We studied in class today about the Cauchy's mean value theorem, but in somewhat more complicated version, and i find it difficult to prove. here the theorem:

Let $f,\ g:\mathbb{R}\rightarrow\mathbb{R}$ be continuous on $[a,\ b]$ and differentiable on$(a,\ b)$. Suppose that $f(b) \neq f(a)$.
1. if $g\left(a\right)\ne g\left(b\right)$ so one of this things happens:
1.1 there is some $x_0\in \left(a,b\right)$ such that $f'\left(x_0\right)\:=\:g'\left(x_0\right)=0$
1.2 exist some $c\:\in \left(a,b\right)$ such that $\frac{g'\left(c\right)}{f'\left(c\right)}\:=\:\frac{g\left(b\right)-g\left(a\right)}{f\left(b\right)-f\left(a\right)}$
2.if $g\left(a\right)=g\left(b\right)$ then exist $x_0\in \left(a,b\right)$ such that $g'\left(x_0\right)\:=\:0$

Now, i don't undestand 1.1. from what it derives? What is the magic trick that we need to show here?
2 is just Rolle's theorem, and 1.2 i saw the proof here Conditions of Cauchy's Mean Value Theorem


Cauchy's mean value theorem states:

If functions $f$ and $g$ are both continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a, b)$, then there exists some $c ∈ (a,b)$ such that the following equation holds:

A. $(f(b) - f(a))g'(c) = (g(b) - g(a))f'(c)$

Let's suppose $f(b) \neq f(a)$ and $g(b) \neq g(a)$. So now two cases can be distinguished:

  1. $f'(c) = 0$

    Then, since $f(b) \neq f(a)$, it follows from equation A that $g'(c) = 0$, so your proposition 1.1 is true.

  2. $f'(c) \neq 0$

    Now it's possible to divide both equation A members by $f'(c)$, the resulting equation being the one stated in your 1.2 proposition.

This means that, under your hypothesis 1, it's either the case that proposition 1.1 is true or (inclusive) that proposition 1.2 is true.


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