# Intermediate Value Theorem related question

1. Aug 23, 2008

### kehler

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. Aug 23, 2008

### d_leet

What is p(0)? What is p(1/2)? How do these quantities relate to one another?

3. Aug 23, 2008

### kehler

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.

4. Aug 23, 2008

### d_leet

There is also the case that p(0)=p(1)=0, but that is also easy to deal with.

5. Aug 23, 2008

### kehler

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?

6. Aug 23, 2008

### d_leet

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.

7. Aug 23, 2008

### kehler

Oh ok I see what u mean. Thanks :)

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