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Continuous Functions, IVT/EVT?

  1. Oct 25, 2011 #1
    1. The problem statement, all variables and given/known data
    Suppose that f(x) is a continuous function on [0,2] with f(0) = f(2). Show that
    there is a value of x in [0,1] such that f(x) = f(x+1).

    2. Relevant equations
    Intermediate Value Theorem?
    Extreme Value Theorem?
    Periodicity?

    3. The attempt at a solution
    For sure there's an f(x1)=f(x2), where x1≠x2, but I don't know how to prove that they're 1 unit apart. Would it also have something to do with the fact that 1 is the half-width of the interval?
     
  2. jcsd
  3. Oct 25, 2011 #2
    Consider the function
    [tex]g(x)\equiv f(x) - f(x+1)[/tex]
    for [itex]x \in [0,1] [/itex]

    Maybe I don't even need to say more :biggrin:
    What's [itex]g(0)[/itex], what's [itex]g(1)[/itex]?
     
  4. Oct 25, 2011 #3
    Wow I'm a bit stuck today. I'm just going around and around in circles with:

    [itex]g(0) = f(0) - f(1)[/itex]
    [itex]g(1) = f(1) - f(2)[/itex]

    I juggle them around and end up with:

    [itex]f(0) - g(0) = g(1) + f(2)[/itex]
    [itex]f(1) - f(2) = f(1) - f(0)[/itex]

    which brings me nowhere.
     
  5. Oct 25, 2011 #4
    [tex]
    g(0)=f(0)-f(1)[/tex]
    [tex]
    g(1)=f(1)-f(2)=f(1)-f(0)=-(f(0)-f(1))=-g(0)[/tex]
    Now use IVT

    Cheers
     
  6. Oct 25, 2011 #5
    Still confused as to how I incorporate [itex]f(x) = f(x+1)[/itex]
    I have that between [itex](0 , g(0))[/itex] and [itex](1, -g(0))[/itex] there has to be a coordinate [itex](x , 0)[/itex], but then how do I get from that back to [itex]f(x) = f(x+1)[/itex]?
     
  7. Oct 25, 2011 #6
    [tex]g(x)=0\Rightarrow f(x)-f(x+1)=0 \Rightarrow f(x)=f(x+1)[/tex]
     
  8. Oct 25, 2011 #7
    Oh right, duh... Thanks a lot!
     
  9. Oct 25, 2011 #8
    No worries mate ;)
     
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