Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 3 - Section 3.10 - Linear Approximations and Differentials. - 3.10 Exercises - Page 259: 32

Answer

$\frac{1}{4.002} \approx 0.249875$

Work Step by Step

Let $y = \frac{1}{x}$ $\frac{dy}{dx} = -\frac{1}{x^2}$ $dy = -\frac{1}{x^2}~dx$ Let $x = 4$ and let $dx = 0.002$ $dy = -\frac{1}{4^2}~(0.002)$ $dy = -\frac{1}{8000}$ We can find an approximation for $\frac{1}{4.002}$: $\frac{1}{4.002} \approx \frac{1}{4} -\frac{1}{8000}$ $\frac{1}{4.002} \approx \frac{2000}{8000} -\frac{1}{8000}$ $\frac{1}{4.002} \approx \frac{1999}{8000}$ $\frac{1}{4.002} \approx 0.249875$
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