Let xo = 0.5. Given 1 f(x) = -2ex + 4x² ELL f'(x) = 2e-x + x³ 1 1 1 120 x -x5 + 2x 24 1 ƒ"(x) = −2e¯* + 3x²-x f(4)(x) = -2e-* +6-x f(5)(x) = 2e-x-1 f(6)(x) = -2e-* 6 + 2 1 2 f''(x) = 2e-x + 6x - x² 2 x3 a. Find the Taylor Polynomial, T3 (x), of degree at most 3 for f(x) expanded about xo. b. Give the general error formula for f(x) - T3(x) for any x. c. Find the absolute error in using T3 (0.65) to approximate f(0.65). d. Use the error formula to find a bound for the absolute error in approximating f (0.65) with T3 (0.65).

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter5: Inverse, Exponential, And Logarithmic Functions
Section: Chapter Questions
Problem 18T
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Let xo = 0.5. Given
1
f(x) = -2e-x +
f'(x) = 2e-x + x³ x4 + 2
1
4x²
120
1
24
f"(x) = -2e-x + 3x² - ²x³
——
5
x³ + 2x
f''(x) = 2e-x + 6x - 7x²
f(4)(x) = -2e-x + 6-*
f(5) (x) = 2e-x-1
f(6)(x) = -2e-*
a. Find the Taylor Polynomial, T3 (x), of degree at most 3 for f(x) expanded about xo.
b. Give the general error formula for f(x) - T3(x) for any x.
c. Find the absolute error in using T3 (0.65) to approximate f(0.65).
d. Use the error formula to find a bound for the absolute error in approximating f(0.65) with T3 (0.65).
Transcribed Image Text:Let xo = 0.5. Given 1 f(x) = -2e-x + f'(x) = 2e-x + x³ x4 + 2 1 4x² 120 1 24 f"(x) = -2e-x + 3x² - ²x³ —— 5 x³ + 2x f''(x) = 2e-x + 6x - 7x² f(4)(x) = -2e-x + 6-* f(5) (x) = 2e-x-1 f(6)(x) = -2e-* a. Find the Taylor Polynomial, T3 (x), of degree at most 3 for f(x) expanded about xo. b. Give the general error formula for f(x) - T3(x) for any x. c. Find the absolute error in using T3 (0.65) to approximate f(0.65). d. Use the error formula to find a bound for the absolute error in approximating f(0.65) with T3 (0.65).
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