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

### 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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