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Math Help - Points for an inverse relation.

  1. #1
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    Points for an inverse relation.

    I promise, this is the last question for a while.

    I need 4 points for the inverse of this relation:

    y = 2|x|-1

    I dunno how to make absolute values using the MATH button.

    But yeah, I did think about substituting 2, 1, -1, -2 in for X, but is that the same as inverse?
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by mankvill View Post
    I promise, this is the last question for a while.

    I need 4 points for the inverse of this relation:

    y = 2|x|-1

    I dunno how to make absolute values using the MATH button.

    But yeah, I did think about substituting 2, 1, -1, -2 in for X, but is that the same as inverse?
    By inverse relation do you mean inverse function?

    If so then consider that

    \forall{x}<0\quad{y=-2x-1}

    and

    \forall{x}>0\quad{y=2x-1}
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  3. #3
    Senior Member nikhil's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    By inverse relation do you mean inverse function?
    If so then consider that
    you can not consider it inverse function as y=2|x|-1 is not one to one function which is a neccesary condition for a function to have inverse.
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  4. #4
    Senior Member nikhil's Avatar
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    Lightbulb Check this out

    I think you really mean inverse relation
    y=2|x|-1
    y=2x-1 if x>0
    x=(y+1)/2>0
    x=(y+1)/2 y>-1 or
    y=(x+1)/2 x>-1(only interchanged x and y because generally x is taken to represent independent variable)
    so we obtained the inverse relation (or inverse function if we particularly talk about y=2x-1 if x>0 as overall y=2|x|-1 do not have an inverse function) similarly same procedure can be followed for y=-2x-1 if x<0. (every function is a relation but every relation is not a function and every function may not have an inverse function)

    so here is the easy Method
    you said you need 4 points of inverse relation. So put any arbitrary value of x in the equation then interchange the range and domain for example let x=2 then y=2|x|-1
    =2|2|-1=3 so the obtained point for relation is (2,3) now interchanging range and domain, point for inverse relation will be (3,2). Remember (3,-2) will also be an element of inverse relation
    (3,2) and (3,-2) are not valid points points for a function.
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by nikhil View Post
    you can not consider it inverse function as y=2|x|-1 is not one to one function which is a neccesary condition for a function to have inverse.
    They never specified, but based on other problems by this poster, I assumed they accidentally omited a domain on which the function is to be considered, which based on the domain would make the function injective.

    EDIT: But thank you very much, I shouldn't assume.
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