CHAPTER 3 Applications of Differentiation

Transcription

1 CHAPTER Applications of Differentiation Section. Etrema on an Interval Section. Rolle s Theorem and the Mean Value Theorem. 07 Section. Increasing and Decreasing Functions and the First Derivative Test Section. Concavit and the Second Derivative Test.... Section.5 Limits at Infinit Section. A Summar of Curve Sketching Section.7 Optimization Problems Section. Newton s Method Section.9 Differentials Review Eercises Problem Solving

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

3 0 Chapter Applications of Differentiation. f,,. f Left endpoint:, 5 Minimum Right endpoint:, Maimum Critical number: 0, 0 Critical number:, f,, f Left endpoint:, 5 Maimum Critical number: 0, 0 Minimum Right endpoint:, 5. gt t,, 7. t gt t t Left endpoint: Maimum Critical number: 0, 0 Minimum Right endpoint:,, Maimum hs, 0, s hs s Left endpoint: Right endpoint: 0, Maimum, Minimum 9. f cos, 0,. f sin Left endpoint: 0, Maimum Right endpoint: Minimum, tan,, sec sec 0 On the interval,, this equation has no solutions. Thus, there are no critical numbers. Left endpoint:,,. Maimum Right endpoint:, Minimum. (a) Minimum: 0, 5. Maimum:, (b) Minimum: (c) Maimum: 0,, (d) No etrema f (a) Minimum:, Maimum:, (b) Maimum:, (c) Minimum:, (d) Minimum:,

4 Section. Etrema on an Interval f, 0, < Left endpoint: 0, Minimum Right endpoint:, Maimum 9. f,, Right endpoint:, Minimum. (a) 5 (b) (,.7) 0 (0.9,.0) Maimum:,.7 (endpoint) Minimum: 0.9, f , 0, f ± f ± f.7 Maimum (endpoint) f Minimum: 0.9,.0. f, 0, 5. f f, 0, f f f 0 5 Setting f 0, In the interval 0,, choose we have ± 00 0 ± 0 0 ± is the maimum value. f f 9 f 7 7 f 5 0 f 5 50 f 0 5 is the maimum value.

5 0 Chapter Applications of Differentiation 7. f tan f is continuous on 0, but not on 0,. lim tan (a) Yes 5. (a) No 5 5 f (b) No (b) Yes 55. P VI RI I 0.5I, 0 I P 0 when I 0. P 7.5 when I 5. P I 0 Critical number: I amps When I amps, P 7, the maimum output. No, a 0-amp fuse would not increase the power output. P is decreasing for I >. csc sec S hs s cos sin, ds d s csc cot csc s csc cot csc 0 cot arcsec radians s S hs s S hs Sarcsec hs s S is minimum when arcsec radians. 59. (a) a b c a b A B The coordinates of B are 500, 0, and those of A are 500, 5. From the slopes at A and B, 500 9% % a b a b 0.0. Solving these two equations, ou obtain a 0000 and b 00. From the points 500, 0 and 500, 5, ou obtain c c. 00 In both cases, c Thus, CONTINUED

6 Section. Rolle s Theorem and the Mean Value Theorem CONTINUED (b) d For 500 0, d a b c For 0 500, d a b c 0.0. (c) The lowest point on the highwa is 00,, which is not directl over the point where the two hillsides come together.. True. See Eercise 5.. True. Section. Rolle s Theorem and the Mean Value Theorem. Rolle s Theorem does not appl to over 0, since f is not differentiable at. f. f -intercepts:, 0,, 0 f 0 at. 5. f 7. -intercepts:, 0, 0, 0 f f 0 at f, 0, f 0 f 0 f is continuous on 0,. f is differentiable on 0,. Rolle s Theorem applies. f 0 c value: 9. f,,. f f 0 f is continuous on,. f is differentiable on,. Rolle s Theorem applies. c f f 0 ±, c f,, f f f is continuous on,. f is not differentiable on, since f0 does not eist. Rolle s Theorem does not appl.

7 0 Chapter Applications of Differentiation. f,, f f 0 f is continuous on,. (Note: The discontinuit,, is not in the interval.) f is differentiable on (,. Rolle s Theorem applies. f 0 0 c value: 5 ± 5 ± 5 5. f sin, 0, 7. 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 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.

8 Section. Rolle s Theorem and the Mean Value Theorem 09. f tan,, 5. f f 0 f is continuous on,. f is differentiable on,. Rolle s Theorem applies. f sec 0 sec sec ± ± arcsec ±0.5 radian ± arccos f t t t (a) f f (b) v ft must be 0 at some time in,. ft t 0 t seconds c values: ±0.5 radian tangent line (c, f(c )) 9. f, 0, f has a discontinuit at. (a, f(a)) (c, f(c )) a tangent line f (b, f(b)) b secant line. f is continuous on, and differentiable on. f is continuous on 0, and differentiable on,. 0,. f f f f 0 0 f when Therefore,. c. f 7 c 7

