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Math Help - Domain again

  1. #1
    Member M670's Avatar
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    Domain again


    Find the domain of this

     (-infinity,0] U [3,infinity) becuase its to the 4th root we have a restirction unlike an odd root which is all real number
    why am I wrong ?
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  2. #2
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    Re: Domain again

    let's look at it this way:

    let g(x) = \sqrt[4]{x}

    let h(x) = x^2 - 9

    then f = g \circ h

    now h is clearly defined for all x, no problem there.

    but g is only defined for x ≥ 0.

    that means that f is only defined for h(x) ≥ 0.

    so we need to look at what the RANGE of h is, in particular, for which x is h(x) ≥ 0?

    this is precisely when:

    x^2 \geq 9

    that is:

    x \in (-\infty, -3] \cup [3,\infty).

    to see why your answer is wrong, let's pick something in it, and see what happens:

    let's use x = -1, which is in the interval (-∞,0].

    now x2 - 9 = (-1)2 - 9 = 1 - 9 = -8. how are we going to take a 4th root of that?

    EDIT: darn, i left out an x!

    it should be:

    x2 - 9x (is my memory, or my eyesight, going?)

    this is ≥ 0 when either:

    x ≥ 0 and x - 9 ≥ 0, that is x ≥ 9...in this case the "9" controls (since its bigger and we have to have both), so this is the interval [x,∞).

    or:

    x < 0 and x - 9 < 0, that is x < 9...in this case the "0" controls (since it is smaller), so this is the interval (-∞,0].

    therefore, the correct answer is as Soroban said:

    (-∞,0] U [9,∞).
    Last edited by Deveno; January 10th 2013 at 09:51 PM.
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  3. #3
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    Re: Domain again

    Hello, M670!

    \text{Find the domain: }\;f(x) \;=\;\sqrt[4]{x^2-9x}

    We see that x^2 - 9x must not be negative.

    When is x^2-9x greater than or equal to zero?

    Consider the parabola y \:=\:x^2-9x.
    When is it above the x-axis?
    Code:
            |
          * |             *
            |
           *|            *
        ----*-----------*----
           0| .       . 9
            |    . .
            |
    Domain: . (\text{-}\infty,\:0\:\!] \cup [\:\!9,\:\infty)

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