Consider the function $f(x)=\sqrt[3]{x-2}-x+4$ on $[1,3]$. $f$ is certainly continuous on $[1,3]$ and $f(1)=2=f(3)$, but $f$ is not differentiable on $(1,3)$ since $$f^\prime(x)=\frac{1}{3\sqrt[3]{(x-2)^2}}-1$$
is undefined at $x=2$.
So then this particular function fulfills two of the three hypotheses of Rolle's Theorem.
But we can still find $c\in(1,3)$ such that $f^\prime(c)=0$! To be specific, $f^\prime(c)=0$ for $c=2\pm\frac{1}{3\sqrt{3}}\approx1.8075,2.1925$.
Of course the failure of one of the conditions of Rolle's Theorem does not guarantee the nonexistence of $c$ values such that $f^\prime(c)=0$, but I still have questions…
Here $f$ fails to be differentiable at $x=2$ because of a vertical tangent, and not because $f$ has a sharp point. In particular, we should be able to say that graph of $f$ can be given as a smooth curve $\gamma:[a,b]\to\mathbb{R}^2$, correct?
If so, does this "smooth curve property" allow us to extend Rolle's Theorem to the nondifferentiable case where we have a vertical tangent but no sharp point? I've tried to come up with a counterexample, but failed thus far.
Best Answer
You will not be able to find a counterexample. In fact, if we extend the concept of differentiable function in such a way that $f'(x)=\pm\infty$ is allowed, Rolle's theorem is still true. I will not prove that because it is actually the same proof as the proof of the standard version.