University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Practice Exercises - Page 205: 150

Answer

$L(x)= 1-x$

Work Step by Step

The linear approximation of $f(x)$ at $x=0$ is given as: $L(x)=f(0)+x f'(x)$ We are given that $f(x)=\dfrac{1}{1+\tan x}$ and $f(0)=\dfrac{1}{1+\tan(0)}=0$ Also, $f'(x)=\dfrac{-\sec^2 x}{(1+\tan x)^2} \implies f'(0)=\dfrac{-\sec^2 0}{(1+\tan 0)^2}=-1$ Hence, $L(x)=1+x(-1) \implies 1-x$ Hence, the result has been proved.
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