Answer
$L(x)=kx+1$
Work Step by Step
$f(x)=(1+x)^k$,
$L(x)=f(a)+f'(a)(x-a)$
$f'(x)=k(1+x)^{k-1}$
$x_{0}=0$
$f(0)={(1+0)^k}=1$
$f'(0)= k(1+0)^{k-1} = k$
then $L(x)=1+k(x-0)=kx+1$
$L(x)=kx+1$
Thus, the final answer is: $L(x)=kx+1$
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: 15
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