# Thread: Showing the existence of two points

1. ## Showing the existence of two points

let f(x) is continous on [0,1], differentiable on (0,1).
and f(0)=0, f(1)=1.
prove that for any two positive real numbers a and b, there exist two points $0< x_1< x_2< 1\text{ such that }af'(x_1)+bf'(x_2)=a+b$.
Don't know where to start.

2. Originally Posted by Shanks
let f(x) is continous on [0,1], differentiable on (0,1).
and f(0)=0, f(1)=1.
prove that for any two positive real numbers a and b, there exist two points $0< x_1< x_2< 1\text{ such that }af'(x_1)+bf'(x_2)=a+b$.
Don't know where to start.
This is very similar to the problem that you stated in this thread. The only difference is that here you have $f'(x_1)$ and $f'(x_2)$ instead of $1/f'(x_1)$ and $1/f'(x_2)$. You ought to be able to adapt the solution given there so as to deal with the present problem.

3. Sorry! I forgot i have posted a similar problem.
this is a simplized problem in the calculas test, the original problem is to prove the generalized case with finite parameters(point).
prof. said It can be proved by using the intermediate value property of derivative. But I still can't write down the details completely.
We are supposed to find two points whose derivative is closer enough to 1 such that the equality holds,right?