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

10 Section. Rolle s Theorem and the Mean Value Theorem. f,, 9,, 9, m 9 (a) f 9 (b) Secant line: tangent secant 0 (c) f 9 f 9 f c c c c, f c, m f Tangent line: 0. st.9t No. Let f on,. (a) V avg s s msec (b) st is continuous on 0, and differentiable on 0,. Therefore, the Mean Value Theorem applies. f f0 0 and zero is in the interval (, but f f. vt st 9.t.7 msec t seconds 7. Let St be the position function of the plane. If t 0 corresponds to P.M., S0 0, S and the Mean Value Theorem sas that there eists a time t 0, 0 < t 0 < 5.5, such that St 0 vt Appling the Intermediate Value Theorem to the velocit function on the intervals 0, t 0 and t 0, 5.5, ou see that there are at least two times during the flight when the speed was 00 miles per hour. 0 < 00 < 5.5

11 ) Chapter Applications of Differentiation 9. (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,. 5. f is continuous on 5, 5 and does not satisf 5. False. f has a discontinuit at 0. the conditions of the Mean Value Theorem. f is not differentiable on 5, 5. Eample: f 5, 5) f ) ) ) 5, 5) 55. True. A polnomial is continuous and differentiable everwhere. 57. 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. 59. If p A B C, then p A B f b f a b a Ab Bb C Aa Ba C b a Ab a Bb a b a b aab a B b a Ab a B. Thus, A Ab a and b a which is the midpoint of a, b.. f cos differentiable on,. f sin f f < for all real numbers. Thus, from Eercise 0, f has, at most, one fied point. 0.50

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

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

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

15 Chapter Applications of Differentiation 9. f 9 f Critical number: Discontinuities: 0, Test intervals: < < < < 0 0 < < < < Sign of f: f > 0 f f > 0 < 0 f < 0 Conclusion: Increasing Increasing Decreasing Decreasing Increasing on:,,, 0 Decreasing on: 0,,, Relative maimum: 0, 0. f f Critical numbers:, Discontinuit: Test intervals: < < < < < < < < Sign of f: f f > 0 < 0 f f < 0 > 0 Conclusion: Increasing Decreasing Decreasing Increasing Increasing on:,,, Decreasing on:,,, Relative maimum:, Relative minimum:, 0. f cos, 0 < < f sin 0 Critical numbers:, 5 Sign of f: f > 0 Test intervals: 0 < < f f < 0 > 0 Conclusion: Increasing Decreasing Increasing < < 5 5 < < Increasing on: 0, Decreasing on:, 5, 5, Relative maimum: Relative minimum: 5, 5,

16 Section. Increasing and Decreasing Functions and the First Derivative Test 7 5. f sin sin, 0 < < f sin cos cos cos sin 0 Critical numbers:, 7,, Sign of f: f > 0 Test intervals: 0 < < < < 7 7 < < f f < 0 > 0 f f < 0 > 0 Conclusion: Increasing Decreasing Increasing Decreasing Increasing < < < < Increasing on: Decreasing on: 0, Relative minima: Relative maima:, 7, 7 7,,,,,,,,,,, 0 7. f 9,, (a) f 9 9 (c) (b) f 0 f Critical numbers: ± (d) Intervals:, ±,, 0 f < 0 f > 0 f < 0 Decreasing Increasing Decreasing f is increasing when when is negative. f f is positive and decreasing 9. f t t sin t, 0, (a) ft t cos t t sin t (c) tt cos t sin t 0 tt cos t sin t t 0 or t tan t (b) π f f π t t cot t t.9, (graphing utilit) Critical numbers: t.9, t (d) Intervals: 0,.9.9, , ft > 0 ft < 0 ft > 0 Increasing Decreasing Increasing f is increasing when is negative. f f is positive and decreasing when

17 Chapter Applications of Differentiation. f 5 f g for all ±. f, ± f smmetric about origin zeros of f: 0, 0, ±, 0, ± f 0 5 (, ) 5 No relative etrema Holes at, and, (, ) 5. f c is constant f 0 5. f is quadratic f is a line. f f 7. f has positive, but decreasing slope f In Eercises 9 5, f > 0 on,, f < 0 on, and f > 0 on,. 9. g f 5 5. g f 5. g f 0 g f g f g f 0 g0 f0 < 0 g f < 0 g0 f0 > f > 0, undefined, < 0, < f is increasing on,. > f is decreasing on,. Two possibilities for f are given below. (a) (b) 5

18 Section. Increasing and Decreasing Functions and the First Derivative Test The critical numbers are in intervals 0.50, 0.5 and 0.5, 0.50 since the sign of changes in these intervals. f is decreasing on approimatel, 0.0, 0.,, and increasing on 0.0, 0.. Relative minimum when 0.0. Relative maimum when 0.. f 59. f, g sin, 0 < < (a) f g f seems greater than g on 0,. (b) 5 (c) Let h f g sin h cos > 0 on 0,. 0 > sin on 0, Therefore, h is increasing on 0,. Since h0 0, h > 0 on 0,. Thus, sin > 0 > sin f > g on 0,.. 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. 5. (a) (b) B 0.9t.79t t.0t B 0 (c) for t.7, or 9, (. thousand bankruptcies) Actual minimum: 9 (. thousand bankruptcies) 7. (a) Use a cubic polnomial f a a a a 0. (c) The solution is a 0 a 0, a, a : (b) f a a a. f. 0, 0: 0 a 0 f a f0 0 (d) (, ), : a a f 0 a a f 0 (0, 0)

