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Math Help - Interval

  1. #1
    Senior Member
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    Interval

    Hello.

    i know g(0) = 1 and g(x) = 0 , iff x \in ]-\infty, -a] \cup [ a, \infty[

    What do we know about g(x-2) for example?

    It is g(2-2) = 0, but what about g=0?

    Is it g(x-2) = 0, x \in ]-\infty, -a-2] \cup [ a-2, \infty[

    or

    g(x-2) = 0, x \in ]-\infty, -a+2] \cup [ a+2, \infty[ ??

    I am not sure about it.

    Rapha
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  2. #2
    MHF Contributor Calculus26's Avatar
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    Remember g(x-2) is a translation to the right by 2 so


    g(x-2) = 0, ??

    is correct. For eg if x= -a+2 g(-a+2-2) = g(-a) = 0

    if x = a+2 g(a+2-2) = g(a) = 0 as it should

    On the other hand Consider
    g(x-2) = 0,

    Then if x = a-2 you would have g(a-4)= 0 which would not be true for
    a = 4
    Last edited by Calculus26; May 16th 2009 at 08:09 PM.
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  3. #3
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    Alright, thank you, mate
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