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Math Help - Basic domain.

  1. #1
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    [SOLVED]Basic domain.

    y= \sqrt{-x^3+x}

    (-x^3+x)\geq0
    \equiv x(-x^2+1)\geq0
    \equiv x\geq0\cap(x\geq-1\cap~x\leq1)

    Domain = 0\leq~x\leq1


    Is that correct??

    because in my calculator i have a domain =  ]\infty, 1]
    Last edited by Fabio010; October 22nd 2011 at 05:17 AM.
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  2. #2
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    Re: Basic domain.

    No, it is not correct. You calculator is right (assuming you meant ]-\infty, 1] In order that ab> 0, a and b must have the same sign, NOT just "a> 0, b> 0". That is, either a> 0 and b> 0 or a< 0, and b< 0. x> 0 and 1- x^2> 0 gives 0< x< 1. x< 0 and 1-x^2< 1 gives x^2> 0 which is true for any x< 0. Of course, x= 0 and x= 1 are also included.
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  3. #3
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    Re: Basic domain.

    So you are telling me, that i need to assume that x<0 belongs to the function domain.

    But, i solved the domain in incorrectly way.

    I do not know how to solve the domain, too show x<0 in solution.
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  4. #4
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    Re: Basic domain.

    y = \sqrt{x - x^3}

    x - x^3 \ge 0

    x(1 - x^2) \ge 0

    x(1-x)(1+x) \ge 0

    note that the three values of x that make the expression equal to 0 are x = 0 , x = 1 , and x = -1

    these three values break the possible values of x into four intervals ...

    (-\infty,-1] , [-1, 0] , [0, 1] , and [1, \infty)

    pick any value in each interval and check the value of x(1-x)(1+x).
    if that value is positive, then that interval is in the domain.
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  5. #5
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    Re: Basic domain.

    Now i fully understood.
    Thanks.
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  6. #6
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    Re: Basic domain.

    Quote Originally Posted by Fabio010 View Post
    So you are telling me, that i need to assume that x<0 belongs to the function domain.
    No, I didn't say anything of the sort. I said that you cannot assume it does not!

    But, i solved the domain in incorrectly way.

    I do not know how to solve the domain, too show x<0 in solution.
    I thought that was what I just showed you. You have, correctly, that -x^3+ x\ge 0. As I said before, if ab> 0, then either a> 0 and b> 0 or a<0 and b< 0. You must consider both possiblities.
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