Mean Value Theorem in two intervals Suppose that $a&lt;b&lt;c$ for real numbers $a, b, c$. Suppose that $f(x)$ is continuous on $[a, c]$ and differentiable on $(a, c)$. Then prove that there exists $a&lt;\alpha&lt;\beta&lt;c$ such that
$$\frac{f(b)-f(a)}{b-a}=f'(\alpha), \frac{f(c)-f(a)}{c-a}=f'(\beta)$$

I understand that there exists $\alpha$, $\beta$, but how would I prove that $\alpha&lt;\beta$?
Do I have to divide into cases? Can anyone give me any hints?
Define $S(x, y) = \frac {f(y) - f(x)} {y - x}$.

We have $\alpha \in (a, b), \gamma \in (b, c)$ such that $f'(\alpha) = S(a, b)$ and $f'(\gamma) = S(b, c)$.

$S(a, c)$ lies between $S(a, b)$ and $S(b, c)$, so the intermediate value property for derivatives gives us $\beta \in (\alpha, \gamma)$ such that $f'(\beta) = S(a, c)$.
Is your function convex or concave?
For convex function it's proved easily using the second order derivative.
But for concave,I don't know