Given the function $f(x) = 1 - |x|^{2/3}$ which satifies that $f(-1) = f(1) = 0$ show that$\nexists c \in ]-1,1[$ s.t. $f'(c) = 0$.
Furthermore I have to conclude why this does not contradict Rolle's Theorem.
May I ask for help? I calculated $f'(x) = \frac{-2x}{3|x| \cdot \sqrt[3]{x}}$ and thus $f'(-1) = 2/3$ and $f'(1) = -2/3$ but this didn't help me very much. What do I do?
$f $ is continuous at $[-1,1]$
$$f(1)=f(-1)$$
but $f$ is not differentiable at $ (-1,1)$ since it is not differentiable at $x=0$.
in fact,
$$\lim_{x\to 0^+}\frac{f(x)-f(0)}{x-0}=\lim_{x\to 0^+}\frac{-2}{3\sqrt[3]{x}}=-\infty$$
thus, we cannot apply Rolle's theorem.
For $x\ne 0$, $$f'(x)=\frac{\pm 2}{3\sqrt[3]{x}}\ne 0$$