# Mean Value Theorem/Proof 1

## Theorem

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

Then:

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

## Proof

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.

$\blacksquare$