Thread: Proving a point in an interval

1. 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.

2. 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 $g$ ?

3. what is g?

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

4. Yes...

$f$ is defined on $[0,1]$ so $g(x)=f(x)-f(x+1/2)$ is defined on $[0,1/2]$ because both $x$ and $x+1/2$ have to lie in $[0,1]$.

We are told that $f(0)=f(1)$, can you use this to get some information on $g(0)$ and $g(1/2)$ ?

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

6. 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 $g(x)=f(x){\color{red}-}f(x+1/2)$, one has $g(0)=f(0)-f(1/2)$ and $g(1/2)=f(1/2)-f(1)=f(1/2)-f(0)$ hence $g(0)=-g(1/2)$. In other words, if $g(0)$ is positive then $g(1/2)$ is negative and if $g(0)$ is negative then $g(1/2)$ is positive.

7. 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?

8. 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 ?

9. 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.

10. 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.