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: 27



Work Step by Step

Since \begin{array}{ll} {f(x)=x^{-\frac{2}{3}},} & {f(1)=1} \\ {f^{\prime}(x)=-\frac{2}{3} x^{-\frac{5}{3}},} & {f^{\prime}(1)=-\frac{2}{3}} \\ {f^{\prime \prime}(x)=\frac{10}{9} x^{-\frac{8}{3}},} & {f^{\prime \prime}(1)=\frac{10}{9}} \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-\frac{2}{3}(x-1)+\frac{5}{9}(x-1)^{2} \end{aligned} and $$\left|f(1.2)-T_{2}(1.2)\right|=|0.885549-0.888889|=0.00334$$
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