I want to prove Cauchy's mean value theorem without using Rolle's theorem and only using Lagrange's mean value theorem. I tried to prove this by assuming $\frac{f'(x)}{g'(x)}$ but I proceed further please help me out with this.
Answer
If $f$ and $g$ are continuous real functions on $[a,b]$ which are differentiable on $(a,b)$. We could construct a function $F(x)=[f(b)-f(a)]g(x)-[g(b)-g(a)]f(x).$ Obviously, $F(x)$ is continuous on $[a,b]$ and differentiable on $(a,b).$ Applying Lagrange's mean value theorem to $F(x)$, we obtain a point $\epsilon \in (a,b)$ such that
$$
F(b)-F(a)=F'(\epsilon)(b-a).
$$
Noting that $F(b)-F(a)=0$, we have
$$
F'(\epsilon)=[f(b)-f(a)]g(\epsilon)-[g(b)-g(a)]f(\epsilon)=0.
$$
That is,
$$
[f(b)-f(a)]g(\epsilon)=[g(b)-g(a)]f(\epsilon).
$$
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