Answer
$L(x)=x-\pi$
Work Step by Step
Step 1. $f(x)=tan(x)$, $f'(x)=sec^2(x)$,
Step 2. $f(\pi)=tan(\pi)=0$, $f'(\pi)=sec^2(\pi)=1$,
Step 3. The linearization at $a=\pi$ is $L(x)=f(\pi)+f'(\pi)(x-\pi)=0+(x-\pi)=x-\pi$
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0-32187-896-5
ISBN 13:
978-0-32187-896-0
Chapter 3: Derivatives - Section 3.9 - Linearization and Differentials - Exercises 3.9 - Page 174: 5
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