assume $f:[a,b] \to \mathbb R$ such that $f '(a)=f '(b)$ how prove $\exists t\in(a,b) : f(t)-f(a)=f '(t)(t-a)$? let $f:[a,b] \to \mathbb R$ such that $f '(a)=f '(b)$ how prove $$\exists t\in(a,b) : f(t)-f(a)=f '(t)(t-a)$$
Thanks in advance 
 A: I found an idea and a partial solution. I assume that $f'$ is continuous (perhaps the idea can be more easily finished if $f\in C^2$ is assumed). 
Let $g$ denote $f'$, then $f(t)-f(a)=\int_a^tg$, and the statement means that there exist a $t$ such that $g(t)$ is the average of all $g(x)$ for $x\in [a,t]$.
Consider the function $G(t):=\displaystyle\frac{\int_a^tg}{t-a}$ if $t>a$ and $G(a):=g(a)$. This is continuous, and therefore so is $G-g$ on $[a,b]$. Intuitively, we could perhaps restrict it to an interval $[a,b_1]$ where either $g\ge g(a)$ or $g\le g(a)$, and $g(b_1)=g(a)$ again. Then, by checking the signs of $G-g$ and using the mean value theorem, we're done.
However, this arguement as it stands don't apply to continuous functions like $g(x)=x\cdot \sin(1/x)$ (e.g. on $[0,1]$).
For that, we might build up a sequence, with $t_0:=b$, we have $G(b)=g(t_1)$ for some $t_1\in (a,b)$ by Lagrange's intermediate value thm, and so on, ($G(t_1)=g(t_2)$ for some $t_2\in (a,t_1)$...) Though $t_i$ must be convergent, I can't see why can't we have $t_i\to a$, so it makes me uncertain..
