CHAPTER 3 Applications of Differentiation


 Edith Snow
 1 years ago
 Views:
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 OddNumbered 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 0amp 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 nonincreasing 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 crosssectional 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.
CHAPTER 11 VectorValued Functions
CHAPTER VectorValued Functions Section. VectorValued Functions...................... 9 Section. Differentiation and Integration of VectorValued Functions.... Section. Velocit and Acceleration.....................
More informationCHAPTER 5 Logarithmic, Exponential, and Other Transcendental Functions
CHAPTER 5 Logarithmic, Eponential, and Other Transcendental Functions Section 5. The Natural Logarithmic Function: Differentiation.... 9 Section 5. The Natural Logarithmic Function: Integration...... 98
More informationSHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.
Chapter Practice Dicsclaimer: The actual eam is different. On the actual eam ou must show the correct reasoning to receive credit for the question. SHORT ANSWER. Write the word or phrase that best completes
More informationInfinite Limits. Let f be the function given by. f x 3 x 2.
0_005.qd //0 :07 PM Page 8 SECTION.5 Infinite Limits 8, as Section.5, as + f() = f increases and decreases without bound as approaches. Figure.9 Infinite Limits Determine infinite its from the left and
More informationOne of the most common applications of Calculus involves determining maximum or minimum values.
8 LESSON 5 MAX/MIN APPLICATIONS (OPTIMIZATION) One of the most common applications of Calculus involves determining maimum or minimum values. Procedure:. Choose variables and/or draw a labeled figure..
More informationQuick Review 4.1 (For help, go to Sections 1.2, 2.1, 3.5, and 3.6.)
Section 4. Etreme Values of Functions 93 EXPLORATION Finding Etreme Values Let f,.. Determine graphicall the etreme values of f and where the occur. Find f at these values of.. Graph f and f or NDER f,,
More informationCHAPTER P Preparation for Calculus
CHAPTER P Preparation for Calculus Section P. Graphs and Models...................... Section P. Linear Models and Rates of Change............ Section P. Functions and Their Graphs................. Section
More informationFind the volume of the solid generated by revolving the shaded region about the given axis. Use the disc/washer method 1) About the xaxis
Final eam practice for Math 6 Disclaimer: The actual eam is different Find the volume of the solid generated b revolving the shaded region about the given ais. Use the disc/washer method ) About the ais
More informationTangent Line Approximations. y f c f c x c. y f c f c x c. Find the tangent line approximation of. f x 1 sin x
SECTION 9 Differentials 5 Section 9 EXPLORATION Tangent Line Approimation Use a graphing utilit to graph f In the same viewing window, graph the tangent line to the graph of f at the point, Zoom in twice
More informationAnswer Key 1973 BC 1969 BC 24. A 14. A 24. C 25. A 26. C 27. C 28. D 29. C 30. D 31. C 13. C 12. D 12. E 3. A 32. B 27. E 34. C 14. D 25. B 26.
Answer Key 969 BC 97 BC. C. E. B. D 5. E 6. B 7. D 8. C 9. D. A. B. E. C. D 5. B 6. B 7. B 8. E 9. C. A. B. E. D. C 5. A 6. C 7. C 8. D 9. C. D. C. B. A. D 5. A 6. B 7. D 8. A 9. D. E. D. B. E. E 5. E.
More informationMath 121. Practice Questions Chapters 2 and 3 Fall Find the other endpoint of the line segment that has the given endpoint and midpoint.
Math 11. Practice Questions Chapters and 3 Fall 01 1. Find the other endpoint of the line segment that has the given endpoint and midpoint. Endpoint ( 7, ), Midpoint (, ). Solution: Let (, ) denote the
More informationCHAPTER 1 Functions and Their Graphs
PART I CHAPTER Functions and Their Graphs Section. Lines in the Plane....................... Section. Functions........................... Section. Graphs of Functions..................... Section. Shifting,
More informationCHAPTER 2: Partial Derivatives. 2.2 Increments and Differential
CHAPTER : Partial Derivatives.1 Definition of a Partial Derivative. Increments and Differential.3 Chain Rules.4 Local Etrema.5 Absolute Etrema 1 Chapter : Partial Derivatives.1 Definition of a Partial
More information1969 AP Calculus BC: Section I
969 AP Calculus BC: Section I 9 Minutes No Calculator Note: In this eamination, ln denotes the natural logarithm of (that is, logarithm to the base e).. t The asymptotes of the graph of the parametric
More informationCalculus  Chapter 2 Solutions
Calculus  Chapter Solutions. a. See graph at right. b. The velocity is decreasing over the entire interval. It is changing fastest at the beginning and slowest at the end. c. A = (95 + 85)(5) = 450 feet
More information1. The cost (in dollars) of producing x units of a certain commodity is C(x) = x x 2.
