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

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