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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) ?

P..
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Applied mathematician
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3 Answers3

86

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!

P..
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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 06 '13 at 11:00
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    This is what I meant by intuition. Thanks – Applied mathematician Feb 06 '13 at 12:26
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    Best example I've seen yet. Thank you @P.. – Mr.Fry Jul 10 '14 at 00:23
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    @P.. This is the most creative answer I've seen on calculus here – nice job applying the mean value theorem into the real world! – Toby Mak Feb 26 '18 at 11:26
  • I don't see how g'(c) is the average speed of Powell.. – TheNicanova Oct 05 '18 at 04:28
  • If you want time rather than position in space to be an argument to $f$ and $g$, then, you have to state something to the effect that, at the precise instant Bolt passed the finish line, Powell had covered, say, 77% of the distance. Then, the theorem says there is a point in time $c$ such that $\frac{f'(c)}{g'(c)}$ equals $1.0/0.77$. Otherwise, you would have to consider $g'$ and $f'$ to be the rate of change of intermediate average speed with respect to distance. This said, I love this example, as it very clearly relates the generalized mean value theorem to the simple one. – Papa Smurf Jan 31 '22 at 15:33
  • This doesn't make sense. Variables $a$ and $b$ represent time and $b-a$ isn't the same for Bolt and Powell. – Tony Häger Feb 09 '23 at 16:40
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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.

Community
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pitonist
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    Wouldn't it be better to just measure the distance Asafa had travelled when Bolt finished? That way $f_A(0) = f_B(0) = 0$, $f_B(9.63) = 100$, $f_A(9.63) = x$ where $x \in (0, 100)$. – user76284 May 02 '18 at 20:34
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    An example that would work is by considering a match where two runners have to run as fast as they can in some interval $[0,T]$. – Sha Vuklia Apr 20 '19 at 16:16
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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.

awllower
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