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Continuous not bounded above function

  1. Feb 12, 2012 #1
    1. The problem statement, all variables and given/known data

    Let f : R -> R be a continuous function such that f(0) = 0. If S := {f(x) | x in R} is not
    bounded above, prove that [0, infinity) ⊆ S (that is, S contains all non-negative real numbers).

    Then find an appropriate value for a in the Intermediate Value theorem.

    2. Relevant equations

    3. The attempt at a solution

    If y > 0, then since S is not bounded above, there exists b in R such that f(b) > y.

    Then because y>0 , f(b) >0 and [0, infinity) ⊆ S ????

    Applying the intermediate value theorem to this continuous function, it follows that there exists a real number a such that f(a) = y. ????
  2. jcsd
  3. Feb 12, 2012 #2
    you should want to apply the IVT to the closed intervals contained in S of the continuous function.
  4. Feb 14, 2012 #3
    There is no given function. I think its more general question
  5. Feb 14, 2012 #4
    perhaps i am misunderstanding a bit. first show that S is an interval (by continuity and IVT), then show by contradiction this interval has no upper bound.
  6. Feb 14, 2012 #5


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    That's true.
    No, those two facts aren't enough, or even particularly relevant.
    That last statement is true but you haven't given a convincing argument. Explain how your correct statement in red above combines with the IV theorem to get your last statement.
  7. Feb 15, 2012 #6
    how I am going to use the IVT with this statement?
  8. Feb 15, 2012 #7


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    You are trying to show ##y\in S##. You have b > 0 with f(b) > y and f(0)=0. So...
  9. Feb 15, 2012 #8


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    f is continuous on the closed interval [0,b].

    f(0) = 0

    f(b) > y.

    now ask yourself: can f get from f(0) to f(b) without crossing the horizontal line at y?

    try drawing a picture, and re-read the statement of the IVT very carefully....
  10. Feb 16, 2012 #9
    so by IVT f(0)<=y<=f(b). then how we have to find the value for a using IVT
  11. Feb 16, 2012 #10


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    what, exactly, does the IVT say? isn't it an existence theorem of some sort? what does it say exists?

    (you are looking for a phrase such as: there is a c in [a,b] such that....)
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