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Math Help - [SOLVED] Domain of square root

  1. #1
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    [SOLVED] Domain of square root

    So I have to find f of g which turns out to be

    square root of (1 - (square root 1+ x)^2)

    I can simplify this to square root of -x I believe.

    Original f(x) = square root 1-x^2
    Original g(x) = square root 1+x

    Domain of f(x) = x less than or equal to 1
    Domain of g(x) = x greater or equal to -1
    Domain of f of g = x less than or equal to 0

    The domain for square root of -x should then be [0,1] right?
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  2. #2
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    Quote Originally Posted by thekrown View Post
    So I have to find f of g which turns out to be
    square root of (1 - (square root 1+ x)^2)

    Original f(x) = square root 1-x^2
    Original g(x) = square root 1+x
    The domain for square root of -x should then be [0,1] right?
    Surely this is a counter-example: f \circ g( - 0.36) = \sqrt {1 - \left( {\sqrt {1 - 0.36} } \right)^2 } which is defined.

    Here is a post script hint: \text{dom}_{f \circ g}  = \text{dom}_f  \cap \text{Rng}_g
    Last edited by Plato; February 18th 2010 at 03:49 PM.
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  3. #3
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    I didn't understand that post at all.
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  4. #4
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    Quote Originally Posted by thekrown View Post
    I didn't understand that post at all.
    Then do you understand this probem as as given at all?
    You need a live-sit-down with a tutor.
    You are simply lacking the very basics of this material.
    I truly hope you will try to get the basics.
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  5. #5
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    If you're saying the domain is simply the range of it's inverse, I get that but I wanted to try and solve it using the teacher's methods.

    Anyways, wish me luck, my test is in a few hours.
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