# Thread: Problem concerning differentiation

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 $x_1$ < $x_2$ in (0,1) such that
$\frac{a}{f'(x_1)}+\frac{b}{f'(x_2)}=a+b$.
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:

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

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

6. Sorry, I forgot the condition that $x_1.
here I made a complement.
Now how to prove it? help!!!

7. Edit: Reworking