Mean Value Theorem/Proof 1
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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
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.
$\blacksquare$
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
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 11.6$