# Prove that the function

Question:

Prove that the function $f$ given by $f(x)=\log \cos x$ is strictly decreasing on $\left(0, \frac{\pi}{2}\right)$ and strictly increasing on $\left(\frac{\pi}{2}, \pi\right)$.

Solution:

We have,

$f(x)=\log \cos x$

$\therefore f^{\prime}(x)=\frac{1}{\cos x}(-\sin x)=-\tan x$

In interval $\left(0, \frac{\pi}{2}\right), \tan x>0 \Rightarrow-\tan x<0$

$\therefore f^{\prime}(x)<0$ on $\left(0, \frac{\pi}{2}\right)$

$\therefore f$ is strictly decreasing on $\left(0, \frac{\pi}{2}\right)$.

In interval $\left(\frac{\pi}{2}, \pi\right), \tan x<0 \Rightarrow-\tan x>0$

$\therefore f^{\prime}(x)>0$ on $\left(\frac{\pi}{2}, \pi\right)$

$\therefore f$ is strictly increasing on $\left(\frac{\pi}{2}, \pi\right)$.