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 198: 4



Work Step by Step

Given: $f(x)=(x)^{\frac{1}{3}}$, $L(x)=f(a)+f'(a)(x-a)$ $f'(x)=\frac{1}{3}{(x)^{\frac{-2}{3}}}$ $a=-8$ $f(-8)=(-8)^{\frac{1}{3}}=-2$ $f'(-8)=\frac{1}{3}{(-8)^{\frac{-2}{3}}}=\frac{1}{12}$ then $L(x)=-2+\frac{1}{12}(x+8)=\frac{x}{12}-\frac{4}{3}$ $L(x)=\frac{x}{12}-\frac{4}{3}$ Thus, the final answer is: $L(x)=\frac{x}{12}-\frac{4}{3}$
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