Slide 1.8- 1 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

OBJECTIVES

Equations and Inequalities Involving Absolute Value

Learn to solve equations involving absolute value.

Learn to solve inequalities involving absolute value.

SECTION 1.8

1

2


ABSOLUTE VALUE

a a if a 0 and a a if a 0.


THE SOLUTIONS OF

u a is equivalent to u a or u a.

u a, a 0

u a has no solution when a 0.


EXAMPLE 1 Solving an Equation Involving Absolute Value

Solve each equation.

The solution set is {–3}.

a. x 3 0 b. 2x 3 5 8

SolutionLet u x 3, u 0 has only one solution: u 0.

x 3 0x 3 0

x 3

Check the solution.x 3 0

3 3 0

0 0

?


EXAMPLE 1 Solving an Equation Involving Absolute Value

The solution set is {–5, 8}.

b. 2x 3 5 8Solution continued

Let u 2x 3, u 13 is equivalent tou 13 or u 13.

We leave it to you to check the solutions.

2x 3 5 5 8 5 2x 3 13Isolate the absolute value.

2x 3 132x 16x 8

2x 3 132x 10x 5


EXAMPLE 2 Solving an Equation of the Form

If |u| = |v|, then either u is equal to |v| or u is equal to –|v|. Since |v| = ± v in every case, we have u = v or u = –v. Thus,

Solutionx 1 x 5 .

u v

Solve

u v is equivalent to u v or u v.x 1 x 5 is equivalent to

x 1 x 5 1 5 (False)

x 1 x 5 x 2

The solution set is {–2}. We leave the check to you.


RULES FOR SOLVING ABSOLUTE VALUE INEQUALITIES

If a > 0, and u is an algebraic expression, then

1. u a is equivalent to a u a.2. u a is equivalent to a u a.3. u a is equivalent to u a or u a.4. u a is equivalent to u a or u a.


EXAMPLE 3Solving an Inequality Involving an Absolute Value

Solve the inequality 4x 1 9 and graph thethe solution set.

Rule 2 applies here, with u = 4x – 1 and a = 9.Solution

4x 1 9 9 4x 1 9

1 9 4x 11 9 1 8 4x 10

84

4x4

104

2 x 52



Solution continued

x 2 x 52

;The solution set is that is,

the solution set is the closed interval 2,52

.

2 x 52

, or 2,52

1 20–1–2][

3–3 52


EXAMPLE 4

In the introduction to this section, we wanted to find the possible search range (in miles) for a search plane that has 30 gallons of fuel and uses 10 gallons of fuel per hour. We were told that the search plane normally averages 110 miles per hour, but that weather conditions could affect the average speed by as much as 15 miles per hour (either slower or faster). How do we find the possible search range?

Finding the Search Range of an Aircraft


EXAMPLE 4

To find distance, we need both speed and time.Let x = actual speed in mphWe know actual speed is within 15 mph of average speed, 110 mph. That is,|Actual speed – Average speed| ≤ 15 mph


Solution

x 110 15 15 x 110 15

110 15 x 110 1595 x 125

The actual speed of the search plane is between 95 and 125 mph.


EXAMPLE 4

The plane uses 10 g of fuel per hour. It has 30 g, so it can fly for 3 hr. So the actual number of miles the search plane can fly is 3x.


Solution continued

95 x 1253 95 3x 3 125 285 x 375

The search plane’s range is between 285 and 375 miles.



Solve the inequality 2x 8 4 and graph thethe solution set.

Rule 4 applies here, with u = 2x – 8 and a = 9.Solution

2x 8 4 is equivalent to2x 8 4 or 2x 8 4

2x 8 8 4 82x2

42

x 2

2x 8 8 4 82x2

122

x 6



The solution set is {x | x ≤ 2 or x ≥ 6};

Solution continued

x ≤ 2 or x ≥ 6(–∞, 2] U [6, ∞)

[]4 51 76320

in interval notation it is (–∞, 2] U [6, ∞).


EXAMPLE 6Solving Special Cases of Absolute Value Inequalities

Solve each inequality. a. 3x 2 5

a. The absolute value is always nonnegative, so |3x – 2| > –5 is true for all real numbers x. The solution set is all real numbers, or (–∞, ∞).

Solutionb. 5x 3 2

b. There is no real number with absolute value ≤ –2. The solution set is the empty set, or .

Slide 1.8- 1 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.

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