# Thread: Prove f has at most one fixed point in I.

1. ## Prove f has at most one fixed point in I.

A number c is called a fixed point of f if f(x) = c. Prove that if f is differentiable on an interval I and f'(x) < 1 for all x $\in$I, then f has at most one fixed point in I. HINT: From g(x) = f(x) - x.
Assum that I is an open interval (why must we assume this?)

2. Notice $g'(x)<0$ for all $x\in I$ and $g(c)=0$. If there was another number $d\in I$ (assume $c) such that $g(d)=0$ we would have by Rolle's theorem that there is a point $x_0 \in (c,d)$ such that $g'(x_0)=0$ which is a contradiction. I don't see where you need $I$ to be open though.

3. Originally Posted by Jose27
Notice $g'(x)<0$ for all $x\in I$ and $g(c)=0$. If there was another number $d\in I$ (assume $c) such that $g(d)=0$ we would have by Rolle's theorem that there is a point $x_0 \in (c,d)$ such that $g'(x_0)=0$ which is a contradiction. I don't see where you need $I$ to be open though.

should we apply g(x) = f(x) - x in the proof, I don't get the idea

4. Originally Posted by 450081592
should we apply g(x) = f(x) - x in the proof
Yes.

What is it that you don't understand?

5. Originally Posted by Jose27
Yes.

What is it that you don't understand?
where did you use g(x)= f(x)-x here, I have a hard time to show the entire solution

and I don't know what is the contradictin for, what is your hypothesis?