Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - 7.1 Derivative of f(x)=bx and the Number e - Exercises - Page 327: 61

Answer

$$L(x)= x$$

Work Step by Step

Given $$f(x) =e^{-2x}\sin x ,\ \ a=0$$ Since $f(a) =0$ and \begin{align*} f'(x)&= \frac{d}{dx}\left(e^{-2x}\right)\sin \left(x\right)+\frac{d}{dx}\left(\sin \left(x\right)\right)e^{-2x}\\ &= -2e^{-2x}\sin \left(x\right)+e^{-2x}\cos \left(x\right)\\ f'(0)&= 1 \end{align*} Then the linearization is given by \begin{align*} L(x)&=f^{\prime}(a)(x-a)+f(a)\\ &=x \end{align*}
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