APPM 1350 Review #2 Summer 2014 1. The cost (in dollars) of producing units of a certain commodity is C() 5000 + 10 + 0.05 2. (a) Find the average rate of change of C with respect to when the production
More informationFunctions. Introduction CHAPTER OUTLINE
Functions,00 P,000 00 0 970 97 980 98 990 99 000 00 00 Figure Standard and Poor s Inde with dividends reinvested (credit "bull": modification of work b Praitno Hadinata; credit "graph": modification of
More informationFall 2009 Instructor: Dr. Firoz Updated on October 4, 2009 MAT 265
Fall 9 Instructor: Dr. Firoz Updated on October 4, 9 MAT 65 Basic Algebra and Trigonometr Review Algebra Factoring quadratic epressions Eamples Case : Leading coefficient. Factor 5 6 ( )( ), as ,  are
More informationSolutions to the Exercises of Chapter 8
8A Domains of Functions Solutions to the Eercises of Chapter 8 1 For 7 to make sense, we need 7 0or7 So the domain of f() is{ 7} For + 5 to make sense, +5 0 So the domain of g() is{ 5} For h() to make
More informationNote: Unless otherwise specified, the domain of a function f is assumed to be the set of all real numbers x for which f (x) is a real number.
997 AP Calculus BC: Section I, Part A 5 Minutes No Calculator Note: Unless otherwise specified, the domain of a function f is assumed to be the set of all real numbers for which f () is a real number..
More informationReview sheet Final Exam Math 140 Calculus I Fall 2015 UMass Boston
Review sheet Final Eam Math Calculus I Fall 5 UMass Boston The eam is closed tetbook NO CALCULATORS OR ELECTRONIC DEVICES ARE ALLOWED DURING THE EXAM The final eam will contain problems of types similar
More informationAnswer Explanations. The SAT Subject Tests. Mathematics Level 1 & 2 TO PRACTICE QUESTIONS FROM THE SAT SUBJECT TESTS STUDENT GUIDE
The SAT Subject Tests Answer Eplanations TO PRACTICE QUESTIONS FROM THE SAT SUBJECT TESTS STUDENT GUIDE Mathematics Level & Visit sat.org/stpractice to get more practice and stud tips for the Subject Test
More informationProperties of Limits
33460_003qd //04 :3 PM Page 59 SECTION 3 Evaluating Limits Analticall 59 Section 3 Evaluating Limits Analticall Evaluate a it using properties of its Develop and use a strateg for finding its Evaluate
More informationCalculus 1: A Large and In Charge Review Solutions
Calculus : A Large and n Charge Review Solutions use the symbol which is shorthand for the phrase there eists.. We use the formula that Average Rate of Change is given by f(b) f(a) b a (a) (b) 5 = 3 =
More informationMath 2413 General Review for Calculus Last Updated 02/23/2016
Math 243 General Review for Calculus Last Updated 02/23/206 Find the average velocity of the function over the given interval.. y = 6x 35x 28, [8, ] Find the slope of the curve for the given value of
More information1 (C) 1 e. Q.3 The angle between the tangent lines to the graph of the function f (x) = ( 2t 5)dt at the points where (C) (A) 0 (B) 1/2 (C) 1 (D) 3
[STRAIGHT OBJECTIVE TYPE] Q. Point 'A' lies on the curve y e and has the coordinate (, ) where > 0. Point B has the coordinates (, 0). If 'O' is the origin then the maimum area of the triangle AOB is (A)
More informationCHAPTER 1 Functions, Graphs, and Limits
CHAPTER Functions, Graphs, and Limits Section. The Cartesian Plane and the Distance Formula... Section. Graphs of Equations...8 Section. Lines in the Plane and Slope... MidChapter Quiz Solutions... Section.
More informationRolle s Theorem and the Mean Value Theorem. Rolle s Theorem
0_00qd //0 0:50 AM Page 7 7 CHAPTER Applications o Dierentiation Section ROLLE S THEOREM French mathematician Michel Rolle irst published the theorem that bears his name in 9 Beore this time, however,
More informationName Please print your name as it appears on the class roster.
