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

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.

Then from the Combination Theorem for 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 $\displaystyle h = - \frac {f \left({b}\right) - f \left({a}\right)} {b - a}$.

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.