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?
 A: Replacing $f$ by $-f$ if required, we may assume without changing the conclusion that $f''(\xi) > 0$. The following is an instance of such $y = f(x)$ near the point $(\xi, f(\xi))$:
$\hspace{9em}$
This graph suggests how one can find such points $x_1$ and $x_2$. You can simply translate the tangent line little up and choose any two intersecting points. (The assumption on $f''(\xi)$ will guarantee the existence of such points.)
Let us formalize this idea. Since $f$ is $C^2$, there is $\delta > 0$ and $c > 0$ such that
$$ |x - \xi| \leq \delta \quad \Rightarrow \quad f''(x) \geq c.$$
Then Taylor's theorem tells that for $x \in [\xi-\delta,\xi+\delta]$ we have
$$ f(x) \geq f(\xi) + f'(\xi)(x - \xi) + \frac{c}{2}(x - \xi)^2. \tag{*} $$
Now define $g(x)$ by
$$ g(x) = f(x) - f(\xi) - f'(\xi)(x - \xi). $$
By $\text{(*)}$, we know that 
$$ g(\xi \pm \delta) \geq \frac{c\delta^2}{2} \quad \text{and} \quad g(\xi) = 0. $$
So by the intermediate value theorem, there exist $x_1 \in [\xi-\delta, \xi)$ and $x_2 \in (\xi, \xi+\delta]$ such that $g(x_1) = c\delta^2/2 = g(x_2)$. Therefore we have
$$ \frac{f(x_1) - f(x_2)}{x_1 - x_2} - f'(\xi) = \frac{ g(x_1) - g(x_2)}{x_1 - x_2} = 0 $$
and the claim follows.

Without the assumption $f''(\xi) \neq 0$ the conclusion may be false as you can see from the example $f(x) = x^3$ with $\xi = 0$.
A: 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.