Berkele Cit College Practice Problems Math 1 Precalculus  Final Eam Preparation Name Please print our name as it appears on the class roster. SHORT ANSWER. Write the word or phrase that best completes
More informationMATH 115: Final Exam Review. Can you find the distance between two points and the midpoint of a line segment? (1.1)
MATH : Final Eam Review Can ou find the distance between two points and the midpoint of a line segment? (.) () Consider the points A (,) and ( 6, ) B. (a) Find the distance between A and B. (b) Find the
More informationSHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.
Practice for the Final Eam MAC 1 Sullivan Version 1 (2007) SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question. Find the distance d(p1, P2) between the points
More informationLaw of Sines, Law of Cosines, Heron s Formula:
PreAP Math Analsis nd Semester Review Law of Sines, Law of Cosines, Heron s Formula:. Determine how man solutions the triangle has and eplain our reasoning. (FIND YOUR FLOW CHART) a. A = 4, a = 4 ards,
More information1.3 LIMITS AT INFINITY; END BEHAVIOR OF A FUNCTION
. Limits at Infinit; End Behavior of a Function 89. LIMITS AT INFINITY; END BEHAVIOR OF A FUNCTION Up to now we have been concerned with its that describe the behavior of a function f) as approaches some
More information6 = 1 2. The right endpoints of the subintervals are then 2 5, 3, 7 2, 4, 2 9, 5, while the left endpoints are 2, 5 2, 3, 7 2, 4, 9 2.
5 THE ITEGRAL 5. Approimating and Computing Area Preliminar Questions. What are the right and left endpoints if [, 5] is divided into si subintervals? If the interval [, 5] is divided into si subintervals,
More informationMath 125 Practice Problems for Test #3
Math Practice Problems for Test # Also stud the assigned homework problems from the book. Donʹt forget to look over Test # and Test #! Find the derivative of the function. ) Know the derivatives of all
More informationApplications of Derivatives
58_Ch04_pp8660.qd /3/06 :35 PM Page 86 Chapter 4 Applications of Derivatives A n automobile s gas mileage is a function of man variables, including road surface, tire tpe, velocit, fuel octane rating,
More information( 1 3, 3 4 ) Selected Answers SECTION P.2. Quick Review P.1. Exercises P.1. Quick Review P.2. Exercises P.2 SELECTED ANSWERS 895
44_Demana_SEANS_pp8904 /4/06 8:00 AM Page 89 SELECTED ANSWERS 89 Selected Answers Quick Review P.. {,,, 4,, 6}. {,, }. (a) 87.7 (b) 4.7 7. ;.7 9. 0,,,, 4,, 6 Eercises P.. 4.6 (terminating)..6 (repeating).
More information= x. Algebra II Notes Quadratic Functions Unit Graphing Quadratic Functions. Math Background
Algebra II Notes Quadratic Functions Unit 3.1 3. Graphing Quadratic Functions Math Background Previousl, ou Identified and graphed linear functions Applied transformations to parent functions Graphed quadratic
More informationMATH 150/GRACEY EXAM 2 PRACTICE/CHAPTER 2. MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
MATH 0/GRACEY EXAM PRACTICE/CHAPTER Name MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Find the indicated derivative. ) Find if = 8 sin. A) = 8
More information4.3 Exercises. local maximum or minimum. The second derivative is. e 1 x 2x 1. f x x 2 e 1 x 1 x 2 e 1 x 2x x 4
SECTION 4.3 HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH 297 local maimum or minimum. The second derivative is f 2 e 2 e 2 4 e 2 4 Since e and 4, we have f when and when 2 f. So the curve is concave downward
More information1. sin 2. csc 2 3. tan 1 2. Cos 8) Sin 10. sec. Honors PreCalculus Final Exam Review 2 nd semester. TRIGONOMETRY Solve for 0 2
Honors PreCalculus Final Eam Review nd semester Name: TRIGONOMETRY Solve for 0 without using a calculator: 1 1. sin. csc 3. tan 1 4. cos 1) ) 3) 4) Solve for in degrees giving all solutions. 5. sin 1
More informationSection 2.5: Graphs of Functions
Section.5: Graphs of Functions Objectives Upon completion of this lesson, ou will be able to: Sketch the graph of a piecewise function containing an of the librar functions. o Polnomial functions of degree
More informationMATH 122A FINAL EXAM STUDY GUIDE (Spring 2014)
MATH A FINAL EXAM STUDY GUIDE (Spring 0) The final eam for spring 0 will have a multiple choice format. This will allow us to offer the final eam as late in the course as possible, giving more inclass
More informationCHAPTER 3 Graphs and Functions
CHAPTER Graphs and Functions Section. The Rectangular Coordinate Sstem............ Section. Graphs of Equations..................... 7 Section. Slope and Graphs of Linear Equations........... 7 Section.
