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Thread: domain of inverse square root

  1. #1
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    domain of inverse square root

    domain of inverse square root-ii.pngI'm asked to find the domain for this question...
    My ans is (x-1)^2 + (y+1) ^2 >-1 , but the ans given by the author is different , from the sketch , we can know that the author ans is (x-1)^2 + (y+1) ^2 -1 , am i right ?
    Which is correct ? domain of inverse square root-ooo.jpghttp://imgur.com/a/Kn68f
    Attached Thumbnails Attached Thumbnails domain of inverse square root-lll.jpg  
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  2. #2
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    Re: domain of inverse square root

    (1) I'm assuming there's a typo in the original function... isn't is supposed to be
    $\displaystyle f(x, y) = \frac{x}{\sqrt{(x - 1)^2 + (y {\color{red}+} 1)^2 - 1}}$?

    (2) You have this on the second line of your work, which is not correct:
    $\displaystyle (x - 1)^2 + (y {\color{red}-} 1)^2 {\color{red}+} 1 > 0$
    It should be:
    $\displaystyle (x - 1)^2 + (y {\color{red}+} 1)^2 {\color{red}-} 1 > 0$

    (3) The answer should be
    $\displaystyle (x - 1)^2 + (y + 1)^2 > {\color{red}1}$
    which is consistent with the diagram you attached.


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  3. #3
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    Re: domain of inverse square root

    Hey xl5899.

    You should look at your working where the signs should be switched. Try checking your algebra on line 2 going from line 1.

    In terms of your working out you have a circle in the form of (x-a)^2 + (y-b)^2 > 1 which means that your circle is centered at (a,b) corresponding to a co-ordinate of (+1,-1) for that particular equation.

    This corresponds to the answer given in the third attachment.

    As a point of checking you might want to substitute a point that is not in the third attachment but is in the second and see if you get an error where the thing inside the square root is less than zero which will help provide evidence of this.
    Thanks from xl5899
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