Math 241 Homework 7 Solutions

Transcription

1 Math 241 Homework 7 s Section 4.2 Problem 1. Find the value or values c that satisfy the equation = f (c) in the conclusion of the Mean Value Theorem for functions and intervals: f(b) f(a) b a f(x) = x 2 + 2x 1, [0, 1] f(b) f(a) b a = ( ) ( ) 1 0 f (x) = 2x + 2 so we want to solve the equation 2x + 2 = 3 = 3 2x + 2 = 3 2x = 1 x = 1 2 Problem 3. Find the value or values c that satisfy the equation = f (c) in the conclusion of the Mean Value Theorem for functions and intervals: f(b) f(a) b a f(x) = x + 1 x, [1 2, 2] f(b) f(a) b a = (1/2 + 2) (2 + 1/2) 2 1/2 = 0 The solution in the given interval is x = 1. f (x) = 1 1 x 2 = x2 1 x 2 f (x) = 0 x 2 1 = 0 x = ±1 Problem 7. Does f(x) = x(1 x) on [0, 1] satisfy the hypotheses of the Mean Value Theorem? Continuous for x(1 x) 0 0 x 1 and differentiable for x (0, 1) so it does. 1

2 Problem 9. The function x, 0 x < 1 f(x) = 0, x = 1 is zero at x = 0 and x = 1 and differentiable on (0, 1) but its derivative on (0, 1) is never zero. How can this be? Doesn t Rolle s Theorem say the derivative has to be zero somewhere in (0, 1)? Give reasons for your answer. The function is not continuous on [0, 1] because it is not continuous at 1. lim x = 1 and f(1) = 0 x 1 2

3 (a) (ii) (iii) x = 0 Undefined Local min 0 Critical point or endpoint Derivative Extremum Value x =-2 Undefined Local max 0 x =-22 0 Minimum (a) cos 2t + C (b) 2 sin t 2 + (c) cos 2t + 2 sin t 2 + C 39. ƒ(x) = x 2 - x 41. ƒ(x) = s = 4.9t Problem 11. (a) Plot the zeros of each polynomial on a line together with the zeros of its 2 + 5t s = 1 x = 22 0 Maximum 2 first derivative. x = 2 Undefined Local min s = e t + 19t s = sin ( 51. If T(t) is the temperature of the therm y = x Critical point T(0) =-19 C and T(14) = 100 C 4 Derivative Extremum Value (ii) y = x 2 or endpoint Theorem, there exists a 0 6 t x + 15 x = 1 Undefined Minimum C>sec = T (t 0 ), the rate at whic (iii) y = x 3 3x = (x + 1)(x 2) 2 changing at t = t 0 as measured by th 75. (iv) x 3 33x 2 Critical point + 216x = x(x 9)(x 24) Derivative Extremum Value thermometer. or endpoint 53. Because its average speed was appro x =-1 0 Maximum 5 (b) Use Rolle s Theorem to prove that between every two zeros of x n + a n 1 x n 1 by the Mean Value Theorem, it must x = 1 Undefined Local min at least + a 1 once x + during a 0 the trip. there lies a zero of x = 3 0 Maximum The conclusion of the Mean Value Th nx n 1 + (n 1)a n 1 x n a (a) No (b) The derivative is defined and nonzero for x 2. Also, ƒ(2) = 0 and ƒ(x) 7 0 for all x 2. (c) No, because (-q, q) is not a closed interval. (d) The answers are the same as parts (a) and (b), with 2 1 b - 1 a b - a =- 1 c 2 1 c2 a a - b ab b = 61. ƒ(x) must be zero at least once betwee Value Theorem. Now suppose that ƒ(x replaced by a. y b. Then, by the Mean Value Theorem, 79. Yes = 2x at least once between the two zeros of 81. g assumes x 2 a local maximum at -c. since we are given that ƒ (x) 0 on 83. (a) Maximum 4 value = 0is 144 x at = x ±2 = 2. and 2x = x = 0 is zero once and only once between a (b) The largest volume of the box is 144 cubic units, and it occurs when x = ƒ(0.1) 1.1 y = 2x + 8 = 2(x + 4) Section 4.3, pp y 2 0 2g + s 0 2(x + 4) = 0 x = 4 and x 2 1. (a) 0, 1 + 8x + 15 = (x + 5)(x + 3) = 0 x = 3, (b) 5Increasing on (-q, 0) and (1, q 87. Maximum value is 11 at x = 5; minimum value is 5 on the interval [- 3, 2]; local maximum at (- 5, 9). (c) Local maximum at x = 0; local 89. Maximum value is 5 on the interval [3, q); minimum value is 3. (a) -2, 1-5 on the interval (-q, y = 3x -2]. 2 6x = 3x(x 2) (b) Increasing on (-2, 1) and (1, q) 3x(x 2) = 0 x = 0, 2 and (x + 1)(x 2) 2 (c) No local maximum; local minim Section 4.2, pp = 0 x = 1, 25. (a) Critical point at x = 1 (b) Decreasing on (-q, 1), increasi (iv) 1. 1> { 1-4 A y = 3x 2 p { (c) Local (and absolute) minimum a 66x = 3(x ) = 3(x 18)(x 4) 7. (a) 0, , (b) Increasing on (-q, -2) and (1, 3(x 18)(x 9. Does not; 4) ƒ = is 0not differentiable x = 18, 4 and at the x(x interior 9)(x domain 24) point = 0 x = 0, 9, 24and (0, 1) x = 0. The number lines are 11. as Does follows: 13. Does not; ƒ is not differentiable at x = (a) i) x ii) iii) iv) x x x (c) Local minimum at x = 1 9. (a) -2, 2 (b) Increasing on (-q, -2) and (2, and (0, 2) (c) Local maximum at x =-2; loca 11. (a) -2, 0 (b) Increasing on (-q, -2) and (0, (c) Local maximum at x =-2; loca 13. (a) p 2, 2p 3, 4p 3 (b) Let x 1 and x 2 be two zeros of f(x) = x n + a n 1 x n a 1 x + a 0. By Rolle s theorem we have that there exists a c between x 1 and x 2 such that nc n 1 + (n 1)a n 1 c n a 1 = f (c) = f(x 1) f(x 2 ) x 1 x 2 = 0 3

