Mean Value Theorem

Theorem
Let $f$ be a real function which is continuous on the closed interval $\closedint a b$ and differentiable on the open interval $\openint a b$.

Then:
 * $\exists \xi \in \openint a b: \map {f'} \xi = \dfrac {\map f b - \map f a} {b - a}$

Proof
For any constant $h \in \R$ we may construct the real function defined on $\closedint a b$ by:
 * $\map F x = \map f x + h x$

We have that $h x$ is continuous on $\closedint a b$ from Linear Function is Continuous.

From the Sum Rule for Continuous Functions, $F$ is continuous on $\closedint a b$ and differentiable on $\openint a b$.

Let us calculate what the constant $h$ has to be such that $\map F a = \map F b$:

Since $F$ satisfies the conditions for the application of Rolle's Theorem:
 * $\exists \xi \in \openint a b: \map {F'} \xi = 0$

But then:
 * $\map {F'} \xi = \map {f'} \xi + h = 0$

The result follows.

Also see

 * Mean Value Theorem for Integrals