Prove there are $x_1,x_2\in(a,b)$ such that $\frac{f(x_2)-f(x_1)}{x_2-x_1}=f'(\xi)$ Let $f(x) \in C^2$ in $(a,b)$ and $f''(\xi)\not=0, \xi \in (a,b)$. Prove there are $x_1,x_2\in(a,b)$ such that $$\frac{f(x_2)-f(x_1)}{x_2-x_1}=f'(\xi) \tag{1}$$ So I don't really understand what's the dufficulty of that task. Why can't we just say that as $f(x)$ is differentiable on $(a,b)$ so let's just take some $x_1,x_2\in (a,b)$ hence it's continous on $[x_1,x_2]$ hence it satisfies all demands for Lagrange theorem. And $(1)$ is automatically right, isn't it?

Hint: Special case: $f'(\xi) = 0$ and $f''(\xi) >0.$ Because $f$ is $C^2,$ we have $f''>0$ in some neighborhood of $\xi.$ This is enough to give a strict local minimum of $f$ at $\xi.$ Now stare at the picture to see there are $x_1<\xi < x_2$ such that $f(x_1) = f(x_2);$ of course you need to prove this. This gives the conclusion for this case.
