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48. f" (t) = t² + 1/t², t>0, f(2)=3, f' (1)=2

48. f" (t) = t² + 1/t², t>0, f(2)=3, f' (1)=2

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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4.9: #48, please. For #48, find f given the equation.
42. f'(x) = (x+1)/√x, f(1) = 5 43. f' (t)= sec t (sec t + tan t), -T/2 <t</2, f(π/4)= Answer + 44. f' (t) = 3 - 3/t, f(1) = 2, f(-1) = 1 45. f" (x) = -2 + 12x12x², f(0) = 4, f' (0) = 12 Answer 46. f" (x) = 8x³ +5, f(1) = 0, f' (1) = 8 - 47. f" (0) = sin 0 + cos 0, f(0) = 3, f'(0) = 4 Answer 48. f" (t) = t²+1/t², t > 0, f(2)=3, f' (1) = 2 49. f" (x) = 4 + 6x + 24x², f(0) = 3, f(1) = 10 Answer 50. f" (2) = æ® +sinhæ, 51. f" (x)=e- 2 sin x, Answer f (0)= 1, f(0) = 3, a> 0. f (2)=2.6 f(π/2) = 0 52. f" (t) = t-cos t, f(0) = 2, f(1) = 2 53. f" (x) = -². f(1) = 0. f(2)=0
Transcribed Image Text:42. f'(x) = (x+1)/√x, f(1) = 5 43. f' (t)= sec t (sec t + tan t), -T/2 <t</2, f(π/4)= Answer + 44. f' (t) = 3 - 3/t, f(1) = 2, f(-1) = 1 45. f" (x) = -2 + 12x12x², f(0) = 4, f' (0) = 12 Answer 46. f" (x) = 8x³ +5, f(1) = 0, f' (1) = 8 - 47. f" (0) = sin 0 + cos 0, f(0) = 3, f'(0) = 4 Answer 48. f" (t) = t²+1/t², t > 0, f(2)=3, f' (1) = 2 49. f" (x) = 4 + 6x + 24x², f(0) = 3, f(1) = 10 Answer 50. f" (2) = æ® +sinhæ, 51. f" (x)=e- 2 sin x, Answer f (0)= 1, f(0) = 3, a> 0. f (2)=2.6 f(π/2) = 0 52. f" (t) = t-cos t, f(0) = 2, f(1) = 2 53. f" (x) = -². f(1) = 0. f(2)=0
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