Calculus (3rd Edition)

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

Chapter 4 - Applications of the Derivative - 4.2 Extreme Values - Exercises - Page 181: 49


The minimum is $ f(\pi/4) =1-\dfrac{\pi}{2}$ and the maximum is $f( 0) =0$

Work Step by Step

Given $$y=\tan x-2 x, \quad[0,1]$$ Take the derivative \begin{align*} \frac{dy}{dx}&=\sec^2x-2 \end{align*} Find the critical points \begin{align*} f'(\theta)&=0\\ \sec^2x-2 &=0 \end{align*} Then $x=\pi/4 \in [0,1]$ and \begin{align*} f(0)&=0 \\ f(\pi/4)&=1-\frac{\pi}{2}\\ f(1)&=\tan (1)-2 \end{align*} Then the minimum is $ f(\pi/4) =1-\dfrac{\pi}{2}$ and the maximum is $f( 0) =0$
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