# 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 $F \left({a}\right) = F \left({b}\right)$:

 $\displaystyle F \left({a}\right)$ $=$ $\displaystyle F \left({b}\right)$ $\displaystyle \leadsto \ \$ $\displaystyle f \left({a}\right) + h a$ $=$ $\displaystyle f \left({b}\right) + h b$ $\displaystyle \leadsto \ \$ $\displaystyle f \left({a}\right) - f \left({b}\right)$ $=$ $\displaystyle h b - h a$ rearranging $\displaystyle \leadsto \ \$ $\displaystyle f \left({a}\right) - f \left({b}\right)$ $=$ $\displaystyle h \left({b - a}\right)$ Real Multiplication Distributes over Real Addition $\displaystyle \leadsto \ \$ $\displaystyle h$ $=$ $\displaystyle -\dfrac {f \left({b}\right) - f \left({a}\right)} {b - a}$ rearranging

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.

$\blacksquare$