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4. Inequalities

4. Inequalities. 4.1 Solving Linear Inequalities. Problem Basic fee: $20 Per minute: 5 ¢ Budget: $40 How many minutes?: x 20 + 0.05x ≤ 40 x ≤ 400. a. b. a. b. Notations. Closed interval [a, b] = {x | a ≤ x ≤ b} Open interval

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4. Inequalities

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  1. 4. Inequalities

  2. 4.1 Solving Linear Inequalities • Problem • Basic fee: $20 • Per minute: 5¢ • Budget: $40 • How many minutes?: x • 20 + 0.05x ≤ 40 • x ≤ 400

  3. a b a b Notations • Closed interval • [a, b] = {x | a ≤ x ≤ b} • Open interval • (a, b) = {x | a < x < b}

  4. Notations • Infinite interval • [a, ∞) = {x | a ≤ x < ∞} • Infinite interval • (-∞ , b] = {x | -∞ < x ≤ b} a b

  5. Your Turn • Express in set-builder notation. • [a, b) • (- ∞ , b)

  6. Solving Inequalities in One Variable • 0.05x + 20 ≤ 400.05x ≤ 20x ≤ 20/0.05x ≤ 400 • [0, 400] (Interval notation) • {x | x ≤ 400} (set-builder notation)

  7. Properties of Inequalities • Addition • a < b → a + c < b + ca < b → a - c < b – c • Positive Multiplication (c > 0) • a < b → ac < bca < b → a /c < b/c • Negative Multiplication (c < 0) • a < b → ac ≥ bca < b → a /c ≥ b/c

  8. Example 1 • -2x – 4 > x + 5 • -3x > 9(-1/3)(-3x) < (-1/3)9x < -3(- ∞ ,-3)

  9. Example 2 • (x + 3) (x – 2) 1--------- ≥ ---------- + --- 4 3 4 • 3(x + 3) 4(x – 2) 3 ------------ ≥ ----------- + ----- 12 12 12 3x + 9 ≥ 4x – 8 + 3-x ≥ -14 x ≤ 14(-∞, 14]

  10. Special Cases • x > x + 1 • {x | x > x + 1} • What kind of set is this? • x < x + 1 • {x } x < x + 1} • What kind of set is this?

  11. A B A ∩ B 4.2 Compound Inequalities • Intersection of Sets • Given set A and B, intersection A and B,A ∩ B = {x | x ε A AND x ε B}

  12. A A A UB Compound Inequalities • Union of Sets • Given set A and B, union of A and B,A U B = {x | x ε A OR x ε B}

  13. Intersection of Sets • Given: A = {1, 2, 3,5, 9} B = {3, 5, 9, 10, 12} A ∩ B = {5, 9} • Given: A = {x | x ≥ 3} B = {x | x ≤ 10} A ∩ B = {x | x ≥ 3 AND x ≤ 10 } [ ] 3 10

  14. Union and Intersection of Sets • Given: • A = set of all male students at CUH • B = set of all female students at CUH • C = set of all freshman students at CUH • Draw a diagram of: • A ∩ B • A U B • A ∩ C • A U C

  15. Solving Compound Inequality • Given: 2x – 7 > 3 AND 5x – 4 < 6 • What does it mean: Solve the compound inequality? • It means: Find the set of x so that both inequalities are true • Solution Set:{x | 2x – 7 > 3 AND 5x – 4 < 6}

  16. Solving an AND Compound Inequality • 2x – 7 ≥ 3 AND 5x – 4 ≤ 62x ≥ 3 + 7 5x ≤ 6 + 42x ≥ 10 5x ≤ 10x ≥ 10/2 x ≤ 10/5x ≥ 5 x ≤ 2 • Solution Set: { Φ } 2 5

  17. Solving an AND Compound Inequality • -3 < 2x + 1 ≤ 3 • This means: (-3 < 2x + 1 AND 2x + 1 ≤ 3) -3 – 1 < 2x + 1 – 1 ≤ 3 - 1-2 < 2x ≤ 2-1 < x ≤ 1 • Solution Set: { x | -1 < x ≤ 1 } -1 1

