Download 2013_3.6 Abs Value Inequalities

3.6 Absolute Value
Inequalities
Hornick
Warm Up
Solve each inequality and graph the solution.
1.
2.
x+7<4
14x ≥ 28
3. 5 + 2x > 1
x < –3
x≥2
x > –2
–5 –4 –3 –2 –1
0
1
2
3
4
5
–5 –4 –3 –2 –1
0
1
2
3
4
5
–5 –4 –3 –2 –1
0
1
2
3
4
5
Objectives
Solve compound inequalities in one variable involving
absolute-value expressions.
Example 1
Solve the inequality and graph the solutions.
|x|– 3 < –1
|x|– 3 < –1
+3 +3
|x| < 2
Write as a compound inequality.
–2 <x < 2
2 units
–2
–1
2 units
0
1
2
Example 2
Solve the inequality and graph the solutions.
|x – 1| ≤ 2
x – 1 ≥ –2 AND x – 1 ≤ 2
+1 +1
+1 +1
-1≤ x ≤ 3
–3
–2
–1
0
1
2
3
Helpful Hint
Just as you do when solving absolute-value
equations, you first isolate the absolute-value
expression when solving absolute-value
inequalities.
Example 3 AND
Solve the inequality and graph the solutions.
2|x| ≤ 6
2|x| ≤ 6
2
2
|x| ≤ 3
–3 ≤ x ≤ 3
3 units
–3
–2
–1
3 units
0
1
2
3
Example 4 OR
Solve the inequality and graph the solutions.
|x| + 14 ≥ 19
|x| + 14 ≥ 19
– 14 –14
|x|
≥ 5
x ≤ –5 OR x ≥ 5
Write as a compound inequality.
5 units 5 units
–10 –8 –6 –4 –2
0
2
4
6
8 10
Example 5 OR
Solve the inequality and graph the solutions.
3 + |x + 2| > 5
3 + |x + 2| > 5
–3
–3
|x + 2| > 2
x + 2 < –2 OR x + 2 > 2
–2 –2
–2 –2
x
< –4 OR x
>0
–10 –8 –6 –4 –2
0
2
4
6
8 10
Example 6: Application
A pediatrician recommends that a baby’s
bath water be 95°F, but it is acceptable for
the temperature to vary from this amount by
as much as 3°F. Write and solve an absolutevalue inequality to find the range of
acceptable temperatures. Graph the
solutions.
Let t = the actual water temperature.
**The difference between t and the ideal
temperature is at most 3°F.
t – 95
≤
|t – 95| ≤ 3
3
Example 6 Continued
|t – 95| ≤ 3
t – 95 ≥ –3 AND t – 95 ≤
3
+95 +95
+95 +95
t
≥ 92 AND t
≤ 98
90
92
94
96
98
100
The range of acceptable temperature is 92 ≤ t ≤ 98.
Example 6
Solve the inequality.
|x + 4|– 5 > – 8
|x + 4|– 5 > – 8
+5
+5
|x + 4|
>
–3
Absolute-value expressions
are always nonnegative.
Therefore, the statement is
true for all real numbers.
All real numbers are solutions
(infinitely many solutions)
Example 7
Solve the inequality.
|x – 2| + 9 < 7
|x – 2| + 9 < 7
–9 –9
|x – 2|
< –2
Absolute-value expressions
are always nonnegative.
Therefore, the statement is
false for all values of x.
The inequality has no solutions. Be careful!
Remember!
An absolute value represents a distance, and
distance cannot be less than 0.
Work on Practice 3-6 (Multiples of 3
ONLY)
Homework
pg 178 #s 1, 2, 3-27(mult of 3), 29-34
Exit Card
Solve each inequality and graph the solutions.
1. 3|x| > 15
–10
–5
5
0
10
2. |x + 3| + 1 < 3
–6
–5
–4
–3
–2
–1
0
3. A number, n, is no more than 7 units away
from 5. Write and solve an inequality to show
the range of possible values for n.
Solve each inequality.
4. |3x| + 1 < 1
5. |x + 2| – 3 ≥ – 6
Exit Card
Solve each inequality and graph the solutions.
1. 3|x| > 15
–10
–5
x < –5 or x > 5
0
2. |x + 3| + 1 < 3
–6
–5
–4
5
10
–5 < x < –1
–3
–2
–1
0
3. A number, n, is no more than 7 units away
from 5. Write and solve an inequality to show
the range of possible values for n.
|n– 5| ≤ 7; –2 ≤ n ≤ 12
Exit Card
Solve each inequality.
4. |3x| + 1 < 1 no solutions
5. |x + 2| – 3 ≥ – 6
all real numbers