1. Slide 4- 1 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 1- 1 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

2. Inequalities and Problem Solving 9.1Inequalities and Domain 9.2Intersections, Unions, and Compound Inequalities 9.3Absolute-Value Equations and Inequalities 9.4Inequalities in Two Variables 9.5Applications Using Linear Programming 9

3. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Inequalities and Domain Solving Inequalities Graphically Domain Problem Solving 9.1

4. Slide 4- 4 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving Inequalities Graphically An inequality is any sentence containing Any value for a variable that makes an inequality true is called a solution. The set of all solutions is called the solution set. When all solutions of an inequality are found, we say that we have solved the inequality. Examples:

5. Slide 4- 5 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Addition Principle for Inequalities For any real numbers a, b, and c: a < b is equivalent to a + c < b + c; a > b is equivalent to a + c > b + c; Similar statements hold for

6. Slide 4- 6 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Multiplication Principle for Inequalities For any real numbers a, b, and for any positive number c: a < b is equivalent to ac < bc; a > b is equivalent to ac > bc. For any real numbers a, b, and for any negative number c, a bc; a > b is equivalent to ac < bc. Similar statements hold for

7. Slide 4- 7 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve 5x – 2 > x + 10. Example

8. Solve graphically (1, 5)

9. Slide 4- 9 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Domain We know that only nonnegative numbers have square roots that are real numbers. Thus finding the domain of a radical function often involves solving an inequality.

10. Slide 4- 10 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution Find the domain of f if Example

11. Slide 4- 11 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Lazer Line charges $65 plus $45 per hour for copier repair. Jonas remembers being billed less than $150. How many hours was Jonas’ copier worked on? Example

12. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Intersections, Unions, and Compound Inequalities Intersection of Sets and Conjunctions of Sentences Unions of Sets and Disjunctions of Sentences Interval Notation and Domains 9.2

13. Slide 4- 13 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Two inequalities joined by the word “and” or the word “or” are called compound inequalities.

14. Slide 4- 14 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Intersection of Sets and Conjunctions of Sentences The intersection of two sets A and B is the set of all elements that are common to both A and B. We denote the intersection of sets A and B as AB

15. Slide 4- 15 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Find the intersection: Example

16. Slide 4- 16 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley When two or more sentences are joined by the word and to make a compound sentence, the new sentence is called a conjunction of the sentences. The following is a conjunction of inequalities: –1 < x and x < 3. A number is a solution of a conjunction if it is a solution of both of the separate parts. For example, 0 is a solution because it is a solution of –1 < x as well as x < 3.

17. Slide 4- 17 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Note that for a < b, and, equivalently, So 3 < 2x +1 < 7 can be solved as 3 < 2x +1 and 2x + 1 < 7 a < x and x < b can be abbreviated a < x < b; b > x and x > a can be abbreviated b > x > a.

18. Slide 4- 18 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Mathematical Use of the Word “and” The word “and” corresponds to “intersection” and to the symbol Any solution of a conjunction must make each part of the conjunction true.

19. Slide 4- 19 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve and graph: Example

20. Slide 4- 20 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Sometimes there is no way to solve both parts of a conjunction at once. AB In this situation, A and B are said to be disjoint.

21. Slide 4- 21 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve and graph: Example

22. Slide 4- 22 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The union of two sets A and B is the collection of elements belonging to A and/or B. We denote the intersection of sets A and B by Unions of Sets and Disjunctions of Sentences AB

23. Slide 4- 23 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Find the union: Example

24. Slide 4- 24 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley When two or more sentences are joined by the word or to make a compound sentence, the new sentence is called a disjunction of the sentences. x 8 A number is a solution of a disjunction if it is a solution of at least one of the separate parts. For example, x = 12 is a solution since 12 > 8. Example:

25. Slide 4- 25 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Mathematical Use of the Word “or” The word “or” corresponds to “union” and to the symbol For a number to be a solution of a disjunction, it must be in at least one of the solution sets of the individual sentences.

26. Slide 4- 26 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve and graph: Example

27. Slide 4- 27 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Interval Notation and Domains Find the domain of

28. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Absolute-Value Equations and Inequalities Equations with Absolute Value Inequalities with Absolute Value 9.3

29. Slide 4- 29 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Absolute Value (When x is nonnegative, the absolute value of x is x. When x is negative, the absolute value of x is the opposite of x.) The absolute value of x, denoted |x|, is defined as:

30. Slide 4- 30 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution Find the solution set: a) |x| = 6; b) |x| = 0; c) |x| = –2 Example

31. Slide 4- 31 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Absolute-Value Principle for Equations For any positive number p and any algebraic expression X: a)The solutions of |X| = p are those numbers that satisfy X = –p or X = p. b) The equation |X| = 0 is equivalent to the equation X = 0. c) The equation |X| = –p has no solution.

