Cauchy Mean Value Theorem

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Theorem

Let $f$ and $g$ be a real functions which are continuous on the closed interval $\left[{a \,.\,.\, b}\right]$ and differentiable on the open interval $\left({a \,.\,.\, b}\right)$.

Suppose that:

$\forall x \in \left({a \,.\,.\, b}\right): g' \left({x}\right) \ne 0$


Then:

$\exists \xi \in \left({a \,.\,.\, b}\right): \dfrac {f' \left({\xi}\right)} {g' \left({\xi}\right)} = \dfrac {f \left({b}\right) - f \left({a}\right)} {g \left({b}\right) - g \left({a}\right)}$


Proof

Let $F$ be the real function defined on $\left[{a \,.\,.\, b}\right]$ by $F \left({x}\right) = f \left({x}\right) + h g \left({x}\right)$, where $h \in \R$ is a constant.

The plan is to choose the constant $h$ such that $F \left({a}\right) = F \left({b}\right)$ and so apply Rolle's Theorem.

We need to make:

$f \left({a}\right) + h g \left({a}\right) = f \left({b}\right) + h g \left({b}\right)$

We have that:

$\forall x \in \left({a \,.\,.\, b}\right): g' \left({x}\right) \ne 0$

So, by Rolle's Theorem:

$g \left({a}\right) \ne g \left({b}\right)$

Thus we can let:

$h = - \dfrac {f \left({b}\right) - f \left({a}\right)} {g \left({b}\right) - g \left({a}\right)}$

So, by Rolle's Theorem:

$\exists \xi \in \left({a \,.\,.\, b}\right): 0 = F' \left({\xi}\right) = f' \left({\xi}\right) + h g' \left({\xi}\right)$

That is:

$h = - \dfrac {f' \left({\xi}\right)} {g' \left({\xi}\right)}$

Hence the result, from the definition of the derivative.

$\blacksquare$


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 $\map g x$:

[Definition:Continuous Real Function|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:


Cauchy's Mean Value Theorem.png



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:

\(\displaystyle V_a\) \(=\) \(\displaystyle \frac {d \left({t_2}\right) - d \left({t_1}\right)} {t_2 - t_1}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {100 \ \mathrm m} {9.63 \ \mathrm s}\)
\(\displaystyle \) \(=\) \(\displaystyle 10.384 \ \mathrm {m/s}\)
\(\displaystyle \) \(=\) \(\displaystyle 37.38 \ \mathrm {km/h}\)


The Mean Value Theorem gives:

$f' \left({c}\right) = \dfrac {f \left({b}\right) - f \left({a}\right)} {b - a}$

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:

\(\displaystyle \frac {f' \left({c}\right)} {g' \left({c}\right)}\) \(=\) \(\displaystyle \frac {f \left({b}\right) - f \left({a}\right)} {g \left({b}\right) - g \left({a}\right)}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {\frac {f \left({b}\right) - f \left({a}\right)} {b - a} } {\frac {g \left({b}\right) - g \left({a}\right)} {b - a} }\)

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.


Sources