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Math Help - Intermediate Value Theorem

  1. #1
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    Intermediate Value Theorem

    a) Let f :[a,b] --> [a,b] be continuous on [a,b]. Prove that f has a fixed point, i.e., prove that (exist) c in [a,b] such that f(c)=c

    b) let f,g : [a,b]->[a,b] be two continuous function on [a,b] such that f(a)>or equal g(a) and f(b)<or equal g(a).
    Prove that exist c in [a,b] such that f(c) =g(c)
    Plx help
    Last edited by mathbeginner; November 6th 2010 at 02:36 PM.
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  2. #2
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    I see that you have severval other postings.
    You should understand that this is not a homework service nor is it a tutorial service.
    PLease either post some of your own work on this problem or explain what you do not understand about the question.
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  3. #3
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    Sorry,
    I just want to get some idea for how to start the question since I have no idea how to start the question.
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  4. #4
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    Here is a standard hint.

    Define g(x)=f(x)-x, a continuous function.

    Now show that g(a)\ge 0~\&~g(b)\le 0.
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  5. #5
    MHF Contributor Also sprach Zarathustra's Avatar
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    For b)

    Define new function H(x)=f(x)-g(x),

    1. prove that H continuous...
    2. Prove that exist x=t in [a,b] such that H(t)=0...What is the meaning of that?
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  6. #6
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    Quote Originally Posted by Also sprach Zarathustra View Post
    For b)

    Define new function H(x)=f(x)-g(x),

    1. prove that H continuous...
    2. Prove that exist x=t in [a,b] such that H(t)=0...What is the meaning of that?
    I get a problem when I am trying to prove that H(t)=0, how can I say that H(t)=0?
    I think I have to show that f(t)is not > g(t) and also g(t) is not > f(t) then g(t)=f(t) but I have no idea how can I show that is true.
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  7. #7
    MHF Contributor Also sprach Zarathustra's Avatar
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    Try to look at H(a) and H(b) and then use Intermediate Value Theorem.
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  8. #8
    MHF Contributor Drexel28's Avatar
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    My favorite way to prove a)

    Is to suppose that f(x)\ne x for all x\in[a,b] then g:[a,b]\to\{-1,1\}:x\mapsto\frac{f(x)-x}{|f(x)-x|} is continuous and surjective, but this is impossible...why?
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  9. #9
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    Quote Originally Posted by Plato View Post
    Here is a standard hint.

    Define g(x)=f(x)-x, a continuous function.

    Now show that g(a)\ge 0~\&~g(b)\le 0.
    when I when that g(a)\ge 0~\&~g(b)\le 0 f(c) will equal 0. how can I relat to c???
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  10. #10
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    Quote Originally Posted by mathbeginner View Post
    when I when that g(a)\ge 0~\&~g(b)\le 0 f(c) will equal 0. how can I relat to c???
    Go to your textbook and class notes and review examples that apply the Intermediate Value Theorem. Spend more than just a few minutes doing this. Then come back if you still have this question. (You will find that the question you posted is a standard example in many textbooks that cover this material).
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  11. #11
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    Quote Originally Posted by mr fantastic View Post
    Go to your textbook and class notes and review examples that apply the Intermediate Value Theorem. Spend more than just a few minutes doing this. Then come back if you still have this question. (You will find that the question you posted is a standard example in many textbooks that cover this material).
    I find out that what is c depend on how do u set up the question, ex. g(a)<c<g(b) c can be anything ex. 0 or 1 if I put 0 on it then f(c)=0 or if I put 1 on it then f(c)=1

    Am I right? I just want to make sure that am I get it or not. Thanks.
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  12. #12
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    Perhaps you are missing the point of this problem.
    You are not asked to actually find the value of c.
    You are asked to show that one actually exists.
    Of course, it does depend upon the setup.
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  13. #13
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    Quote Originally Posted by Plato View Post
    Perhaps you are missing the point of this problem.
    You are not asked to actually find the value of c.
    You are asked to show that one actually exists.
    Of course, it does depend upon the setup.
    So you are saying that I should prove that f(c)is not > or < c is equal c ?
    Or saying I did find out f(c)=0 since it equal 0 so that is a c exist?

    I am so confused abt this because when I was doing that first I did prove that that f(c)is not > or < c is equal c. Then I thought I get it wrong because what I am doing is proving that IVT is true. That's why I use the other way, to use g(a)<0<g(b) find out that f(c)= 0.

    Thanks
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  14. #14
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    Quote Originally Posted by mathbeginner View Post
    So you are saying that I should prove that f(c)is not > or < c is equal c ?
    Or saying I did find out f(c)=0 since it equal 0 so that is a c exist?

    I am so confused abt this because when I was doing that first I did prove that that f(c)is not > or < c is equal c. Then I thought I get it wrong because what I am doing is proving that IVT is true. That's why I use the other way, to use g(a)<0<g(b) find out that f(c)= 0.

    Thanks
    Your confusion is so profound that I have absolutely nothing more to say.
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  15. #15
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    Quote Originally Posted by mathbeginner View Post
    So you are saying that I should prove that f(c)is not > or < c is equal c ?
    Or saying I did find out f(c)=0 since it equal 0 so that is a c exist?

    I am so confused abt this because when I was doing that first I did prove that that f(c)is not > or < c is equal c. Then I thought I get it wrong because what I am doing is proving that IVT is true. That's why I use the other way, to use g(a)<0<g(b) find out that f(c)= 0.

    Thanks
    Please take the advice given in post #10. Then come back with fresh eyes on this thread.
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