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Math Help - Domain.

  1. #1
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    Domain.

    Can someone please show me how to work out the domain for this function?


     \frac{2}{3} \sqrt{36-p^2}

    I know the answer is 0 < x < 4
    But just can't worked out how they got to that.

    Any help is appreciated!
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  2. #2
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    Quote Originally Posted by el123 View Post
    Can someone please show me how to work out the domain for this function?


     \frac{2}{3} \sqrt{36-p^2}

    I know the answer is 0 < x < 4
    But just can't worked out how they got to that.

    Any help is appreciated!
    The domain for this function comes from the square root expression involved. If the function is a real function, then the argument of a square root can never be negative.

    So what you are searching for, really, is the values of p for which  36 - p^2 \geq 0 .

    You can find these values by solving that inequality.

    You have your answer wrong though. The domain is  -6 \leq p \leq 6
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  3. #3
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    Do you mean 0 < p < 4? (there isn't an x in the term)

    I would think the answer is |p| <= 6.

    I am assuming the question only wants real numbers? If that is so, then you cannot have a negative number in the radical. So 36 minus any number less than 36 (which will always be positive because of the square) is fine, and 36-36 is fine too, since you can have a square root of zero. But 36 - 37 would be bad, since that would be the square root of a negative number, which is nonreal.
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  4. #4
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    and i keep coming up with 0<x<6 but its not the correct answer.
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  5. #5
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    Im using an old exam answer sheet. It says the domain is 0<x<4.

    I get why it can't be negative. But this answer sheet says thats what the answer is.
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  6. #6
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    The answer sheet is wrong. You can simply use a calculator to confirm. Try x = 4, x=5, x=6 and x = 7. If the answer sheet is right, you should get an error for 5, 6, 7. But if you and I are correct, then you should only get an error for 7.
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  7. #7
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    Quote Originally Posted by el123 View Post
    and i keep coming up with 0<x<6 but its not the correct answer.
    The correct method and answer was given in post #2. From the subsequent replies it appears no-one read it.
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