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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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    Can this be proven for dimensions greater than 1? i.e. for $f^{(n)}$2015-06-16

2 Answers 2

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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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    Great proof. Thanks!2012-05-22
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NOTE:

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