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:

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


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

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

$\blacksquare$


Sources