# 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): $\S 11.6$