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Math Help - continuous functions on intervals 2

  1. #1
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    continuous functions on intervals 2

    Let I:=[a,b] and let f: be a continuous function on I such that for each x in I there exists y in I such that . Prove that there exists a point c in I such that f(c)=0
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  2. #2
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    I think you can say something along the lines of take some f(x) in R, and w/ d(0, f(x)) \leq \epsilon. We can find a y_1 such that d(0,f(y_1))\leq \epsilon /2. Suppose there was minimal value c, such that f(c) \geq 0. But we know we can find another value that is even smaller than c, and by the Archimedean Property, it must be 0. This is not a very rigorous proof though, since I didn't really use continuity.
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    Quote Originally Posted by terr13 View Post
    I think you can say something along the lines of take some f(x) in R, and w/ d(0, f(x)) \leq \epsilon. We can find a y_1 such that d(0,f(y_1))\leq \epsilon /2. Suppose there was minimal value c, such that f(c) \geq 0. But we know we can find another value that is even smaller than c, and by the Archimedean Property, it must be 0. This is not a very rigorous proof though, since I didn't really use continuity.
    yeh i think it has to be more rigorous and using continuity. Any idea how to do it that way??
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  4. #4
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    There is a theorem say that “A continuous function on a closed interval has a minimum and a maximum”.
    So \left( {\exists m \in I} \right)\left( {\forall z \in I} \right)\left[ {f(m) \leqslant f(z)} \right].
    So suppose that \left( {\forall z \in I} \right)\left[ {0 \ne f(z)} \right] then from the given f(m) > 0.
    A contradiction follows at once from the minimum.
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  5. #5
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    Quote Originally Posted by Plato View Post
    There is a theorem say that “A continuous function on a closed interval has a minimum and a maximum”.
    So \left( {\exists m \in I} \right)\left( {\forall z \in I} \right)\left[ {f(m) \leqslant f(z)} \right].
    So suppose that \left( {\forall z \in I} \right)\left[ {0 \ne f(z)} \right] then from the given f(m) > 0.
    A contradiction follows at once from the minimum.
    I have been trying to do this problem as well along with the other problem, yet i still cant do it. So frustrated . Being trying to read the book but I just dont understand this material.
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  6. #6
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    You are simply over-thinking all of this.
    If the absolute minimum is f(m) there cannot be any y such that \left| {f(y)} \right| \leqslant \frac{1}{2}\left| {f(m)} \right|
    Otherwise, f(m) would not be the absolute minimum!
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