Let $f(x)$ be a real differentiable funtion on $(a,b)$ and continuous on $[a,b]$. Prove that for any $c \in [a,b]$ not a supremum of $f'(x)$ there is $x_1,x_2 \in [a,b]$ such that $f'(c)=\frac{f(x_1)-f(x_2)}{x_1-x_2}$.
I tried many times but failed this problem. Let $k_1, k_2$ be functions such that $f'(k_1) < f'(c) < f'(k_2)$. It is pretty obvious that one of the $k_1, k_2$ must be $x_1$. I started by looking at the equation of rhe tanngent line that passes through each of the $k_i$'s, but that failed. The hint for this problem is that we must restrict a certain funtion $g$ roughly of the form $\frac{f(b)-f(a)}{b-a}$ to a subset of $[a,b]$ and show that $g$ can take negative and positive values, from which we may conclude by mean value theorem or intermediate value theorem that $g(x)=0$ for some $x$, which would give us the value of $x_1$ and $x_2$. I am not sure how to find such a $g$.
Not exactly the hint, but whatever.
Call $I=(a,b)$. Consider the set $U=\{(x,y)\in\Bbb R^2\,:\, a<x<y<b\}$. It is a connected subset of $\Bbb R^2$. Consider the map $\Delta:U\to\Bbb R$, $\Delta(x,y)=\frac{f(y)-f(x)}{y-x}$. Since $\Delta$ is continuous on $U$, the image is connected. So, it is an interval. By Lagrange's theorem, $\Delta(U)\subseteq f'(I)$ while, by definition of derivative, $\overline{\Delta(U)}\supseteq f'(I)$. Since the closure of an interval is just the interval itself plus its (at most) two extremal points, $f'(I)\setminus\Delta(U)$, being a subset of $\overline{\Delta(U)}\setminus\Delta(U)$, contains at most two points. These two points may only be $\sup_{x\in(a,b)}f'(x)$ or $\inf_{x\in(a,b)}f'(x)$ (though they might possibly be in $\Delta(U)$ in some case). This proves the theorem.