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Math Help - continious function

  1. #1
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    Question continious function

    A function f: A-> R is increasing if f(x)<= f(y) for every x, y in A such that x<=y.


    suppose that f:[a,b] ->R is increasing and that for every Y in [f(a), f(b)] there is some x[0] in [a,b] sucht that f( x[0] ) = Y

    How do i show that fis a continious function?
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  2. #2
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    I have an idea.

    The limit \lim_{x\to c^-}f(x)=\inf \{f(x)|x<c\}=f(c)
    And \lim_{x\to c^+}f(x)=\sup \{f(x)|x>c\}=f(c)
    So the left and right limits exists and are equal to the functional evaluation.

    Because since the function is increasing it places the role of the least upper bound.
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  3. #3
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    Question increasing functions part 2

    Thankz...
    i want to show that the asuumption that f is increasing is essential by
    finding an example of a function f:[-1,1] ->R which has the property that for every Y in [f(-1), f(1)] there is some x[0] in [-1,1] such that f( x[0] ) = Y , bu it is not continious and so fcan't be increasing.

    Does any one know any examples with these values for a=-1 and b=1?

    thankz
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  4. #4
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    Quote Originally Posted by dopi View Post
    Thankz...
    i want to show that the asuumption that f is increasing is essential by
    finding an example of a function f:[-1,1] ->R which has the property that for every Y in [f(-1), f(1)] there is some x[0] in [-1,1] such that f( x[0] ) = Y , bu it is not continious and so fcan't be increasing.
    What you actually are asking is for the converse of the intermediate value theorem. You know that any continous function on a closed interval satisfies the Intermediate property. However you wish to now if a functions satisfies the Intermediate property then it is not necessarly continous.

    Consider a semicircle f:[-1,1]\to \sqrt{1-x^2}
    Now take any point not on endpoints of its extreme value and move it done "slightly" IVT still works but the function has replacable discontinuity.
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