Mean Value Theorem

Theorem
Let $f$ be a real function which is continuous on the closed interval $\left[{a \,.\,.\, b}\right]$ and differentiable on the open interval $\left({a \,.\,.\, b}\right)$.

Then:
 * $\exists \xi \in \left({a \,.\,.\, b}\right): f' \left({\xi}\right) = \dfrac {f \left({b}\right) - f \left({a}\right)} {b - a}$

Proof
For any constant $h \in \R$ we may construct the real function defined on $\left[{a \,.\,.\, b}\right]$ by:
 * $F \left({x}\right) = f \left({x}\right) + h x$

We have that $h x$ is continuous on $\left[{a \,.\,.\, b}\right]$ from Linear Function is Continuous.

From the Sum Rule for Continuous Functions, $F$ is continuous on $\left[{a \,.\,.\, b}\right]$ and differentiable on $\left({a \,.\,.\, b}\right)$.

Let us choose the constant $h$ such that $F \left({a}\right) = F \left({b}\right)$:

Since $F$ satisfies the conditions for the application of Rolle's Theorem:
 * $\exists \xi \in \left({a \,.\,.\, b}\right): F' \left({\xi}\right) = 0$

But then:
 * $F' \left({\xi}\right) = f' \left({\xi}\right) + h = 0$

The result follows.

Also see

 * Mean Value Theorem for Integrals