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Quick domain of function question: Why do you have 2 cases for this function?

  1. Sep 30, 2012 #1
    1. The problem statement, all variables and given/known data

    This question here. Can someone please tell me why we have 2 cases to consider.

    http://archives.math.utk.edu/visual.calculus/0/domain.1/8.html

    I dont understand why it says we have 2 cases to consider.

    8 x - 4 > 0 and x - 3 > 0.
    8 x - 4 < 0 and x - 3 < 0.

    why is one greater than and less than???


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 30, 2012 #2

    ehild

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    The expression under the square root can not be negative. Therefore the sign of both the numerator and denominator have to be same - either positive or negative.


    ehild
     
  4. Sep 30, 2012 #3
    ahhhg. I dont fully understand.
     
  5. Sep 30, 2012 #4
    Well, as ehild said, the total value of the function under the radical cannot be negative in order to produce a real value. Therefore, both the numerator and the denominator under the radical must both be either positive or negative to result in a positive under the radical. The easiest way to solve [itex]\frac{8x-4}{x-3} = 0[/itex] is to solve the numerator such as this:

    8x-4 ≥ 0, if x-3 > 0 and 8x-4 ≤ 0, if x-3 < 0

    For simplicity, use something like this:

    8x-4 = A
    x-3 = B

    Then solve:

    A ≥ 0
    = x ≥ 1/2

    B > 0
    = x > 3

    Then the negative solution,

    A ≤ 0
    = x ≤ 1/2

    B < 0
    = x < 3

    Putting the first two parts and the second sets together, you get the cases for the function:

    A ≥ 1/2 if B > 3,
    A ≤ 1/2 if B < 3

    Hope this helps you!
     
  6. Sep 30, 2012 #5
    ahhh GOT it!

    how about this: (x^2 - 1) / (x + 1)

    would it not just be x must be greater than -1 ??

    why is the answer (-infinity, -1) (-1, infinity)

    ???
     
  7. Sep 30, 2012 #6
    For a standard rational function, it is continuous except when the denominator is equal to 0. Therefore, you solve x + 1 = 0, and x = -1, giving you the only place that the denominator will be zero. However, if it was a radical like the other problem, the solution would be x > -1, since at all other places there will be a negative under the radical.
     
  8. Sep 30, 2012 #7

    SammyS

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    That's not a complete question.

    What is it that you want to do with (x2-1)/(x+1) ?
     
  9. Sep 30, 2012 #8
    so when there is NO radical... all we would do is evaluate the denominator and its always just must not equal to 0 ?
     
  10. Sep 30, 2012 #9
    I dont understand, because lets take this question then.


    1/(x + 5)

    I would just go:

    x + 5 > 0
    x > -5

    But from you are saying, thats wrong!

    I should then just go x CANNOT equal 5, it would be (-∞, -5)(-5, ∞) ?
     
  11. Sep 30, 2012 #10
    Correct, as long as there isn't something else going on with the function that somehow restricts values. That is what happened with the rational function under the square root; since there is not a real solution for a negative under a square root, it it not within the domain. Similar things could happen if there was something such as a logarithm somewhere in the function. As long as there are nothing except integer polynomials in the numerator and denominator, the only restrictions of the domain are values that result in the denominator being equal to 0.

    And yes, the solution to the domain of [itex]\frac{1}{x+5}[/itex] would be (-∞,-5), (-5,∞).
     
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