19 0 Chapter Applications of Differentiation 9. (a) Use a fourth degree polnomial f a a a a a 0. (b) f a a a a (0, 0:, 0:, : 0 a 0 f a f a a a f 0 0 5a a a f 0 a a a f 0 a a a f 0 (c) The solution is a a 0 a 0, a, a,. f (d) 5 (, ) (0, 0) (, 0) 5 7. True Let h f g where f and g are increasing. Then h f g > 0 since f > 0 and g > False Let f, then f and f onl has one critical number. Or, let f, then f has no critical numbers. 75. False. For eample, f does not have a relative etrema at the critical number Assume that f < 0 for all in the interval a, b and let < be an two points in the interval. B the Mean Value Theorem, we know there eists a number c such that < c <, and fc f f. Since fc < 0 and > 0, then f f < 0, which implies that f < f. Thus, f is decreasing on the interval. 79. 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.

20 Section. Concavit and the Second Derivative Test Section. Concavit and the Second Derivative Test.,. Concave upward:, f, Concave upward:,,, Concave downward:, 5. f, 7. Concave upward:,,, Concave downward:, f f f Concave upward:, Concave downward:, 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

21 Chapter Applications of Differentiation 5. 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, 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,,. Point of inflection:, 0 Test interval: 0 < < Sign of f: < < f f < 0 > 0 Conclusion: Concave downward Concave upward

22 Section. Concavit and the Second Derivative Test. 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. 5. 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 f f < 0 > 0.0 < < Conclusion: Concave downward Concave upward Concave downward Concave upward Points of inflection:.,.5,, 0,.,.5 7. f 9. f f f 5 f 5 f Critical numbers: 0, Critical number: 5 However, f0 0, so we must use the First Derivative Test. f < 0 on the intervals, 0 and 0, ; hence, 0, is not an etremum. f > 0 so, 5 is a relative minimum. f5 > 0 Therefore, 5, 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 5 g 5 Critical numbers: g0 > 0 0, 5, Therefore, 0, 0 is a relative minimum. g < 0 Therefore, 5,.7 is a relative minimum. g 0 Test fails b the First Derivative Test,, 0 is not an etremum.

23 Chapter Applications of Differentiation 5. 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 f f Critical numbers: ± f < 0 Therefore,, is a relative maimum. f > 0 Therefore,, is a relative minimum. 9. f cos, 0 f sin 0 Therefore, f is non-increasing and there are no relative etrema.. f 0.,, (a) f 0.5 (c) f f f (b) f0 < 0 0, 0 is a relative maimum. f 5 > 0.,.79 is a relative minimum. f Points of inflection:, 0, 0.5, 0.709,.9, 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 sin 5 sin 5, 0, (a) f cos cos cos 5 (c) f 0 when, 5,. f sin sin 5 sin 5 5 f 0 when,,.7,.95 f π π f f π (b) f < 0 Points of inflection: is a relative maimum.,.5, 0.7,.7, 0.9,.95, 0.97, 5, 0.7 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 Note: 0, 0 and, 0 are not points of inflection since the are endpoints.

24 Section. Concavit and the Second Derivative Test 5 f < 0 f > 0 5. (a) means f decreasing (b) means f increasing f increasing means concave upward f increasing means concave upward 7. Let f. 9. f f f0 0, but 0, 0 is not a point of inflection. f f 5 5. f f f 5. (, 0) (, 0) (, 0) (, 0) 5 f f f is linear. f is quadratic. f is cubic. f concave upwards on,, downward on,.

25 Chapter Applications of Differentiation 59. (a) n : n : n : n : f f f f f f 0 f f f f f f No inflection points No inflection points Inflection point:, 0 No inflection points: Relative minimum:, 0 Relative minimum:, Point of inflection Conclusion: If n and n is odd, then, 0 is an inflection point. If n and n is even, then, 0 is a relative minimum. (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: 5, Point of inflection:, f a b c, f a b f 7a 9b c d 9a b c 9a b c f 5 5a 5b 5c 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 5, d f 5

26 Section. Concavit and the Second Derivative Test 7. f a b c d Maimum:, Minimum: 0, 0 (a) f a b c, f a b f 0 0 d 0 f a b c f 0 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. 5. D 5L L 7. D 5L L 5L L 0 0 or 5L ± L B the Second Derivative Test, the deflection is maimum when 5 L 0.57L. 5 ± L C C C C average cost per unit dc when 00 d B the First Derivative Test, C is minimized when 00 units. 9. S 5000t t St 0,000t t St 0,000 t t St 0 for t.. Sales are increasing at the greatest rate at t. ears. 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.

27 Chapter Applications of Differentiation 7. f, f 0 f, f f0 f, f0 P 5 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 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 True. Let a b c d, a 0. Then when ba, and the concavit changes at this point.

