Rolle theorem proof I'm wondering if we can use the Intermediate Value Theorem to prove Rolle's Theorem.
The hypotheses of Rolle's Theorem are:


*

*The function should be continuous on a closed interval $[a,b]$.

*The function should be differentiable on the open interval $(a,b)$.

*$f(a)=f(b)$
The theorem then shows that there exist $c$ in $(a,b)$ such that  $f'(c)=0$.
For I.V.T we need the function to be continuous and $f'(a).f'(b) \leq 0$.
If the conditions for Rolle's Theorem conditions are achieved, does it mean that $f'$ is continuous?
For the second condition we have that $$ \frac{f(x)-f(a)}{x-a} .\frac{f(x)-f(b)}{x-b}<0$$ because $$ f(x)-f(a)=f(x)-f(b)$$ and $$ 0<x-a<b-a $$
$$ a-b<x-b<0 $$
 A: 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. 
A: +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. 
A: Yes, we can use the IVT to prove Rolle's theorem, but not in the way you expect.
The idea is to construct a sequence of nested intervals, each half as wide as the previous, where the value of $f$ is the same at both endpoints. These will converge to a point where $f'(x)=0$.

Set $a_0=a$ and $b_0=b$. The $n$th interval will be $[a_n,b_n]\subset[a_{n-1},b_{n-1}]$, with length $b_n-a_n=(b_{n-1}-a_{n-1})/2=(b_0-a_0)/2^n$.
(We'll start with $n=0$ and use induction, of course.) Consider the function
$$g(x)=f(x)-f\left(x+\tfrac{b_n-a_n}{2}\right)$$
defined for $a_n\leq x\leq\tfrac{a_n+b_n}{2}$, so $\tfrac{a_n+b_n}{2}\leq x+\tfrac{b_n-a_n}{2}\leq b_n$. Note that $g$ is continuous on this interval, because $f$ is continuous. Also, since $f(a_n)=f(b_n)$,
$$g(a_n)=f(a_n)-f\left(\tfrac{a_n+b_n}{2}\right)=-g\left(\tfrac{a_n+b_n}{2}\right).$$
If this value is $g(a_n)=0$, then $f(a_n)=f\left(\tfrac{a_n+b_n}{2}\right)=f(b_n)$, and we can simply bisect the interval and proceed with either half: $[a_{n+1},b_{n+1}]=\left[a_n,\tfrac{a_n+b_n}{2}\right]$ or $\left[\tfrac{a_n+b_n}{2},b_n\right]$. (If this happens for both $n=0$ and $n=1$, then we should take the interval not containing $a_0$ nor $b_0$, so we don't get convergence to an endpoint. That is, $[a_2,b_2]=\left[\tfrac{3a+b}{4},\tfrac{a+b}{2}\right]$ or $\left[\tfrac{a+b}{2},\tfrac{a+3b}{4}\right]$, instead of $\left[a,\tfrac{3a+b}{4}\right]$ or $\left[\tfrac{a+3b}{4},b\right]$.)
If $g(a_n)\neq0$, then $g$ has opposite signs at the endpoints of its domain, so by IVT, there is some $x\in\left(a_n,\tfrac{a_n+b_n}{2}\right)$ such that $g(x)=0$. Set $[a_{n+1},b_{n+1}]=\left[x,x+\tfrac{b_n-a_n}{2}\right]$, so $f(a_{n+1})=f(b_{n+1})$.
Now we have the desired sequence of nested intervals; they converge to some point $c\in(a,b)$.

We assumed $f$ is differentiable on $(a,b)$, so $f'(c)$ exists.
If $c=a_m$ for some $m\in\mathbb N$, then $a_n=c$ for all $n>m$ (since $c=a_m\leq a_n\leq c$ by the nesting). This implies
$$0=\frac{f(b_n)-f(a_n)}{b_n-a_n}=\frac{f(b_n)-f(c)}{b_n-c}$$
$$0=\lim_{n\to\infty}\frac{f(b_n)-f(c)}{b_n-c}=f'(c).$$
Similarly if $c=b_m$ for some $m$ then $f'(c)=0$. So assume $c\in(a_m,b_m)$ for all $m$. (Thus the following denominators are non-zero.)
$$0=\frac{f(b_n)-f(a_n)}{b_n-a_n}=\frac{f(b_n)-f(c)+f(c)-f(a_n)}{b_n-a_n}$$
$$=\frac{b_n-c}{b_n-a_n}\cdot\frac{f(b_n)-f(c)}{b_n-c}+\frac{c-a_n}{b_n-a_n}\cdot\frac{f(c)-f(a_n)}{c-a_n}.$$
From the definition of the derivative, for any $\varepsilon>0$, there exists $m\in\mathbb N$ such that
$$-\varepsilon<\frac{f(b_n)-f(c)}{b_n-c}-f'(c)<\varepsilon$$
for all $n>m$, and there also exists $l\in\mathbb N$ such that
$$-\varepsilon<\frac{f(a_n)-f(c)}{a_n-c}-f'(c)<\varepsilon$$
for all $n>l$. Therefore, for all $n>\max\{l,m\}$,
$$0=\frac{b_n-c}{b_n-a_n}\cdot\frac{f(b_n)-f(c)}{b_n-c}+\frac{c-a_n}{b_n-a_n}\cdot\frac{f(c)-f(a_n)}{c-a_n}$$
$$<\frac{b_n-c}{b_n-a_n}\cdot\Big(f'(c)+\varepsilon\Big)+\frac{c-a_n}{b_n-a_n}\cdot\Big(f'(c)+\varepsilon\Big)$$
$$=(1)\big(f'(c)+\varepsilon\big)$$
$$0-\varepsilon<f'(c)$$
and similarly
$$0=\frac{b_n-c}{b_n-a_n}\cdot\frac{f(b_n)-f(c)}{b_n-c}+\frac{c-a_n}{b_n-a_n}\cdot\frac{f(c)-f(a_n)}{c-a_n}$$
$$>\frac{b_n-c}{b_n-a_n}\cdot\Big(f'(c)-\varepsilon\Big)+\frac{c-a_n}{b_n-a_n}\cdot\Big(f'(c)-\varepsilon\Big)$$
$$=(1)\big(f'(c)-\varepsilon\big)$$
$$0+\varepsilon>f'(c)$$
so we have $-\varepsilon<f'(c)<\varepsilon$. Since $\varepsilon$ was arbitrary, we conclude that $f'(c)=0$.
