Continuous Functions, IVT/EVT?

  • Thread starter SuspectX
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  • #1
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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?
 

Answers and Replies

  • #2
Consider the function
[tex]g(x)\equiv f(x) - f(x+1)[/tex]
for [itex]x \in [0,1] [/itex]

Maybe I don't even need to say more :biggrin:
What's [itex]g(0)[/itex], what's [itex]g(1)[/itex]?
 
  • #3
7
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Wow I'm a bit stuck today. I'm just going around and around in circles with:

[itex]g(0) = f(0) - f(1)[/itex]
[itex]g(1) = f(1) - f(2)[/itex]

I juggle them around and end up with:

[itex]f(0) - g(0) = g(1) + f(2)[/itex]
[itex]f(1) - f(2) = f(1) - f(0)[/itex]

which brings me nowhere.
 
  • #4
[tex]
g(0)=f(0)-f(1)[/tex]
[tex]
g(1)=f(1)-f(2)=f(1)-f(0)=-(f(0)-f(1))=-g(0)[/tex]
Now use IVT

Cheers
 
  • #5
7
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Still confused as to how I incorporate [itex]f(x) = f(x+1)[/itex]
I have that between [itex](0 , g(0))[/itex] and [itex](1, -g(0))[/itex] there has to be a coordinate [itex](x , 0)[/itex], but then how do I get from that back to [itex]f(x) = f(x+1)[/itex]?
 
  • #6
[tex]g(x)=0\Rightarrow f(x)-f(x+1)=0 \Rightarrow f(x)=f(x+1)[/tex]
 
  • #7
7
0
Oh right, duh... Thanks a lot!
 
  • #8
No worries mate ;)
 

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