Rolle theorem proof

Here is an answer to the wrong question (using MVT to prove Rolle's), followed by an answer to the question I think you were asking.

You can almost certainly use the MVT to prove Rolle's -- indeed, Rolle's is the MVT in the special case where $f(a) = f(b)$. But usually Rolle's is used to prove the MVT, so to make this an "honest" proof, you'd need an alternative proof of the MVT.

NB Actually, having edited the question, I realize OP's asking about the INTERMEDIATE value theorem, not the MEAN value theorem.

To answer one of the questions asked: if the conditions of Rolle's theorem are achieved, does that mean that $f'$ is continuous? The answer is no. Let $$ f(x) =\begin{cases} 0 & x = 0 \\ x^2 \sin(\frac{1}{x}) & \text{else}\end{cases}. $$ Then $f$ is differentiable everywhere, has $f(-1/\pi) = f(1/\pi) = 0$, but $f'$ is not continuous at $x = 0$.

Because we cannot assume that $f'$ is continuous, your proof of Rolle via IVT doesn't seem like it's going to work, no.

+1 for a nice question. It is rather strange that the question's concerns were not addressed fully for such a long time and no one noticed it (perhaps due to the fact that it was marked accepted).

The basic premise of the question $f'(a)f'(b) \leq 0$ is wrong. First point is that differentiability of $f$ at $a, b$ is not given so one can't talk of $f'(a) $ and $f'(b) $. Secondly even if that is allowed by modifying the hypotheses (ie assume $f'$ exists in $[a, b] $) it does not follow that $f'(a) f'(b) \leq 0$.

The inequality in question should be $$\frac{f(x) - f(a)} {x-a} \cdot\frac{f(x) - f(b)} {x-b} \leq 0$$ but from here we can't go to $f'(a) f'(b) \leq 0$ via limiting procedure as we can't take $x\to a$ and $x\to b$ simultaneously.

The conclusion does not hold in general. However if $f$ is not constant on $[a, b] $ we can apply one of the proofs of IVT and get a subinterval $[p, q] $ such that $f(p) =f(q) =f(a) $ and $f(x) \neq f(a) $ for $x\in(p, q) $. And then one can prove easily that $f'(p) f'(q) \leq 0$.

Also another curious point is that the derivative need not be continuous (as shown in another answer here) and asker's intent is to use IVT via continuity of $f'$ to get an $f'(c) =0$. Well the derivative may not be continuous but it satisfies intermediate value property via Darboux theorem and we indeed get a point $c\in[p, q] $ with $f'(c) =0$.

The proof of Rolle's theorem as well as Darboux theorem are based on the same two ideas:

• A continuous function on a closed interval takes its minimum and maximum values.
• The sign of derivative at a point gives us information about the increasing/decreasing nature of function at a point (this is an immediate consequence of definition of derivative) so that derivative at interior extremum points must vanish.

So in essence the approach suggested in the question is a roundabout way of proving Rolle's theorem. Its best to rely on the usual proof.