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. 
A: Continuing from John Ma's answer, the approximate solutions are (have a look here)
$$x_n = q -\frac 1 q - \frac{2}{3q^3}  - \frac{13}{15q^5}  - \frac{146}{105q^7} +\cdots $$ where $q=\left(n+\frac{1}{2}\right)\pi$, odd $n$ corresponding to minimum values and even $n$ to maximum values of the function.
A: Proof of the fact that minimum is not attained anywhere outside the intervals $(\pi , 2\pi )$ and $(-2\pi, -\pi )$: we can concentrate on the positive side since the given function is even. Let $x >2\pi $. Then $|\frac {\sin x} x| <\frac 1 {2\pi }$ so $\frac {\sin x } x >-\frac 1 {2\pi }$. Now consider $\frac {\sin (\pi +\pi /4)} {\pi +\pi /4}=-\frac {\cos (\pi /4)} {\pi +\pi /4}=-\frac 1 {\sqrt 2} \frac 1 {{\pi +\pi /4}}<-\frac 1 {2\pi }< \frac {\sin x } x$. Hence the function does not attain its minimum at any point of $(2\pi , \infty )$. Obviously, the minimum value is negative so minumum is not attained on $(0,\pi )$. 
A: When $\sin x/ x $ is an extremum, by differentiation we get a transcendental equation 
$$ \tan x = x $$
The graphs of $\tan x, x $ intersect at infinitely many points, supplying maximum and minimum values of its roots in alternate order as shown:
WAgraph 
Only the first minima ( near$\approx 4.49341..$ ) are required. They can be found by e.g., Newton-Raphson iteration.
