2 Jan 1999

efinition

The absolute value or modulus of a real number x, written |x| is defined as

|x| =  ì
í
î		 x
 
-x    		if x ³ 0,
 
if x < 0 (or £ 0).

The graph of y = |x| is shown below.

Geometrically, |x| is the distance of x from the origin.

Example 1:

|3 - x| =  ì
í
î		 3 - x
 
-(3 - x)    		if 3 - x ³ 0,
 
if 3 - x < 0;
 
= 
 ì
í
î		 3 - x
 
x -3		 if x £ 3,
 
if x > 3.

roperties

  1. |-a| = |a|
    2.
  2. Ö(a2) = |a|
    3.
  3. |ab| = |a||b|
    4.
  4. |a/b| = |a|/|b|  provided b ¹ 0
    5.
  5. |a + b| £ |a| + |b|
    6.
  6. |a - b| ³ |a| - |b|
    7.
  7. For k ³ 0, |a| = k  Þ  a = ±k
    8.
  8. For any k, |a| = |k|  Þ  a = ±k
    9.
  9. For k > 0, |a| < k  Û  -k < a < k
    10.
  10. For k > 0, |a| > k  Û  x < -k or x > k
    11.

raphs of Functions Involving Modulus

Since |f(x)| =  ì
í
î		 f(x)
 
-f(x)    		if f(x) ³ 0,

if f(x) < 0;

the graph of y = |f(x)| is obtained from the graph of y = f(x) by reflecting the negative part (ie below the x-axis) about the x-axis.

Example 2:  Sketch y = |x2 - 1|.

Example 3:  Sketch y = |x| + |2x - 3|.

The values 0 and 3/2 divide the number line into three intervals.

 

  x £ 0 0 £ x £ 3/2 x ³ 3/2
|x| -x x x
|2x - 3| -2x + 3 -2x + 3 2x - 3
y = |x| + |2x - 3| -3x + 3 -x + 3 3x - 3

Example 6 will make use of this graph.

nequalities Involving Modulus

Example 4:  Solve |3x - 2| < 3 - 2x.

Solution:

First of all, we must have
3 - 2x > 0  Þ  x < 3/2.

Next, we 'removed the modulus sign'

|3x - 2| < 3 - 2x
Þ  3x - 2  <  3 - 2x     and    3x - 2  >  -(3 - 2x)
x
  <  1
and
x
  >  -1
\  -1 < x < 1
 
Taking the intersection, we have
\  -1 < x < 1

Example 5:  Solve |3x - 2| > |3 - 2x|.

Solution:

Squaring both sides,

9x2 - 12x + 4  >  9 - 12x + 4x2
x2
  >  1
|x|
  >  1
\  x > 1 or x < -1

Example 6:  Find the solution set of |x| + |2x - 3| ³ 4x.

Solution:

We have sketch the graph in Example 3.

We will sketch an addition graph of y = 4x.

At the point of intersection,

-x + 3
  = 4x
x
  = 3/5

The solution set is {x Î R : x £ 3/5}.

Example 7:  Solve 
|x| + 1
¾¾¾
|x| - 1
  < 3.

Solution:

|x| + 1 - 3|x| + 3
¾¾¾¾¾¾¾
|x| - 1
 < 0
 
2|x| - 4
¾¾¾
|x| - 1
 > 0
 
Let y = |x|.  Since 2 is positive,
we may ignored it.
y - 2
¾¾¾
y - 1
 > 0

Thus  y < 1   or  y > 2
Þ    |x| < 1
or
 |x| > 2
-1 < x < 1
or
 x < -2 or x > 2
\  x < -2, -1 < x < 1 or x > 2

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