More information9.1 Practice A. Name Date sin θ = and cot θ = to sketch and label the triangle. Then evaluate. the other four trigonometric functions of θ.
.1 Practice A In Eercises 1 and, evaluate the si trigonometric functions of the angle. 1.. 8 1. Let be an acute angle of a right triangle. Use the two trigonometric functions 10 sin = and cot = to sketch
More informationUniversity of Toronto Mississauga
Surname: First Name: Student Number: Tutorial: Universit of Toronto Mississauga Mathematical and Computational Sciences MATY5Y Term Test Duration  0 minutes No Aids Permitted This eam contains pages (including
More informationMULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Eam Name MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Determine from the graph whether the function has an absolute etreme values on the interval
More informationMat 267 Engineering Calculus III Updated on 9/19/2010
Chapter 11 Partial Derivatives Section 11.1 Functions o Several Variables Deinition: A unction o two variables is a rule that assigns to each ordered pair o real numbers (, ) in a set D a unique real number
More informationSyllabus Objective: 2.9 The student will sketch the graph of a polynomial, radical, or rational function.
Precalculus Notes: Unit Polynomial Functions Syllabus Objective:.9 The student will sketch the graph o a polynomial, radical, or rational unction. Polynomial Function: a unction that can be written in
More informationAdditional Topics in Differential Equations
0537_cop6.qd 0/8/08 :6 PM Page 3 6 Additional Topics in Differential Equations In Chapter 6, ou studied differential equations. In this chapter, ou will learn additional techniques for solving differential
More informationK. Function Analysis. ). This is commonly called the first derivative test. f ( x) is concave down for values of k such that f " ( k) < 0.
K. Function Analysis What you are finding: You have a function f ( ). You want to find intervals where f ( ) is increasing and decreasing, concave up and concave down. You also want to find values of where
More informationReview Problems for Test 2
Review Problems for Test 2 Math 603/06 0 0/ 2007 These problems are provided to help you study. The fact that a problem occurs here does not mean that there will be a similar problem on the test. And
More informationFunctions of Several Variables
Chapter 1 Functions of Several Variables 1.1 Introduction A real valued function of n variables is a function f : R, where the domain is a subset of R n. So: for each ( 1,,..., n ) in, the value of f is
More informationMath 2250 Final Exam Practice Problem Solutions. f(x) = ln x x. 1 x. lim. lim. x x = lim. = lim 2
Math 5 Final Eam Practice Problem Solutions. What are the domain and range of the function f() = ln? Answer: is only defined for, and ln is only defined for >. Hence, the domain of the function is >. Notice
More informationKEY IDEAS. Chapter 1 Function Transformations. 1.1 Horizontal and Vertical Translations PreCalculus 12 Student Workbook MHR 1
Chapter Function Transformations. Horizontal and Vertical Translations A translation can move the graph of a function up or down (vertical translation) and right or left (horizontal translation). A translation
More informationReview Sheet for Second Midterm Mathematics 1300, Calculus 1
Review Sheet for Second Midterm Mathematics 300, Calculus. For what values of is the graph of y = 5 5 both increasing and concave up? >. 2. Where does the tangent line to y = 2 through (0, ) intersect
More information1.6 CONTINUITY OF TRIGONOMETRIC, EXPONENTIAL, AND INVERSE FUNCTIONS
.6 Continuit of Trigonometric, Eponential, and Inverse Functions.6 CONTINUITY OF TRIGONOMETRIC, EXPONENTIAL, AND INVERSE FUNCTIONS In this section we will discuss the continuit properties of trigonometric
More informationPreCalculus Mathematics Limit Process Calculus
NOTES : LIMITS AND DERIVATIVES Name: Date: Period: Mrs. Nguyen s Initial: LESSON.1 THE TANGENT AND VELOCITY PROBLEMS PreCalculus Mathematics Limit Process Calculus The type of it that is used to find
More informationReview of Prerequisite Skills, p. 350 C( 2, 0, 1) B( 3, 2, 0) y A(0, 1, 0) D(0, 2, 3) j! k! 2k! Section 7.1, pp
. 5. a. a a b a a b. Case If and are collinear, then b is also collinear with both and. But is perpendicular to and c c c b 9 b c, so a a b b is perpendicular to. Case If b and c b c are not collinear,
More information9 (0, 3) and solve equations to earn full credit.
