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Math Help - Give example and prove it

  1. #1
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    Give example and prove it

    An open subinterval of the closed interval [a, b] is an open interval (c, d)
    such that (c, d) subset of [a, b] (i.e. a <= c < d <= b). Give an example of numbers a < b and of a function f : [a, b] --> R which is increasing on
    every open subinterval of [a, b] and is not increasing on [a, b]. Be sure
    to justify your example (Hint: Choose f to be discontinuous.)


    I am really bad at something that contains with intervals
    Last edited by 450081592; November 15th 2009 at 01:38 AM.
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  2. #2
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    The first thing I would notice is that (a, b) is an open subinterval of [a,b]. So your example function must be increasing on (a, b) (and there is a very easy example of such a function). And then you would need to pick values for f(a) and f(b) so that the function on [a,b] is NOT increasing. Again, fairly easy.
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    The first thing I would notice is that (a, b) is an open subinterval of [a,b]. So your example function must be increasing on (a, b) (and there is a very easy example of such a function). And then you would need to pick values for f(a) and f(b) so that the function on [a,b] is NOT increasing. Again, fairly easy.

    I have got an example
    f(x) = 1 if x = 0,
    f(x) = x if 0<x<2,
    f(x) = 1.5 if x = 2

    This function is continuous and increasing in every open subinterval of [0,2]
    But it is not increasing in [0,2].

    but I dont know how to prove it, can you show me the proof?
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