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.
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
- 1961: David V. Widder: Advanced Calculus (2nd ed.) ... (previous) ... (next): $1$ Partial Differentiation: $\S 2$. Functions of One Variable: $2.4$ Law of the Mean
- 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): mean-value theorem
- 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): mean-value theorem
- 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): mean value theorem