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Math Help - Domain, Range, and Function Problems

  1. #1
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    Domain, Range, and Function Problems

    1.) Please help me find the domain and range of <br />
h(x)=-\sqrt{x^2-36}<br />

    2.) For f(x) = 4x- \sqrt{x} and g(x) = (3x+2)^2, please find f(g(x)).

    Thanks in advance for your assistance.
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  2. #2
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    Did you graph these two functiona?
    That will tell you a great deal.
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  3. #3
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    Quote Originally Posted by currypuff View Post
    1.) Please help me find the domain and range of <br />
h(x)=-\sqrt{x^2-36}<br />
    Domain: We must have x^2 - 36\geq 0 \implies x\geq 6 \mbox{ or }x\leq -6.

    Range: It is all real y such that the equation y = - \sqrt{x^2 - 36} has a solution for x in the domain. Thus, it means we want to solve, -y = \sqrt{x^2 - 36} and RHS is always non-negative thus LHS must be non-negative and so we have a constraint that -y \geq 0 \implies y\leq 0. Square both sides, y^2 = x^2 - 36 \implies x^2 = y^2 + 36 since y^2 + 36 \geq 0 a square root exists and so the solution is  x = \pm \sqrt{y^2 + 36} but we want that x\geq 6 \mbox{ or }x\leq -6. If x\geq 6 then \sqrt{y^2 + 36} \geq 6 \implies y^2 + 36 \geq 36 \implies y^2\geq 0 which is always true. And similarly with -\sqrt{y^2 + 36} \leq -6. Thus, the range is y\leq 0.
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  4. #4
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    Quote Originally Posted by currypuff View Post
    1.) Please help me find the domain and range of <br />
h(x)=-\sqrt{x^2-36}<br />
    This one is actually quite simple, you can set what is inside of the square root equal to zero and solve for x.

    -1(x^2-36) \geq0

    -x^2 + 36 \geq 0

    -x^2 \geq -36

     x^2 \leq 36

    \sqrt{x^2} \leq \sqrt{36}

    x \leq 6

    There's your domain. Not sure how to find the range, but I'm sure someone here can help you on that

    EDIT: Dag nabbit! Hacker beat me to it again!!!
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  5. #5
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    Quote Originally Posted by mathgeek777 View Post
    x \leq 6

    There's your domain. Not sure how to find the range, but I'm sure someone here can help you on that
    That would be
    -6 \leq x \leq 6
    actually.

    -Dan
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