28 Section.5 Limits at Infinit 9. False. f sin cos f cos sin cos sin 0 Critical number: cos sin tan tan. False. Concavit is determined b f. ftan.0555 is the maimum value of. Section.5 Limits at Infinit. f. f 5. f sin No vertical asmptotes No vertical asmptotes No vertical asmptotes Horizontal asmptote: Horizontal asmptote: 0 Horizontal asmptotes: 0 Matches (f) Matches (d) Matches (b) 7. f f lim f f f lim f. f f lim f 5

29 0 Chapter Applications of Differentiation. (a) (b) (c) h f lim h (Limit does not eist) h f lim h 5 h f lim h 0 5. (a) (b) (c) lim 0 lim lim (Limit does not eist) 7. (a) 5 lim lim lim 0 (b) 5 lim (c) 5 lim (Limit does not eist). lim lim lim lim Limit does not eist lim lim lim, for < 0 we have 7. lim for < 0, lim lim ( 9. 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

30 . (a) f lim lim Therefore, and are both horizontal asmptotes. Section.5 Limits at Infinit = = 5. lim sin Let t. sin t lim t 0 t 7. lim 9. lim lim lim lim lim lim f lim lim lim lim f Let t. lim sin sint lim t 0 t lim sint t 0 t

31 Chapter Applications of Differentiation 5. (a) f 7. Yes. For eample, let f. (b) lim f lim f 0 (c) Since lim f, the graph approaches that of a horizontal line, lim f Intercepts:, 0, 0, Smmetr: none 5 Horizontal asmptote: since lim lim. 5 Discontinuit: (Vertical asmptote) Intercept: 0, 0 Intercept: 0, 0 Smmetr: origin Smmetr: -ais Horizontal asmptote: 0 Horizontal asmptote: since Vertical asmptote: ± lim lim Relative minimum: 0, 0

32 Section.5 Limits at Infinit 55. Intercept: 0, 0 Smmetr: -ais Horizontal asmptote: Vertical asmptote: ± 57. Domain: > 0 Intercepts: none Smmetr: -ais Horizontal asmptote: 0 since Discontinuit: 0 (Vertical asmptote) lim 5 0 lim. 59. Intercept: 0, 0. Intercepts: ±, 0 Smmetr: none Smmetr: -ais Horizontal asmptote: since Horizontal asmptote: since lim Discontinuit: (Vertical asmptote) lim. lim lim. Discontinuit: 0 (Vertical asmptote) 5 5. Intercept: 0, 0 Smmetr: none Horizontal asmptote: Vertical asmptote: 0

33 Chapter Applications of Differentiation 5. Domain:,,, Intercepts: none Smmetr: origin Horizontal asmptote: none Vertical asmptotes: ± (discontinuities) f 5 5 Domain:, 0, 0, f No relative etrema f No points of inflection Vertical asmptote: 0 Horizontal asmptote: 5 7 = 5 = 0 9. f f 5 0 for an in the domain of f. f 0 when 0. = = Since f > 0 on, 0 and f < 0 on 0,, then 0, 0 is a point of inflection. Vertical asmptotes: ± Horizontal asmptote: 0 7. f f = 0 f when. Since f > 0 on, and f < 0 on,, then, 0 is a point of inflection. = = Vertical asmptote:, Horizontal asmptote: 0

34 Section.5 Limits at Infinit 5 7. f f No relative etrema = = f 5 0 when 0. Point of inflection: 0, 0 Horizontal asmptotes: ± No vertical asmptotes 75. g sin, < < g cos. π (, π ) = sin() Horizontal asmptote: Relative maimum: No vertical asmptotes f, g (a) (c) 70 f = g (b) f g The graph appears as the slant asmptote. 79. C C C C lim

35 Chapter Applications of Differentiation. line: m 0. (a) T t 0.00t 0.77t.5 (b) 90 5 T = m + (, ) (c) 90 (a) (b) d A B C A B m m 7 m m 0 T T t 5 t (d) T 0. T (c) lim dm lim dm m m The line approaches the vertical line 0. Hence, the distance approaches. (e) lim T t (f) The limiting temperature is. T has no horizontal asmptote. 5. Answers will var. See page False. Let f (See Eercise.). Section. A Summar of Curve Sketching. f has constant negative slope. Matches (D). The slope is periodic, and zero at 0. Matches (A) 5. (a) f 0(c) for and is increasing on 0,. f > 0 f is negative for < < (decreasing function). f f is positive for > and < (increasing function). (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.

36 Section. A Summar of Curve Sketching when 0. 0 when ±., (0, 0) ) =, Horizontal asmptote: < < < < < < < < Conclusion Decreasing, concave down 0 Point of inflection Decreasing, concave up 0 0 Relative minimum Increasing, concave up 0 Point of inflection Increasing, concave down 9.. 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 asmptote: ± Horizontal asmptote: 0 Smmetr with respect to the origin 7, 0 0 0, 7 (0, 0)

37 ) Chapter Applications of Differentiation. g g g 0 when 0.9,.05 0 when ± 0,),.. 7, 0) ) 0. 9, 0. )., 577 ) 0. 5,. 7) ) g0.9 < 0, therefore, 0.9,.0 is relative maimum. g.05 > 0, therefore,.05,.7 is a relative minimum. Points of inflection:,.,,.577 Intercepts: 0,,.7, 0 Slant asmptote: 5. f f 0 when ±. (, ) = f 0 (, ) Relative maimum:, = 0 Relative minimum:, Vertical asmptote: 0 Slant asmptote: 7. 0 when,. (0, ) (, ) (, ) 0 < 0 when. Therefore,, is a relative maimum. > 0 when. Therefore,, is a relative minimum. Vertical asmptote: Slant asmptote:

