How prove that there exists $\xi\in(a,b)$ with $f'(\xi)=\frac{f(\xi)-f(a)}{b-a}$ 
Let $f(x)$ be continuous on $[a,b]$, differentiable on $(a,b)$, and with some $c\in(a,b)$ such that $f'(c)=0$. Show: There exists $\xi\in(a,b)$ such that
  $$
f'(\xi)=\dfrac{f(\xi)-f(a)}{b-a}
$$

My idea: I know this: Let
$$F(x)=e^{-\dfrac{x}{b-a}}[f(x)-f(a)]$$
$$\Longrightarrow F'(x)=-\dfrac{1}{b-a}e^{-\dfrac{x}{b-a}}[f(x)-f(a)]+e^{-\dfrac{x}{b-a}}f'(x)$$
$$\Longrightarrow F'(c)=e^{-\dfrac{c}{b-a}}[f'(c)-\dfrac{f(c)-f(a)}{b-a}]=-\dfrac{F(c)}{b-a}$$
If $f'(x)$ is continuous, I can prove this problem, 
But this problem condition can't have $f'(x)$ have continuous, so I can't continue
 A: Consider the function
$$
g(x)=\exp\left(\frac{b-x}{b-a}\right)\Big[f(x)-f(a)\Big]\tag{1}
$$
Then
$$
g'(x)=-\frac1{b-a}\exp\left(\frac{b-x}{b-a}\right)\Big[f(x)-f(a)-(b-a)f'(x)\Big]\tag{2}
$$
If $f'(c)=0$, then
$$
g(a)=0\tag{3}
$$
and
$$
g(c)=\exp\left(\frac{b-c}{b-a}\right)\Big[f(c)-f(a)\Big]\tag{4}
$$
and
$$
g'(c)=-\frac1{b-a}\exp\left(\frac{b-c}{b-a}\right)\Big[f(c)-f(a)\Big]\tag{5}
$$
If $g(c)=0$ there is a $\xi\in(a,c)$ so that $g'(\xi)=0$.
If $g(c)\gt0$, then $g'(c)\lt0$. Thus, $g$ has a maximum at $\xi\in(a,c)$ so that $g'(\xi)=0$.
If $g(c)\lt0$, then $g'(c)\gt0$. Thus, $g$ has a minimum at $\xi\in(a,c)$ so that $g'(\xi)=0$.
Therefore, there is a $\xi\in(a,c)$ so that
$$
0=g'(\xi)=-\frac1{b-a}\exp\left(\frac{b-\xi}{b-a}\right)\Big[f(\xi)-f(a)-(b-a)f'(\xi)\Big]\tag{6}
$$
and thus,
$$
\frac{f(\xi)-f(a)}{b-a}=f'(\xi)\tag{7}
$$
A: 
${\bf Lemma}$. Let $g:[0,1]\to \Bbb{R}$ be a continuous function which
  is has a derivative on $(0,1)$. Suppose that $g(0)=0$, and that there
  is a $c\in(0,1)$ with $g'(c)=0$. Then, there is
  $\xi\in(0,1)$ with $g'(\xi)=g(\xi)$.

${\it Proof.}$ consider $h:[0,1]\to\Bbb{R}$ defined by $h(x)=g(x)e^{-x}$. We consider two cases:


*

*Case 1: There is a $\beta\in(0,1]$ with $g(\beta)=0$. It follows that $h(0)=h(\beta)$, and using Rolle's Theorem, there is a $\xi\in(0,1)$ with $h'(\xi)=0$, which is equivalent to $g'(\xi)=g(\xi)$.

*Case 2: $g(x)\ne0$ for every $x\in(0,1]$. Thus, by the Intermediate  Value Theorem $g$ keeps a constant sign on the interval $ (0,1]$. Replacing $g$ by $-g$ if necessary, we may suppose that $g(x)>0$ for every $x$ in $(0,1]$.
Now, from $h'(c)=-g(c)e^{-c}<0$ we conclude that there $\alpha\in (0,c)$ such that
$$
\forall\,x\in[\alpha,c),\quad\frac{h(x)-h(c)}{x-c}<0
$$
that is $h(\alpha)>h(c)$. Now the function $x\mapsto h(x)-h(c)$ changes its sign on the interval $[0,\alpha]$, So there is a $\gamma\in(0,\alpha)$ such that $h(\gamma)=h(c)$. Again, using Rolle's Theorem, there is a $\xi\in(\gamma,c)$ with $h'(\xi)=0$, which is equivalent to $g'(\xi)=g(\xi)$.


This concludes the proof of the Lemma.$\qquad\square$
Now, the general case of the proposed question follows by applying the lemma
to the function $g(x)=f(a+x(b-a))-f(a)$.
A: Let
$$F(x)=e^{\frac{x}{a-b}}\cdot(f(x)-f(a)),\ x\in[a,b],$$
then
$$F'(x)=e^{\frac{x}{a-b}}\cdot\left(f'(x)-\frac{f(x)-f(a)}{b-a}\right),\ x\in(a,b).$$
We only need to prove there exists $\xi\in(a,b)$ such that $F'(\xi)=0$.
Otherwise, by Darboux's theorem, $F'(x)>0$ or $F'(x)<0$ when $x\in(a,b)$.
Without loss of generality, we assume that $F'(x)>0$. So $F$ is strictly increasing on $[a,b]$, and $F(c)>0=F(a)$ which implies $f(c)>f(a)$. Consequently
$$F'(c)=e^{\frac{c}{a-b}}\cdot\frac{f(c)-f(a)}{a-b}<0.$$
This is in contradiction with $F'(x)>0$.
