Let $f:[a,b]\rightarrow \mathbb{R} $ be differentiable with $f'(a) = f'(b)$. There exist a $c\in(a,b)$ such that $f'(c) = \frac{f(c) - f(a) }{c -a}$. Can someone help me with the following problem?

Let $f:[a,b]\rightarrow \mathbb{R} $ be differentiable and suppose that $f'(a) = f'(b)$. Show that there exist a $c\in(a,b)$ such that $f'(c) = \frac{f(c) - f(a) }{c -a}$.

My idea is to take first the case $f'(a) = f'(b) = 0$. In this case, consider 
$$\phi(x) = \frac{f(x)-f(a)}{x-a}, \quad \forall x \in (a,b],$$
and $\phi(a) = 0$. Is obvious that $\phi$ is continuous in $(a,b]$. While in $a$, the continuity of $\phi(a)$ follows from the fact $$\lim_{x\to a^{+}} \phi(x) = f'(a) = 0 = \phi(a).$$
Therefore, $\phi$ is continuous in $[a,b]$. So, $\phi$ attains its maximum and its minimum in $[a,b]$. If one of them is in $(a,b)$, let $c$ be this point. So, since $\phi$ is differentiable in $(a,b)$, I know that we must have $\phi'(c) = 0$. But
$$\phi'(x) = \frac{f'(x)}{x-a} - \frac{f(x)-f(a)}{(x-a)^2}.$$
So
$$\phi'(c) = 0 \Rightarrow f'(c) = \frac{f(c) - f(a) }{c -a}.$$
The problem is that I can't ensure that a point of maximum or minimum must be in $(a,b)$. 
Note that, once I solve this case ($f'(a) = 0$), the general case is solved just considering $g(x) = f(x) - f'(a) x$. Because $g'(a) = g'(b) = 0$. And then if there exists a $c\in(a,b)$ with 
$$ g'(c) = \frac{g(c)-g(a)}{c-a},$$
then
$$f'(c) - f'(a) = \frac{f(c) - f'(a)c -f(a) + f'(a)a}{c-a} = \frac{f(c) -f(a)}{c-a} - f'(a).$$
Therefore,
$$f'(c) = \frac{f(c) - f(a) }{c -a}.$$
 A: Assume that the only extremum points of $\phi$ are $a$ and $b$. Without loss of generality, we assume that $\phi(a) \leq \phi(x) \leq \phi(b)$ for all $x \in [a,b]$. The second inequality means that
$$
f(x) \leq f(a) + (x-a) \phi(b)
$$
for all $x \in [a,b]$. This yields, for any $x \in [a,b)$,
$$
\frac{f(b)-f(x)}{b-x} \geq \frac{f(b)-f(a)-(x-a)\phi(b)}{b-x} = \frac{f(b)-f(a)}{b-a}.
$$
Letting $x \to b^{-}$, we deduce $f'(b) \geq \phi(b)$. Since $f'(b)=f'(a)$, we have $f'(a) \geq \phi(b)$, so $\phi(a) \geq \phi(b)$. This implies that $\phi$ is constant, and this is a trivial case.
Hence either $\phi$ is constant, or $\phi$ has an interior critical point. In both cases, your proof is complete.
A: Thanks, guys, for the help! I solved of the following way:
First, we have that $\phi'(b)$ exist:
$$\begin{array}{rcl}
\frac{\phi(x) - \phi(b)}{x-b} &=& \frac{\frac{f(x)-f(a)}{x-a} - \frac{f(b)-f(a)}{b-a}}{x-b}\\
&=& \frac{1}{x-a}\left[\frac{f(x)-f(b)}{x-b}\right] + \frac{1}{x-a}\left[\frac{f(b)-f(a)}{x-b}\right] - \frac{1}{b-a}\left[\frac{f(b)-f(a)}{x-b}\right] \\
&=& \frac{1}{x-a}\left[\frac{f(x)-f(b)}{x-b}\right] + \left(\frac{1}{x-a} - \frac{1}{b-a}\right)\left[\frac{f(b)-f(a)}{x-b}\right] \\
&=& \frac{1}{x-a}\left[\frac{f(x)-f(b)}{x-b}\right] + \left(\frac{b-x}{(x-a)(b-a)}\right)\left[\frac{f(b)-f(a)}{x-b}\right] \\
&=& \frac{1}{x-a}\left[\frac{f(x)-f(b)}{x-b}\right] - \frac{1}{x-a}\left[\frac{f(b)-f(a)}{b-a}\right] \\
&\xrightarrow[x \to b^{-}]{} & \frac{f'(b) -\phi(b)}{b-a} = -\frac{\phi(b)}{b-a}
\end{array}$$
Where in the last equality I use that $f'(b) = 0$.
Therefore,
$$\phi'(b) = -\frac{\phi(b)}{b-a}.$$
Using the mean value theorem, we have $d\in(a,b)$ such that
$$\phi'(d) = \frac{\phi(b)-\phi(a)}{b-a} = \frac{\phi(b)}{b-a} = -\phi'(b).$$
So, by Darboux's theorem (the intermediate value theorem for the derivative, which does not require $\phi'$ continuos), there exist a $c^{*}\in [d,b]$ with $\phi'(c^{*}) = 0$. If $c^{*} \neq b$, we take $c = c^{*}$. If $c^{*} = b$, then $\phi'(d) = -\phi'(b) = 0$ and so we take $c = d$. And this way we can ensure that $c \in (a,b)$.
A: For the purpose of finding a contradiction, we may investigate $\phi'$ near $a$. If $c$ indeed happens to be $a$, then $\phi'(a)=0$. So, for the moment let's suppose $f$ is twice diferentiable at $a$, 
$$\lim_{x\rightarrow a}\phi'(x)=\lim_{x\rightarrow a}\left( \frac{f'(x)}{x-a}-\frac{f(x)-f(a)}{(x-a)^2}\right)=\lim_{x\rightarrow a}\left( \frac{f'(x)(x-a)-(f(x)-f(a))}{(x-a)^2}\right)$$
By the continuity and differentiability of $f$ at $a$ we get $0/0$ and so we can apply L'Hopital
$$\lim_{x\rightarrow a}\phi'(x)=\lim_{x\rightarrow a}\frac{f'(x)+f''(x)(x-a)-f'(x)}{2(x-a)}=\lim_{x\rightarrow a}\frac{f''(x)}{2}=\frac{1}{2}f''(a)$$
OK, so if $\phi'$ is continuous at $a$ we get
$$\phi'(a)=\frac{1}{2}f''(a)$$
This tells us that if $a$ is a critical point of $\phi$ then $f''(a)=0$, which is not in the hypothesis.
For the other endpoint, 
$$\phi'(b)=0 \implies f'(b)=\frac{f(b)-f(a)}{b-a}$$
Again, is not necessarily true. So, by the counter-positive, your problem is solved for all $f:[a,b]\rightarrow \mathbb{R}$ differentiable with $f'(a)=f'(b)$ also satisfying:
twice differentiable at $a$,
$$f''(a)\neq 0$$
and
$$f'(b)\neq\frac{f(b)-f(a)}{b-a}$$
My next step would be assuming each of those things and showing it still can't be done, but I ran out of ideas. Hope this can help you out.
