If $f''(x)$ exist in $(a, b)$ then there exist $\xi \in (a, b)$ so that $f''(\xi)(b-a)=f'(b)-f'(a)$ Let $f$ be differentiable on $[a, b]$ and $f'$ is differentiable on $(a, b)$. How can we prove that there exists $\xi \in (a, b)$, such that $f'(b)-f'(a)=f''(\xi)(b-a)$ ?
We cannot use the mean value theorem because $f'(x)$ may have discontinuities at $a$ or $b$.
I came up with this: If $f''(\xi_1)>\frac{f'(b)-f'(a)}{b-a}$ and $f''(\xi_2)<\frac{f'(b)-f'(a)}{b-a}$ for some $\xi_1, \xi_2$, then by Darboux's theorem there will be $\xi$, such that $f''(\xi)=\frac{f'(b)-f'(a)}{b-a}$. Therefore we can assume $f''(x)>\frac{f'(b)-f'(a)}{b-a}$ or $f''(x)<\frac{f'(b)-f'(a)}{b-a}$ for all $x$.
Am I going the right direction?
 A: Replacing $f$ with the function $g(x)=f(x)+cx^2$ for appropriately chosen $c$, we may assume that $f'(a)=f'(b)$ (adding $cx^2$ to $f(x)$ does not change whether the equation $f'(b)-f'(a)=f''(\xi)(b-a)$ is true since it just adds $2c(b-a)$ to both sides).  So, assuming $f'(a)=f'(b)$, we want to show there exists $\xi\in(a,b)$ such that $f''(\xi)=0$. [This step is similar to how you reduce the mean value theorem to Rolle's theorem by adding a linear function.]
As you have said, using Darboux's theorem, we may assume that either $f''(x)>0$ or $f''(x)<0$ for all $x\in(a,b)$; let's say $f''(x)>0$ (the other case is similar).  This means $f'$ is strictly increasing on $(a,b)$.  But together with $f'(a)=f'(b)$, this violates Darboux's theorem: picking any $d\in (a,b)$, suppose $f'(d)>f'(a)$.  Then $f'(d)>f'(b)$ so by Darboux there exists $e\in (d,b)$ such that $f'(d)>f'(e)>f'(b)$, contradicting that $f'$ is increasing on $(a,b)$.  Similarly, if $f'(d)<f'(a)$, there is $e\in (a,d)$ such that $f'(a)>f'(e)>f'(d)$ which is again a contradiction.  So we must have $f'(d)=f'(a)$ for all $d\in (a,b)$, but that again contradicts the fact that $f'$ is strictly increasing on $(a,b)$.