32. Slide 4- 32 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Find the solution set: a) |2x +1| = 5; b) |3 – 4x| = –10 Example

33. Slide 4- 33 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Given that f (x) = 3|x+5| – 4, find all x for which f (x) = 11. Example

34. Slide 4- 34 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Sometimes an equation has two absolute- value expressions. Consider |a| = |b|. This means that a and b are the same distance from zero. If a and b are the same distance from zero, then either they are the same number or they are opposites.

35. Slide 4- 35 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley To solve |3x – 5| = |8 + 4x| we would consider the two cases Example

36. Slide 4- 36 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Inequalities with Absolute Value Our methods for solving equations with absolute value can be adapted for solving inequalities. Inequalities of this sort arise regularly in more advanced courses.

37. Slide 4- 37 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve |x| < 3. Then graph. Example

38. Slide 4- 38 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example

39. Slide 4- 39 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Principles for Solving Absolute-Value Problems For any positive number p and any expression X: a) The solutions of |X| = p are those numbers that satisfy X = –p or X = p. –p p

40. Slide 4- 40 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley b) The solutions of |X| < p are those numbers that satisfy –p < X < p. c) The solutions of |X| > p are those numbers that satisfy X < –p or p < X. –p p ( ) ( )

41. Slide 4- 41 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve |3x + 7| < 8. Then graph. Example

42. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Inequalities in Two Variables Graphs of Linear Inequalities Systems of Linear Inequalities 9.4

43. Slide 4- 43 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Graphs of Linear Inequalities When the equal sign in a linear equation is replaced with an inequality sign, a linear inequality is formed. Solutions of linear inequalities are ordered pairs.

44. Slide 4- 44 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Determine whether (1, 5) and (6, –2) are solutions of the inequality 3x – y < 5. Example

45. Slide 4- 45 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The graph of a linear equation is a straight line. The graph of a linear inequality is a half-plane, with a boundary that is a straight line. To find the equation of the boundary line, we simply replace the inequality sign with an equals sign.

46. Slide 4- 46 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Graph x y -5 -4 -3 -2 -1 1 2 3 4 5 -3 2 -2 3 1 6 5 4 y = x -4 -5 Example

47. Slide 4- 47 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Graph x y -5 -4 -3 -2 -1 1 2 3 4 5 -3 2 -2 3 1 6 5 4 y = 3 – 8x -4 -5 (3, 1) Example

48. Slide 4- 48 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Steps for Graphing Linear Inequalities 1. Replace the inequality sign with an equals sign and graph this line as the boundary. If the inequality symbol is, draw the line dashed. If the inequality symbol is, draw the line solid. 2. The graph of the inequality consists of a half-plane on one side of the line and, if the line is solid, the line as well.

49. Slide 4- 49 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Steps for Graphing Linear Inequalities a) If the inequality is of the form y < mx + b or shade below the line. If the inequality is of the form y > mx + b or shade above the line. b) If y is not isolated, either solve for y and graph as in part (a) or simply graph the boundary and use a test point. If the test point is a solution, shade the half-plane containing the point. If it is not a solution, shade the other half-plane.

50. Slide 4- 50 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution Graph x y -5 -4 -3 -2 -1 1 2 3 4 5 -3 2 -2 3 1 6 5 4 -4 -5 Example

51. Slide 4- 51 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution Graph The graph consists of the line x = –3 and the half-plane to the right of the line x = –3. x y -5 -4 -3 -2 -1 1 2 3 4 5 -3 2 -2 3 1 6 5 4 -4 -5 Example

52. Slide 4- 52 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Systems of Linear Inequalities To graph a system of equations, we graph the individual equations and then find the intersection of the individual graphs. We do the same thing for a system of inequalities, that is, we graph each inequality and find the intersection of the individual graphs.

53. Slide 4- 53 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution Graph the system x y -5 -4 -3 -2 -1 1 2 3 4 5 -3 2 -2 3 1 6 5 4 -4 -5 Example x y -5 -4 -3 -2 -1 1 2 3 4 5 -3 2 -2 3 1 6 5 4 -4 -5

54. Slide 4- 54 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution Graph x y -5 -4 -3 -2 -1 1 2 3 4 5 -3 2 -2 3 1 6 5 4 -4 -5 Example

55. Slide 4- 55 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley A system of inequalities may have a graph that consists of a polygon and its interior. In Section 4.5 we will have use for the corners, or vertices (singular vertex), of such a graph.

