Exercise on a fixed end Lagrange's MVT Given a function f with derivative on all $[a;b]$ with $f'(a) = f'(b)$, show that there exists $c \in (a;b)$ such that $f'(c) = \frac{f(c)-f(a)}{c-a}$.
This is some kind of MVT with constraint. I have a proof but it uses Darboux's theorem. Can you prove it without using it?

Sketch of the proof I have using Darboux
I suppose first that $f'(a) = 0$ (it's easy to get back to this case with an affine transform) and I set $g(x) = \frac{f(x)-f(a)}{x-a}$. $g'(b) = - \frac{g(b)-g(a)}{b-a} = - g'(d)$ for some $d \in (a;b)$ by Lagrange's MVT. Then $g'(b)$ and $g'(d)$ are either null or have different signs in which case we can find $e \in (d;b)$ such that $g'(e) = 0$. In all cases $g'$ has a zero which is the $c$ we need to find.

This paper https://arxiv.org/pdf/1309.5715.pdf contains two nice proofs of Flett's MVT, which is a statement of your question. Of course, Riedel and Sahoo book I pointed in the comment is also good.