4 Problem 17. Show that g(t) = t t 4 has exactly one zero in (0, ) g(1) = < 0 and g(5) = > 0 Thus by the Intermediate Value Theorem we have that there is at least on zero on (0, ). Assume there are two zeroes. Then by Rolle s Theorem there is a c in (0, ) such that g (c) = 0 f (c) = 0 g (t) = 1 t + = = t 1 + t + t t + t t t 1 + t t 1 t + t 2 ( 1 + t t) which is impossible thus there cannot be two solutions. 1 c + c 2 ( 1 + c c) = 0 1 = 0 Problem 27. Find all possible functions with the given derivative: y = x (ii) y = x 2 (iii) y = x 3 x2 2 has derivative x so the answer is x2 2 + C (ii) x3 3 has derivative x2 so the answer is x3 3 + C (iii) x4 4 has derivative x3 so the answer is x4 4 + C 4

5 Problem 35. Find the function with the given derivative whose graph passes through the point P : r (θ) = 8 csc 2 θ, ( π 4, 0) 8θ + cot θ has derivative 8 csc 2 θ so we have the general form is 8θ + cot θ + C. We also want it to pass through P so we have 0 = 8 π 4 + cot π 4 + C 0 = 2π C C = 1 2π r(θ) = 8θ + cot θ 1 2π Problem 47. Classical accounts tell us that a 170-oar trireme (ancient Greek or Roman warship) once covered 184 sea miles in 24 hours. Explain why at some point during this feat, the trireme s speed exceeded 7.5 knots (sea miles per hour). The average speed is Thus by the Mean Value Theorem there has to be a point where the boat is going 7.7 knots which exceeds 7.5 knots. 5

6 Problem 54. Construct a polynomial f(x) that has zeros at x = 2, 1, 0, 1, and 2 (ii) Graph f and its derivative f together. How is what you see related to Rolle s Theorem? (iii) Do g(x) = sin x and its derivative g illustrate the same phenomenon? We have that f(x) looks like x(x + 2)(x + 1)(x 1)(x 2) (ii) Graphing this we have (iii) Yes Which shows that the derivative is zero (i.e. intercepts the x-axis) between every two zeros of f Problem 55. Assume that f is continuous on [a, b] and differentiable on (a, b). Also assume that f(a) and f(b) have opposite signs and that f 0 between a and b. Show that f(x) = 0 exactly once between a and b. By the Intermediate Value Theorem, there must be at least one zero between a and b. If there were more than one, then by Rolle s Theorem, there would have to be a point between them such that the derivative f is zero. However, we are given that the derivative is never zero between a and b and thus there can only be exactly one solution. 6

