Find Number of Critical Points for Modulus Function in Given Interval

To find the number of critical points of the function $f(x)$ in the interval $(-2\pi, 2\pi)$, we first recall the definition of a critical point. A point $x_0$ in the domain of a function $f(x)$ is called a critical point if either $f'(x_0) = 0$ or $f'(x_0)$ does not exist.

The function is defined as: $f(x) = \begin{cases} \left|\frac{\sin x}{x}\right| &, x \neq 0 \\ 1 &, x=0 \end{cases}$

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Step 1: Differentiability at $x = 0$

For $x \in (-\pi, \pi) \setminus \{0\}$, we have $\frac{\sin x}{x} > 0$, which implies: $f(x) = \frac{\sin x}{x}$

Using the limit definition of the derivative at $x = 0$: $f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{\frac{\sin h}{h} - 1}{h} = \lim_{h \to 0} \frac{\sin h - h}{h^2}$

Using the Taylor series expansion $\sin h = h - \frac{h^3}{6} + O(h^5)$: $f'(0) = \lim_{h \to 0} \frac{-\frac{h^3}{6} + O(h^5)}{h^2} = 0$

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

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Step 2: Points where $f(x)$ is non-differentiable

The absolute value function $\left|\frac{\sin x}{x}\right|$ introduces potential non-differentiable points where $\sin x = 0$ for $x \neq 0$.

In the open interval $(-2\pi, 2\pi)$, $\sin x = 0$ occurs at $x = \pi$ and $x = -\pi$.

Let's evaluate the one-sided derivatives at $x = \pi$:

• Left-hand derivative ($f'_ -(\pi)$): For $x \in (0, \pi)$, $\sin x > 0$, so $f(x) = \frac{\sin x}{x}$. $f'_{-}(\pi) = \left. \frac{x \cos x - \sin x}{x^2} \right|_{x=\pi} = \frac{\pi (-1) - 0}{\pi^2} = -\frac{1}{\pi}$

• Right-hand derivative ($f'_ +(\pi)$): For $x \in (\pi, 2\pi)$, $\sin x < 0$, so $f(x) = -\frac{\sin x}{x}$. $f'_{+}(\pi) = \left. -\frac{x \cos x - \sin x}{x^2} \right|_{x=\pi} = \frac{1}{\pi}$

Since $f'_{-}(\pi) \neq f'_{+}(\pi)$, the derivative at $x = \pi$ does not exist. By symmetry, $f'(x)$ also does not exist at $x = -\pi$.

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

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Step 3: Points where $f'(x) = 0$ (Stationary Points)

For $x \in (-2\pi, 2\pi) \setminus \{0, -\pi, \pi\}$, the derivative of $f(x) = \pm \frac{\sin x}{x}$ is given by: $f'(x) = \pm \frac{x \cos x - \sin x}{x^2}$

Setting $f'(x) = 0$ yields: $x \cos x - \sin x = 0 \implies \tan x = x$

We count the non-zero solutions to $\tan x = x$ in $(-2\pi, 2\pi)$:

1. For $x \in (0, \pi)$:

   • On $\left(0, \frac{\pi}{2}\right)$, $\tan x > x > 0$.
   • On $\left(\frac{\pi}{2}, \pi\right)$, $\tan x < 0 < x$.
   • Thus, no roots exist in $(0, \pi)$.

2. For $x \in (\pi, 2\pi)$:

   • On $\left(\pi, \frac{3\pi}{2}\right)$, $\tan x$ increases continuously from $\tan \pi = 0$ to $+\infty$, while $x$ takes values in $\left(\pi, \frac{3\pi}{2}\right)$. Since $\tan \pi < \pi$, there is exactly one root in $\left(\pi, \frac{3\pi}{2}\right)$.
   • On $\left(\frac{3\pi}{2}, 2\pi\right)$, $\tan x < 0 < x$, so no roots exist.

3. For $x \in (-2\pi, 0)$:

   • By odd symmetry of the equation $\tan x = x$, there is exactly one root in $\left(-\frac{3\pi}{2}, -\pi\right)$.

Thus, there are 2 stationary points where $f'(x) = 0$ (excluding $x = 0$).

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Conclusion

Summing up all the critical points in the interval $(-2\pi, 2\pi)$:

1. $x = 0$ (where $f'(0) = 0$) $\rightarrow$ 1 point
2. $x = \pm \pi$ (where $f'(x)$ does not exist) $\rightarrow$ 2 points
3. Non-zero solutions of $\tan x = x$ (where $f'(x) = 0$) $\rightarrow$ 2 points

$\text{Total number of critical points} = 1 + 2 + 2 = 5$

Hence, the correct option is C.