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: 33

Answer

$\sqrt[3] {1001} \approx 10.0033$

Work Step by Step

Let $y = \sqrt[3] x$ $\frac{dy}{dx} = \frac{1}{3}x^{-2/3}$ $\frac{dy}{dx} = \frac{1}{3~x^{2/3}}$ $dy = \frac{1}{3~x^{2/3}}~dx$ Let $x = 1000$ and let $dx = 1$ $dy = \frac{1}{3~(1000^{2/3})}~(1)$ $dy = \frac{1}{300}$ We can find an approximation for $\sqrt[3] {1001}$: $\sqrt[3] {1001} \approx \sqrt[3] {1000}+\frac{1}{300}$ $\sqrt[3] {1001} \approx 10+\frac{1}{300}$ $\sqrt[3] {1001} \approx 10.0033$
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