Precalculus: Absolute Value

1. Definitions

|a b| = |a| |b|

|a/b| = |a|/|b|

|a + b| <= |a| + |b|


a >= 0

|f(x)| = a
→ f(x) = + a OR f(x) = - a
|f(x)| > = a
→ f(x) <= - a OR f(x) >= + a
|f(x)| < = a
→ - a <= f(x) <= a

a < 0

|f(x)| = a
has no sense, there are no solutions for this equality.
|f(x)| < a
has no sense, there are no solutions for this inequality.
|f(x)| > a
is true for any real x.

2. Worked examples

a) |+ 5|= + 5

b) |- 4| = - (- 4)

c) |2 x - 3| = 2 x - 3 if 2 x - 3 > 0 that is x >= 3/2
- (2 x - 3) = - 2 x + 3 if 2 x - 3 <0 that is x < 3/2

d)| 2 x - 1| = + 3 → 2 x - 1 = + 3 OR 2 x - 1 = - 3
That is:
x = + 2 OR x = - 1

e) | 4 x - 7| = - 12 has no sense.

f) | 4 x - 7| < = - 5 has no sense.

g) |3 x - 4| > 4 → 3 x - 4 <= - 4 OR 3 x - 4 >= + 4
That is:
x <= 0 OR x >= + 8/3

h) |2 x - 10| < = 12 → - 12 <= 2 x - 10 <= + 12
That is:
- 1 <= x <= + 11

3. Exercises

Give the following expressions without absolute value bars:

|- 19|

|4 x + 5|

Solve for x the following:

| 3 x - 4| = 3

| 2 x - 5| < - 6

| 2 x + 6| > - 7

| 3 x - 5| <= 2

|3 x - 8| >= 10