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

• Nov 15th 2009, 04:43 PM
450081592
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?)
• Nov 15th 2009, 05:12 PM
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.
• Nov 15th 2009, 07:36 PM
450081592
Quote:

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(Clapping)
• Nov 15th 2009, 07:54 PM
Jose27
Quote:

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

Yes.

What is it that you don't understand?
• Nov 16th 2009, 04:33 AM
450081592
Quote:

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?