Answer
$L(x) = xln(2) + 1$
Work Step by Step
$f(x) = 2^x$ , $a = 0$
$f(a) = 2^0$
$f(a) = 1$
$f'(x) = 2^x ln(2)$
$f'(2) = 2^0 ln(2)$
$f'(2) = 1 ln(2)$
$L(x) = f(a) + f'(a)(x-a)$
$L(x) = 1 + ln 2(x-0)$
$L(x) = 1 + xln2 - 0$
$L(x) = xln(2) + 1$
Calculus: Early Transcendentals 8th Edition
by Stewart, James
Published by
Cengage Learning
ISBN 10:
1285741552
ISBN 13:
978-1-28574-155-0
Chapter 3 - Section 3.10 - Linear Approximations and Differentials. - 3.10 Exercises - Page 256: 4
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