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Domain of the function

  1. Jul 11, 2013 #1
    1. The problem statement, all variables and given/known data

    f(x)= √[(2x+1)/x3-3x2+2x]

    2. Relevant equations



    3. The attempt at a solution

    here
    For f(x) to exist
    x3-3x2+2x >0

    and 2x+1≥0
    by solving this i found that x [itex]\in[/itex] [-[itex]\frac{1}{2}[/itex],-∞][itex]\bigcup[/itex][2,∞]
    but the answer given is (-∞,-1/2][itex]\bigcup[/itex](0,1)(2,∞)


    so please anybody can explain why that is so


    thanks
     
  2. jcsd
  3. Jul 11, 2013 #2

    mfb

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    2016 Award

    Staff: Mentor

    What about 2x+1 <= 0, x3-3x2+2x < 0?

    There are brackets missing in the problem statement.
     
  4. Jul 11, 2013 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    I think you mean √[(2x+1)/(x3-3x2+2x)]

    No, the numerator and denominator do NOT have to both be positive. What is true is that the fraction cannot be negative which means that, as long as the numerator is not 0, they must have the same sign. That is, either x3-3x2+2x >0 and 2x+1≥0 OR x3-3x2+2x <0 and 2x+1< 0.

     
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