Answer
$L(x)=x-\pi$
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: $L(x)=x-\pi$
University Calculus: Early Transcendentals (3rd Edition)
by Hass, Joel R.; Weir, Maurice D.; Thomas Jr., George B.
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
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