Math 0 Intermediate Algebra II Final Eam Review Page of Instructions: (6, ) Use our own paper for the review questions. For the final eam, show all work on the eam. (6, ) This is an algebra class do not
More informationTO THE STUDENT: To best prepare for Test 4, do all the problems on separate paper. The answers are given at the end of the review sheet.
MATH TEST 4 REVIEW TO THE STUDENT: To best prepare for Test 4, do all the problems on separate paper. The answers are given at the end of the review sheet. PART NONCALCULATOR DIRECTIONS: The problems
More informationWorksheets for MA 113
Worksheet # : Review Worksheets for MA 3 http://www.math.uky.edu/~ma3/s.3/packet.pdf Worksheet # 2: Functions, Logarithms, and Intro to Limits Worksheet # 3: Limits: A Numerical and Graphical Approach
More informationSection 1.1: A Preview of Calculus When you finish your homework, you should be able to
Section 1.1: A Preview of Calculus When you finish your homework, you should be able to π Understand what calculus is and how it compares with precalculus π Understand that the tangent line problem is
More informationProperties of Derivatives
6 CHAPTER Properties of Derivatives To investigate derivatives using first principles, we will look at the slope of f ( ) = at the point P (,9 ). Let Q1, Q, Q, Q4, be a sequence of points on the curve
More informationLesson 9.1 Using the Distance Formula
Lesson. Using the Distance Formula. Find the eact distance between each pair of points. a. (0, 0) and (, ) b. (0, 0) and (7, ) c. (, 8) and (, ) d. (, ) and (, 7) e. (, 7) and (8, ) f. (8, ) and (, 0)
More informationThird Annual NCMATYC Math Competition November 16, Calculus Test
Third Annual NCMATYC Math Competition November 6, 0 Calculus Test Please do NOT open this booklet until given the signal to begin. You have 90 minutes to complete this 0question multiple choice calculus
More informationMath Honors Calculus I Final Examination, Fall Semester, 2013
Math 2  Honors Calculus I Final Eamination, Fall Semester, 2 Time Allowed: 2.5 Hours Total Marks:. (2 Marks) Find the following: ( (a) 2 ) sin 2. (b) + (ln 2)/(+ln ). (c) The 2th Taylor polynomial centered
More informationCALCULUS I. Practice Problems. Paul Dawkins
CALCULUS I Practice Problems Paul Dawkins Table of Contents Preface... iii Outline... iii Review... Introduction... Review : Functions... Review : Inverse Functions... 6 Review : Trig Functions... 6 Review
More information24. ; Graph the function and observe where it is discontinuous. 2xy f x, y x 2 y 3 x 3 y xy 35. 6x 3 y 2x 4 y 4 36.
SECTION. LIMITS AND CONTINUITY 877. EXERCISES. Suppose that, l 3, f, 6. What can ou sa about the value of f 3,? What if f is continuous?. Eplain wh each function is continuous or discontinuous. (a) The
More informationFinding Limits Graphically and Numerically. An Introduction to Limits
60_00.qd //0 :05 PM Page 8 8 CHAPTER Limits and Their Properties Section. Finding Limits Graphicall and Numericall Estimate a it using a numerical or graphical approach. Learn different was that a it can
More informationSEE and DISCUSS the pictures on pages in your text. Key picture:
Math 6 Notes 1.1 A PREVIEW OF CALCULUS There are main problems in calculus: 1. Finding a tangent line to a curve though a point on the curve.. Finding the area under a curve on some interval. SEE and DISCUSS
More informationReview Sheet for Exam 1 SOLUTIONS
Math b Review Sheet for Eam SOLUTIONS The first Math b midterm will be Tuesday, February 8th, 7 9 p.m. Location: Schwartz Auditorium Room ) The eam will cover: Section 3.6: Inverse Trig Appendi F: Sigma
More informationTangent Line Approximations
60_009.qd //0 :8 PM Page SECTION.9 Dierentials Section.9 EXPLORATION Tangent Line Approimation Use a graphing utilit to graph. In the same viewing window, graph the tangent line to the graph o at the point,.