38 Section. A Summar of Curve Sketching 9 9., Domain:, 0 when and undefined when. 0 when and undefined when. (, (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: h when ± ± h 7 0 when 0 9 Relative maimum:, 9 Relative minimum:, 9 Intercepts: 0, 0, ±, 0 Smmetric with respect to the origin Point of inflection: 0, 0 ( ), 9 5 (, 0) (0, 0) (, 0) ( ), when and undefined when 0. < 0 when 0. (, 0 0) (, ) 7 (, 0 ) 5 < < < < < < Conclusion Decreasing, concave down 0 Undefined Undefined Relative minimum Increasing, concave down 0 Relative maimum Decreasing, concave down

39 0 Chapter Applications of Differentiation 5. 0 when 0, 0 when < < 0 Conclusion Increasing, concave down 0 0 Relative maimum 0 < < Decreasing, concave down 0 Point of inflection < < Decreasing, concave up 0 Relative minimum < < Increasing, concave up ( 0.79, 0) (0, ) (, ) (.5, 0) (, (.7, 0) ) 7. 5 No critical numbers 0 when 0. < < < < Conclusion Decreasing, concave up 0 Point of inflection Decreasing, concave down (0, ) (, 0) 9. f 9 (.75, 0) f when ±) (, 7 f 0 when 0 < < < < < < < < f f f Conclusion Increasing, concave down 7 0 Relative maimum Decreasing, concave down 0 Point of inflection Decreasing, concave up 5 0 Relative minimum Increasing, concave up (, 0 ) (.7, 0) (, 5) (0., 0)

40 ( Section. A Summar of Curve Sketching. 0 when 0,. 0 when 0,. < < Conclusion Decreasing, concave up 0 Relative minimum (, 0 ) (0, 0) (, ) (, 7 < < Increasing, concave up 7 0 Point of inflection < < < < Increasing, concave down Point of inflection Increasing, concave up. f f 0 when,. f 0 when 0,. f f f Conclusion < < Decreasing, concave up 0 Relative minimum < < 0 Increasing, concave up Point of inflection 0 < < Increasing, concave down 0 0 Point of inflection < < Increasing, concave up (, ) (0, 0) (, ) (.79, 0) when ±. 0 0 when 0. Conclusion Increasing, concave down 0 Relative maimum < < 0 Decreasing, concave down Point of inflection 0 < < Decreasing, concave up 0 Relative minimum < < Increasing, concave up < < ( 5, 0) (, ) (0, 0) (, ) ( ) 5, 0

41 Chapter Applications of Differentiation 7. 0 undefined at. < < Conclusion Decreasing (0, ), 0 0 Undefined Relative minimum < < Increasing 9. sin sin, 0 cos sin Relative maimum: Relative minimum: Inflection points: cos 0 when,. sin 0 when 0,, 5,, 7,., 9, 9, 9, 5, 9,, 0, 7, 9,, 9 π π π. sec tan 0 when 0. Relative maimum: Relative minimum:, Inflection point: 0, 0 Vertical asmptotes: ± tan, < <. sec 0 when ±,. csc sec, 0 < < sec tan csc cot 0 Relative minimum: Vertical asmptotes:, 0, π π π

42 Section. A Summar of Curve Sketching 5. g tan, < < g g sin cos cos cos sin cos Vertical asmptotes:,,, Intercepts:, 0, 0, 0,, 0 Smmetric with respect to -ais. Increasing on 0, and, Points of inflection: ±.0, 0 0 when 0 7. f vertical asmptote 0 horizontal asmptote Minimum:.0, 9.05 Maimum:.0, 9.05 Points of inflection:., 7.,., 7. π 0 0 π π π f is cubic. f f is quadratic. is linear. f f 0, 0 point of inflection ± horizontal asmptotes f 5. f f (an vertical translate of f will do)

43 Chapter Applications of Differentiation 55. f f (an vertical translate of f will do) 57. Since the slope is negative, the function is decreasing on,, and hence f > f f 5 Vertical asmptote: none Horizontal asmptote: 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.. h. f, Undefined, if if The rational function is not reduced to lowest terms. The graph appears to approach the slant asmptote. hole at, 5. f cos, 0, (a).5 (b) On 0, there seem to be 7 critical numbers: 0.5,.0,.5,.0,.5,.0,.5 f cos cos sin 0 Critical numbers, 0.97,,.9, 5,.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

44 Section.7 Optimization Problems 5 7. Vertical asmptote: 5 9. Vertical asmptote: 5 Horizontal asmptote: 0 Slant asmptote: 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 5: 5. (e) n 0 5 M 0 N (a) (b) When t 0, N0 bacteria. (c) N is a maimum when t 7. (seventh da). (d) Nt 0 for t. (e) lim Nt,50 9. t 7 Section.7 Optimization Problems. (a) First Number, Second Number Product, P CONTINUED

45 Chapter Applications of Differentiation. CONTINUED (b) First Number, Second Number Product, P The maimum is attained near 50 and 0. (c) P 0 0 (d) 500 (55, 05) (e) dp 0 0 when 55. d d P d < The solution appears to be 55. P is a maimum when The two numbers are 55 and 55.. Let and be two positive numbers such that 9. S 9 ds 9 0 when 9. d d S d > 0 when 9. S is a minimum when Let be a positive number. S ds d 0 when. d S d > 0 when. The sum is a minimum when and. 7. Let be the length and the width of the rectangle A 50 da 50 0 when 5. d d A d < 0 when 5. A is maimum when 5 meters. 9. Let be the length and the width of the rectangle. dp d P 0 when. d P 5 > 0 when. d P is minimum when feet.