56. Slide 4- 56 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The graph of the system is shown and the vertices are labeled. These vertices can be found by finding the intersection points of the pairs of lines. Red Blue Green x y -5 -4 -3 -2 -1 1 2 3 4 5 -3 2 -2 3 1 6 5 4 -4 -5 (3, 5) (3, –3 ) (–1, 1 ) Example

57. Slide 4- 57 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Type Example Solution Linear equations 2x – 8 = 3(x + 5) A number in one variable Graph Let’s look at 6 different types of problems that we have solved, along with illustrations of each type.

58. Slide 4- 58 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Type Example Solution Linear inequalities –3x + 5 > 2 A set of numbers; in one variable an interval Graph -2012 )

59. Slide 4- 59 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Type Example Solution Linear equations 2x + y = 7 A set of ordered in two variables pairs; a line Graph

60. Slide 4- 60 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Type Example Solution Linear inequalities x + y ≥ 4 A set of ordered in two variables pairs; a half-plane Graph

61. Slide 4- 61 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Type Example Solution System of x + y = 3, An ordered pair or equations in 5x – y = –27 a (possibly empty) two variables set of ordered pairs Graph

62. Slide 4- 62 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Type Example Solution System of 6x – 2y ≤ 12, A set of ordered inequalities in y – 3 ≤ 0, pairs; a region two variables x + y ≥ 0 of a plane Graph

63. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Applications Using Linear Programming Objective Functions and Constraints Linear Programming 9.5

64. Slide 4- 64 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Function Many real world situations require finding a greatest value (a maximum) or a least value (a minimum). Often a quantity we wish to maximize depends on two or more quantities.

65. Slide 4- 65 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley For example, a small refreshment stand’s profits, P, might depend on the number of cans of soda s and the number of bottles of water w sold. If the stand makes $0.25 profit from each can of soda and $0.45 profit from each bottle of water, the total profit, in dollars, is given by the objective function P = 0.25s + 0.45w.

66. Slide 4- 66 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Constraints The operator of the stand might be tempted to sell only bottles of water since they yield the largest profit. This would be a good idea if it were not for the fact that the number of cans of soda and the number of bottles of water – and thus the total profit – is subject to the demands, or constraints, of the situation. For example, because of the size of the cooler, the stand may be limited to no more than a total of 30 bottles and cans. Thus the objective function would be subject to the constraint

67. From past history the operator of the stand knows that at least 5 bottles of water will be requested. This leads to the second constraint Finally, the operators of the stand have been told to spend no more than $10.00 on the drinks. If the cans of soda cost $0.25 each and the bottles of water cost $0.50 each this leads to the third constraint Cost of sodaCost of water Cannot exceed $10.00

68. In short, the operator of the stand wishes to maximize the objective function P = 0.25s + 0.45w, subject to the constraints The constraints form a system of linear inequalities that can be graphed.

69. Slide 4- 69 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Linear Programming The refreshment stand operator’s problem is “How many cans of soda and bottles of water should be sold, subject to the constraints listed, in order to maximize profit?” To solve such a problem, we use an important result from a branch of mathematics known as linear programming.

70. Slide 4- 70 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Corner Principle Suppose that an objective function F = ax + by + c depends on x and y (with a, b, and c constant). Suppose also that F is subject to constraints on x and y, which form a system of linear inequalities. If F has a minimum or a maximum value, then it can be found as follows: 1. Graph the system of inequalities and find the vertices.

71. Slide 4- 71 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Corner Principle 2. Find the value of the objective function at each vertex. The greatest and least of those values are the maximum and the minimum values of the function, respectively. 3. The ordered pair at which the maximum and minimum occurs indicates the choice of (x, y) for which that maximum or minimum occurs.

72. Slide 4- 72 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley We graph the system. The portion of the graph that is shaded represents all pairs that satisfy the constraints. 35 5 30 20 25 15 10 0 5 10 15 20 25 30 35 40 45 50 s w It is sometimes called the feasible region.

73. Slide 4- 73 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 35 5 30 20 25 15 10 0 5 10 15 20 25 30 35 40 45 50 s w (0, 20) (20, 10) (25, 5) (0, 5) According to the corner principle, P is maximized at one of the vertices of the shaded region. Determine the coordinates of the vertices.

74. Slide 4- 74 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Vertex (s, w) Profit P = 0.25s + 0.45w (0, 5) (0, 20) (20, 10) (25, 5) 0.25(0) + 0.45(5) = 2.25 0.25(0) + 0.45(20) = 9.00 0.25(20) + 0.45(10) = 9.50 0.25(25) + 0.45(5) = 8.50 We now find the value of P at each vertex. The greatest value of P occurs at (20, 10). Thus profit is maximized at $9.50 if the operator of the stand stocks 20 cans of soda and 10 bottles of water to sell. Incidentally, we have also shown that profit is minimized at $2.25 if 5 bottles of water and no cans of soda are stocked. Max. Min.