Mean Value Theorem

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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

Mean-value-theorem.png

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.

Some sources hyphenate: Mean-Value Theorem.


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

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


Also see


Sources