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 - 9.4 Taylor Polynomials - Exercises - Page 493: 25


$T_{2}=1+x+\frac{1}{2} x^{2}$ $0.0185$

Work Step by Step

Since \begin{array}{ll} {f(x)=e^{x},} & {f(0)=1} \\ {f^{\prime}(x)=e^{x},} & {f^{\prime}(0)=1} \\ {f^{\prime \prime}(x)=e^{x},} & {f^{\prime \prime}(0)=1} \end{array} Then \begin{aligned} T_{2}(x) &=f(a)+\frac{f^{\prime}(a)}{1 !}(x-a)+\frac{f^{\prime \prime}(a)}{2 !}(x-a)^{2} \\ &=1+(x-0)+\frac{1}{2}(x-0)^{2} \\ &=1+x+\frac{1}{2} x^{2} \end{aligned} and $$\left|f(-0.5)-T_{2}(-0.5)\right|=|0.606531-0.625|=0.0185$$
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