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I face some difficulties to find the domain

  1. Oct 11, 2009 #1
    I face some difficulties to find the domain

    Now i put my questions and please help me to answer it

    i try solve it


    answer 1.JPG

    answer 2.JPG

  2. jcsd
  3. Oct 11, 2009 #2


    Staff: Mentor

    It can sometimes take several hours for attachments to be approved, so for quicker responses, it's best to type your questions out so that we can see them right away.
  4. Oct 11, 2009 #3
    I think attachments are approved now >> :smile:
  5. Oct 11, 2009 #4
    The domain of problem 46 is not all real numbers. For √x, domain is x≥0. Use that for √(-1-s).

    The domain of 47 is larger than just w≥4; perhaps you were still thinking about domains with square roots. Try thinking about the domain of 1/x, or where it is undefined, and then think about the domain of (2x+3)/(w-4) again.

    I can't quite make out your answer for the domain of 51, but it doesn't look right. But let's look at what you have with (y + 1)(y - 3)≥0. You know where the parabola crosses the x-axis and that it opens upward since the leading coefficient is positive. What part(s) of the graph will be above the x-axis, or ≥0?
    Last edited: Oct 11, 2009
  6. Oct 11, 2009 #5
    As Bohrok said, if the domain was all real numbers for √(-1-s), it would be incorrect. Say s is equal to 2. -1-2 = -3. √-3 is an imaginary number. You want to make sure whatever is under the square root bracket is equal to 0 or more.
  7. Oct 11, 2009 #6
    53. x^2-4x+5 and x^2+2x+3 can't be factor the way you did.
  8. Oct 11, 2009 #7


    Staff: Mentor

    The domain for f(x) = √x is x >= 0.
  9. Oct 11, 2009 #8
    I hit the wrong two-key combination instead of Ctrl-V. :redface: Corrected.
  10. Oct 12, 2009 #9
    46 )
    1-s ≥ 0
    -s ≥ - 1

    47) i think domain all real numer

    51 ) please i want the answer

    help me .>
  11. Oct 12, 2009 #10


    Staff: Mentor

    No. Start with -1 -s ≥ 0
    You should end up with an inequality that involves s, not -s. Review the rules for working with inequalities if needed.
    No. There is one real number that isn't in the domain of this function.
    For this function to be defined, it must be that y2 -2y - 3 ≥ 0.
    Factor the left hand side into two linear factors. For the product of the two factors to be ≥ 0, both factors have to be ≥ 0, or both have to be <= 0. Each case will produce its own set of numbers.
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