Q:

If $x_1|f(x_2)|.$

Problem
Denote $f(x)=\dfrac{\sin x}{x},x>0.$ Let $x_1,x_2 $ be any two local extremum points of $f(x)$. Prove that if $x_1<x_2$, then $|f(x_1)|>|f(x_2)|.$
Note
I have obtained that：
1. $|f(x)|=\dfrac{|\sin x|}{x} \leq \dfrac{1}{x};$
2. There exists one and only one local extremum point over every interval $(k\pi,(k+1)\pi)$, where $k=1,2,\cdots.$
3. $f'(x)=\dfrac{x\cos x-\sin x}{x^2}.$ If $x_0$ is the local extremum point, then $x_0\cos x_0=\sin x_0$. Thus, $f(x_0)=\dfrac{\sin x_0}{x_0}=\cos x_0.$
This will  help?

A:

If $f(x) = \dfrac{\sin(x)}{x}$ has a local extremum at $x_0> 0$ then
$$ 
f'(x_0) = 0 \implies \tan(x_0) = x_0 \, .
$$
(Idea: Use this relation to get an expression for $|f(x_0|$ 
that does not involve trigonometric functions anymore.)
Using the trigonometric identity
$$
 \sin(x) = \pm \frac{\tan(x)}{\sqrt{1 + \tan(x)^2}}
$$
we get that at a the local extremum $x_0$
$$
 \sin(x_0) = \pm \frac{x_0}{\sqrt{1 + x_0^2}}
$$
 and therefore
$$
  |f(x_0)| = \frac{1}{\sqrt{1 + x_0^2}}  \, .
$$
The desired conclusion now follows because the right-hand side
is a (strictly) decreasing function of $x_0$.

