# Homework Help: Continuous Functions, IVT/EVT?

1. Oct 25, 2011

### SuspectX

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. Oct 25, 2011

### susskind_leon

Consider the function
$$g(x)\equiv f(x) - f(x+1)$$
for $x \in [0,1]$

Maybe I don't even need to say more
What's $g(0)$, what's $g(1)$?

3. Oct 25, 2011

### SuspectX

Wow I'm a bit stuck today. I'm just going around and around in circles with:

$g(0) = f(0) - f(1)$
$g(1) = f(1) - f(2)$

I juggle them around and end up with:

$f(0) - g(0) = g(1) + f(2)$
$f(1) - f(2) = f(1) - f(0)$

which brings me nowhere.

4. Oct 25, 2011

### susskind_leon

$$g(0)=f(0)-f(1)$$
$$g(1)=f(1)-f(2)=f(1)-f(0)=-(f(0)-f(1))=-g(0)$$
Now use IVT

Cheers

5. Oct 25, 2011

### SuspectX

Still confused as to how I incorporate $f(x) = f(x+1)$
I have that between $(0 , g(0))$ and $(1, -g(0))$ there has to be a coordinate $(x , 0)$, but then how do I get from that back to $f(x) = f(x+1)$?

6. Oct 25, 2011

### susskind_leon

$$g(x)=0\Rightarrow f(x)-f(x+1)=0 \Rightarrow f(x)=f(x+1)$$

7. Oct 25, 2011

### SuspectX

Oh right, duh... Thanks a lot!

8. Oct 25, 2011

### susskind_leon

No worries mate ;)