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Finding the range of rational functions algebraically

  1. Oct 11, 2012 #1
    how to find the range of rational functions like f(x) = [itex]\frac{1}{{x}^{2}-4}[/itex] algebraically , i graphed it and seen that (-1/4,0] can not be in range . generally i am interested in how to find the range of functions and rational functions in particular
     
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  3. Oct 12, 2012 #2

    haruspex

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    How do you think it might relate to the maxima and minima of the function? Where are those for your example?
     
  4. Oct 12, 2012 #3

    yes i got those points by differentiating but can't we find it without using derivatives ? i mean any algebraic method ?
     
  5. Oct 12, 2012 #4

    Mentallic

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    I can't think of any algebraic methods, and if I were to use logical reasoning, it would still involve some kind of crude link to limits and derivatives.
     
  6. Oct 12, 2012 #5

    mathman

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    By inspection |x| = 2 is critical and x = 0 is a local minimum.
     
  7. Oct 12, 2012 #6
    minimum ? it's max. there is no minmimum for this function .but how can i find the range though ? i mean how it's useful to find out the range exactly ?
     
    Last edited: Oct 13, 2012
  8. Oct 13, 2012 #7

    mathman

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    You're right. It is local max (-1/4). As |x| -> 2 from below, f(x) -> -∞. However |x| -> 2 from above, f -> +∞. Finally as |x| -> ∞, f -> 0.

    Net result: range has two parts (-∞,-1/4) for |x| < 2, and (0,+∞) for |x| > 2.
     
  9. Oct 13, 2012 #8

    Mentallic

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    I took a little while to think about it, and yes, there is an algebraic solution to the problem.

    For the function

    [tex]y=\frac{1}{x^2-4}[/tex]

    let the range be denoted R, which is also the y-value, so we have

    [tex]R=\frac{1}{x^2-4}[/tex]

    Now, we want to solve for x:

    [tex]Rx^2-4R=1[/tex]

    [tex]x^2=\frac{1+4R}{R}[/tex]

    [tex]x=\pm\sqrt{\frac{1+4R}{R}}[/tex]

    Now, x exists (and thus a correspondent range exist) whenever

    [tex]\frac{1+4R}{R}\geq 0[/tex]

    and clearly the opposite of that is, if

    [tex]\frac{1+4R}{R}< 0[/tex]

    then the domain does not exist, thus the range does not exist.

    Now, if we solve this inequality, for R>0 (we can see that [itex]R\neq 0[/itex])

    [tex]1+4R<0[/tex]

    [tex]R<\frac{-1}{4}[/tex]

    And we obviously can't have that both R>0 and [itex]R<\frac{-1}{4}[/itex] so we scrap that. Now if R<0

    [tex]1+4R>0[/tex]

    [tex]R>\frac{-1}{4}[/tex]

    Which gives us the intersection [tex]\frac{-1}{4}<R<0[/tex] as required.
     
  10. Oct 14, 2012 #9

    mathman

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    There are no values of x where -1/4 < y < 0. As I have indicated, the range is everything but this interval.
     
  11. Oct 14, 2012 #10

    Mentallic

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    In my solution I wrote

     
  12. Oct 15, 2013 #11

    can u plz explain to me how did u reach the conclusion that R will be greater than zero? solving the inequality only yields R>-1/4
     
  13. Oct 15, 2013 #12

    Mentallic

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    I had written that precisely a year ago :biggrin:

    Since [itex]-1/4 < R < 0[/itex] is where the range of the function does not exist, everything else is where the range does exist.

    So the range is
    [itex]y> 0 \cup y\leq -1/4[/itex]

    (edit: [itex]y\neq 0[/itex] because we can't divide by 0)

    If there's something in my earlier solution that you don't understand, just point it out.
     
    Last edited: Oct 15, 2013
  14. Oct 15, 2013 #13
    lol well i'm glad u're still alive :P, yes there is something i do not get, the range of that function would be the domain of the inverse function correct? 1+4R/R > or equal to 0 will give us all the possible range correct?

    solving this only gives us range bigger than -1/4, and range does not equal to 0, but as u already know this function has range above than zero, the question is how did u deduce that from the problem/inequality?
     
  15. Oct 15, 2013 #14

    Mentallic

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    Yes and yes.

    Oh you're right about excluding R=0 from the solution, I missed that. You're not solving the inequality correctly though.

    Remember that when you multiply through by R, if R<0 then you need to change the reverse the sign of the inequality

    So for R>0
    [itex]1+4R\geq 0[/itex]

    but for R<0
    [itex]1+4R\leq 0[/itex]

    Now when you solve these two inequalities, you only keep the intersections that makes sense. For example, if we assume R<0 but then solve the inequality and find that R>1, there are no solutions for R such that both of these can hold true at the same time (we say that the intersection of the sets is the empty set). If however we assume R<0 and solve the inequality to find R>-1, then our solution set is -1<R<0.
     
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