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Math Help - Domain of Square Root Function

  1. #1
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    Domain of Square Root Function

    How do you find the domain and range of square root functions?

    (1) Find the domain of y = sqrt{2x + 3}

    (2) Find the domain of f(x) = sqrt{2x - x^2}
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  2. #2
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    Well since we're dealing with real numbers, we aren't allowed to take the square-root of a negative number. So find out for what x-values the expression inside the square-root is less than 0.
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  3. #3
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    Quote Originally Posted by Jameson View Post
    Well since we're dealing with real numbers, we aren't allowed to take the square-root of a negative number. So find out for what x-values the expression inside the square-root is less than 0.
    Just to expand a bit, the domains are given by the solution to the following inequations:

    (1) Solve 2x + 3 \geq 0.

    (2) Solve 2x - x^2 \geq 0.
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  4. #4
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    yes but......

    Quote Originally Posted by mr fantastic View Post
    Just to expand a bit, the domains are given by the solution to the following inequations:

    (1) Solve 2x + 3 \geq 0.

    (2) Solve 2x - x^2 \geq 0.
    Yes, but why? I am looking for the why of things not just how to find the answer.
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  5. #5
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    Well, it's not that hard to find the "why."

    Let's use our brain: having f(x)=\sqrt x. For what values of x, \sqrt x is well defined? Yes, we require that x\ge0, 'cause x can be zero, or positive, but it can't be negative, since square roots of negative numbers don't belong to the Real numbers, that's why we require x\ge0.

    Now, by applyin' this to your problems, you'll see this does make sense.
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  6. #6
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    ok

    Quote Originally Posted by Krizalid View Post
    Well, it's not that hard to find the "why."

    Let's use our brain: having f(x)=\sqrt x. For what values of x, \sqrt x is well defined? Yes, we require that x\ge0, 'cause x can be zero, or positive, but it can't be negative, since square roots of negative numbers don't belong to the Real numbers, that's why we require x\ge0.

    Now, by applyin' this to your problems, you'll see this does make sense.
    ok...now I get it.
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