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 $$h \in \mathbb{R}$$ 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 $$h = - \frac {f \left({b}\right) - f \left({a}\right)} {b - a}$$.

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.