Find the maximum and minimum value of this function. f(x)= x+ ^2 … - Vidyadip

Question

Find the maximum and minimum value of this function.
$f(x)=\sec x+\log \cos ^2 x, 0 < x < 2 \pi $

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Answer

Given
$f(x) =\sec x+\log \cos ^2 x$
$ =\sec x+2 \log \cos x$
$\Rightarrow f^{\prime}(x) =\sec x \tan x-2 \tan x$
$ =\tan x(\sec x-2)$
$\text { for finding critical point } f^{\prime}(x)=0$
$\Rightarrow \tan x(\sec x-2)=0$
$ \tan x=0 \text { or } \sec x=2$
$\Rightarrow \tan x=0 \text { or } \sec x=\frac{1}{2}$
$\Rightarrow x=\pi, x=\frac{\pi}{3}, \frac{5 \pi}{3} \quad[\because 0 < x <2 \pi]$
$ \text { now } f^{\prime}(x)=\tan x(\sec x-2)$
$\Rightarrow f^{\prime \prime}(x)=\sec ^2 x(\sec x-2)+\tan ^2 x \sec x$
$=\sec ^2 x(\sec x-2)+\sec x\left(\sec ^2 x-1\right)$
$\Rightarrow f^{\prime \prime}(x)=2 \sec ^3 x-2 \sec ^2 x-\sec x$
$\text { at } x=\frac{\pi}{3}, f^{\prime \prime}\left(\frac{\pi}{3}\right)=2 \sec ^3 \frac{\pi}{3}-2 \sec ^2 \frac{\pi}{3}-\sec \frac{\pi}{3}$
$=2 \times 8-2 \times 4-2=6>0$
hence minimum point is at $x=\frac{\pi}{3}$ and minimum value$
f\left(\frac{\pi}{3}\right) =\sec \frac{\pi}{3}+\log \cos ^2 \frac{\pi}{3}=2+\log \frac{1}{4}$
$ =2-2 \log 2$
at $x=\pi, f^{\prime \prime}(\pi)=2 \sec ^3 \pi-2 \sec ^2 \pi-\sec \pi$
$=-2-2+1=-3<0$
hence, maximum point is $x=\pi$ and maximum value
$f(\pi) =\sec \pi+\log \cos ^2 \pi$
$ =-1+\log (1)=-1$
$\text { at } x=\frac{5 \pi}{3}, f^{\prime \prime}\left(\frac{5 \pi}{3}\right)=2 \sec ^3\left(\frac{5 \pi}{3}\right)-2 \sec ^2\left(\frac{5 \pi}{3}\right)$$
-\sec \left(\frac{5 \pi}{3}\right)$
$=2(2)^3-2(2)^2+2=10>0$
hence minimum point is at $x=\frac{5 \pi}{3}$ and minimum value$
f\left(\frac{5 \pi}{3}\right) =\sec \frac{5 \pi}{3}+\log \cos ^2 \frac{5 \pi}{3}=2+\log \left(\frac{1}{4}\right)$
$ =2-2 \log 2$

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