University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.11 - Linearization and Differentials - Exercises - Page 199: 43


a) $\frac{-1}{3}$ b) $\frac{-2}{5}$ c) $\frac{1}{15}$

Work Step by Step

We are given: $f(x)=x^{-1}$ $\delta f=f(0.5+0.1)-f(0.5)$ or $\delta f=f(0.6)-f(0.5)=0.6^{-1}-0.5^{-1}=\frac{5}{3}-2=\frac{-1}{3}$ or $\delta f=\frac{-1}{3}$ differentiation of f(x): $f'(x)=-x^{-2}$ $df=f'(0.5)\times0.1$ $df=(-0.5^{-2})\times0.1=-0.4$ $df=-\frac{2}{5}$ so $|\delta f-df|=|\frac{-1}{3}-\frac{-2}{5}|=|\frac{1}{15}|=\frac{1}{15}$ The final answer is: $a)\frac{-1}{3}, b)\frac{-2}{5}, c)\frac{1}{15}$
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