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Range of rational functions


Date: Thu, 01 Dec 1994 09:48:40 -0500 (EST)
From: Anonymous
Subject: pre calc question

Hello, this is our first attempt to use your service.

We are a senior class in upstate NY who is struggling with rational 
functions.

Given the following rational function:  f(x)= (x-1)/(x^2+3x-1)

How do you find the range?  Our book gives the answer of:
(neg infinity to 0.12] U [0.65 to pos infinity)

Thank You in advance.

John Mehrman and Danielle Nowotenski
Keene Central School,  Keene Valley New York


Date: Thu, 1 Dec 1994 15:58:37 -0500 (EST) 
From: Dr. Ethan
Subject: Re: pre calc question

     Well your book is right.  That is the range. I think that you said
that this was a pre calc. class so I don't know how much calculus I can 
use but I will try to use as little as possible.  First of all do you have an
idea what this graph looks like?  I guess that is kinda what we are trying
to do.   

     Okay. To start with, we know that some of the points we need to
watch out for are the roots of the equation on the bottom.  Those are the
points where we get the indeterminant form k/0 where k is a constant.  You
probably know that those points are problems.  So at those points on both
sides we need to figure out if it is going to positive and negative infinity.  

     Next take the limits as x goes to positive and negative infinity.  You
hopefully will notice that since the power of x is bigger on the top
than the bottom, then the limit will be zero in both directions. However
in the positive direction it is coming in from the positive that means
that it will have to have a maximum point between the root (x=1) and
positive infinity (that point is where they get the .12). How to find this
point without calculus is very hard but using the first derivative it is
relatively straightforward (the calculations are a bit nasty on this one).

     Similarly, the thing in the middle of the two asymptotic lines (do you
know what these are?) looks like a parabola and you can see that it is going
to have a mininimum.  Again to find this min is pretty straightforward using
calculus but hard otherwise.  Then you have a pretty clear drawing of the
thing and the range falls right out.

     I hope that helps; if not then right back and somebody else can take
a shot.
                Ethan, Doctor On Call
    
Associated Topics:
High School Calculus

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