Mean Value Theorem/Proof 1

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Let $f$ be a real function which is continuous on the closed interval $\closedint a b$ and differentiable on the open interval $\openint a b$.


$\exists \xi \in \openint a b: \map {f'} \xi = \dfrac {\map f b - \map f a} {b - a}$


For any constant $h \in \R$ we may construct the real function defined on $\closedint a b$ by:

$\map F x = \map f x + h x$

We have that $h x$ is continuous on $\closedint a b$ from Linear Function is Continuous.

From the Sum Rule for Continuous Real Functions, $F$ is continuous on $\closedint a b$ and differentiable on $\openint a b$.

Let us calculate what the constant $h$ has to be such that $\map F a = \map F b$:

\(\ds \map F a\) \(=\) \(\ds \map F b\)
\(\ds \leadsto \ \ \) \(\ds \map f a + h a\) \(=\) \(\ds \map f b + h b\)
\(\ds \leadsto \ \ \) \(\ds \map f a - \map f b\) \(=\) \(\ds h b - h a\) rearranging
\(\ds \leadsto \ \ \) \(\ds \map f a - \map f b\) \(=\) \(\ds h \paren {b - a}\) Real Multiplication Distributes over Real Addition
\(\ds \leadsto \ \ \) \(\ds h\) \(=\) \(\ds -\dfrac {\map f b - \map f a} {b - a}\) rearranging

Since $F$ satisfies the conditions for the application of Rolle's Theorem:

$\exists \xi \in \openint a b: \map {F'} \xi = 0$

But then:

$\map {F'} \xi = \map {f'} \xi + h = 0$

The result follows.