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Math Help - domain & range

  1. #1
    fuzzyorama
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    domain & range

    what is the domain, range for these equations? kindly include explaination.

    1. z = 4x + y
    2. z = √(x + y - 16)
    3. z = √(4 - y)
    4. z = 1/xy
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  2. #2
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    Quote Originally Posted by fuzzyorama View Post
    what is the domain, range for these equations? kindly include explaination.

    1. z = 4x + y
    2. z = √(x + y - 16)
    I start your off.

    1)The domain is all RxR. The range is >=0 because x^2>=0 and y^2>=0.

    2)We require that x^2+y^2>=16. And the range is >=0 because sqrt( ) function can get arbitrarily big.
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  3. #3
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    Hello, fuzzyorama!

    What is the domain, range for these equations?

    1)\; z\:=\:4x^2 + y^2
    x can be any real value.
    y can be any real value.
    . . Domain: . x,y \in (-\infty,\,\infty)

    Since x^2 and y^2 are nonnegative, z will be nonnegative.
    . . Range: . z \,\geq\,0



    2)\; z \:= \:\sqrt{x^2 + y^2 - 16}

    The radicand must be nonnegative.
    . . Domain: . x^2+y^2\,\geq\,16

    z will be nonnegative.
    . . Range: . z \,\geq\,0



    3)\; z \:= \:\sqrt{4 - y^2}

    The radicand must be nonegative. . 4-y^2\:\geq\:0
    . . Domain: . -2\:\leq \:y\:\leq\:2

    z will be nonnegative.
    . . Range: . z\,\geq\,0



    4)\; z \:= \:\frac{1}{x^2y^2}

    x,\,y cannot be zero.
    . . Domain: . x,y\:\neq\:0

    Since x^2y^2 is always positive, z is positive.
    . . Range: . z\,>\,0

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