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Intermediate Value Theorem related question

  1. Aug 23, 2008 #1
    1. The problem statement, all variables and given/known data
    Suppose the g:[0,1] -> R is continuouse, with g(0)=g(1). Show that there exists an x in [0,1/2] such that g(x) = g(x+1/2)


    2. Relevant equations
    The intermediate value theorem is used in there somewhere. IVT:
    If f(x) is continuous on the interval [a,b] and f(a)<q<f(b), then there exists an x in [a,b] such that f(x)=q.


    3. The attempt at a solution
    I tried creating another function, p(x) = g(x) - g(x+1/2). I want to show p(x)=0 at some point in [0,1/2] by finding a positive number in the interval and a negative one, then quote the IVT. But I can't seem to do that. I tried substituting the endpoint values in but they didn't help much. And I assume that g(0)=g(1) must come into play somewhere?
     
  2. jcsd
  3. Aug 23, 2008 #2
    What is p(0)? What is p(1/2)? How do these quantities relate to one another?
     
  4. Aug 23, 2008 #3
    Hmm...
    p(0) = g(0) - g(1/2)
    p(1/2) = g(1/2) - g(1) = g(1/2) - g(0) = -p(0)

    Ooh. So I just say that they're negative of each other? And if one's negative and the other's positive then there must be a zero in between. Is that it? :D
    Silly me.
     
  5. Aug 23, 2008 #4
    There is also the case that p(0)=p(1)=0, but that is also easy to deal with.
     
  6. Aug 23, 2008 #5
    Can I do it for p(1) if we want x in [0,1/2]? :S
    Anyway, thanks for the hint. Someone else told me to consider the three cases p(0)=0, p(0)>0 and p(0)<0 so I've been sitting here for ages trying to figure that out! Any idea how to do it with that method?
     
  7. Aug 23, 2008 #6
    Sorry I meant p(0)=p(1/2)=0. That method is basically what you've done, you've already figured out the cases when p(0)>0 and p(0)<0 since you know p(1/2)=-p(0). That only leaves p(0)=0 which is the case above p(0)=p(1/2)=0.
     
  8. Aug 23, 2008 #7
    Oh ok I see what u mean. Thanks :)
     
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