In many textbooks, the former two have been proved with the help of Rolle's theorem. However my teacher(and many sites as well) say that Rolle's theorem is a special case of LMVT and Cauchy's is a generalization.
Can we prove Cauchy's MVT and LMVT without using Rolle's theorem? If not, should we admit that these are just applications of Rolle's theorem and hence yield no extra result?
$\mathcal{Remark}$
I found an analogy which could be useful:
Suppose a car is travelling at an average speed of 40 miles/hr. In the course of the travel, it has to, at some point, travel at exactly 40 miles an hour.
This is exactly what LMVT has to say.
Best Answer
Here is a direct proof of the mean value theorem. The arguments are closed to those given in E. Goulart's Cours d'analyst mathématique, Tome I. The argument same argument can be used to prove Rolle's theorem.
Suppose $f$ is continous in $[a,b]$ and differentiable in $(a,b)$. Define $\phi(x):=f(x)-\frac{f(b)-f(a)}{b-a}x$
It is clear that $\phi$ is continuous in $[a,b]$ and so, it attains its minimum and maximum values in $[a,b]$.
If $\phi$ attains its maximum and minimum at the endpoints, then $\phi$ is constant in $[a,b]$ since $\phi(a)=\phi(b)=\frac{f(a)b-f(b)a}{b-a}$. Since $\phi$ is differentiable in $(a,b)$, $\phi'(x)\equiv0$.
If $\phi$ achieves its wither its maximum or its minimum at some $c\in (a,b)$. Since $\phi$ is differentiable in $(a,b)$, it follows that $\phi'(c)=0$.
All this means that
$$ f'(c)=\frac{f(b)-f(a)}{b-a} $$ for some $c\in (a,b)$.