# Mean Value Theorem

## 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 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.

$\blacksquare$

## 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:

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

$\blacksquare$

## Proof 3

Recall the Cauchy Mean Value Theorem:

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

Suppose:

$\forall x \in \openint a b: \map {g'} x \ne 0$

Then:

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

The result follows by setting $\map g x = x$ for all $x \in \R$.

$\blacksquare$

## Also presented as

The Mean Value Theorem can also be presented in the form:

$\map f {c + h} - \map f c = h \map {f'} {c + \theta h}$

for some $\theta \in \openint 0 1$.

## Also known as

The Mean Value Theorem is also known as the law of the mean.

## Examples

### Example: $x^3$: Formulation $1$

Let $f$ be the real function defined as:

$\map f x = x^3$

Let:

$a = 1$, $b = 2$

Then when $\xi = \sqrt {\dfrac 7 3}$:

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

### Example: $x^3$: Formulation $2$

Let $f$ be the real function defined as:

$\map f x = x^3$

Let:

$c = 2$, $h = -1$

Then when $\theta = 2 - \sqrt {\dfrac 7 3}$:

$\map {f'} {c + \theta h} = \dfrac {\map f {c + h} - \map f c} h$