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Math Help - Prove f has at most one fixed point in I.

  1. #1
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    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 \inI, 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?)
    Last edited by 450081592; November 15th 2009 at 04:54 PM.
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    Notice g'(x)<0 for all x\in I and g(c)=0. If there was another number d\in I (assume c<d) 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.
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    Quote Originally Posted by Jose27 View Post
    Notice g'(x)<0 for all x\in I and g(c)=0. If there was another number d\in I (assume c<d) 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
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    Quote Originally Posted by 450081592 View Post
    should we apply g(x) = f(x) - x in the proof
    Yes.

    What is it that you don't understand?
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  5. #5
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    Quote Originally Posted by Jose27 View Post
    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?
    Last edited by 450081592; November 16th 2009 at 09:08 PM.
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