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: 5



Work Step by Step

$f(x)=\tan{x} $, $L(x)=f(a)+f'(a)(x-a)$ $f'(x)=sec^2{x}$ $a=\pi$ $f(\pi)=\tan{\pi}=0$ $f'(\pi)=\sec^2{\pi}=(-1)^2=1$ then $L(x)=0+(x-\pi)=x-\pi$ $L(x)=x-\pi$ Thus, the final answer is: $L(x)=x-\pi$
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