Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 9 - Further Applications of the Integral and Taylor Polynomials - Chapter Review Exercises - Page 497: 27

Answer

$$T_4=x-x^3$$

Work Step by Step

Given $$f(x)=x e^{-x^{2}}, \quad T_{4}, \quad a=0$$ Since \begin{array}{rlrl} {f(x)} & {=x e^{-x^{2}}} & {f(0)} & {=0} \\ {f^{\prime}(x)} & {=-\left(2 x^{2}-1\right) e^{-x^{2}}} & {f^{\prime}(0)} & {=1} \\ {f^{\prime \prime}(x)} & {=\left(4 x^{3}-6 x\right) e^{-x^{2}}} & {f^{\prime \prime}(0)} & {=0} \\ {f^{\prime \prime \prime}(x)} & {=-\left(8 x^{4}-24 x^{2}+6\right) e^{-x^{2}}} & {f^{\prime \prime \prime}(0)} & {=-6} \\ {f^{(4)}(x)} & {=\left(16 x^{5}-80 x^{3}+60 x\right) e^{-x^{2}}} & {f^{(4)}(0)} & {=0} \end{array} Then \begin{aligned} T_{4}(x)&= f(a)+f^{\prime}(a)(x-a)+\frac{f^{\prime \prime}(a)}{2 !}(x-a)^{2}+\frac{f^{\prime \prime \prime}(a)}{3 !}(x-a)^{3}++\frac{f^{(4)}(a)}{4 !}(x-a)^{4} \\ &= x-\frac{6}{3!}x^3=x-x^3 \end{aligned}
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