More informationAdditional Topics in Differential Equations
6 Additional Topics in Differential Equations 6. Eact FirstOrder Equations 6. SecondOrder Homogeneous Linear Equations 6.3 SecondOrder Nonhomogeneous Linear Equations 6.4 Series Solutions of Differential
More informationRolle s Theorem, the Mean Value Theorem, and L Hôpital s Rule
Rolle s Theorem, the Mean Value Theorem, and L Hôpital s Rule 5. Rolle s Theorem In the following problems (a) Verify that the three conditions of Rolle s theorem have been met. (b) Find all values z that
More informationFitting Integrands to Basic Rules
6_8.qd // : PM Page 8 8 CHAPTER 8 Integration Techniques, L Hôpital s Rule, and Improper Integrals Section 8. Basic Integration Rules Review proceres for fitting an integrand to one of the basic integration
More informationFind the distance between the pair of points. 2) (7, 7) and (3, 5) A) 12 3 units B) 2 5 units C) 6 units D) 12 units B) 8 C) 63 2
Sample Departmental Final  Math 9 Write the first five terms of the sequence whose general term is given. 1) a n = n 2  n 0, 2,, 12, 20 B) 2,, 12, 20, 30 C) 0, 3, 8, 1, 2 D) 1,, 9, 1, 2 Find the distance
More informationDirectional derivatives and gradient vectors (Sect. 14.5). Directional derivative of functions of two variables.
Directional derivatives and gradient vectors (Sect. 14.5). Directional derivative of functions of two variables. Partial derivatives and directional derivatives. Directional derivative of functions of
More informationAP Calculus BC Summer Review
AP Calculus BC 0708 Summer Review Due September, 07 Name: All students entering AP Calculus BC are epected to be proficient in PreCalculus skills. To enhance your chances for success in this class, it
More information( ) 2 3x=0 3x(x 3 1)=0 x=0 x=1
Stewart Calculus ET 5e 05497;4. Partial Derivatives; 4.7 Maimum and Minimum Values. (a) First we compute D(,)= f (,) f (,) [ f (,)] =(4)() () =7. Since D(,)>0 and f (,)>0, f has a local minimum at (,)
More informationGraphing Calculator Computations 2
Graphing Calculator Computations A) Write the graphing calculator notation and B) Evaluate each epression. 4 1) 15 43 8 e) 15  4 * 3^ + 8 ^ 4/  1) ) 5 ) 8 3 3) 3 4 1 8 3) 7 9 4) 1 3 5 4) 5) 5 5) 6)
More informationCHAPTER 3 Derivatives and Their Applications
CHAPTER Derivatives and Their Applications Review of Prerequisite Skills, pp. 6 7. a. b. c. d. 0 6 6 0 6 6 6 0 6 6 6 0 6 6 6 6 Calculus and Vectors Solutions Manual e. f. 8 6 0 8 8 6 0 8. a. ( ) ( ) 6
More informationC) x m A) 260 sq. m B) 26 sq. m C) 40 sq. m D) 364 sq. m. 7) x x  (6x + 24) = 4 A) 0 B) all real numbers C) 4 D) no solution
Sample Departmental Final  Math 46 Perform the indicated operation. Simplif if possible. 1) 7   22 + 3 2  A) +  2 B)  + 42 C) + 42 D)  +  2 Solve the problem. 2) The sum of a number and its
More information90 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions. Name Class. (a) (b) ln x (c) (a) (b) (c) 1 x. y e (a) 0 (b) y.
90 Chapter 5 Logarithmic, Eponential, and Other Transcendental Functions Test Form A Chapter 5 Name Class Date Section. Find the derivative: f ln. 6. Differentiate: y. ln y y y y. Find dy d if ey y. y
More informationDifferential Calculus I   : Fundamentals
Differential Calculus I   : Fundamentals Assessment statements 6. Informal ideas of limits and convergence. Limit notation. Definition of derivative from first principles f9() 5 lim f ( h) f () h ( h
More informationMath 2413 Final Exam Review 1. Evaluate, giving exact values when possible.
