1. Copyright © 2011 Pearson Education, Inc. More on Inequalities, Absolute Value, and Functions CHAPTER 8.1Compound Inequalities 8.2Equations Involving Absolute Value 8.3Inequalities Involving Absolute Value 8.4 Functions and Graphing 8.5Function Operations 8

2. Copyright © 2011 Pearson Education, Inc. Compound Inequalities 8.1 1.Solve compound inequalities involving and. 2.Solve compound inequalities involving or.

3. Slide 8 - 3 Copyright © 2011 Pearson Education, Inc. Compound Inequality: Two inequalities joined by either and or or. Examples:x > 3 and x  8  2  x or x > 4 Intersection: For two sets A and B, the intersection of A and B, symbolized by A  B, is a set containing only elements that are in both A and B.

4. Slide 8 - 4 Copyright © 2011 Pearson Education, Inc. Example 1 For the compound inequality x >  5 and x < 2, graph the solution set and write the compound inequality without “and” if possible. Then write in set-builder notation and in interval notation. Solution The set is the region of intersection. x >  5 x < 2 x >  5 and x < 2 ( ) ( )

5. Slide 8 - 5 Copyright © 2011 Pearson Education, Inc. continued x >  5 and x < 2 Without “and”:  5 < x < 2 Set-builder notation: {x|  5 < x < 2} Interval notation: (  5, 2) Warning: Be careful not to confuse the interval notation with an ordered pair.

6. Slide 8 - 6 Copyright © 2011 Pearson Education, Inc. Example 2 For the inequality graph the solution set. Then write the solution set in set-builder notation and in interval notation. Solution Solve each inequality in the compound inequality. and

7. Slide 8 - 7 Copyright © 2011 Pearson Education, Inc. continued [ ) [ ) Without “and”:  2  x < 4 Set-builder notation: {x|  2  x < 4} Interval notation: [  2, 4)

8. Slide 8 - 8 Copyright © 2011 Pearson Education, Inc. Solving Compound Inequalities Involving and To solve a compound inequality involving and, 1. Solve each inequality in the compound inequality. 2. The solution set will be the intersection of the individual solution sets.

9. Slide 8 - 9 Copyright © 2011 Pearson Education, Inc. Example 3a Solution For the compound inequality, graph the solution set. Then write the solution set in set-builder notation and in interval notation. ]( Set-builder notation: {x| 2 < x  6} Interval notation: (2, 6]

10. Slide 8 - 10 Copyright © 2011 Pearson Education, Inc. Example 3b Solution For the compound inequality, graph the solution set. Then write the solution set in set-builder notation and in interval notation.

11. Slide 8 - 11 Copyright © 2011 Pearson Education, Inc. continued ( [ Solution set: [ Set-builder notation: {x|x  2} Interval notation: [2, ∞)

12. Slide 8 - 12 Copyright © 2011 Pearson Education, Inc. Solution For the compound inequality, graph the solution set. Then write the solution set in set-builder notation and in interval notation. Since no number is greater than 5 and less than 1, the solution set is the empty set Example 3c Set builder notation: { } or Interval notation: We do not write interval notation because there are no values in the solution set.

13. Slide 8 - 13 Copyright © 2011 Pearson Education, Inc. Union: For two sets A and B, the union of A and B, symbolized by A  B, is a set containing every element in A or in B. Solving Compound Inequalities Involving or To solve a compound inequality involving or, 1. Solve each inequality in the compound inequality. 2. The solution set will be the union of the individual solution sets.

