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(D) For $x > 0$,the derivative is given by $f'(x) = \sin \frac{\pi}{x} - \frac{\pi}{x} \cos \frac{\pi}{x}$.We want to find the number of points $c \in

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infinite

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(D) For $x > 0$,the derivative is given by $f'(x) = \sin \frac{\pi}{x} - \frac{\pi}{x} \cos \frac{\pi}{x}$.We want to find the number of points $c \in (0, 1)$ such that $f'(c) = 0$.This implies $\sin \frac{\pi}{c} = \frac{\pi}{c} \cos \frac{\pi}{c}$,or $	an \frac{\pi}{c} = \frac{\pi}{c}$.Consider the intervals $I_k = \left[ \frac{1}{k+1}, \frac{1}{k} ight]$ for $k = 1, 2, 3, \dots$.At the endpoints of these intervals,$f\left(\frac{1}{k+1}ight) = \frac{1}{k+1} \sin((k+1)\pi) = 0$ and $f\left(\frac{1}{k}ight) = \frac{1}{k} \sin(k\pi) = 0$.Since $f(x)$ is continuous on $[0, 1]$ and differentiable on $(0, 1)$,by Rolle's Theorem,there exists at least one point $c_k \in \left( \frac{1}{k+1}, \frac{1}{k} ight)$ such that $f'(c_k) = 0$ for every $k \in \mathbb{N}$.Since there are infinitely many such intervals $I_k$ contained in $(0, 1)$,there are infinitely many points where $f'(x) = 0$.

Consider the function $f(x) = \begin{cases} x \sin \frac{\pi}{x} & \text{for } x > 0 \\ 0 & \text{for } x = 0 \end{cases}$. Then the number of points in $(0, 1)$ where the derivative $f'(x)$ vanishes is:

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