# Generalized mean value theorem

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) ?

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I think the tag of calculus is more appropriate for this question, rather than real analysis. –  awllower Feb 6 at 10:02
See here en.wikipedia.org/wiki/… for the theorem together with a proof –  sonystarmap Feb 6 at 10:04
Possible duplicate –  awllower Feb 6 at 10:27

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 Bolt was actually running at the average speed of $37.38$ km/h.

Powell Asafa was participating in that race, with a time $11.99=1.245\times9.63$ seconds, so that 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!

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I know you must think that a proof is very easy. But per chance it is better to include a proof? Thanks in any case. –  awllower Feb 6 at 11:00
This is what I meant by intuition. Thanks –  MSKfdaswplwq Feb 6 at 12:26
$h(x)=(g(b)-g(a))f(x)-(f(b)-f(a))g(x)$.
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.
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.