Calculus (3rd Edition)

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

Chapter 4 - Applications of the Derivative - Chapter Review Exercises - Page 221: 2


$$-6.25\times 10^{-3},\ \ \ -6.13520\times 10^{-3} ,\ \ \ 0.0001148$$

Work Step by Step

Given $$\frac{1}{\sqrt{4.1}}-\frac{1}{2} $$ Consider $f(x)=x^{-1 / 2}, a=4,$ and $\Delta x=0.1$, since \begin{align*} f^{\prime}(x)&=\frac{-1}{2}x^{-3/2}\\ f^{\prime}(4)&=\frac{-1}{16} \end{align*} Then \begin{align*} \Delta f&=f (a+\Delta x)-f(a)\\ &\approx f'(a)\Delta x\\ &\approx \frac{-0.1}{16} \\ &\approx -6.25\times 10^{-3} \end{align*} By using a calculator, $\frac{1}{\sqrt{4.1}}-\frac{1}{2}=-6.13520\times 10^{-3}$ and the error in the linear approximation is$$ |6.13520\times 10^{-3}-6.25\times 10^{-3}|=0.0001148 $$
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