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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 '15 at 17:36
Can this be proven for dimensions greater than 1? i.e. for $f^{(n)}$ – kathystehl Jun 16 '15 at 17:02
up vote 9 down vote accepted

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. – EsX_Raptor 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


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

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