46 Section.7 Optimization Problems 7. d 0. 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. 7 f 7. f d 7 Since d is smallest when the epression inside the radical is smallest, ou need onl find the critical numbers of f 7. f 0 B the First Derivative Test, the point nearest to, is,. (, ) (, ) d d (, (, 0) ( 5. dq d kq 0 kq 0 k 7. 0,000 (see figure) d Q d kq 0 k kq 0 0 when Q 0. d Q d k < 0 when Q 0. dqd is maimum when Q 0. S 0,000 of fence needed. d S 70,000 d > 0 when 00. where S is the length ds 0,000 0 when 00. d S is a minimum when 00 meters and 00 meters. 9. (a) A area of side area of Top (b) V lengthwidthheight (a) A 50 square inches (a) V 99 cubic inches (b) A square inches (b) V cubic inches (c) A.5 50 square inches (c) V.5 7 cubic inches (c) S V 75 V 75 0 ±5 B the First Derivative Test, 5 ields the maimum volume. Dimensions: (A cube!)

47 Chapter Applications of Differentiation. (a) V s, 0 < < s dv s s d s s 0 when s, s s is not in the domain. d V s d d V d < 0 when s. V s 7 is maimum when 5. (b) If the length is doubled, V 7s. Volume is increased b a factor of. 7s. A da d 0 when d A d. < 0 when The area is maimum when feet and feet. 5. (a) 0 0 L CONTINUED, > (b) (.57,.) 0 L is minimum when.57 and L..

48 ( Section.7 Optimization Problems 9 5. CONTINUED (c) Area A A ± 0, (select ) Then and A. Vertices: 0, 0,, 0, 0, 0 7. A 5 see figure da d 5 5 (, when.5. B the First Derivative Test, the inscribed rectangle of maimum area has vertices ±5, 0, ±5, 5. Width: 5 Length: 5 ; A 0 da d see figure when 0. B the First Derivative Test, the dimensions b are 0 b 0 (approimatel 7.77 b 7.77). These dimensions ield a minimum area V r h cubic inches or h r (a) Radius, r Height Surface Area CONTINUED

49 50 Chapter Applications of Differentiation. CONTINUED (b) (c) S r rh Radius, r 0. Height 0. Surface Area (d) rr h r r r r r (.5,.) (e) 0 The minimum seems to be. for r.5. ds dr r r 0 when r.5 in. h.0 in. r Note: Notice that h r r The minimum seems to be about. for r... 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.

50 ( Section.7 Optimization Problems 5 5. V h r r see figure dv d r r r r rr 0 r r rr 0 (0, r) rr r r r 9 r r 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. 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 V r r h. Let be the length of a side of the square and the length of a side of the triangle. h d S dr r r r r S r rh r r r r r r r r r ds dr r r 0 when r 9. cm. r > 0 when r 9 cm. The surface area is minimum when r 9 cm and h 0. The resulting solid is a sphere of radius r. cm. A is minimum when A d A 9 d da d 0 0 > 0 h r 0 9 and 0 9.

51 5 Chapter Applications of Differentiation. Let S be the strength and k the constant of proportionalit. Given h w, h w, S kwh S kw57 w k57w w ds dw k57 w 0 when w, h. d S kw < 0 when w. dw 5. dr d d R d R v 0 sin g v 0 g cos 0 when v 0 g sin < 0 when B the Second Derivative Test, R is maimum when.,.. These values ield a maimum. 7. sin h s s h, 0 < < sin tan h h tan s tan sec sin di k sin sin cos cos cos d I k sin k sin k s sec sin cos k cos cos sin α h ft s α k cos sin Since 0 when,, or when sin ±. is acute, we have sin h tan feet. Since d Id k sin 9 sin 7 < 0 when sin, this ields a maimum. 9. S, L Time T 0 dt d 0 0 S = + L = + ( ( Q You need to find the roots of this equation in the interval 0,. B using a computer or graphics calculator, ou can determine that this equation has onl one root in this interval. Testing at this value and at the endpoints, ou see that ields the minimum time. Thus, the man should row to a point mile from the nearest point on the coast.

52 Section.7 Optimization Problems 5 5. T v 0 v dt d v v 0 0 Since sin and 0 sin θ θ Q we have sin v sin v 0 sin v sin v. Since d T d v v 0 > 0 this condition ields a minimum time. 5. f sin (a) Distance from origin to -intercept is. Distance from origin to -intercept is.57. (b) d sin π π π (0.797, ) Minimum distance at (c) Let f d sin. f sin cos Setting f 0, ou obtain 0.797, which corresponds to d F cos kw F sin Since F df d k cos sin k tan arctan k 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. k + θ k

