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Math Help - Proof of intermediate value theorem for a particular case

  1. #1
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    Proof of intermediate value theorem for a particular case

    Let
    f be a continuous function on the closed interval [0; 1], and suppose that f(0) < 0
    and that
    f(1) > 0. By considering the set S= {x E [0,1] :f(x)_<0) prove that there exists
    a x' such that f(x')=0

    S is non empty as x=0 is a member and bounded above by 1, so by the completness axiom Sup(S) exists. Sup(S)_> x E S and Sup(s) _< 1. Not sure where to go from here. Thanks in advance.

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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Denote c = sup S and suppose f ( c ) < 0 or f ( c ) > 0 . In both cases you'll get a contradiction. So, necessarily f ( c ) = 0 .
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  3. #3
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    Quote Originally Posted by poirot View Post
    [LEFT]Let f be a continuous function on the closed interval [0; 1], and suppose that f(0) < 0
    and that
    f(1) > 0. By considering the set S= {x E [0,1] :f(x)_<0) prove that there exists a x' such that f(x')=0
    This proof relies upon the following theorem.
    If f is continuous and then there is a neighbored of c on which f has the same sign as f(c).
    If you take and suppose that then you can find a contradiction.
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  4. #4
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    let c=Sup(S)
    Assume F(c) < 0. I don't see a contradiction
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  5. #5
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    Quote Originally Posted by poirot View Post
    I don't see it
    If then .
    Can't you find a contradiction there???
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    I don't know if c is a member of the set. If it is then surely sigma will be 1-c.
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  7. #7
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    ok I'm in a better maths mood today. I got frustated by this yesterday.
    assume F(c)<0 which implies the theorem.
    c<_1 so F(c)<0 implies c is a member of the set
    So for all m E [0,1] with m > c implies that F(m) > 0 but get stuck here. Its sort of intuitive that it can't be because of the density of reals but I don't know how to prove it convincingly.
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  8. #8
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    Quote Originally Posted by Plato View Post
    If then .
    Can't you find a contradiction there???
    It follows that that means .
    That is a contradiction because c is the greatest number with that property
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    this is only true if c+ sigma/2 <1
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  10. #10
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    Quote Originally Posted by poirot View Post
    this is only true if c+ sigma/2 <1
    Well of course. But you have to pick delta to make that true.
    Look, I am not doing this question for you.
    If you do not understand this approach, please do not use it.
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  11. #11
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    sorry but you have put 'there exists a delta >0 '. So you have to show such a delta does not exist, not picking a counter example. Thats just bad logic. I am close to understanding.
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  12. #12
    MHF Contributor FernandoRevilla's Avatar
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    poirot: I don't understand which is the problem. According to Plato's answer (#5):



    and this, evidently contradicts the choice of c.
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