14. Slide 8 - 14 Copyright © 2011 Pearson Education, Inc. Example 4a For the compound inequality, graph the solution set. Then write the solution set in set-builder notation and in interval notation. Solution or [ ) [ )

15. Slide 8 - 15 Copyright © 2011 Pearson Education, Inc. continued Solution set: Set-builder notation: {x|x <  1 or x  1} Interval notation: ( ,  1)  [1,  )

16. Slide 8 - 16 Copyright © 2011 Pearson Education, Inc. Example 4b For the compound inequality, graph the solution set. Then write the solution set in set-builder notation and in interval notation. Solution or ] ] ] Solution set:

17. Slide 8 - 17 Copyright © 2011 Pearson Education, Inc. continued Solution set: Set-builder notation: {x|x ≤ 5} Interval notation: ( , 5]

18. Slide 8 - 18 Copyright © 2011 Pearson Education, Inc. Example 4c For the compound inequality, graph the solution set. Then write the solution set in set-builder notation and in interval notation. Solution or ] ( Solution set:

19. Slide 8 - 19 Copyright © 2011 Pearson Education, Inc. continued Solution set: Set-builder notation: {x|x is a real number}, or Interval notation: ( ,  )

20. Slide 2 - 20 Copyright © 2011 Pearson Education, Inc. Solve: a) b) c) d) 8.1

21. Slide 2 - 21 Copyright © 2011 Pearson Education, Inc. Solve: a) b) c) d) 8.1

22. Slide 2 - 22 Copyright © 2011 Pearson Education, Inc. Solve: a) b) c) d) 8.1

23. Slide 2 - 23 Copyright © 2011 Pearson Education, Inc. Solve: a) b) c) d) 8.1

24. Copyright © 2011 Pearson Education, Inc. Equations Involving Absolute Value 8.2 1.Solve equations involving absolute value.

25. Slide 8 - 25 Copyright © 2011 Pearson Education, Inc. Absolute Value Property If |x| = a, where x is a variable or an expression and a  0, then x = a or x =  a.

26. Slide 8 - 26 Copyright © 2011 Pearson Education, Inc. Example 1 Solve. a. |2x +1| = 5 b. |3 – 4x| = –10 Solution a. b. |3 – 4x| = –10 The solution are –3 and 2. x = –3 or x = 2 2x = –6 or 2x = 4 2x +1 = –5 or 2x +1 = 5 This equation has the absolute value equal to a negative number. Because the absolute value of every real number is a positive number or zero, this equation has no solution.

27. Slide 8 - 27 Copyright © 2011 Pearson Education, Inc. Example 2 Solve. |3x + 4| + 3 = 11 Solution |3x + 4| + 3 = 11 Subtract 3 from both sides to isolate the absolute value. |3x + 4| = 8 3x + 4 = 8 or 3x + 4 = –8 3x = 4 or 3x = –12 x = 4/3 or x = –4 The solutions are 4/3 and  4.

28. Slide 8 - 28 Copyright © 2011 Pearson Education, Inc. Solving Equations Containing a Single Absolute Value To solve an equation containing a single absolute value, 1. Isolate the absolute value so that the equation is in the form |ax + b| = c. If c > 0, proceed to Steps 2 and 3. If c < 0, the equation has no solution. 2. Separate the absolute value into two equations, ax + b = c and ax + b =  c. 3. Solve both equations.

29. Slide 8 - 29 Copyright © 2011 Pearson Education, Inc. Solving Equations in the Form |ax + b| = |cx + d| To solve an equation in the form |ax + b| = |cx + d|, 1. Separate the absolute value equation into two equations: ax + b = cx + d and ax + b =  (cx + d). 2. Solve both equations.

30. Slide 8 - 30 Copyright © 2011 Pearson Education, Inc. Example 3a Solve: |3x – 5| = |8 + 4x|. Solution 3x – 5 = 8 + 4x or 3x – 5 =  (8 + 4x) –13 + 3x = 4x –13 = x The solutions are  13 and  3/7.

