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Range of quadratic equation.

  1. Feb 21, 2012 #1

    PrincePhoenix

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    The problem statement, all variables and given/known data

    The following equation defines a rational function.Find its Range
    f(x) = x2/(x-1)

    The attempt at a solution

    Let y = x2/(x-1)

    =>y(x-1) = x2
    =>xy-y = x2
    =>x2-xy+y = 0

    comparing with the general form of quadratic equation ax2+bx+c = 0,
    a=1, b=-y, c=y.

    Putting the values in quadratic formula,

    x = y±√(y2 - 4y)/2

    It is clear that f(x) will only be real when the term y2-4y is greater than or equal to zero. So,

    y2-4y ≥ 0

    y(y-4) ≥ 0 ---------(i)

    divide both sides by y,
    y-4 ≥ 0
    y ≥ 4
    ---------------------------------------------------------------------------------
    However just by looking at the discriminant it is clear that y ≤ 0. How do I get that? When I divide y-4 on both sides of (i), I get y≥0.
     
    Last edited: Feb 21, 2012
  2. jcsd
  3. Feb 21, 2012 #2
    Can y-4 be negative?
     
  4. Feb 21, 2012 #3

    PrincePhoenix

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    I don't get what you're trying to say.

    What I mean is that along with y-4 ≥ 0, there is another interval for values of range which is supposed to be y ≤ 0. How do I get that?
     
  5. Feb 21, 2012 #4
    What happens when you multiply (or divide) both sides of an inequality like [itex]a \le b[/itex], what sign goes here [itex]-a\ ?\ -b[/itex]
     
  6. Feb 21, 2012 #5

    PrincePhoenix

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    -a≥-b?

    But isn't that because we multiplied '-' on both sides of the inequality?? :confused:
     
  7. Feb 21, 2012 #6

    PrincePhoenix

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    Do you mean this is the way to do it??

    y2-4y ≥ 0

    y(y-4) ≥ 0 ---------(i)

    multiply both sides by '-1'

    -(y-4)y ≤ 0
    divide both sides by -(y-4)

    y≤0?
     
  8. Feb 21, 2012 #7
    From here, take different cases;
    y(y-4) ≥ 0

    Suppose y-4=0 then;
    Suppose y-4<0 then;
    Suppose y-4>0 then;
     
  9. Feb 21, 2012 #8

    PrincePhoenix

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    y-4 = 0 => y=4
    y-4 > 0 => y>4
    y-4 < 0 => y<4?

    Still I don't get you. I take the three different possible cases. What do I do with them? Do I have to check each to see which are correct? Or somehow use them to get y ≤ 0?
     
  10. Feb 21, 2012 #9
    If (y-4) < 0, then (y-4) is negative

    What happens now when you divide

    y(y-4) ≥ 0

    by the negative number (y-4)?
     
  11. Feb 21, 2012 #10

    PrincePhoenix

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    y ≤ 0. :smile:
    Thank you.
     
  12. Feb 21, 2012 #11

    PrincePhoenix

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    Just a couple of questions to clear it up. When all the three possibilities were tried, how should I have known that the answer I got by y-4<0 one of the two intervals?
    And can I find the second interval that I had found in my original solution using this method?
     
  13. Feb 21, 2012 #12
    You know that it is correct because you're just checking equalities at all points, you're free to pick a value for y (or set of values for y) and check each of them.
    When you set y-4<0 you are checking every y that's less than 4
    When you set y-4=0 you are checking for y=0
    When you set y-4>0 you are checking every y that's greater than 4
    By doing this you have checked every possible value for y.

    Using this method you will find ALL solutions

    y(y-4) ≥ 0

    Suppose y-4>0 then
    y ≥ 0
    Which just returns y>4

    Suppose y-4=0 then
    y*0=0
    Which is obviously true, so y=4 is valid, combined with the first result we get
    y ≥ 4

    Suppose y-4<0 then
    0 ≥ y

    And so we arrive at the two sets of solutions using this method
    0 ≥ y
    y ≥ 4
     
  14. Feb 21, 2012 #13

    PrincePhoenix

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    Thanks a lot genericusername. :smile:. That cleared it up.
     
  15. Feb 21, 2012 #14
    No problem buddy! :biggrin:
     
  16. Feb 21, 2012 #15
    It must be that [tex]y(y - 4) \ge 0.[/tex]


    Get critical numbers by setting y = 0 and y - 4 = 0 and solving for y in each respective equation.

    So the critical numbers are y = 0, -4.


    Test numbers to the left, in between, and to the right to see which
    regions satisfy the inequality.

    Get to your desired range from there.
     
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