Generalized mean value theorem

Question

I know and understand the mean value theorem. But at the moment I don't have the intuition to understand the generalized mean value theorem

If $f$ and $g$ are continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)$, then there exists a point $c\in(a,b)$ where$$[f(b)-f(a)]g'(c)=[g(b)-g(a)]f'(c).$$If $g'$ is never zero on $(a,b)$, then the conclusion can be stated as$$\frac{f'(c)}{g'(c)}=\frac{f(b)-f(a)}{g(b)-g(a)}.$$

What is the intuition? And how can you prove this (generally) ?

Answer 1

Note: Except some technicality issues the following example gives a good intuition behind the mean value theorems.

In the 2012 Olympics Usain Bolt won the 100 metres gold medal with a time of 9.63 seconds.
His average speed was total distance, $d(t_2)-d(t_1)$, over total time, $t_2-t_1$: $$V_a=\dfrac{d(t_2)-d(t_1)}{t_2-t_1}=\dfrac{100}{9.63}=10.384 \ \text{m/s}=37.38 \ \text{km/h}.$$ Mean value theorem $$ f'(c)=\dfrac{f(b)-f(a)}{b-a} $$ says that at some point (which is $c$ seconds) Bolt was actually running at the average speed of $37.38$ km/h.

Powell Asafa was participating in that race also, with a time $11.99=1.245\times9.63$ seconds, so Bolt's average speed was $1.245$ times the average speed of Powell. Generalized mean value theorem: $$\dfrac{f'(c)}{g'(c)}=\dfrac{f(b)-f(a)}{g(b)-g(a)}=\dfrac{\frac{f(b)-f(a)}{b-a}}{\frac{g(b)-g(a)}{b-a}},$$ says that at some point Bolt was actually running at a speed exactly $1.245$ times of Powell's speed!

Answer 2

The example of P.. does not satisfy the conditions of generalized mean value theorem. Let me briefly explain.

$f$ is the position of Bolt while $g$ is the position of Powell. Domains of these functions are time. $f$ is defined for $[0,9.63]$ and $g$ is defined for $[0,11.99]$. So the domains are not equal. In this case Generalized Mean Value Theorem will not work. Think about this unrealistic scenario where Powell has waited for the first 9.63 second where Bolt has finished the race and started running at t=9.63, finishing 100m in the remaining 2.36 seconds. In this case there is no instant at which Bolt was running 1.245 times Powell's speed.

Answer 3

Consider the function:
$h(x)=(g(b)-g(a))f(x)-(f(b)-f(a))g(x)$.
Then clearly $h(b)-h(a)=0$. Hence, by the ordinary mean-value theorem, $h'(c)=0$ for some $c$ in between $b$ and $a$. And that is what we wanted.
As for the intuition, it could be thought of as requiring a tangent line to a parameterised curve on the plane, as in the Wiki-article. Also it is in fact equivalent to the ordinary mean-value theorem, as the proof shows.
P.S. Suppose $g'(x) \neq 0$, then there is an inverse of $g$. If we substitute $y=g^{-1}(x)$, then the theorem is nothing but the ordinary mean-value theorem. Thus, in some sense, this "generalised" mean-value theorem could even be deemed as a "specialized" version of mean-value theorem.

Answer 4

[This is from Shilov’s Real and Complex Analysis book]

Thm: Let functions ${ f, g : [a, b] \to \mathbb{R} }$ be continuous on ${ [a, b] }$ and differentiable on ${ (a, b) . }$ Further say ${ g ^{’} }$ is never ${ 0 }$ on ${ (a, b). }$ Then

$${ \frac{f(b) - f(a)}{g(b) - g(a) } = \frac{f ^{’} (c) }{ g ^{’} (c) } }$$

at some ${ c \in (a, b). }$

Informally, ratio of average rates of change is the ratio of instantaneous rates of change at some point.

Pf: Given the above functions ${ f }$ and ${ g , }$ it is natural to apply Rolles theorem to the linear combination

$${ \varphi(x) := f(x) + \lambda g(x) }$$

where ${ \lambda }$ is such that

$${ \varphi(a) = \varphi(b) }$$

that is

$${ f(a) + \lambda g(a) = f(b) + \lambda g(b) }$$

that is

$${ \lambda := - \left( \frac{f(b) - f(a)}{g(b) - g(a)} \right) . }$$

Now by Rolles theorem, there is a ${ c \in (a, b) }$ such that

$${ \varphi ^{’}(c) = 0 }$$

that is

$${ f ^{’}(c) = - \lambda g ^{’} (c) }$$

that is

$${ \frac{f ^{’} (c)}{g ^{’} (c)} = \frac{f(b) - f(a)}{g(b) - g(a) } }$$

as needed. ${ \blacksquare }$