31. Slide 8 - 31 Copyright © 2011 Pearson Education, Inc. Example 3b Solve: |2x – 9| = |3 − 2x|. Solution 2x – 9 = 3 − 2x or 2x – 9 =  (3 − 2x) 4x = 12 2x – 9 =  3 + 2x x = 3 The absolute value equation has only one solution, 3. – 9 =  3

32. Slide 2 - 32 Copyright © 2011 Pearson Education, Inc. Solve: a) b) c) d) 8.2

33. Slide 2 - 33 Copyright © 2011 Pearson Education, Inc. Solve: a) b) c) d) 8.2

34. Slide 2 - 34 Copyright © 2011 Pearson Education, Inc. Solve: a) b) c) all real numbers d) no solution 8.2

35. Slide 2 - 35 Copyright © 2011 Pearson Education, Inc. Solve: a) b) c) all real numbers d) no solution 8.2

36. Copyright © 2011 Pearson Education, Inc. Inequalities Involving Absolute Value 8.3 1.Solve absolute value inequalities involving less than. 2.Solve absolute value inequalities involving greater than.

37. Slide 8 - 37 Copyright © 2011 Pearson Education, Inc. Solving Inequalities in the Form |x| 0 To solve an inequality in the form |x| 0, 1. Rewrite the inequality as a compound inequality involving and: x >  a and x < a (or use  a < x < a). 2. Solve the compound inequality. Similarly, to solve |x|  a, we write x   a and x  a (or  a  x  a).

38. Slide 8 - 38 Copyright © 2011 Pearson Education, Inc. Example 1a For the inequality, solve, graph the solution set, and write the solution set in both set-builder and interval notation. |x| < 9. Solution x > −9 and x < 9 So our graph is as follows: Set-builder notation: {x|  9 < x < 9} Interval notation (  9, 9) ( )

39. Slide 8 - 39 Copyright © 2011 Pearson Education, Inc. Example 1b For the inequality, solve, graph the solution set, and write the solution set in both set-builder and interval notation. |x − 4| ≤ 5 Solution A number line solution: Set-builder notation: {x|  1 ≤ x ≤ 9} Interval notation [  1, 9] ][ x – 4  −5 and x – 4 ≤ 5 x  −1 and x ≤ 9

40. Slide 8 - 40 Copyright © 2011 Pearson Education, Inc. Example 1c For the inequality, solve, graph the solution set, and write the solution set in both set-builder and interval notation. |x – 3| < 6 Solution |x – 3| < 6  6 < x – 3 < 6 Rewrite as a compound inequality.  3 < x < 9 Add 3 to each part of the inequality. A number line solution: Set-builder notation: {x|  3 < x < 9} Interval notation (  3, 9). ) (

41. Slide 8 - 41 Copyright © 2011 Pearson Education, Inc. Example 1e For the inequality, solve, graph the solution set, and write the solution set in both set-builder and interval notation. |2x – 3| + 8 < 5. Solution Isolate the absolute value. |2x – 3| + 8 < 5 |2x – 3| < –3 Since the absolute value cannot be less than a negative number, this inequality has no solution: . Set builder notation: { } or Interval notation: We do not write interval notation because there are no values in the solution set.

42. Slide 8 - 42 Copyright © 2011 Pearson Education, Inc. Solving Inequalities in the Form |x| > a, where a > 0 To solve the inequality in the form x|  a, where a > 0, 1. Rewrite the inequality as a compound inequality involving or: x a. 2. Solve the compound inequality. Similarly, to solve |x|  a, we would write x   a or x  a.

43. Slide 8 - 43 Copyright © 2011 Pearson Education, Inc. Example 2a For the inequality, solve, graph the solution set, and write the solution set in both set-builder and interval notation. |x + 7| > 5. Solution We convert to a compound inequality and solve each. |x + 7| > 5 x + 7 5 x <  12 A number line solution: Set-builder notation: {x| x  2} Interval notation: ( ,  12)  (  2,  ). 5-14-12-10-8-6-4-2024-15-13-9-53-15-71-115-3 ) (

