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: 4


\begin{aligned} T_{2}(x) &=\frac{1}{2}+\frac{1}{2}(x+1)+\frac{1}{4}(x+1)^{2} \\ T_{3}(x) &=\frac{1}{2}+\frac{1}{2}(x+1)+\frac{1}{4}(x+1)^{2} \end{aligned}

Work Step by Step

Since \begin{array}{ll} {f(x)=\frac{1}{1+x^{2}},} & {f(-1)=\frac{1}{2}} \\ {f^{\prime}(x)=-\frac{2 x}{\left(1+x^{2}\right)^{2}},} & {f^{\prime}(-1)=\frac{1}{2}} \\ {f^{\prime \prime}(x)=\frac{2\left(x^{2}-1\right)}{\left(1+x^{2}\right)^{3}},} & {f^{\prime \prime}(-1)=\frac{1}{2}} \\ {f^{\prime \prime \prime}(x)=-\frac{24 x\left(x^{2}-1\right)}{\left(1+x^{2}\right)^{4}},} & {f^{\prime \prime \prime}(-1)=0} \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}{2}+\frac{1}{2}(x+1)+\frac{1}{4}(x+1)^{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}{2}+\frac{1}{2}(x+1)+\frac{1}{4}(x+1)^{2}+(0)(x+1)^{3} \\ &=\frac{1}{2}+\frac{1}{2}(x+1)+\frac{1}{4}(x+1)^{2} \end{aligned}
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