7 Section 4.3 Problem 1. Answer the following questions about the function whose derivative is given by f (x) = x(x 1): What are the critical points of f? (ii) On what intervals is f increasing or decreasing? (iii) At what points, if any, does f assume local maximum and minimum values? f (x) = 0 x = 0, 1 Thus the critical points occur at x = 0 and x = 1 (ii) Plotting the critical points we get 0 1 Thus f is increasing on (, 0), (1, ) and decreasing on (0, 1) (iii) There is a local max at x = 0 and a local min at x = 1 7

8 Problem 7. Answer the following questions about the function whose derivative is given by f (x) = x 1/3 (x + 2): What are the critical points of f? (ii) On what intervals is f increasing or decreasing? (iii) At what points, if any, does f assume local maximum and minimum values? f (x) = x + 2 x 1/3 f is undefined at x = 0 and zero at x = 2 so the critical points are x = 0 and x = 2 (ii) Testing the intervals we have -2 0 Thus f is increasing on (, 2), (0, ) and decreasing on ( 2, 0) (iii) There is a local min at x = 0 and a local max at x = 2 8

9 Problem 9. Let g(t) = t 2 3t + 3 Find the intervals on which the function is increasing and decreasing. (ii) Then identify the function s local extreme values, if any, saying where they are taken on. (iii) Which, if any, of the extreme value are absolute? (iv) Support your findings with a graphing calculator or computer grapher. Testing the intervals we get g (t) = 0 2t 3 = 0 t = 3 2-3/2 (ii) Thus g is increasing on (, 3 /2) and decreasing on ( 3 /2, ) There is a local max of 21 /4 at x = 3 /2 f ( 3 2 ) = ( 3 2 ) 2 3 ( 3 2 ) + 3 = = 21 4 (iii) Since the function has only one local max, 21/4 at x = 3 /2 is an absolute max. (iv) The graph looks like 9

10 Problem 15. Let f(r) = 3r r Find the intervals on which the function is increasing and decreasing. (ii) Then identify the function s local extreme values, if any, saying where they are taken on. (iii) Which, if any, of the extreme value are absolute? (iv) Support your findings with a graphing calculator or computer grapher. f (r) = 9r > 0 Thus f is increasing on (, ) (ii) There are no local extrema (iii) There are no absolute extrema (iv) The graph looks like 10

11 Problem 19. Let H(t) = 3 2 t4 t 6 Find the intervals on which the function is increasing and decreasing. (ii) Then identify the function s local extreme values, if any, saying where they are taken on. (iii) Which, if any, of the extreme value are absolute? (iv) Support your findings with a graphing calculator or computer grapher. H (t) = 6t 3 6t 5 = 6t 3 (1 t 2 ) = 6t 3 (1 t)(1 + t) H (t) = 0 t = ±1, 0 Plotting these critical points and testing the intervals we have Thus H is increasing on (, 1), (0, 1) and decreasing on ( 1, 0), (1, ) (ii) H( 1) = 1 2, H(0) = 0, H(1) = 1 2 Thus there are local maxes of 1 /2 at x = 1 and x = 1 and a local min of 0 at x = 0 (iii) Since the graph points downward at its ends there is no local minimum and the two local maxes are absolute (iv) The graph looks like 11

12 Problem 35. Let h(x) = x3 3 2x2 + 4x, 0 x < Identify the function s local extreme values in the given domain, and say where they are assumed. (ii) Which, if any, of the extreme value are absolute? (iii) Support your findings with a graphing calculator or computer grapher. h (x) = x 2 4x + 4 = (x 2) 2 and h (x) = 0 x = 2 Since h (x) > 0 we have that h is increasing on [0, ) Thus there is a local minimum of 0 at x = 0 (ii) Since the local minimum is attained at the endpoint and the function is always increasing, the local minimum is an absolute minimum (iii) The graph is 12