# College Algebra Fifth Edition James Stewart  Lothar Redlin  Saleem Watson - PowerPoint PPT Presentation Download Presentation College Algebra Fifth Edition James Stewart  Lothar Redlin  Saleem Watson

College Algebra Fifth Edition James Stewart  Lothar Redlin  Saleem Watson Download Presentation ## College Algebra Fifth Edition James Stewart  Lothar Redlin  Saleem Watson

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
##### Presentation Transcript

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

2. Absolute Value of a Number • Recall from Section P.3 that the absolute value of a number a is given by

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

4. 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.

5. Absolute Value Equations

6. E.g. 1—Solving an Absolute Value Equation • Solve the inequality |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.

7. 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.

8. Absolute Value Inequalities

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

10. 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.

11. 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.

12. Solution 1 E.g. 3—Absolute Value Inequality • 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).

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

14. Solution 2 E.g. 3—Absolute Value Inequality • From the figure, we see that this is the interval (3, 7).

15. 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)

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

17. 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.

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

19. 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

20. 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.