# Rolle's Theorem and vertical tangents

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.

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.