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 492: 3


\begin{aligned} T_{2}(x) &=\frac{1}{3}-\frac{1}{9}(x-2)+\frac{1}{27}(x-2)^{2} \\ T_{3}(x) &= \frac{1}{3}-\frac{1}{9}(x-2)+\frac{1}{27}(x-2)^{2}-\frac{1}{81}(x-2)^{3} \end{aligned}

Work Step by Step

Since \begin{array}{ll} {f(x)=\frac{1}{1+x},} & {f(2)=\frac{1}{3}} \\ {f^{\prime}(x)=-\frac{1}{(1+x)^{2}},} & {f^{\prime}(2)=-\frac{1}{9}} \\ {f^{\prime \prime}(x)=\frac{2}{(1+x)^{3}},} & {f^{\prime \prime}(2)=\frac{2}{27}} \\ {f^{\prime \prime \prime}(x)=-\frac{6}{(1+x)^{4}},} & {f^{\prime \prime \prime}(2)=-\frac{2}{27}} \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} \\ &=\frac{1}{3}-\frac{1}{9}(x-2)+\frac{1}{27}(x-2)^{2} \\ T_{3}(x) &=f(a)+\frac{f^{\prime}(a)}{1 !}(x-a)+\frac{f^{\prime \prime}(a)}{2 !}(x-a)^{2}+\frac{f^{\prime \prime \prime}(a)}{3 !}(x-a)^{3} \\ &=\frac{1}{3}-\frac{1}{9}(x-2)+\frac{1}{27}(x-2)^{2}-\frac{1}{81}(x-2)^{3} \end{aligned}
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