Let's focus on the following version of Cauchy's Mean Value Theorem:
Cauchy's Mean Value Theorem: Let $f, g$ be functions defined on closed interval $[a, b]$ such that
1) Both $f, g$ are continuous on $[a, b]$
2) Both $f, g$ are differentiable on $(a, b)$
3) $g'$ does not vanish at any point in $(a, b)$
then there is a point $c \in (a, b)$ such that $$\frac{f(b) – f(a)}{g(b) – g(a)} = \frac{f'(c)}{g'(c)}$$
In most good textbooks it is mentioned that this theorem can't be derived from the usual Mean Value Theorem. Using MVT we can get $$\frac{f(b) – f(a)}{g(b) – g(a)} = \frac{(f(b) – f(a))/(b – a)}{(g(b) – g(a))/(b – a)} = \frac{f'(c_{1})}{g'(c_{2})}$$ and we can't guarantee that $c_{1} = c_{2}$ and hence the above strategy to prove Cauchy's theorem fails.
However I beg to differ and I provide the following argument. The geometrical interpretation of MVT is that if we take any two points on a continuous and smooth curve (with some further restrictions on nature of curve) then between this portion of the curve there is a point at which the tangent to the curve is parallel to the chord joining these two points. Cauchy's theorem is obtained at once if we use the parametric equation of the curve with $x = g(t), y = f(t), dy/dx = f'(t)/g'(t)$ for $t \in (a, b) $ and the endpoints of the curve are $(g(a), f(a)), (g(b), f(b))$. Hence Cauchy's Theorem is derivable from MVT.
The rigorous proof proceeds as follows. Since $g'$ does not vanish it is clear from Rolle's theorem that $g$ is one one function so that $g(a) \neq g(b)$. If $g(a) < g(b)$ then we can show that $g$ is strictly increasing. Let $a \leq c < d \leq b$. Clearly $g(c) \neq g(d) \neq g(a)$. Let us suppose $g(d) < g(a)$ so that $g(d) < g(a) < g(b)$ and hence by IVT there will be a point $x$ in $(d, b)$ with $g(x) = g(a)$ contrary to the fact that $g$ is one one. So we must have $g(a) < g(d)$. Similarly $g(c) < g(b)$. Now suppose that $g(c) > g(d)$ so that $g(a) < g(d) < g(c)$ and by IVT there is an $x \in (a, c)$ with $g(x) = g(d)$ contrary to the fact that $f$ is one one. Hence $g(c) < g(d)$. We have thus shown that $g$ is strictly increasing. If $g(a) > g(b)$ then we can show in same manner that $g$ is strictly decreasing.
Hence by inverse function theorem $h = g^{-1}$ exists and is continuous and differentiable. Suppose $g$ is increasing and let $A = g(a), B = g(b)$ so that $a = h(A), b = h(B)$ and consider the function $F(x) = f(h(x))$ on $[A, B]$. By MVT there is a $C \in (A, B)$ such that $$F(B) – F(A) = (B – A)F'(C)$$ or $$f(h(B)) – f(h(A)) = (g(b) – g(a))f'(h(C))h'(C)$$ If $h(C) = c$ then $c \in (a, b)$ and $h'(C) = 1/g'(c)$ and we get $$\frac{f(b) – f(a)}{g(b) – g(a)} = \frac{f'(c)}{g'(c)}$$
I would like to get feedback on this proof above. I haven't found any issue with the argument and think it is perfectly valid.
Best Answer
Or we could just consider the function:
$F(x)=f(x)-f(a)-\frac{f(b)-f(a)}{g(b)-g(a)}(g(x)-g(a))$
Clearly $F(x)$ is defined iff $g(b) $ is different from $g(a)$ , otherwise if $g(a)=g(b)$ then $F(x)$ would satisfy Role's theorem's conditions hence there would exist a point $c$ $ \epsilon$ $(a,b)$ such that $g'(c)=0$ this is a contradiction to our third condition :
Then from the MVT we have :
There exists a point $c$ $\epsilon$ $(a,b)$ such that $F'(c)=0$
thus we get :
$f'(c)=\frac{f(b)-f(a)}{g(b)-g(a)}g'(c)$
or
$\frac{f'(c)}{g'(c)}=\frac{f(b)-f(a)}{g(b)-g(a)}$