The number of critical points of the function f(x) = cases |(...

The critical points of a function are the points in its domain where the derivative is zero or does not exist.

First, consider $x = 0$. The function is continuous at $x = 0$ since $\lim_{x \to 0} \left|\dfrac{\sin x}{x}\right| = 1 = f(0)$.

For $x \in (-\pi, \pi)$, $\dfrac{\sin x}{x} > 0$, so $f(x) = \dfrac{\sin x}{x}$.

The derivative at $x = 0$ is given by $f'(0) = \lim_{x \to 0} \dfrac{\dfrac{\sin x}{x} - 1}{x} = \lim_{x \to 0} \dfrac{\sin x - x}{x^2} = 0$.

Since $f'(0) = 0$, $x = 0$ is a critical point.

Next, consider the points where $f(x) = 0$, which occurs at $x = \pi$ and $x = -\pi$ in the interval $(-2\pi, 2\pi)$.

At $x = \pi$, the inner function $g(x) = \dfrac{\sin x}{x}$ has $g(\pi) = 0$ and $g'(\pi) = -\dfrac{1}{\pi} \neq 0$. Since the derivative of the inner function is non-zero, the absolute value function $f(x) = |g(x)|$ is not differentiable at $x = \pi$. By symmetry, $f(x)$ is also not differentiable at $x = -\pi$.

Thus, $x = \pi$ and $x = -\pi$ are critical points.

For $x \neq 0, \pi, -\pi$, the function is differentiable and $f'(x) = \pm \dfrac{x \cos x - \sin x}{x^2}$.

Setting $f'(x) = 0$ gives $x \cos x - \sin x = 0 \Rightarrow \tan x = x$.

We analyze the roots of $\tan x = x$ in $(-2\pi, 2\pi)$:

In $(0, \pi)$, $\tan x > x$ for $x \in (0, \pi/2)$ and $\tan x < 0 < x$ for $x \in (\pi/2, \pi)$. There are no roots in this interval.

In $(\pi, 2\pi)$, $\tan x$ increases from $0$ to $\infty$ on $(\pi, 3\pi/2)$. Since $\tan \pi = 0 < \pi$ and $\tan x \to \infty$ as $x \to 3\pi/2^{-}$, there is exactly one root in $(\pi, 3\pi/2)$. In $(3\pi/2, 2\pi)$, $\tan x < 0 < x$, yielding no roots.

Since $\tan x = x$ is an odd equation, there is exactly one symmetric root in $(-2\pi, -\pi)$.

This gives $2$ additional critical points where $f'(x) = 0$.

The total number of critical points in $(-2\pi, 2\pi)$ is $1$ (at $x=0$) $+ 2$ (at $x=\pm \pi$) $+ 2$ (roots of $\tan x = x$) $= 5$.

Answer: $5$