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Identity at Some Point

  1. Dec 12, 2005 #1
    "Let f[a,b]->[a,b] be continuous. Prove that there exists at least 1 x in [a,b] such that f(x)=x."

    This seems simple geometrically since if we consider the identity function g(x)=x, if f(x) is continuous, then if you "draw" the graw of f, it must intersect g at some point. At that point, f(x)=x. But I have no idea how to translate this intuition into analytical lingo.
  2. jcsd
  3. Dec 12, 2005 #2
    Use the intermediate value theorem.
  4. Dec 12, 2005 #3
    and what can you say about g(x) = f(x)-x on [a,b]?
  5. Dec 12, 2005 #4


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    Staff Emeritus
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    In particular, what is f(a)- a? What is f(b)-b?
    (Remember that f(a) and f(b) is in [a,b]?)
  6. Dec 13, 2005 #5
    I got it! Thanks for the help.
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