Q:

How to find the minimum value of $f(x)=\dfrac{\sin{x}}{x},x \neq 0.$?

Problem
Find the minimum value of $$f(x)=\dfrac{\sin{x}}{x},(x \neq 0).$$
Note
Maybe, you would consider finding its derivative. But, in fact, there exist infinitely many extremum points over its domain. This way can't work. By graphing, we can obviously see that there exist a minimum value point locating over $(\pi,2\pi)$ and the symmetric interval $(-2\pi,-\pi).$
How to show this fact?

A:

We consider only $x\ge 0$. Note $f\left( \frac{3\pi}{2}\right) = -\frac{2}{3\pi}$ and 
$$f(x) = \frac{\sin x}{x} \ge -\frac{1}{x} \ge -\frac{1}{3\pi}> -\frac{2}{3\pi}$$
when $x\ge 3\pi$. So the global minimum occurs in the interval $(\pi, 2\pi)$. Taking derivative: 
$$ f'(x) = \frac{\cos x}{x} - \frac{\sin x}{x^2}$$
So the global minimum $x_m$ should satisfy 
$$ x = \tan x, \ \ \ x\in (\pi, 2\pi).$$
It seems you have to find that numerically. 

