Continuous Functions, IVT/EVT?

[SuspectX]

Homework Statement

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).

Homework Equations

Intermediate Value Theorem?
Extreme Value Theorem?
Periodicity?

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?

 

[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)?

 

[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.

 

[susskind_leon]

<br /> g(0)=f(0)-f(1)
<br /> g(1)=f(1)-f(2)=f(1)-f(0)=-(f(0)-f(1))=-g(0)
Now use IVT

Cheers

 

[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)?

 

[susskind_leon]

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

 

[SuspectX]

Oh right, duh... Thanks a lot!

 

[susskind_leon]

No worries mate ;)