Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 3 - Applications of Differentiation - 3.3 How Derivatives Affect the Shape of a Graph - 3.3 Exercises - Page 230: 67

Answer

Prove that $f(x)$ is increasing on $\left(0,\frac{\pi}{2}\right)$

Work Step by Step

$f(x)$ = $\tan x-x$ $f'(x)$ = $\sec^{2}x-1$ $\gt$ $0$ for $0$ $\lt$ $x$ $\lt$ $\frac{\pi}{2}$ since $\sec^{2}x$ $\gt$ $1$ for $0$ $\lt$ $x$ $\lt$ $\frac{\pi}{2}$, so $f$ is increasing on $\left(0,\frac{\pi}{2}\right)$. Thus $f(x)$ $\gt$ $f(0)$ = $0$ for $0$ $\lt x\lt\frac{\pi}{2}$ $\tan x-x\gt0$ $\tan x\gt x$ for $0\lt x\lt \frac{\pi}{2}$
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