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f(x) e (2x+4) = 9 (i) find the third derivative, f"", of f.

f(x) e (2x+4) = 9 (i) find the third derivative, f"", of f.

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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(part 1 of 5) When f is the function f(x) = e(2x+4) (i) find the third derivative, f"", of f. 1. f'(x) = 3! e(2x+4) 2. f'(x) = 6 · 2³ e(2x+4) 3. f(x) = 2³ e(2x+4) 4. f(x) = -172³3e(²x+4) 3! 5. f(x) = e(2x+4) (part 2 of 5) (ii) Find the nth-derivative, f(n), of f. 1. f(n)(x) 2ne2x+4 = 2. f(n)(x) = n! 2n 2x+4 е 3. f(n)(x) = ²x+4 1 4. f(n)(x) = -—=—7 2² ²x+4 n! 5. f(n)(x) = n! e²x+4
Transcribed Image Text:(part 1 of 5) When f is the function f(x) = e(2x+4) (i) find the third derivative, f"", of f. 1. f'(x) = 3! e(2x+4) 2. f'(x) = 6 · 2³ e(2x+4) 3. f(x) = 2³ e(2x+4) 4. f(x) = -172³3e(²x+4) 3! 5. f(x) = e(2x+4) (part 2 of 5) (ii) Find the nth-derivative, f(n), of f. 1. f(n)(x) 2ne2x+4 = 2. f(n)(x) = n! 2n 2x+4 е 3. f(n)(x) = ²x+4 1 4. f(n)(x) = -—=—7 2² ²x+4 n! 5. f(n)(x) = n! e²x+4
3. 1/7 f(n) (0) = 27e4 n! 4. —ƒ(¹) (0) 5. f() (0) = 1. p3(x) = e 2. p3(x) = (part 4 of 5) (iv) Determine the degree 3 Taylor polynomial, P3, centered at x 0 for f. = 3. p3(x) = e 6. p3(x) 2n n = 1 -4 5. p3(x) = е¯ -e4 n! 4 -4 4 (1 1+ 2x + 2x² + e¹(1+2x - 2x² e¹ (1 4. p3(x) = e¹(1 − 2x + 2x² -4 ¹(1 2x + 2x² (122:²³) 72³) 4 172³) col Acol 1 + 2x 2x² ² + 1/32²³) 4 e¹ (1 + 2x + 2x² + ²32³3) (part 3 of 5) (iii) Compute the value of f(n) (0). 1. 1- f(n) (0) = 2" . e4 2. 1-7 f(n) (0) = 2n Teln е n!
Transcribed Image Text:3. 1/7 f(n) (0) = 27e4 n! 4. —ƒ(¹) (0) 5. f() (0) = 1. p3(x) = e 2. p3(x) = (part 4 of 5) (iv) Determine the degree 3 Taylor polynomial, P3, centered at x 0 for f. = 3. p3(x) = e 6. p3(x) 2n n = 1 -4 5. p3(x) = е¯ -e4 n! 4 -4 4 (1 1+ 2x + 2x² + e¹(1+2x - 2x² e¹ (1 4. p3(x) = e¹(1 − 2x + 2x² -4 ¹(1 2x + 2x² (122:²³) 72³) 4 172³) col Acol 1 + 2x 2x² ² + 1/32²³) 4 e¹ (1 + 2x + 2x² + ²32³3) (part 3 of 5) (iii) Compute the value of f(n) (0). 1. 1- f(n) (0) = 2" . e4 2. 1-7 f(n) (0) = 2n Teln е n!
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