Prove that the function $f$ given by $f(x) = \log |\cos x|$ is decreasing on $\left(0, \frac{\pi}{2}\right)$ and increasing on $\left(\frac{3\pi}{2}, 2\pi\right)$.
Given $f(x) = \log |\cos x|$.
First,we find the derivative $f'(x)$:
$f'(x) = \frac{1}{\cos x} \cdot (-\sin x) = -\tan x$.
Case $1$: For $x \in \left(0, \frac{\pi}{2}\right)$:
In the first quadrant,$\tan x > 0$.
Therefore,$f'(x) = -\tan x < 0$.
Since $f'(x) < 0$ for all $x \in \left(0, \frac{\pi}{2}\right)$,the function $f(x)$ is strictly decreasing on $\left(0, \frac{\pi}{2}\right)$.
Case $2$: For $x \in \left(\frac{3\pi}{2}, 2\pi\right)$:
In the fourth quadrant,$\tan x < 0$.
Therefore,$f'(x) = -\tan x > 0$.
Since $f'(x) > 0$ for all $x \in \left(\frac{3\pi}{2}, 2\pi\right)$,the function $f(x)$ is strictly increasing on $\left(\frac{3\pi}{2}, 2\pi\right)$.
First,we find the derivative $f'(x)$:
$f'(x) = \frac{1}{\cos x} \cdot (-\sin x) = -\tan x$.
Case $1$: For $x \in \left(0, \frac{\pi}{2}\right)$:
In the first quadrant,$\tan x > 0$.
Therefore,$f'(x) = -\tan x < 0$.
Since $f'(x) < 0$ for all $x \in \left(0, \frac{\pi}{2}\right)$,the function $f(x)$ is strictly decreasing on $\left(0, \frac{\pi}{2}\right)$.
Case $2$: For $x \in \left(\frac{3\pi}{2}, 2\pi\right)$:
In the fourth quadrant,$\tan x < 0$.
Therefore,$f'(x) = -\tan x > 0$.
Since $f'(x) > 0$ for all $x \in \left(\frac{3\pi}{2}, 2\pi\right)$,the function $f(x)$ is strictly increasing on $\left(\frac{3\pi}{2}, 2\pi\right)$.
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