College Algebra Sixth Edition James Stewart  Lothar Redlin  Saleem Watson

1. College Algebra Sixth Edition James StewartLothar RedlinSaleem Watson

2. Equations and Inequalities 1

3. Solving Absolute Value Equations and Inequalities 1.9

4. Absolute Value of a Number Recall from Section P.2 that the absolute value of a number a is given by

5. Absolute Value of a Number It represents the distance from a to the origin on the real number line

6. Absolute Value of a Number More generally, |x – a| is the distance between x and a on the real number line. • The figure illustrates the fact that the distance between 2 and 5 is 3.

7. Absolute Value Equations

8. E.g. 1—Solving an Absolute Value Equation Solve the equation |2x – 5| = 3 • The equation |2x – 5| = 3 is equivalent to two equations: 2x – 5 = 3 or 2x – 5 = –3 2x = 8 or 2x = 2x = 4 or x = 1 • The solutions are 1 and 4.

9. E.g. 2—Solving an Absolute Value Equation Solve the inequality 3|x – 7| + 5 = 14 • First, we isolate the absolute value on one sideof the equal sign. 3|x – 7| + 5 = 14 3|x – 7| = 9 |x – 7| = 3x – 7 = 3 or x – 7 = –3x = 10 or x = 4 • The solutions are 4 and 10.

10. Absolute Value Inequalities

11. Properties of Absolute Value Inequalities We use these properties to solve inequalities that involve absolute value.

12. Properties of Absolute Value Inequalities These properties can be proved using the definition of absolute value. • For example, to prove Property 1, the inequality |x| < c says that the distance from x to 0 is less than c.

13. Properties of Absolute Value Inequalities From the figure, you can see that this is true if and only if x is between -c and c.

14. Solution 1 E.g. 3—Solving an Absolute Value Equation Solve the inequality |x – 5| < 2 • The inequality |x – 5| < 2 is equivalent to –2 < x – 5 < 2 (Property 1) 3 < x < 7 (Add 5) • The solution set is the open interval (3, 7).

15. Solution 2 E.g. 3—Solving an Absolute Value Equation Geometrically, the solution set consists of: • All numbers x whose distance from 5 is less than 2.

16. Solution 2 E.g. 3—Solving an Absolute Value Equation From the figure, we see that this is the interval (3, 7).

17. E.g. 4—Solving an Absolute Value Inequality Solve the inequality |3x + 2| ≥ 4 • By Property 4, the inequality |3x + 2| ≥ 4 is equivalent to 3x + 2 ≥ 4 or 3x + 2 ≤ –4 3x≥ 2 3x≤ –6 (Subtract 2) x≥ 2/3 x≤ –2 (Divide by 3)

18. E.g. 4—Solving an Absolute Value Inequality So, the solution set is {x | x≤ –2 or x ≥ 2/3} = (– ∞, – 2] [2/3, ∞)

19. E.g. 5—Piston Tolerances The specifications for a car engine indicate that the pistons have diameter 3.8745 in. with a tolerance of 0.0015 in. • This means that the diameters can vary from the indicated specification by as much as 0.0015 in. and still be acceptable.

20. E.g. 5—Piston Tolerances • Find an inequality involving absolute values that describes the range of possible diameters for the pistons. • Solve the inequality.

21. Example (a) E.g. 5—Piston Tolerances Let d represent the actual diameter of a piston. • The difference between the actualdiameter (d) and the specified diameter (3.8745) is less than 0.0015. • So, we have |d – 3.8745| ≤ 0.0015

22. Example (b) E.g. 5—Piston Tolerances The inequality is equivalent to –0.0015 ≤ d – 3.8745 ≤ 0.0015 3.8730 ≤ d ≤ 3.8760 • So, acceptable piston diameters may varybetween 3.8730 in. and 3.8760 in.