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Computing the range for a rational function involving absolute value

  1. Nov 3, 2006 #1
    Hi. I need help computing a range.

    The question is : Find the domain and range of [tex]y=\frac{|x+2|}{x}[/tex].

    The domain is obvious, x can't be 0, (-inf,0,) U (0,inf). But how do I find the range?? Can someone help me out? I have tried messing around with the definition of absolute value... if x>0 then |x| = x and if x<0 then |x| = -(x) but it just adds to the confusion, it probably doesn't help that I am running on like 4 cups of coffee. Anyways, thanks for the help :approve:
     
  2. jcsd
  3. Nov 3, 2006 #2

    nazzard

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    Hello Checkfate,

    you can continue to use the definition of absolute value and rewrite the function for both cases. For [itex]x\geq-2[/itex] you would get
    [tex]y=\frac{|x+2|}{x}=\frac{x+2}{x}=1+\frac{2}{x}[/tex]
    and you can discuss the range of this function for [itex]x\in[-2,0)[/itex] and [itex]x\in(0,\infty)[/itex].

    (similar for the case of [itex]x<-2[/itex])

    Regards,

    nazzard
     
    Last edited: Nov 3, 2006
  4. Nov 3, 2006 #3
    Okay, I came to the right answer.... this is how I did it

    I started as you suggested, by first working with the function as defined when [itex]x\geq-2[/itex] and got the equation [tex]y=1+\frac{2}{x}[/tex]. I then had to do some thinking... in the interval [-2,0) y starts out at a maximum of 0 and then quickly declines from that point on as [tex]\frac{2}{x}[/tex] gets larger and larger, in a negative fashion. Then after 0, [tex]\frac{2}{x}[/tex] starts out infinitely large and comes down and tends to a minimum of 1 as x gets larger. So the range for that interval is [tex](-\inf,0] U (1,\inf)[/tex]... Then I look at when [tex]x<-2[/tex] and note that as x decreases, [tex]\frac{-2}{x}[/tex] gets closer to zero and this the function tends towards -1. So the range in this interval is [tex](-1,0}[/tex]... This gives a combined range of [tex](-\inf,0] U (1,\inf)[/tex].

    Thanks alot :)
     
    Last edited: Nov 3, 2006
  5. Nov 3, 2006 #4

    arildno

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    Why is the domain obvious?
    I don't get that at all.
    Please tell me.
     
  6. Nov 4, 2006 #5
    Because with |x+2|/x, the only value of x that will mess it up is 0... Division by zero is undefined :P Everything else is fair game.
     
  7. Nov 4, 2006 #6

    arildno

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    Well, why can't its domain be [2,4], then?
    Or the set of points [itex]\{-1,3.14,57\}[/itex]?

    It is meaningless to supply a function without specifying its domain.
    What you have specified, is known as the MAXIMAL domain of the function within the real number set.
     
    Last edited: Nov 4, 2006
  8. Nov 4, 2006 #7
    But isn't agreeable that when you are asked to specify the domain of a function, what is being asked for is all the possible x's that can be plugged into the function? Unless you are supplied with a graph of the function, then you could only go on what you see.

    I do see your point though :)
     
  9. Nov 5, 2006 #8

    arildno

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    It is obvious that that is what the dumb exercise maker MEANT.
    However, that does not excuse the exercise maker for making an improper question!

    He should have asked something like:
    "Determine the greatest set of real numbers that can serve as the domain of the function"
     
  10. Mar 13, 2010 #9
    i have to do a portfolio on rational functions, so i am supposed to figure out everuthing about asymptotes, range, domain, etc. all by my self. so far i have been succesful in figuring out the domain but still dont know how to find the range of a rational function. for ex. 1/x+3
     
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