53 5 Chapter Applications of Differentiation 57. (a) (b) Base Base Altitude Area cos 0 cos 0 cos 0 cos 0 cos 50 sin 0 sin 0 sin 0 sin 0 sin cos 0 sin 0. Base Base cos 0 cos 0 cos 0 cos 0 cos 50 cos 0 cos 70 cos 0 Altitude sin 0 sin 0 sin 0 sin 0 sin 50 sin 0 sin 70 sin 0 Area 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 C , C Approimation: 0.5 units, or 05 units. S m 5m 0m ds m 5m 5 0m 0 m 0 when m dm. Line: S mi. S m m 5m m 0m m Using a graphing utilit, ou can see that the minimum occurs when 0.. Line: 0. S mi. S (0.,.5) m

54 Section. Newton s Method 55 Section. Newton s Method. f f.7 n n f n f n f n f n n f n f n f sin f cos n n f n f n f n f n n f n f n 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 f f Approimation of the zero of f is.. Similarl, the other zero is approimatel 7.5. f f Approimation of the zero of f is.. n n f n f n f n f n n f n f n n n f n f n f n f n n f n f n f f n n Approimation of the zero of f is CONTINUED f n f n f n f n n f n f n

55 5 Chapter Applications of Differentiation. CONTINUED n n f n f n Approimation of the zero of f is.00. n n f n f n 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 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 h f g h Point of intersection of the graphs of f and g occurs when n n h n h n h n h n n h n h n h f g tan h sec Point of intersection of the graphs of f and g occurs when.9. n n h n h n h n h n n h n h n f a 0. f i i i a i i i a i i a i i a i i i 7 i i i i i i i i

56 Section. Newton s Method f cos f sin Approimation of the zero:. n n f n f n f n f n n f n f n f 9. 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. 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

57 ) 5 Chapter Applications of Differentiation 5. f, f (a) (b) (c) 5 (d) f. f Continuing, the zero is.7. f f.05 f Continuing, the zero is.5. (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 The -intercepts correspond to the values resulting from the first iteration of Newton s Method. 7. f a 0 f n n n a n n n n a n n n a n n a n a n 9. f cos, 0, f sin cos , 0.5) 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 Approimation to the critical number: 0.0

58 Section. Newton s Method 59. f,, 0 d d is minimized when D 7 7 is a minimum. n g D g n g n g n Point closest to, 0 is.99, 0.0. g n g n n g n g n 5 (.99, 0.0) (, 0). Minimize: T T Distance rowed Rate rowed T Distance walked Rate walked Let f 7 and f. Since f 00 and f 5, the solution is in the interval,. n n f n Approimation:.5 miles f n f n f n n f n f n 5.,500, , ,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 The zero occurs when.5 which corresponds to $,5. n f n f n

59 0 Chapter Applications of Differentiation 7. 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. 9. True 5. f f Let. n n f n f n f n f n n f n f n Approimation:.0 Section.9 Differentials. f f f Tangent line at, : f f T f f 5 f Tangent line at, : f f 0 T f sin f cos f sin Tangent line at, sin : f f T cos sin sin cos cos sin 7. f, f,, d 0. f f f. f 0.05 d fd f

60 Section.9 Differentials 9. f, f,, d 0.0 f f f 0.99 f d f d f d d d d d d d cot d cot csc d cot cot d 9. cos d sin d. (a) (b) f.9 f 0. f f0.. (a) f.0 f 0.0 f f0.0 (b) f.9 f 0. f f f.0 f 0.0 f f (a) (b) g.9 g 0.07 g g (a) g. g 0. g g0. (b) g.9 g 0.07 g g g. g 0. g g A. d ± da d A da ± ± square inches A r r r dr ± A da r dr ± ±7 square inches

61 Chapter Applications of Differentiation. (a) 5 centimeter 5. d ±0.05 centimeters r inches r dr ±0.0 inches (b) A da d 5±0.05 Percentage error: da A ±.5 square centimeters da A ± % d d 0.05 d % (a) (b) V r dv r dr ±0.0 ±. cubic inches S r ds r dr ±0.0 ±0.9 square inches (c) Relative error: Relative error: dv V r dr dr r r % ds S r dr dr r r % 7. V r h 0r, r 5 cm, h 0 cm, dr 0. cm V dv 0r dr cm 9. (a) T Lg(b) dt glg dl Relative error: dt T dlglg Lg dl L relative error in L seconds. minutes Percentage error: dt T % % d ±5 ±0.5 (a) h 9.5 csc dh 9.5 csc cot d dh h cot d dh cot h Converting to radians, cot % in radians. θ h b 9.5 (b) dh cot d 0.0 h d d 0.0 cot 0.0 tan 0.0 tan tan % in radians

62 Review Eercises for Chapter. r v 0 sin 5. Let f, 00, d 0.. v 0 00 ftsec changes from 0 to dr 00 cos d 0 d 0 r dr feet 0 cos 0 9 feet 00 f f fd d f Using a calculator: Let f, 5, d. f f f d d f Let f,, d 0.0, f. Then f.0 f fd Using a calculator, In general, when 0, d approaches. 5. True 55. True Review Eercises for Chapter. A number c in the domain of f is a critical number if fc 0 or is undefined at c. f f ( c) is undefined. f ( c) = 0. g 5 cos, 0, g 5 sin (., 7.57) 0 when sin 5. (.7, 0.) Critical numbers: 0.,.7 Left endpoint: 0, 5 Critical number: 0., 5. Critical number:.7, 0. Minimum Right endpoint:, 7.57 Maimum

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

64 Review Eercises for Chapter 5 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. 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 5 5. (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 < < < < 5. g g g Critical numbers: 0, ± (, ) (0, 0) (, ) g0 > 0 g ± < 0 Relative minimum at 0, 0 Relative maimums at ±, 7. 7 (5, f(5)) 5 (, f()) (, 0) (0, 0) The first derivative is positive and the second derivative is negative. The graph is increasing and is concave down.

