Using Interval Notation to Express Answers

Date: 12/03/96 at 05:10:55
From: Juanita Johnston
Subject: Algebra, complex fractions, and inequalities

Solve for x and graph: x/(x+3) > or = 1/(x-1).  The answer is also 
supposed to be given in interval notation.

I've tried everything I can think of, and I just don't know where to 
start.  I tried subracting the 1/(x-1) side, to get the x on the same 
side, and then combining the fractions  I got it to look like:
                         
(x-3)(x+1)               
---------  > or = 0  
(x+3)(x-1)
 
I don't know where to go from there, or even if I'm on the right 
track!

Date: 12/03/96 at 18:10:33
From: Doctor Mike
Subject: Re: Algebra, complex fractions, and inequalities

Juanita,
    
Hey, you didn't leave very much for me to do!  So far, it's perfect.
Next you need to see that the numerator is zero for x=3 and x=-1, so  
the expression is equal to 0 for exactly those values.  Also, the 
denominator is zero for x=-3 and x=1, so the expression is undefined
for those values. (These values make the original version of the 
problem undefined, too.)  If we stay away from those values, then none
of the numerator or denominator factors can get to zero and cross over
to the other sign.   
  
Let's draw a number line with these four special numbers marked: 
    
   -x <------o----o----o----o----0----o----o----o----o------> +x 
                 -3        -1         1         3        
  
The trick is to look at the sign (+ or -) of your expression in each
one of the intervals where nothing changes sign from + to - or back.
These intervals are "up to but not including -3", "strictly between
-3 and -1", etc. There are 5 of them.  

As an example, let's see what happens between -1 and +1. The numerator 
has one positive factor and one negative factor. The same is true for 
the denominator. So both numerator and denominator are negative, 
making the fraction > 0. 

Now we know that it is >= 0 in the interval (-1,+1). Remember the 
fraction is 0 for x=-1 and undefined for +1, so we write [-1,+1); the 
square cornered bracket on the left shows -1 IS IN the interval we are 
talking about where the fraction is >= zero. With me so far?  

Now do this kind of analysis about the 4 other sub-intervals to see 
what happens. As for the graph for such a problem, just draw on the 
number line to show what x values are included. There are two ways to 
show that an end point is or isn't included in the graph. Some books 
use "[" and "(" for this and some use a filled-dot and hollow-dot for 
it. Look in your book.
  
I hope this helps.  Keep up the good work.  

-Doctor Mike,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/