44. Slide 8 - 44 Copyright © 2011 Pearson Education, Inc. Example 2d For the inequality, solve, graph the solution set, and write the solution set in both set-builder and interval notation. |4x + 7|  9 >  12. Solution Isolate the absolute value|4x + 7|  9 >  12 |4x + 7| >  3 This inequality indicates that the absolute value is greater than a negative number. Since the absolute value of every real number is either positive or 0, the solution set is. Set-builder notation: {x|x is a real number} or Interval notation: ( ,  ). 5-14-12-10-8-6-4-2024-15-13-9-53-15-71-115-3

45. Slide 2 - 45 Copyright © 2011 Pearson Education, Inc. Solve: a) b) c) d) 8.3

46. Slide 2 - 46 Copyright © 2011 Pearson Education, Inc. Solve: a) b) c) d) 8.3

47. Slide 2 - 47 Copyright © 2011 Pearson Education, Inc. Solve: a) b) c) d) 8.3

48. Slide 2 - 48 Copyright © 2011 Pearson Education, Inc. Solve: a) b) c) d) 8.3

49. Copyright © 2011 Pearson Education, Inc. Functions and Graphing 8.4 1.Identify the domain and range of a relation and determine if the relation is a function. 2.Find the value of a function. 3.Graph functions.

50. Slide 8 - 50 Copyright © 2011 Pearson Education, Inc. Relation: A set of ordered pairs. Domain: The set of all input values (x-values) for a relation. Range: The set of all output values (y-values) for a relation. Function: A relation in which every value in the domain is paired with exactly one value in the range.

51. Slide 8- 51 Copyright © 2011 Pearson Education, Inc. Example 1a Identify the domain and range of the relation, then determine whether it is a function. BirthdateFamily member March 1Donna April 17Dennis Sept. 3Catherine October 9Denise Nancy The relation is not a function because an element in the domain, Sept. 3, is assigned to two names in the range. Domain: {March 1, April 17, Sept 3, Oct 9} Range: {Donna, Dennis, Catherine, Denise, Nancy}

52. Slide 8- 52 Copyright © 2011 Pearson Education, Inc. Example 1b Identify the domain and range of the relation, then determine whether it is a function. {(−4, −1), (−2, 1), (0, 0), (2, −1), (4, 2)} The relation is a function because every value in the domain is paired with only one value in the range. Domain: {−4, −2, 0, 2, 4} Range: {−1, 1, 0, −1, 2} Solution

53. Slide 8- 53 Copyright © 2011 Pearson Education, Inc. continued For each graph, identify the domain and range. Then state whether each relation is a function. c.d. Domain: {x|x  1} Range: all real numbers Not a function Domain: all real numbers Range: {y   1} Function

54. Slide 8- 54 Copyright © 2011 Pearson Education, Inc. Example 2 For the function f(x) = 3x – 5, find the following. a. f(2)b. f(  4) Solution a. f(2) = 3x – 5 = 3(2) – 5 = 6 – 5 = 1 b. f(  4) = 3x – 5 = 3(  4) – 5 =  12 – 5 =  17

55. Slide 8- 55 Copyright © 2011 Pearson Education, Inc. Example 3 Graph: Solution We use (0, 2) as one ordered pair and then use the slope,, to find the second ordered pair. Recall that slope indicates the “rise” and “run” from any point on the line to another point on the line.

56. Slide 8- 56 Copyright © 2011 Pearson Education, Inc. Example 4 Graph. f(x) = 2x 2 Solution We create a table of ordered pairs, plot the points, and connect with a smooth curve. xf(x)f(x) 22 8 11 2 00 12 28

57. Slide 8- 57 Copyright © 2011 Pearson Education, Inc. Example 5b Graph. f(x) = |x| + 2 Solution We create a table of ordered pairs, plot the points, and connect the points. xf(x)f(x) 44 6 22 4 02 24 46

58. Slide 8- 58 Copyright © 2011 Pearson Education, Inc. Example 5c Graph. f(x) = |x − 3| + 2 Solution We create a table of ordered pairs, plot the points, and connect the points. xf(x)f(x) 05 23 32 43 65

