2. ## oh..

i m actually not meaning the first part...

3. 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).

4. ## dotdot

is there any connection between mean value theorem and the thing we asked to prove ?

i mean the f(b') - f(a') / ... = k

5. is there any connection between mean value theorem and the thing we asked to prove ?
Yes. The mean value theorem states that there exist a $c$ in $(a,b)$ such as $\frac{f(b)-f(a)}{b-a}=f'(c)$. In your example, $g$ and $h$ are defined and continuous on $(a,b)$, you must find a subinterval of $(a,b)$ in which the mean value theorem can be applied.

6. ## cant see

i still cant see any obvious connection...
anyone who can do it for me please?

7. ## be careful

be careful. hav u seen that f'(a) < k < f'(b)

that is f', not f

8. The intermediate value theorem for derivatives states that if $f:I\mapsto \mathbb{R}$ is a differenciable function then $f'$ has the intermediate value property on $I$, an open interval. Meaning if $a,b\in I$ with $a and $c$ lies (strictly) between $f'(a)$ and $f'(b)$ there exists $x_0\in (a,b)$ so that $f'(x_0) = c$.

Without lose of generality we will assume that $f'(a) < c < f'(b)$. Now define the function $g:[a,b]\mapsto \mathbb{R}$ as $g(x) = f(x) - cx$. Since $g$ is continous on $[a,b]$ it assumes a minimum value. We will now show that $g$ does not take a minimum value at $a$. Note that $g'(a) = f'(a) - c < 0$. But since $g'(a) = \lim_{x\to a^+}\tfrac{g(x)-g(a)}{x-a} < 0$ it follows that $g(x) - g(a)<0$ for $a sufficiently close to $x$. Thus, there exists $y\in (a,b)$ so that $g(y) - g(a) < 0$, which is means $g(y) < g(a)$, in particular $a$ is not a minimum for $g$. Using a similar argument we can show $b$ is not a minimum for $g$. Thus, $g$ has a minimum on $(a,b)$ and so there is a point $x_0 \in (a,b)$ so that $g'(x_0) = 0\implies f'(x_0) = c$.