Thread: Proving a point in an interval

Let f be continuous on the interval [0,1] to Reals and such that f(0) = f(1). Prove that there exists a point c in [0,1/2] such that f(c) = f(c + (1/2)).
[Hint: Consider g(x) = f(x) - f(x + (1/2))]

I'm not really sure how to go about proving this. Any help would be much appreciated.

Hi,

Originally Posted by jkru

Let f be continuous on the interval [0,1] to Reals and such that f(0) = f(1). Prove that there exists a point c in [0,1/2] such that f(c) = f(c + (1/2)).
[Hint: Consider g(x) = f(x) - f(x + (1/2))]

What can be said about ?

g(x) is going to be the difference between a point x and (x + 1/2) when applied to the function f(x).

Yes...

is defined on so is defined on because both and have to lie in .

We are told that , can you use this to get some information on and ?

So, we can say that g(0) = g(1/2)?

Originally Posted by jkru

So, we can say that g(0) = g(1/2)?

There was a typo in my post, I've corrected it, sorry.

As , one has and hence . In other words, if is positive then is negative and if is negative then is positive.

Ohh, that makes sense. But, how does this help us prove a point c in [0,1/2]? Are we supposed to say that g(0) and g(1/2) are opposite in sign that there exists a point such that g(c) = 0?

Originally Posted by jkru

Are we supposed to say that g(0) and g(1/2) are opposite in sign that there exists a point such that g(c) = 0?

Yes ! Can you justify this ?

I think there's a theorem in my book that I'm allowed to use. I think it's something along the lines of:

With a continuous function the two points are connected, so the function itself must cross the axis, thus resulting in an intermediate value.

Is that correct? Not sure if I explained it right.

Originally Posted by jkru

With a continuous function the two points are connected, so the function itself must cross the axis, thus resulting in an intermediate value.

Is that correct?

Yes it is . This theorem is called the intermediate value theorem.

Thanks for your help!