Extreme points of $f(x)= 2x+\cot x$

Question

I want to find the extreme points of the function $f(x) =2x+\cot x$.
I began by observing that the function is defined on $D=\mathbb{R} - \{k\pi | k\in \mathbb{Z} \} $. Now, $f'(x) =2-\frac{1}{\sin^2 x} $ $\forall x \in D$. I tried to study $f'$'s sign, but I don' t know how to get around the points where the function is not defined.

Answer 1

$D_f = \mathbb{R} - \{k\pi, k\in \mathbb{Z}\}$

$f(x) = 2x+\cot x $

$f'(x) = 2 - \csc^2x$

At extreme points, $f'(x) = 0$

$2 = \csc^2x \implies \sin^2x = \frac{1}{2} = \sin^2\frac{\pi}{4}$

$$x = 2k\pi\pm\pi/4$$

These are the extreme points.

It'd be clear from the graph below of $f(x)$