# Mean Value Theorem

## Contents

## 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 1

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 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 $\map F a = \map F b$:

\(\displaystyle \map F a\) | \(=\) | \(\displaystyle \map F b\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \map f a + h a\) | \(=\) | \(\displaystyle \map f b + h b\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \map f a - \map f b\) | \(=\) | \(\displaystyle h b - h a\) | rearranging | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \map f a - \map f b\) | \(=\) | \(\displaystyle h \paren {b - a}\) | Real Multiplication Distributes over Real Addition | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle h\) | \(=\) | \(\displaystyle -\dfrac {\map f b - \map f a} {b - a}\) | 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.

$\blacksquare$

## Proof 2

Let $g : \closedint a b \to \R$ be a real function with:

- $\map g x = x$

for all $x \in \closedint a b$.

By Power Rule for Derivatives, we have:

- $g$ is differentiable with $\map {g'} x = 1$ for all $x \in \closedint a b$.

Note that in particular:

- $\map {g'} x \ne 0$ for all $x \in \openint a b$.

Since $f$ is continuous on $\closedint a b$ and differentiable on $\openint a b$, we can apply the Cauchy Mean Value Theorem.

We therefore have that there exists $\xi \in \openint a b$ such that:

- $\dfrac {\map {f'} \xi} {\map {g'} \xi } = \dfrac {\map f b - \map f a} {\map g b - \map g a}$

Note that:

- $\map {g'} \xi = 1$

and:

- $\map g b - \map g a = b - a$

so this can be rewritten:

- $\map {f'} \xi = \dfrac {\map f b - \map f a} {b - a}$

$\blacksquare$

## Also see

## Sources

- 1977: K.G. Binmore:
*Mathematical Analysis: A Straightforward Approach*... (previous) ... (next): $\S 11.6$ - 2014: Christopher Clapham and James Nicholson:
*The Concise Oxford Dictionary of Mathematics*(5th ed.) ... (previous) ... (next): Entry:**mean value theorem**