Continuous Functions, IVT/EVT?

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Homework Help Overview

The problem involves a continuous function f(x) defined on the interval [0,2] with the condition that f(0) = f(2). The objective is to demonstrate that there exists a value of x in the interval [0,1] such that f(x) = f(x+1).

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the implications of the Intermediate Value Theorem (IVT) and the Extreme Value Theorem (EVT) in relation to the problem. There are attempts to define a new function g(x) = f(x) - f(x+1) and evaluate its values at the endpoints of the interval. Questions arise about how to prove the existence of x such that f(x) = f(x+1) and how to relate this back to the original function.

Discussion Status

The discussion is ongoing, with participants exploring various approaches and expressing confusion about the connections between their findings and the original problem statement. Some guidance has been offered regarding the use of the IVT, but no consensus has been reached on the complete solution.

Contextual Notes

Participants are working under the constraints of the problem's conditions, including the continuity of f(x) and the specific values at the endpoints of the interval. There is an emphasis on understanding the relationship between g(x) and the original function without arriving at a definitive conclusion.

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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?
 
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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]?
 
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.
 
[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
 
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]?
 
[tex]g(x)=0\Rightarrow f(x)-f(x+1)=0 \Rightarrow f(x)=f(x+1)[/tex]
 
Oh right, duh... Thanks a lot!
 
No worries mate ;)
 

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