Vtyjkyuk

If $h(x) = (fcirc g)(x), xin [a,b]$, $f(x)$ and $g(x)$ are continuous over $[a,b]$ and differentiable over $(a,b)$. And $g(a) = b, g(b) = a$. We need to show that there exists at least one $c in (a,b)$ such that
$$h(a)-h(b) = f'(c)(b-a)$$

I tried to use the mean value theorem and I got
$$h(b)-h(a) = h'(c)(b-a)$$

Do you have any suggestions please

As I commented, if we assume $g([a,b]) subseteq [a,b]$, then $h$ would be well-defined. Now use the condition about $g$, we write the equation we want to prove as
$$
frac h(a) - h(b) g(a) - g(b) = f'(c).
$$
Now consider Cauchy Mean Value Theorem. If $g'neq 0$ on $[a,b]$ is also presumed, then the equation is easily deduced. Otherwise this question would be much complicated.

It was easier than I expected

I have to apply the Mean Value Theorem on $f(x)$, then

there exists at least one $c in (a,b)$ such that

$$f'(c) = fracf(b) - f(a)b-a$$

Now, since $h(a) = (f circ g) (a) = f(g(a)) = f(b)$

$h(b) = (f circ g) (b) = f(g(b)) = f(a)$

then

$$f'(c) = frach(a) - h(b)b-a$$