# Proving Cauchy's Generalized Mean Value Theorem

This is an exercise from Stephen Abbott's Understanding Analysis. The hint it gives on how to solve it is not very clear, in my opinion, so I would like for a fresh set of eyes to go over it with me:

pp 143 Exercise 5.3.4. (a) Supply the details for the proof of Cauchy's Generalized Mean Value Theorem (Theorem 5.3.5.).

Theorem 5.3.5. (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$$\frac{f'(c)}{g'(c)}=\frac{f(b)-f(a)}{g(b)-g(a)}.$$

*Hint: This result follows by applying the Mean Value Theorem to the function*$$h(x)=[f(b)-f(a)]g(x)-[g(b)-g(a)]f(x)$$

First of all, I know that the Mean Value Theorem (MVT) states that if $f:[a,b]\to\mathbb{R}$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists a point $c\in(a,b)$ where$$f'(c)=\frac{f(b)-f(a)}{b-a}.$$

If we assume that $h$ has the above properties, then applying the MVT to it, for some $c\in(a,b)$, would yield$$h'(c)=\frac{h(b)-h(a)}{b-a}=$$

$$\frac{[f(b)-f(a)]g(b)-[g(b)-g(a)]f(b) \quad - \quad [f(b)-f(a)]g(a)+[g(b)-g(a)]f(a)}{b-a}=$$

$$[f(b)-f(a)]\left(\frac{g(b)-g(a)}{b-a}\right) \quad - \quad[g(b)-g(a)]\left(\frac{f(b)-f(a)}{b-a}\right)=$$

$$[f(b)-f(a)]g'(c) \quad - \quad [g(b)-g(a)]f'(c).$$This is the best I could achieve; I have no clue on how to reach the second equation in the above theorem.

Do you guys have any ideas? Thanks in advance!

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By the way, I really appreciate your great formatting and that you've shown your work. –  mixedmath Feb 29 '12 at 4:30
How do you know that $\frac{g(b)-g(a)}{b-a}$ gives you g'(c),i.e c is the same c for which $\frac{h(b)-h(a)}{b-a}$ is h'(c)? –  p.s Dec 5 '12 at 18:59
Very interesting exercise, you can do the same thing for $f,g: \bar{\mathbb{R}} \to \bar{\mathbb{R}}$ with $g(a),g(b),f(a),f(b)$ finite. –  Maman Jun 6 at 17:36
Can this be proven for dimensions greater than 1? i.e. for $f^{(n)}$ –  kathystehl Jun 16 at 17:02

Note that $$\begin{eqnarray}h(a)&=&[f(b)-f(a)]g(a)-[g(b)-g(a)]f(a)\\ &=&f(b)g(a)-g(b)f(a)\\ &=&[f(b)-f(a)]g(b)-[g(b)-g(a)]f(b)\\ &=&h(b)\end{eqnarray}$$ and so $h'(c)=0$ for some point $c\in (a,b)$. Then differentiate $h$ normally and note that this makes $c$ the desired point.

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That was it! It follows from Rolle's theorem that $h'(c)=0$ if $h(a)=h(b)$. How could I've forgotten! Thank you very much. –  Josué Feb 29 '12 at 4:22
Glad to help, and to get a chance to review myself. –  Alex Becker Feb 29 '12 at 4:27
Great proof. Thanks! –  Double AA May 22 '12 at 5:43

NOTE:

You should calculate out $h(b) - h(a)$. You'll immediately see that you are done.

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