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: 14



Work Step by Step

$f(x)=\sin^{-1}x$, $L(x)=f(a)+f'(a)(x-a)$ $f'(x)=\frac{1}{\sqrt{(1-x^2)}}$ $x_{0}=0$ $f(0)={\sin^{-1}0}=0$ $f'(0)= \frac{1}{1} = 1$ then $L(x)=0+1(x-0)=x$ $L(x)=x$ Thus, the final answer is: $L(x)=x$
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