65 Chapter Applications of Differentiation. (a) (b) D 0.00t 0.5t.9t 0.t (c) Maimum at.9, Minimum at., (d) Outlas increasing at greatest rate at the point of inflection 9., lim lim cos 5. lim 0, since 5 cos h 9. f Discontinuit: Discontinuit: 0 lim lim lim Vertical asmptote: Horizontal asmptote: Vertical asmptote: 0 Horizontal asmptote:. f. f Relative minimum:, 0 Relative maimum:, 0 Relative minimum: 0.55,.077 Relative maimum:.55, Vertical asmptote: 0. Horizontal asmptote: 0 5. f Domain:, ; Range:, 5, ) ) f 0 when. f Therefore,, is a relative maimum. Intercepts: 0, 0,, 0 5

66 Review Eercises for Chapter 7 7. f, Domain:,, Range:, 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 9. f Domain:, ; Range:, f 5 0 when, 5,. f 5 0 when, ±. 5 f > 0 Therefore,, 0 is a relative minimum. f 5 < 0 Therefore, 5, 5 5 is a relative maimum. Points of inflection:, 0, 5 Intercepts: 0, 9,, 0,, 0, 0.0, 5, 0. (, 0),. 5 (.9, 0.) (, 0) (.7, 0.0) ( ( 5. f Domain:, ; Range:, f f 0 when and undefined when, 0. is undefined when 0,. 5 B the First Derivative Test, 0 is a relative maimum and is a relative minimum. 0, 0 is a point of inflection., 5, 0) ) ),.59) 0, 0) ) Intercepts:, 0, 0, 0

67 Chapter Applications of Differentiation 5. f Domain:,,, ; Range:,,, f < 0 if. f Horizontal asmptote: Vertical asmptote: Intercepts:, 0, 0, 55. f Domain:, ; Range: 0, f 0 when 0. f 0 when ±. (, ( 5 (0, ), ( ( f0 < 0 Therefore, 0, is a relative maimum. Points of inflection: ±, Intercept: 0, Smmetric to the -ais Horizontal asmptote: f 0 Domain:, 0, 0, ; Range:,,, 5 (, ) f 0 when ±. (, ) 5 0 f 0 f < 0 Therefore,, is a relative maimum. f > 0 Therefore,, is a relative minimum. Vertical asmptote: 0 Smmetric with respect to origin

68 ) Review Eercises for Chapter f 9 Domain:, ; Range: 0, f 9 9 f 9 is undefined at ±. 9 f0 < 0 Therefore, 0, 9 is a relative maimum. 0 when 0 and is undefined when ±. 0 5, 0) 0, 9) ) ), 0) Relative minima: ±, 0 Points of inflection: ±, 0 Intercepts: ±, 0, 0, 9 Smmetric to the -ais. f cos, ) Domain: 0, ; Range:, f sin 0, f is increasing. f cos 0 when,. Points of inflection: Intercept: 0,,,,, ) 0, ), ). 0 (a) The graph is an ellipse: Maimum:, Minimum:, (b) 0 d d d d 0 d d d d (, ) (, ) The critical numbers are and. These correspond to the points,,,,,, and,. Hence, the maimum is, and the minimum is,.

69 70 Chapter Applications of Differentiation 5. Let t 0 at noon. L d 00 t 0t 0,000 00t t dl t 0 when t.9 hr. dt Ship A at 0.9, 0; Ship B at 0, 9. d 0,000 00t t 09. when t.9 :55 P.M.. d km (0, 0) B d (0, 0 t) (00 t, 0) A (00, 0) 7. We have points 0,,, 0, and,. Thus, Let m 0 f L. 0 or f 0. 0 when 0, 5 minimum. Vertices of triangle: 0, 0, 5, 0, 0, (0, ) (, ) (, 0) 0 9. A Average of basesheight s s s see figure da d s s s s s s s s 0 when s. s s A is a maimum when s. s s s+ s s s s 7. You can form a right triangle with vertices 0, 0,, 0 and 0,. Assume that the hpotenuse of length L passes through,. Let m 0 0 or f L. f 7 0 when 0 or. L.05 feet 0

70 Review Eercises for Chapter 7 7. csc csc L L L L csc 9 csc dl csc cot 9 sec tan 0 d L or L csc see figure 9 or L 9 csc L π θ ( θ 9 csc 9 sec L θ ( tan tan sec csc tan sec tan L 9 Compare to Eercise 7 using a 9 and b. ft.07 ft 75. Total cost Cost per hournumber of hours T v v 550 v 0 v dt 550 dv 0 v v,000 0v 0 when v mph. d T 00 so this value ields a minimum. dv v > 0 when v f From the graph ou can see that f has three real zeros. f n n f n f n f n f n n f n f n n n f n f n f n f n n f n f n n n f n f n f n f n n f n f n The three real zeros of f are.5, 0.7, and.79.