# Cauchy Mean Value Theorem

## 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:

- $\exists \xi \in \openint a b: \dfrac {\map {f'} \xi} {\map {g'} \xi} = \dfrac {\map f b - \map f a} {\map g b - \map g a}$

## Proof

First we check $\map g a \ne \map g b$.

Aiming for a contradiction, suppose $\map g a = \map g b$.

From Rolle's Theorem:

- $\exists \xi \in \openint a b: \map {g'} \xi = 0$.

This contradicts $\forall x \in \openint a b: \map {g'} x \ne 0$.

Thus by Proof by Contradiction $\map g a \ne \map g b$.

Let $h = \dfrac {\map f b - \map f a} {\map g b - \map g a}$.

Let $F$ be the real function defined on $\closedint a b$ by:

- $\map F x = \map f x - h \map g x$.

Then:

\(\ds \map F b - \map F a\) | \(=\) | \(\ds \paren {\map f b - h \map g b} - \paren {\map f a - h \map g a}\) | as $\map F x = \map f x - h \map g x$ | |||||||||||

\(\ds \) | \(=\) | \(\ds \paren {\map f b - \map f a} - h \paren {\map g b - \map g a}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds 0\) | ||||||||||||

\(\ds \leadsto \ \ \) | \(\ds \map F a\) | \(=\) | \(\ds \map F b\) | |||||||||||

\(\ds \leadsto \ \ \) | \(\ds \exists \xi \in \openint a b: \, \) | \(\ds \map {F'} \xi\) | \(=\) | \(\ds \map {f'} \xi - h \map {g'} \xi\) | Sum Rule for Derivatives, Derivative of Constant Multiple | |||||||||

\(\ds \) | \(=\) | \(\ds 0\) | Rolle's Theorem | |||||||||||

\(\ds \leadsto \ \ \) | \(\ds \exists \xi \in \openint a b: \, \) | \(\ds \frac {\map {f'} \xi} {\map {g'} \xi}\) | \(=\) | \(\ds h\) | $\forall x \in \openint a b: \map {g'} x \ne 0$ | |||||||||

\(\ds \) | \(=\) | \(\ds \frac {\map f b - \map f a} {\map g b - \map g a}\) |

$\blacksquare$

## Also known as

The **Cauchy mean value theorem** is also known as the **generalized mean value theorem**.

## Geometrical Interpretation

Consider two functions $\map f x$ and $\map g x$:

- continuous on the closed interval $\closedint a b$
- differentiable on $\openint a b$.

For every $x \in \closedint a b$, we consider the point $\tuple {\map f x, \map g x}$.

If we trace out the points $\tuple {\map f x, \map g x}$ over every $x \in \closedint a b$, we get a curve in two dimensions, as shown in the graph:

This article, or a section of it, needs explaining.An explanation is needed as to the coordinate frame used to depict this.You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Explain}}` from the code. |

In the drawing, the slope of the red line is $\dfrac {\map g b - \map g a} {\map f b - \map f a}$.

This is because:

- $\dfrac {\Delta y} {\Delta x} = \dfrac {\map g b - \map g a} {\map f b - \map f a}$

assuming that the vertical axis, which contains the value of $\map f x$, is the $y$-axis.

The slope of the green line is $\dfrac {\map {g'} c} {\map {f'} c}$.

This is because:

- $\valueat {\dfrac {\d g} {\d f} } {x \mathop = c} = \valueat {\dfrac {\d g / \d x} {\d f / \d x} } {x \mathop = c} = \dfrac {\map {g'} c} {\map {f'} c}$

The drawing illustrates that for the value of $c$ chosen in the pictures, the slopes of the red line and green line are the same.

That is:

- $\dfrac {\map g b - \map g a} {\map f b - \map f a} = \dfrac {\map {g'} c} {\map {f'} c}$

## Example

In the 2012 Olympics Usain Bolt won the 100 metres gold medal with a time of $9.63$ seconds.

By definition, his average speed was the total distance travelled divided by the total time it took:

\(\ds V_a\) | \(=\) | \(\ds \frac {\map d {t_2} - \map d {t_1} } {t_2 - t_1}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \frac {100 \ \mathrm m} {9.63 \ \mathrm s}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds 10.384 \ \mathrm {m/s}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds 37.38 \ \mathrm {km/h}\) |

The Mean Value Theorem gives:

- $\map {f'} c = \dfrac {\map f b - \map f a} {b - a}$

Hence, at some point Bolt was *actually running* at the average speed of $37.38 \ \mathrm {km/h}$

Asafa Powell was participating in that same race.

He achieved a time of $11.99 \ \mathrm s = 1.245 \times 9.63 \ \mathrm s$.

So Bolt's average speed was $1.245$ times the average speed of Powell.

The Cauchy Mean Value Theorem gives:

\(\ds \frac {\map {f'} c} {\map {g'} c}\) | \(=\) | \(\ds \frac {\map f b - \map f a} {\map g b - \map g a}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \frac {\dfrac {\map f b - \map f a} {b - a} } {\dfrac {\map g b - \map g a} {b - a} }\) |

Hence, at some point, Bolt was *actually running* at a speed exactly $1.245$ times that of Powell's.

## Source of Name

This entry was named for Augustin Louis Cauchy.