Mean Value Theorem

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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