Calculus (3rd Edition)

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

Chapter 4 - Applications of the Derivative - 4.1 Linear Approximation and Applications - Exercises - Page 174: 54

Answer

$$L(x) \approx \frac{2 \sqrt{2}(\pi-4) }{\pi^{2}}x-\frac{\sqrt{2}\left(\frac{\pi}{2}-4\right)}{\pi}$$

Work Step by Step

Given $$y=\frac{\sin x}{x}, \quad a=\frac{\pi}{4}$$ Since $y(\pi/4)=\dfrac{2 \sqrt{2}}{\pi}$ and \begin{align*} y'(x)& =\frac{x \cos x-\sin x}{x^{2}}\\ y'(\pi/4)&=\frac{2 \sqrt{2}(\pi-4)}{\pi^{2}} \end{align*} Then the linearization at $a=\pi/4$ is given by \begin{align*} L(x)&\approx y^{\prime}(a)(x-a)+y(a)\\ &= y^{\prime}(\pi/2)(x-\pi/4)+y(\pi/4)\\ &=\frac{2 \sqrt{2}(\pi-4)}{\pi^{2}}(x-\pi/4)+\frac{2 \sqrt{2}}{\pi}\\ &=\frac{2 \sqrt{2}(\pi-4) }{\pi^{2}}x-\frac{\sqrt{2}\left(\frac{\pi}{2}-4\right)}{\pi} \end{align*}
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