Continuous Functions, IVT/EVT?

In summary, the conversation discusses the problem of showing that there is a value of x in the interval [0,1] where f(x) is equal to f(x+1), given that f(x) is a continuous function on [0,2] with f(0) = f(2). The Intermediate Value Theorem and the function g(x) = f(x) - f(x+1) are suggested as potential solutions. It is eventually determined that using the Intermediate Value Theorem on g(x) = 0, it can be shown that f(x) = f(x+1) at a certain value of x in the interval.
  • #1
SuspectX
7
0

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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  • #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
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
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
Oh right, duh... Thanks a lot!
 
  • #8
No worries mate ;)
 

Related to Continuous Functions, IVT/EVT?

1. What is a continuous function?

A continuous function is a mathematical function that has no abrupt changes or breaks in its graph. This means that the function can be drawn without lifting your pencil from the paper, and there are no gaps or holes in the graph.

2. What is the Intermediate Value Theorem (IVT)?

The Intermediate Value Theorem states that if a continuous function f(x) is defined on an interval [a, b], and if k is any number between f(a) and f(b), then there exists at least one number c in the interval [a, b] such that f(c) = k. In other words, the function takes on every value between f(a) and f(b) somewhere on the interval [a, b].

3. What is the Extreme Value Theorem (EVT)?

The Extreme Value Theorem states that if a continuous function f(x) is defined on a closed interval [a, b], then f(x) must have a minimum and a maximum value on that interval. In other words, the function must reach its highest and lowest points somewhere on the interval [a, b].

4. How are IVT and EVT related?

IVT and EVT are related because they both deal with the behavior of continuous functions. IVT guarantees that a continuous function will take on every value between two points, while EVT guarantees that a continuous function will have a maximum and minimum value on a given interval.

5. Why are continuous functions important?

Continuous functions are important because they accurately model many real-world phenomena, such as motion, sound, and temperature. They also allow for the use of powerful mathematical tools, such as the Intermediate Value Theorem and the Extreme Value Theorem, which can be used to solve problems and prove the existence of solutions.

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