Calculus: Early Transcendentals 8th Edition

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: 4

Answer

$L(x) = xln(2) + 1$

Work Step by Step

$f(x) = 2^x$ , $a = 0$ $f(a) = 2^0$ $f(a) = 1$ $f'(x) = 2^x ln(2)$ $f'(2) = 2^0 ln(2)$ $f'(2) = 1 ln(2)$ $L(x) = f(a) + f'(a)(x-a)$ $L(x) = 1 + ln 2(x-0)$ $L(x) = 1 + xln2 - 0$ $L(x) = xln(2) + 1$
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