Ok. Next time be more precise then . Take a look at Mean value theorem - Wikipedia, the free encyclopedia. You must prove this theorem... called the mean value theorem. (or something very close to it).
Yes. The mean value theorem states that there exist a $\displaystyle c$ in $\displaystyle (a,b)$ such as $\displaystyle \frac{f(b)-f(a)}{b-a}=f'(c)$. In your example, $\displaystyle g$ and $\displaystyle h$ are defined and continuous on $\displaystyle (a,b)$, you must find a subinterval of $\displaystyle (a,b)$ in which the mean value theorem can be applied.is there any connection between mean value theorem and the thing we asked to prove ?
The intermediate value theorem for derivatives states that if $\displaystyle f:I\mapsto \mathbb{R}$ is a differenciable function then $\displaystyle f'$ has the intermediate value property on $\displaystyle I$, an open interval. Meaning if $\displaystyle a,b\in I$ with $\displaystyle a<b$ and $\displaystyle c$ lies (strictly) between $\displaystyle f'(a)$ and $\displaystyle f'(b)$ there exists $\displaystyle x_0\in (a,b)$ so that $\displaystyle f'(x_0) = c$.
Without lose of generality we will assume that $\displaystyle f'(a) < c < f'(b)$. Now define the function $\displaystyle g:[a,b]\mapsto \mathbb{R}$ as $\displaystyle g(x) = f(x) - cx$. Since $\displaystyle g$ is continous on $\displaystyle [a,b]$ it assumes a minimum value. We will now show that $\displaystyle g$ does not take a minimum value at $\displaystyle a$. Note that $\displaystyle g'(a) = f'(a) - c < 0$. But since $\displaystyle g'(a) = \lim_{x\to a^+}\tfrac{g(x)-g(a)}{x-a} < 0$ it follows that $\displaystyle g(x) - g(a)<0$ for $\displaystyle a<x$ sufficiently close to $\displaystyle x$. Thus, there exists $\displaystyle y\in (a,b)$ so that $\displaystyle g(y) - g(a) < 0$, which is means $\displaystyle g(y) < g(a)$, in particular $\displaystyle a$ is not a minimum for $\displaystyle g$. Using a similar argument we can show $\displaystyle b$ is not a minimum for $\displaystyle g$. Thus, $\displaystyle g$ has a minimum on $\displaystyle (a,b)$ and so there is a point $\displaystyle x_0 \in (a,b)$ so that $\displaystyle g'(x_0) = 0\implies f'(x_0) = c$.