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 + 2 1) (0 + 0 1) 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
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
(a) (ii) (iii) 71. 5 25 x = 0 Undefined Local min 0 Critical point or endpoint Derivative Extremum Value x =-2 Undefined Local max 0 x =-22 0 Minimum 2 37. (a) - 1 2 cos 2t + C (b) 2 sin t 2 + (c) - 1 2 cos 2t + 2 sin t 2 + C 39. ƒ(x) = x 2 - x 41. ƒ(x) = 1 + 43. s = 4.9t Problem 11. (a) Plot the zeros of each polynomial on a line together with the zeros of its 2 + 5t + 10 45. s = 1 x = 22 0 Maximum 2 first derivative. x = 2 Undefined Local min 0 47. s = e t + 19t + 4 49. s = sin ( 51. If T(t) is the temperature of the therm y = x 2 73. 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 0 6 14 + 8x + 15 x = 1 Undefined Minimum 2 8.5 C>sec = T (t 0 ), the rate at whic (iii) y = x 3 3x 2 + 4 = (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 1 +... at least + a 1 once x + during a 0 the trip. there lies a zero of x = 3 0 Maximum 5 57. The conclusion of the Mean Value Th nx n 1 + (n 1)a n 1 x n 2 +... + a 1 77. (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 = 2. 71. 1.09999 ƒ(0.1) 1.1 y = 2x + 8 = 2(x + 4) Section 4.3, pp. 228 230 85. 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. 222 224 = 0 x = 1, 25. (a) Critical point at x = 1 (b) Decreasing on (-q, 1), increasi (iv) 1. 1>2 3. 1 5. { 1-4 A y = 3x 2 p {0.771 2 (c) Local (and absolute) minimum a 66x + 216 = 3(x 2 22 + 72) = 3(x 18)(x 4) 7. (a) 0, 1 7. 1 3 11 + 272 1.22, 1 11-272 -0.549 3 (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 =-1. 17. (a) i) x 2 0 2 ii) iii) iv) 5 4 3 1 0 2 0 4 9 18 24 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 1 +... + 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 2 +... + a 1 = f (c) = f(x 1) f(x 2 ) x 1 x 2 = 0 3
Problem 17. Show that g(t) = t + 1 + t 4 has exactly one zero in (0, ) g(1) = 1 + 2 4 < 0 and g(5) = 5 + 6 4 > 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 + = = 1 1 + t 1 + t + t t + t 2 1 + 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
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 + 1 + 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 184 24 7.7 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
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
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
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
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 = 9 4 + 18 4 + 12 4 = 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
Problem 15. Let f(r) = 3r 3 + 16r 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 2 + 16 > 0 Thus f is increasing on (, ) (ii) There are no local extrema (iii) There are no absolute extrema (iv) The graph looks like 10
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 -1 0 1 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
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