Calculus (3rd Edition)

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

Chapter 4 - Applications of the Derivative - Chapter Review Exercises - Page 222: 42


$x= \dfrac{(2n+1)\pi}{2}$ Where $n$ is an integer.

Work Step by Step

Given $$y=x-2 \cos x $$ Since \begin{align*} y'&=1+2\sin x \\ y''&=2\cos x \end{align*} Then $f''(x)=0$ for $x=\dfrac{(2n+1)\pi}{2}$ and $$ y''>0\ \ \text{ for } \ x\in \left(\frac{(4n-1)\pi}{2},\frac{(4n+1)\pi}{2}\right) \\ y''<0\ \ \text{ for } \ x\in \left(\frac{(4n+1)\pi}{2},\frac{(4n+3)\pi}{2}\right)$$ Hence there exists an inflection point at $x= \dfrac{(2n+1)\pi}{2}$ Where $n$ is an integer.
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