Let f(x) = (x - 1)^(2/3) on [0, 2], where the real cube root is used…

Step 1. The ends. $f(0) = (-1)^{2/3} = 1$ and $f(2) = 1^{2/3} = 1$. They are equal, and $f$ is continuous on $[0, 2]$.

Step 2. Differentiability at $1$. $\dfrac{f(1 + h) - f(1)}{h} = \dfrac{h^{2/3}}{h} = h^{-1/3}$. It grows without bound in size, so $f$ is not differentiable at $1$. The answer is $0$.

Step 3. The slope elsewhere. For $x \ne 1$, $f'(x) = \dfrac{2}{3}(x - 1)^{-1/3}$, which is never zero.

Step 4. The lesson. The conditions of Rolle's theorem fail at $x = 1$, and the conclusion fails too. There are $0$ points with $f'(x) = 0$.