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Math Help - Finding the Domain Problems:

  1. #1
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    Finding the Domain Problems:

    Please help me find the domain of the following problems:

    1. 5throot(-4+3x)

    The function is defined on the interval from _________ to __________

    2. 6throot(-4+3x)

    The function is defined on the interval from _________ to __________

    3. The domain of the function sqrt(x(x-4)) in interval notation is _________

    Thanks so much!!
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    Quote Originally Posted by qbkr21 View Post
    Please help me find the domain of the following problems:

    1. 5throot(-4+3x)

    The function is defined on the interval from _________ to __________

    2. 6throot(-4+3x)

    The function is defined on the interval from _________ to __________

    3. The domain of the function sqrt(x(x-4)) in interval notation is _________

    Thanks so much!!
    For an "even" root we require that the argument is non-negative. For an "odd" root we don't have that restriction. So the answer to 1 is all real numbers and the answer to 2 is  [0, \infty).

    3 is a tad more difficult, but all we need do is determine where x(x-4) \geq 0. This happens for ( - \infty, 0) \cup (4, \infty) as you can verify by several means.

    -Dan
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    Except for #3 you must use bracket, also I am unable to figure out how you got #2 could you please explain a bit more?
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by qbkr21 View Post
    Except for #3 you must use bracket, also I am unable to figure out how you got #2 could you please explain a bit more?
    Yes, the answer for 3 should be ( - \infty, 0] \cup [4, \infty) because of the \geq relation. Thank you.

    For 2 we require that the argument under the 6th root be non-negative for the same reason we require the same for the square root: x^{1/6} is not defined for negative x. (Or if you wish to speak more loosely, a negative x will return an imaginary value.)

    So we see that I goofed again. The argument of the 6th root is -4 + 3x so we require that -4 + 3x \geq 0 or the solution for x is the interval: [4/3, \infty ).

    (I was obviously doing this problem too fast. My apologies!)

    -Dan
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