Intermediate value theorem problem

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Homework Help Overview

The discussion revolves around the Intermediate Value Theorem (IVT) and its application to proving the existence of a fixed point for a continuous function F defined on the interval [0,1]. Participants explore the implications of F being bijective and continuous, and the meaning of the equation F(c) = c.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the significance of the fixed point condition F(c) = c, with some suggesting the use of the IVT. Questions arise about the implications of continuity and the behavior of F at the endpoints of the interval. There is also a consideration of whether a fixed point can exist if F is not continuous.

Discussion Status

The conversation is ongoing, with various interpretations being explored. Some participants have offered guidance on visualizing the problem and applying the IVT, while others have raised questions about the assumptions regarding continuity and the selection of points within the interval.

Contextual Notes

There is an emphasis on the continuity of F and its bijective nature, with participants questioning the implications of these properties on the existence of fixed points. The discussion also touches on the necessity of understanding the function's behavior at the endpoints of the interval.

shrug
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1. How can you prove if F is continuous, then there exists a fixed point of F in [0,1]?



I know F:[0,1] ---> [0,1] an bijective, but what is f(c)=c mean?
 
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Have you tried drawing a picture? It might be enlightening.
 


I think it means more or less exactly what it says. There is a c in [0,1] such that F(c)=c. Use the intermediate value theorem. If F is bijective there is an a such that F(a)=0 and a b such that F(b)=1. What happens in between? Apply the IVT to F(x)-x on the interval [a,b].
 


Dick said:
I think it means more or less exactly what it says. There is a c in [0,1] such that F(c)=c. Use the intermediate value theorem. If F is bijective there is an a such that F(a)=0 and a b such that F(b)=1. What happens in between? Apply the IVT to F(x)-x on the interval [a,b].

Does that mean I need to pick a point like .5, which is between [0,1].

Also, would there be any point where F isn't continuous and a fixed point may not exist.

Thanks guys
 


I don't think you really understood what I wrote. As morphism said, draw a picture. Look up the IVT.
 


No, you can't "pick" a point. And you are told that the function is continuous. Why are you asking if it isn't?

If F is from [0, 1] what can you say about F(0)? What can you say about F(1)?

What can you say about 0- F(0) and 1- F(1)?
 

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