. 7. A see figure da d (, 0 when.. B the First Derivative Test, the inscribed rectangle of maimum area has vertices ±, 0, ±,. Width: Length: ; 9. 0 0 A 0 da d 0 0 0 0 see figure 0 0 0 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. + +. (a) P r 00 00 00 CONTINUED

( Section.7 Optimization Problems. CONTINUED (b) Length, Width, Area, 0 0 0 0 0 00 0 00 0 00 0 00 0 00 0 0 00 0 0 00 0 0 00 0 7 0 00 0 09 0 00 0 7 0 00 0 0 00 0 9 (c) A 00 00 (d) A 00. when 0. Maimum when length 0 m and width 00. (e) 000 (0, 9.) 0 00 0 A 0 Maimum area is approimatel 9. m 0 m. The maimum area of the rectangle is approimatel 9 m.. Let be the sides of the square ends and the length of the package. P 0 0 V 0 0 dv d 0 when. d V d < 0 when. The volume is maimum when inches and 0 inches.. 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 0 9 r 9 r 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 r 9 r r cubic units.

Chapter Applications of Differentiation 7. 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 0. 9. V r r h h S r rh r r r r ds dr r r 0 when r 9. cm. d S dr r r r r r > 0 when r 9 cm. r r r r r The surface area is minimum when r 9 cm and h 0. The resulting solid is a sphere of radius r. cm. r h. Let be the length of a side of the square and the length of a side of the triangle. 0 0 9 0 A 0 da d 0 0 0 9 d A 9 d > 0. Let S be the strength and k the constant of proportionalit. Given h w, h w, S kwh S kw7 w k7w w ds dw k7 w 0 when w, h. d S dw kw < 0 when w. These values ield a maimum. A is minimum when 0 9 and 0 9.. 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.,..

Section.7 Optimization Problems 7. sin h s s h, 0 < < sin tan h h tan s tan sec sin h s I k sin k sin k s sec sin cos α α di k sin sin cos cos cos d ft k cos cos sin 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. 9. S, L Time T 0 dt d 0 0 S = + L = + ( ( Q 9 0 9 0 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.. 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.

Chapter Applications of Differentiation. f sin. F cos kw F sin k + θ k π (a) Distance from origin to -intercept is. Distance from origin to -intercept is.7. (b) d sin π π (0.797, 0.979) Minimum distance 0.979 at 0.797. (c) Let f d sin. f sin cos df d k cos sin k tan arctan k Since F cos k sin the minimum force is F kw cos k sin kwk cos sin cos k sin 0 k k k k, kw kw cos k sin k. Setting f 0, ou obtain 0.797, which corresponds to d 0.979. 7. (a) (b) Base Base Altitude Area cos 0 cos 0 cos 0 cos 0 cos 0 sin 0 sin 0 sin 0 sin 0 sin 0.. 9.7 7.7 0. cos 0 sin 0. Base Base cos 0 cos 0 cos 0 cos 0 cos 0 cos 0 cos 70 cos 0 Altitude sin 0 sin 0 sin 0 sin 0 sin 0 sin 0 sin 70 sin 0 Area.. 9.7 7.7 0.. 0.7 7.0 cos 90 sin 90.0 The maimum cross-sectional area is approimatel. square feet. (c) (e) A a b h cos sin cos sin, 0 < 00 (0,.) < 90 (d) da cos cos sin sin d cos cos sin cos cos cos cos 0 when 0, 0, 00. The maimum occurs when 0. 0 90 0

Section. Newton s Method 7 9. C 00 00 0, C 00 00 0 0 Approimation: 0. units, or 0 units. S m m 0m ds m m 0m 0 m 0 when m dm. Line: S 0 0 0. mi. S m m m m 0m m Using a graphing utilit, ou can see that the minimum occurs when 0.. Line: 0. S 0. 0. 00. 0.. mi. S 0 0 0 (0.,.) m. f ; 9 0 Hence, or. f f is increasing on, and,. Hence, f is increasing on, and,. f, f. The maimum value of f is. Section. Newton s Method. f f.7 n n f n f n f n f n n f n f n.7000 0.00.000 0.0.7.7 0.00. 0.000.7. f sin f cos n n f n f n f n f n n f n f n.0000 0. 0.9900 0... 0.0009.0000 0.0009.

