1. ## Problem concerning differentiation

f(x) is continous in [0,1], differentiable in (0,1), and f(0)=0,f(1)=1.
prove that for any positive numbers a and b, there exist $\displaystyle x_1$ < $\displaystyle x_2$ in (0,1) such that
$\displaystyle \frac{a}{f'(x_1)}+\frac{b}{f'(x_2)}=a+b$.
how to prove this kind of problem about " there exist ..."?
I know there is three theorems in the textbook may be helpful.
But here there are two parameters.
Any suggestion, hint or solution is greatly appreciated!

2. We can try the Mean Value Theorem:

$\displaystyle \mbox{There exists a }c, 0< c<1,\mbox{ such that }\frac{f(1)-f(0)}{1-0}=f'(c).$

3. I know there exist a number c in (0,1) such that $\displaystyle f'(c)=1$.
that is obviously true by the theorem. And then.....
we still didn't get closer to the result.

4. Any body help me, PLZ!

5. Originally Posted by Shanks
I know there exist a number c in (0,1) such that $\displaystyle f'(c)=1$.
that is obviously true by the theorem. And then.....
we still didn't get closer to the result.
This is the result: $\displaystyle x_1=x_2=c$.

6. Sorry, I forgot the condition that $\displaystyle x_1<x_2$.