Consider the function $\mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{l}\mathrm{xsin}\frac{\mathrm{\pi }}{\mathrm{x}}, \mathrm{for} \mathrm{x}>0\\ 0, \mathrm{for} \mathrm{x}=0\end{array}\right.$ Then, the number of points in (0, 1) where the derivative f '(x) vanishes is

f (x) vanishes at points where

$\begin{array}{l} \sin \frac{\mathrm{\pi }}{\mathrm{x}}=0, \mathrm{i}.\mathrm{e}., \frac{\mathrm{\pi }}{\mathrm{x}}=\mathrm{k\pi }, \mathrm{k}=1,2,3,4.....\\ \mathrm{Hence},\mathrm{x}=\frac{1}{\mathrm{k}}.\\ \mathrm{Also}, \mathrm{f}\text{'}\left(\mathrm{x}\right)=\sin \frac{\mathrm{\pi }}{\mathrm{x}}-\frac{\mathrm{\pi }}{\mathrm{x}} \cos \frac{\mathrm{\pi }}{\mathrm{x}} \mathrm{if} \mathrm{x}\ne 0.\end{array}$

Since the function has a derivative at any interior point of the interval (0,1), it is continuous in [0,1], and f(0)=f(1), Rolle's theorem is applicable to any one of the intervals

      $\left[\frac{1}{2},1\right],\left[\frac{1}{3},\frac{1}{2}\right]=,...., \left[\frac{1}{\mathrm{k}+1},\frac{1}{\mathrm{k}}\right]$

Hence, there exists at least one c in each of these intervals where $\mathrm{f}\text{'}\left(\mathrm{c}\right)=0$. Therefore, there are infinite points.