Chapter Applications of Differentiation. f f Approimation of the zero of f is 0.. n n f n f n f n f n n f n f n 0.000 0.70.700 0. 0.7 0.7 0.07.07 0.0 0. 0. 0.00.00 0.0009 0. 7. f f Approimation of the zero of f is.. Similarl, the other zero is approimatel 7.. n n f n f n f n f n n f n f n.000 0.. 0.00.9.9 0.0.0 0.000.. 0.000.9 0.000.9 9. f f Approimation of the zero of f is.97. n n f n f n f n f n n f n f n.000 0.70.700 0.0.. 0.0.9 0.00.. 0.0000.07 0.00000.. f.9.79. f 7..79 n n f n f n 0.000 0.0.00 0.09 0.709 0.709 0.09 0.7 0.77 0. 0. 0.0 0.07 0.07 0.799 0.799 0.00 0.9 0.0 0.90 0.90 0.000 0.0 0.000 0.9000 0.9000 0.0000 0.000 0.0000 0.9000 Approimation of the zero of f is 0.900. n n f n f n Approimation of the zero of f is.00. Approimation of the zero of f is.900. f n f n f n f n n f n f n n f n f n. 0.0000 0.00 0.0000.000 n n f n f n f n f n n f n f n.9 0.0000 0.000 0.0000.9000. f sin f cos Approimation of the zero of f is 0.9. n n f n f n f n f n n f n f n 0.000 0.00.77 0.00 0.90 0.90 0.0000.7 0.0000 0.90

Section. Newton s Method 9. h f g h Point of intersection of the graphs of f and g occurs when 0.9. n n h n h n h n h n n h n h n 0.000 0.0.79 0.0 0.7 0.7 0.0000.7 0.0000 0.7 7. h f g tan h sec Point of intersection of the graphs of f and g occurs when.9. Note: f and g tan intersect infinitel often. n n h n h n h n h n n h n h n.000 0.7.0 0.00.9.9 0.009 0.7 0.000.9 9. (a) f a, a > 0 (b) : n n n. f n n f n f n n n... n n a n For eample, given, n a n 9... 7: n n 7 n, n n.7.77.. 7.. f. f f 0; therefore, the method fails. n n f n f n 0 0 f 0 f and so on. Fails to converge. Answers will var. See page. Newton s Method uses tangent lines to approimate c such that f c 0. First, estimate an initial close to c (see graph). Then determine b f f. Calculate a third estimate b f f. Continue this process until n n is within the desired accurac. Let n be the final approimation of c. a c f() b

) 0 Chapter Applications of Differentiation 7. Let g f cos g sin. The fied point is approimatel 0.7. n n g n g n g n g n n g n g n.0000 0.97. 0.9 0.70 0.70 0.090.9 0.0 0.79 0.79 0.0000.7 0.0000 0.79 9. f, f (a) (d) f (b) f. f Continuing, the zero is.7. (c) f.0 f Continuing, the zero is.... 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.. f a 0 f n n n a n n n n a n n n a n n a n a n. f cos, 0, f sin cos 0 0. 0, 0.) Letting F f, we can use Newton s Method as follows. F sin cos n n F n F n F n F n n F n F n 0.9000 0.0. 0.09 0.0 0.0 0.000.077 0.000 0.0 Approimation to the critical number: 0.0

Section. Newton s Method. f,, 0 d 0 7 7 d is minimized when D 7 7 is a minimum. g D g (.99, 0.0) (, 0) n n g n g n g n g n n g n g n.0000.0000.0000 0.0.9.9 0.00.9 0.007.9.9 0.00.09 0.0000.9.99 Point closest to, 0 is.99, 0.0. 7. Minimize: T T 0 0 9 7 0 Distance rowed Rate rowed T Distance walked Rate walked 0 0 0 Let f 7 and f. Since f 00 and f, the solution is in the interval,. n n f n.7000 9.7.0 0....00 0.70 0.0070.. 0.00 9.9 0.0000. Approimation:. miles f n f n f n n f n f n 9.,00,000 7 0 0,000 7 0,0,000 0 Let f 7 0,0,000 f 90. From the graph, choose 0. n n f n f n f n f n n f n f n 0.0000 000.0000 00.0000.070 7.90 7.90 77.09.09 0.00.0.0 70.09. 0.0...0. 0.000. The zero occurs when. which corresponds to $,. (The larger zero is.07

Chapter Applications of Differentiation. False. Let f. is a discontinuit. It is not a zero of f. This statement would be true if f pq is given in reduced form.. True. f sin f cos Let 0, 0, sin 0 be a point on the graph of f. If ( 0, 0 is a point of tangenc, then cos 0 0 0 0 0 0 0 sin 0 0. Thus, 0 tan 0. 0.9 Slope cos 0 0.7 You can verif this answer b graphing sin and the tangent line 0.7. Section.9 Differentials. f.9.99.0. f f.00.90.00.00 Tangent line at, : f f T.000.900.000.000. f.9.99.0. f f.70.00.00 0.0 Tangent line at, : f f 0 T 0.0000.000.000 0.0000 0. f sin.9.99.0. f cos f sin 0.9 0.9 0.909 0.90 0. Tangent line at, sin : f f T cos sin 0.909 0.9 0.909 0.90 0.77 sin cos cos sin

