# 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:

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

## Proof

Let $F$ be the real function defined on $\left[{a \,.\,.\, b}\right]$ by $F \left({x}\right) = f \left({x}\right) + h x$, where $h \in \R$ is a constant.

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

Then 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 $F \left({a}\right) = F \left({b}\right)$.

Then:

- $f \left({a}\right) + h a = f \left({b}\right) + h b$

Hence:

- $h = - \dfrac {f \left({b}\right) - f \left({a}\right)} {b - a}$

Since $F$ satisfies the conditions for the application of Rolle's Theorem:

- $\exists \xi \in \left({a \,.\,.\, b}\right): F^{\prime} \left({\xi}\right) = 0$

But then:

- $F^{\prime} \left({\xi}\right) = f^{\prime} \left({\xi}\right) + h = 0$

The result follows.

$\blacksquare$

## Sources

- K.G. Binmore:
*Mathematical Analysis: A Straightforward Approach*(1977)... (previous)... (next): $\S 11.6$