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: 50

Answer

The maximum is $f(-\pi/6) \approx2.488$ and the minimum is $f(0)=0$

Work Step by Step

Given $$ y =\sec ^{2} x-2 \tan x \quad \text { on }\left[-\frac{\pi}{6}, \frac{\pi}{3}\right]$$ Since \begin{aligned} y'&=2 \sec x \cdot \sec x \tan x-2 \sec ^{2} x\\ &=2 \sec ^{2} x(\tan x-1) \end{aligned} Then $f(x)$ has critical points when \begin{align*} f'(x)&=0\\ 2 \sec ^{2} x(\tan x-1)&=0 \end{align*} Then $x=\pi/4\in\left[-\frac{\pi}{6}, \frac{\pi}{3}\right] $, since \begin{align*} f(-\pi/6)&\approx2.488\\ f(\pi/4)&=0\\ f(\pi/3)&\approx 0.5359\ \end{align*} Hence the maximum is $f(-\pi/6) \approx2.488$ and the minimum is $f(0)=0$
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