Section.9 Differentials 7. f, f,, d 0. f f f. f 0.0 d f d f0. 0. 0. 9. f, f,, d 0.0 f f f 0.99 f 0.99 0.09 d f d f0.0 0.0 0.0.. d d d d. 7. d d d cot d cot csc d cot cot d 9. cos. (a) d sin d f.9 f 0. f f0. 0. 0.9 (b) f.0 f 0.0 f f0.0 0.0.0. (a) (b) f.9 f 0. f f0.. (a) 0..0 f.0 f 0.0 f f0.0 (b) 0.0 0.9 g.9 g 0.07 g g0.07 0.07.0 g. g 0. g g0. 0. 7.9 7. A 9. d ± da d A da ± ± square inches A r r r dr ± A da r dr ± ±7 square inches. (a) centimeters d ±0.0 centimeters A da d ±0.0 (b) da A d d 0.0 d 0.0 0.0.% ±. square centimeters Percentage error: da A. 0.00... %

Chapter Applications of Differentiation. r inches r dr ±0.0 inches (a) V r (c) Relative error: dv V r dr dr r r (b) dv r dr ±0.0 ±. cubic inches S r ds r dr ±0.0 ±0.9 square inches Relative error: 0.0 0.0 % ds S r dr dr r r 0.0 0.00... %. V r h 0r, r cm, h 0 cm, dr 0. cm V dv 0r dr 00. 0 cm 7. (a) (b) T Lg9. dt glg dl Relative error: dt T dlglg Lg dl L relative error in L 0.00 0.00 Percentage error: dt T 00 0.% % 0.0000 seconds. minutes.7 d ± ±0. (a) (b) dh cot.70. h Converting to radians, cot 0.90.00 0.007 0.7% in radians. dh h cot d 0.0 d d h 9. csc dh 9. csc cot d dh h cot d 0.0 cot 0.0 tan 0.0 tan.7.7 θ 0.0 tan 0.9 0.9 0.0.% in radians h b 9.. r v 0 sin. Let f, 00, d 0.. v 0 00 ftsec changes from 0 to. dr 00 cos d 0 d 0 r dr 0 00 9 feet 0 cos 0 00 f f f d d f 99. 00 0. 9.97 00 Using a calculator: 99. 9.999

Review Eercises for Chapter. Let f,, d. f f f d d f.99 00 7. Let f,, d 0.0, f. Then f.0 f f d.0 0.0 0.0. Using a calculator,.990. 9. f. f tan f f sec f0 0 At 0,, f0, f0 f0 Tangent line: 0 Tangent line at 0, 0: 0 0 f (0, 0) (0, ) f. In general, when 0, d. True 7. 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 cos, 0, g sin 0 (, 7.7) 0 when sin. (.7, 0.) Critical numbers: 0.,.7 Left endpoint: 0, Critical number: 0.,. Critical number:.7, 0. Minimum Right endpoint:, 7.7 Maimum 0 0

Chapter Applications of Differentiation. Yes. f f 0. f is continuous on,, differentiable on,. f 0 for. c satisfies fc 0. 7. f 9. (a) 0 f b f a b a f, f 7 f f 7 0 (b) f is not differentiable at. fc c 7 c 9 7.7 79. 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,. 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: 9. ht t t ht t 0 when t. Relative minimum:, Test Interval: < t < < t < Sign of ht: ht < 0 ht > 0 Conclusion: Decreasing Increasing

Review Eercises for Chapter 7. cost sint v sint cost (a) When t inch and v inches/second., (b) sint cost 0 when sint Therefore, sint and cost The maimum displacement is. (c) Period: inch. Frequenc: cost tant.. 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 < < < <. g g g Critical numbers: 0, ± (, ) (0, 0) (, ) g0 > 0 g ± < 0 Relative minimum at 0, 0 Relative maimums at ±, 7. 7 (, f()) (, f()) (, 0) (0, 0) 7 9. The first derivative is positive and the second derivative is negative. The graph is increasing and is concave down.. (a) D 0.00t 0.t.9t 0.t 9.0 (b) 9 0 0 9 (c) Maimum at.9, 9. 99 Minimum at., 9. 97 (d) Outlas increasing at greatest rate at the point of inflection 9., 7.7 979

Chapter Applications of Differentiation. lim lim. lim cos 7. lim 0, since cos. 9. lim cos. h. f Discontinuit: Discontinuit: 0 lim lim lim Vertical asmptote: Horizontal asmptote: Vertical asmptote: 0 Horizontal asmptote:. f 7. f Relative minimum:, 0 Relative maimum:, 0 Relative minimum: 0.,.077 Relative maimum:., 0.077 00 0. 00 Vertical asmptote: 0. Horizontal asmptote: 0 9. f Domain:, ; Range:,, ) ) f 0 when. f Therefore,, is a relative maimum. Intercepts: 0, 0,, 0. f Domain:, ; Range:, f 0 when ± and undefined when ±. f, (, 0) (, 0) (0, 0), f > 0 Therefore,, is a relative minimum. f < 0 Therefore,, is a relative maimum. Point of inflection: 0, 0 Intercepts:, 0, 0, 0,, 0 Smmetr with respect to origin