Absolute Value (Real Numbers: Rational)

Date: 02/02/99 at 05:11:50
From: Laura Noack
Subject: Absolute Value (Real Numbers: Rational)

|1/(x-2)| >= 4

If x > 0

 1/(x-2) >= 4
       1 >= 4x-8
       9 >= 4x
     9/4 >= x

This part is correct.

If x < 0,

1/-(x-2) >= 4
     1/4 <= -(x-2)
       1 <= -4x + 8
      -7 <= -4x
     7/4 >= x


This is what I got for x < 0, but the answer is supposed to be 
7/4 <= x, not 7/4 >= x. Can you also help me display the answer on a 
real number line?

It should look like this:

       (*)---------(*)
      _________________
       7/4         9/4

Date: 02/02/99 at 12:11:35
From: Doctor Rick
Subject: Re: Absolute Value (Real Numbers: Rational)

Do you know what you mean by x > 0 ? The correct condition is:

  1/(x-2) > 0

that is, if what's between the absolute value signs is greater than 
zero, you can remove the absolute value, since it has no effect in this 
case. Further, note that you multiply both sides by (x-2) and the 
direction of the inequality does not change. This is correct, but only 
because 

  1/(x-2) > 0 => x > 2

so that the quantity you multiply by, (x-2), is positive.

From what I said about the first part, can you see where you went wrong 
in the second part? The condition should be 
 
 1/(x-2) < 0

which implies that x < 2. In this case, as you say, you can replace the 
absolute value with multiplication by -1. But the quantity that you 
multiply by, -(x-2), is POSITIVE (check out that condition), and once 
again you shouldn't change the direction of the inequality!

In case you're not clear on this, let me spell out the complete 
solution.

  (x > 2 and x <= 9/4) or (x > 2 and x >= 7/4)

The first () is what you get from the first part, and the condition 
x > 2 (so that the quantity in the absolute value is positive) must be 
kept as part of the solution. Likewise in the second (), the condition 
that the quantity in the absolute value is negative must be kept as 
part of the solution.

What happens at x = 2?

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/