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Math Help - Tricky Domain and Range Question

  1. #1
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    Tricky Domain and Range Question

    Hello. I have encountered a question that begins simply enough and then ends up confusing me.
    Here it is:

    Given f(x) = (sqrt)(x^2 - 9) and g(x) = 1/(x-4), find the domain and range of the following functions:
    a) f(x)
    b) f(g(x))
    c. f(f(x))

    I solved "a" with a domain of (- infinity, to -3) (3, infinity) and range of all real #'s.

    Part "b" I get stuck though. I know f(g(x)) = (sqrt)((1/(x-4))^2 - 9) but then I do not know where to go. Help Please!

    Also, I do not get part "c" at all. f(f((x)) = (sqrt)((sqrt)(x^2 -9))^2 -9) ??
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  2. #2
    Junior Member bondesan's Avatar
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    For b) you have f(g(x))=\sqrt{\left(\frac{1}{x-4}\right)^2 - 9}

    The square root function only is defined when the argument is positive, so all you have to do is find x which makes

    \left(\frac{1}{x-4}\right)^2 - 9>0 true.

    Expand the denominator, multiply both sides by it then you'll end with a polynomial - find its roots and the interval in which it is positive. This is the domain of f(g(x)).

    For c) you use the same approach.
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  3. #3
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    OK so I did that and got:

    (1/(x-2))^2 - 9 > 0
    1 / (x^2 -8x +16) - 9 > 0
    -9x^2 + 72x -144 > x^2 - 8x + 16
    10x^2 - 80x + 160
    10(x-4)^2
    Isn't it always positive?
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  4. #4
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    note that y = \sqrt{x^2 - 9} has domain x \le -3 and x \ge 3

    for y = \sqrt{[g(x)]^2 - 9} ... domain is g(x) \le -3 and g(x) \ge 3

    substitute \frac{1}{x-4} in for g(x) and finish solving the inequalities.
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