Math 4 Final Eam Review. Evaluate, giving eact values when possible. sin cos cos sin y. Evaluate the epression. loglog 5 5ln e. Solve for. 4 6 e 4. Use the given graph of f to answer the following: y f
More informationPRECALCULUS FINAL EXAM REVIEW
PRECALCULUS FINAL EXAM REVIEW Evaluate the function at the indicated value of. Round our result to three decimal places.. f () 4(5 ); 0.8. f () e ; 0.78 Use the graph of f to describe the transformation
More informationAP Calculus FreeResponse Questions 1969present AB
AP Calculus FreeResponse Questions 1969present AB 1969 1. Consider the following functions defined for all x: f 1 (x) = x, f (x) = xcos x, f 3 (x) = 3e x, f 4 (x) = x  x. Answer the following questions
More informationFitting Integrands to Basic Rules. x x 2 9 dx. Solution a. Use the Arctangent Rule and let u x and a dx arctan x 3 C. 2 du u.
58 CHAPTER 8 Integration Techniques, L Hôpital s Rule, and Improper Integrals Section 8 Basic Integration Rules Review proceres for fitting an integrand to one of the basic integration rules Fitting Integrands
More informationSTANDARD FORM is a QUADRATIC FUNCTION and its graph is a PARABOLA. The domain of a quadratic function is the set of all real numbers.
EXERCISE 23 Things to remember: 1. QUADRATIC FUNCTION If a, b, and c are real numbers with a 0, then the function f() = a 2 + b + c STANDARD FORM is a QUADRATIC FUNCTION and its graph is a PARABOLA. The
More information3.1 ANALYSIS OF FUNCTIONS I INCREASE, DECREASE, AND CONCAVITY
MATH00 (Calculus).1 ANALYSIS OF FUNCTIONS I INCREASE, DECREASE, AND CONCAVITY Name Group No. KEYWORD: increasing, decreasing, constant, concave up, concave down, and inflection point Eample 1. Match the
More informationSkills Practice Skills Practice for Lesson 1.1
Skills Practice Skills Practice for Lesson. Name Date Lots and Projectiles Introduction to Quadratic Functions Vocabular Give an eample of each term.. quadratic function 9 0. vertical motion equation s
More information2. Use your graphing calculator to graph each of the functions below over the interval 2, 2
MSLC Math 0 Final Eam Review Disclaimer: This should NOT be used as your only guide for what to study. if 0. Use the piecewise defined function f( ) to answer the following: if a. Compute f(0), f(), f(),
More informationCHAPTER 3: DERIVATIVES
(Exercises for Section 3.1: Derivatives, Tangent Lines, and Rates of Change) E.3.1 CHAPTER 3: DERIVATIVES SECTION 3.1: DERIVATIVES, TANGENT LINES, and RATES OF CHANGE In these Exercises, use a version
More informationPreCalculus Final Exam Review Revised Spring 2014
PreCalculus Final Eam Review Revised Spring 0. f() is a function that generates the ordered pairs (0,0), (,) and (,). a. If f () is an odd function, what are the coordinates of two other points found
More informationFind the slope of the curve at the given point P and an equation of the tangent line at P. 1) y = x2 + 11x  15, P(1, 3)
Final Exam Review AP Calculus AB Find the slope of the curve at the given point P and an equation of the tangent line at P. 1) y = x2 + 11x  15, P(1, 3) Use the graph to evaluate the limit. 2) lim x
More informationCHAPTER 4 Integration
CHAPTER Itegratio Sectio. Atierivatives a Iefiite Itegratio......... Sectio. Area............................. Sectio. Riema Sums a Defiite Itegrals........... Sectio. The Fuametal Theorem of Calculus..........
More information2.1 Intercepts; Symmetry; Graphing Key Equations
Ch. Graphs.1 Intercepts; Smmetr; Graphing Ke Equations 1 Find Intercepts from an Equation MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. List the
More information( ) 2 + 2x 3! ( x x ) 2
Review for The Final Math 195 1. Rewrite as a single simplified fraction: 1. Rewrite as a single simplified fraction:. + 1 + + 1! 3. Rewrite as a single simplified fraction:! 4! 4 + 3 3 + + 5! 3 3! 4!
More informationSOLUTIONS 1 (27) 2 (18) 3 (18) 4 (15) 5 (22) TOTAL (100) PROBLEM NUMBER SCORE MIDTERM 2. Form A. Recitation Instructor : Recitation Time :
Math 5 March 8, 206 Form A Page of 8 Name : OSU Name.# : Lecturer:: Recitation Instructor : SOLUTIONS Recitation Time : SHOW ALL WORK in problems, 2, and 3. Incorrect answers with work shown may receive
More information