Proving Bounded Function and Continuity: Intermediate Value Theorem Explained

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In summary, the problem at hand is to prove that for a function f(x) that is bounded and continuous on [0,1], there exists at least one point x where f(x)=x. This can be done using the Intermediate Value Theorem by considering the function g(x)=f(x)-x and showing that there is a zero for g(x) between the points g(0) and g(1). It is important to note that the interval [0,1] is inclusive of its endpoints, as using (0,1) as the interval can lead to counterexamples.
  • #1
lepton123
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Homework Statement



I have to prove that if a function is bounded for x between (0,1) and f(x) is bounded also between (0,1) and f(x) is continuos that there exists some x such that f(x)=x


Homework Equations



Intermediate value theorem?


The Attempt at a Solution



I know that this problem makes intuitive sense, and I've drawn out bounds, and I know obviously, that this statement will hold true, and the proof probably involves the intermediate theorem law, but I am just not sure how to construct my proof for it. Can I just say that for f(0)<x<f(1) then x=f(x) for some x (and repeat the same thing for my other two cases, when f(1)>f(0) and f(1)=f(0))?

Any help would be great
 
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  • #2
lepton123 said:

Homework Statement



I have to prove that if a function is bounded for x between (0,1) and f(x) is bounded also between (0,1) and f(x) is continuos that there exists some x such that f(x)=x


Homework Equations



Intermediate value theorem?


The Attempt at a Solution



I know that this problem makes intuitive sense, and I've drawn out bounds, and I know obviously, that this statement will hold true, and the proof probably involves the intermediate theorem law, but I am just not sure how to construct my proof for it. Can I just say that for f(0)<x<f(1) then x=f(x) for some x (and repeat the same thing for my other two cases, when f(1)>f(0) and f(1)=f(0))?

Any help would be great

First off, you need the conditions you've listed to hold on ##[0,1]##. A counterexample on ##(0,1)## would be ##f(x)=x^2##.

You are correct that the Intermediate Value Theorem might prove useful. Think about the function ##g(x)=f(x)-x##.
 
  • #3
Okay, I considered doing that; would I try to now show that g(x)=0 for some x?
 
  • #4
lepton123 said:
Okay, I considered doing that; would I try to now show that g(x)=0 for some x?

Yes.
 
  • #5
Okay, I think I got it now, thanks!
I just considered two points; g(1) and g(0), and showed how between those points, there always exists a zero as g(1) can be between 0 and -1 and g(0) is between 0 and 1, so by the intermediate value theorem, there has to be a zero between these two points for any g(x) that satisfy the properties of f(x)
 
  • #6
That sounds like the right idea. You'll definitely need to beef up the rigor a bit to make it a proper proof.
 
  • #7
lepton123 said:
so by the intermediate value theorem, there has to be a zero between these two points for any g(x) that satisfy the properties of f(x)
It's important to remember that that "between" is inclusive of rather than exclusive of the end points. You used (0,1) in the problem statement rather than [0,1]. gopher_p's counterexample of f(x)=x2 satisfies the conditions of the opening post, but there is no point in (0,1) where f(x)=x. There is such a point (two points in this case) if one looks at [0,1] instead.
 

1. What is the Intermediate Value Theorem?

The Intermediate Value Theorem (IVT) is a mathematical theorem that states that if a continuous function has values of opposite signs at two points, then it must have at least one root (zero) between those points.

2. How does the Intermediate Value Theorem relate to bounded functions?

The IVT is used to prove that a bounded function must have a maximum and minimum value on a given interval. This is because if the function is continuous and bounded, it must cross every horizontal line between the maximum and minimum value, which implies that it must have a root for every value in between.

3. Can the Intermediate Value Theorem be used to prove continuity?

Yes, the IVT is often used as a tool to prove continuity. This is because the theorem states that a function is continuous if it has no breaks or jumps, and thus must cross every horizontal line between two points. If a function satisfies this condition, then it is continuous.

4. What are the conditions for the Intermediate Value Theorem to be applicable?

The IVT is applicable when the function is continuous on a closed interval [a,b] and has values of opposite signs at the endpoints of the interval.

5. Can the Intermediate Value Theorem be applied to all types of functions?

No, the IVT is only applicable to continuous functions. This means that the function must have no breaks or jumps and can be drawn without lifting the pen from the paper.

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