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Homework Help: F(x)=x for some x

  1. Oct 14, 2013 #1
    1. The problem statement, all variables and given/known data

    I have to prove that if a function is bounded for x between (0,1) and f(x) is bounded also between (0,1) and f(x) is continuos that there exists some x such that f(x)=x

    2. Relevant equations

    Intermediate value theorem?

    3. The attempt at a solution

    I know that this problem makes intuitive sense, and I've drawn out bounds, and I know obviously, that this statement will hold true, and the proof probably involves the intermediate theorem law, but I am just not sure how to construct my proof for it. Can I just say that for f(0)<x<f(1) then x=f(x) for some x (and repeat the same thing for my other two cases, when f(1)>f(0) and f(1)=f(0))?

    Any help would be great
  2. jcsd
  3. Oct 14, 2013 #2
    First off, you need the conditions you've listed to hold on ##[0,1]##. A counterexample on ##(0,1)## would be ##f(x)=x^2##.

    You are correct that the Intermediate Value Theorem might prove useful. Think about the function ##g(x)=f(x)-x##.
  4. Oct 14, 2013 #3
    Okay, I considered doing that; would I try to now show that g(x)=0 for some x?
  5. Oct 14, 2013 #4
  6. Oct 14, 2013 #5
    Okay, I think I got it now, thanks!
    I just considered two points; g(1) and g(0), and showed how between those points, there always exists a zero as g(1) can be between 0 and -1 and g(0) is between 0 and 1, so by the intermediate value theorem, there has to be a zero between these two points for any g(x) that satisfy the properties of f(x)
  7. Oct 14, 2013 #6
    That sounds like the right idea. You'll definitely need to beef up the rigor a bit to make it a proper proof.
  8. Oct 14, 2013 #7

    D H

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    It's important to remember that that "between" is inclusive of rather than exclusive of the end points. You used (0,1) in the problem statement rather than [0,1]. gopher_p's counterexample of f(x)=x2 satisfies the conditions of the opening post, but there is no point in (0,1) where f(x)=x. There is such a point (two points in this case) if one looks at [0,1] instead.
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