 # Find where f(x)=4x-\tan x, \ -\pi/2<x<\pi/2 is increasing or decreasing and find it's maximum and minimum values Dottie Parra 2021-09-15 Answered

Find where is increasing or decreasing and find it's maximum and minimum values.

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differentiating f(x) with respect to x, ${f}^{\prime }\left(x\right)=4-{\mathrm{sec}}^{2}x$
now, $,{f}^{\prime }\left(x\right)=4-{\mathrm{sec}}^{2}x=0$
$⇒4-{\mathrm{sec}}^{2}x=0$
$⇒{\mathrm{sec}}^{2}x={2}^{2}$
$⇒\mathrm{sec}x=±2$
$⇒x=±\frac{\pi }{3}$
so, there are three intervals.$\left(-\frac{\pi }{2},\frac{\pi }{3}\right),\left(-\frac{\pi }{3},\frac{\pi }{3}\right),\left(\frac{\pi }{3},\frac{\pi }{2}\right)$ let's check f'(x)>0 in which intervals. $-\frac{\pi }{3}\mathrm{sec}x<2$ so, ${f}^{\prime }\left(x\right)=4-{\mathrm{sec}}^{2}x>0$

you will get ${f}^{\prime }\left(x\right)<0$ from this intervals

again differentiate with respect to x, $f{}^{″}\left(x\right)=0-2{\mathrm{sec}}^{2}x.\mathrm{tan}x=-2{\mathrm{sec}}^{2}x.\mathrm{tan}x$
$atx=\frac{\pi }{3},f\frac{\pi }{3}\right)=2{\mathrm{sec}}^{2}\left(\frac{\pi }{3}\right).\mathrm{tan}\left(\frac{\pi }{3}\right)<0$