59. Slide 3- 59 Copyright © 2011 Pearson Education, Inc. How would you graph the function y = 3x – 2? a) Plot (0, –2), down 2, right 3 b) Plot (0, –2), down 3, right 1 c) Plot (0, 2), up 3, left 2 d) Plot (0, –2), up 3, right 1 8.4

60. Slide 3- 60 Copyright © 2011 Pearson Education, Inc. How would you graph the function y = 3x – 2? a) Plot (0, –2), down 2, right 3 b) Plot (0, –2), down 3, right 1 c) Plot (0, 2), up 3, left 2 d) Plot (0, –2), up 3, right 1 8.4

61. Slide 3- 61 Copyright © 2011 Pearson Education, Inc. Given f(x) = −3x + 5, find f(−2). a) −1 b) 2 c) 11 d) 13 8.4

62. Slide 3- 62 Copyright © 2011 Pearson Education, Inc. Given f(x) = −3x + 5, find f(−2). a) −1 b) 2 c) 11 d) 13 8.4

63. Copyright © 2011 Pearson Education, Inc. Function Operations 8.5 1.Add or subtract functions. 2.Multiply functions. 3.Divide functions.

64. Slide 8 - 64 Copyright © 2011 Pearson Education, Inc. Adding or Subtracting Functions The sum of two functions, f + g, is founded by (f + g)(x) = f(x) + g(x). The difference of two functions, f – g, is founded by (f – g)(x) = f(x) – g(x).

65. Slide 8- 65 Copyright © 2011 Pearson Education, Inc. Example 1 Given f(x) = 3x + 1 and g(x) = 5x + 2, find the following. a. f + gb. f  gc. (f – g)(−2) Solution a. f + g = f(x) + g(x) = (3x + 1) + (5x + 2) = 8x + 3 b. f − g = f(x) − g(x) = (3x + 1) − (5x + 2) = − 2x − 1 c.Replace x with −2. (f − g)(−2) = −2(−2) + 1 = 4 + 1 = 5

66. Slide 8 - 66 Copyright © 2011 Pearson Education, Inc. Multiplying Functions The product of two functions, f g, is founded by (f g)(x) = f(x)g(x).

67. Slide 8 - 67 Copyright © 2011 Pearson Education, Inc. Example 2 Given f(x) = 2x + 7 and g(x) = x − 4, find f g. Solution f g = f(x)g(x) = (2x + 7)(x − 4) = 2x 2 − 8x + 7x – 28 = 2x 2 − x – 28

68. Slide 8 - 68 Copyright © 2011 Pearson Education, Inc. Dividing Functions The quotient of two functions, f / g, is founded by

69. Slide 8 - 69 Copyright © 2011 Pearson Education, Inc. Example 3 Given f(x) = 16x 3 −12x 2 + 8x and g(x) = 4x, find f / g. Solution

70. Slide 3- 70 Copyright © 2011 Pearson Education, Inc. Given f(x) = 4x – 1 and g(x) = 5x + 2, what is f + g? a) x + 4 b) x − 4 c) 9x + 1 d) 9x – 1 8.5

71. Slide 3- 71 Copyright © 2011 Pearson Education, Inc. Given f(x) = 4x – 1 and g(x) = 5x + 2, what is f + g? a) x + 4 b) x − 4 c) 9x + 1 d) 9x – 1 8.5

72. Slide 3- 72 Copyright © 2011 Pearson Education, Inc. Given f(x) = 3x – 2 and g(x) = 5x – 1, what is f g? a) 15x 2 − 13x + 2 b) 15x 2 − 13x − 2 c) 15x 2 − 7x + 2 d) 15x 2 − 7x − 2 8.5

73. Slide 3- 73 Copyright © 2011 Pearson Education, Inc. Given f(x) = 3x – 2 and g(x) = 5x – 1, what is f g? a) 15x 2 − 13x + 2 b) 15x 2 − 13x − 2 c) 15x 2 − 7x + 2 d) 15x 2 − 7x − 2 8.5