  18. Solving an OR Compound Inequality • Given:2x – 3 < 7 OR 35 – 4x ≤ 32x < 4 OR -4x ≤ -32x < 4 OR x ≥ 8 • Take the union of solution sets{x | x < 4 U x ≥ 8}= {x | x < 4 or x ≥ 8} 4 8

  19. Solving an OR Compound Inequality • Given:3x – 5 ≤ 13 OR 5x + 2 > -33x ≤ 18 OR 5x > -5x ≤ 6 OR x > -1 • Take the union of solution sets{x | x ≤ 6 U x > -1}= {x | x ≤ 6 or x > -1} = R -1 6

  20. Your Turn • Find the following sets • {a, b, c, d, e} ∩ {b, c, 2, 3, x, y} • {a, b, c, d, e} U {b, c, 2, 3, x, y} • Solve the following • 3 ≤ 4x – 3 < 19 • 3x < 3 or 2x > 10

  21. -c 0 c |A| |A| A A 4.3 Equations & Inequalities Involving Absolute Values • Absolute value of A -- |A| -- where A is any algebraic expression: • |A| = c  A = c or A = -c, where c > 0 • |2x – 3| = 112x – 3 = 11 or 2x – 3 = -11

  22. Solving Equation Involving Absolute Value • Solve for x: |2x – 3| = 112x – 3 = 11 or 2x – 3 = -11 2x = 14 2x = -8x = 7 x = - 4 {4, 7} • Solve for x: 5|1 - 4x| -15 = 0 |1 – 4x| = 15/5 = 3 1 – 4x = 3 or 1 – 4x = -3 -4x = 2 -4x = -4x = -1/2 x = 1 {-1/2, 1}

  23. Equation With 2 Absolute Values • Solve for x: |2x – 7| = |x + 3|2x – 7 = (x + 3) or 2x – 7 = -(x + 3) x = 10 2x – 7 = -x – 3 3x = 4 x = 4/3 {4/3, 10}

  24. Solving Absolute Value Inequality (Using Boundary Points) • Solve and graph: |2x + 3| ≥ 5 • Solve the equation2x + 3 = 5 or 2x + 3 = -5 x = 1 x = -4 • Locate the boundary points • Choose a test value in each interval and substitute in the inequality -4 1

  25. - 4 1 • Solve and graph: |2x + 3| ≥ 5 • Locate the boundary points • Choose a test value in each interval and substitute in inequality

  26. - 4 1 - 4 1 • Solve and graph: |2x + 3| ≥ 5 • Write the solution set. Check for boundaries.Preliminary Solution: (-∞ , -4) U (1, ∞) Because |2x + 3| = 5, we need to include the solution set of this equation (i.e., boundaries): x = -4, 1. (This was found in step 1.)Final Solution: (-∞ , -4] U [1, ∞)

  27. Using Boundary Points • Solve and graph: |2x -5| ≥ 3 • Solve the equation2x – 5 = 3 or 2x – 5 = -3 x = 4 x = 1 • Locate the boundary points • Choose a test value in each interval and substitute in inequality 1 4

  28. 1 4 • Solve and graph: |2x - 5| ≥ 3 • Locate the boundary points • Choose a test value in each interval and substitute in inequality

  29. 1 4 1 4 • Solve and graph: |2x – 5| ≥ 3 • Write the solution set. Check for boundaries.Preliminary Solution: (-∞ , 1) U (4, ∞) Because |2x - 5| = 3, we need to include the solution set of this equation (i.e., boundaries): x = 1, 4 (This was found in step 1.)Final Solution: (-∞ , 1] U [4, ∞)

  30. Solving Absolute Value Inequality (Using Compound Inequalities) • Note: • Solution set of |x| < 2 is (-2, 2)(-2, 2)  -2 < x < 2 • Solution set of |x| > 2 is (-∞, -2) U (2, ∞) -2 0 2 -2 0 2 (-∞, -2) U (2, ∞)  x < -2 or x > 2

  31. Solving Absolute Value Inequality (Using Compound Inequalities) • Solve: |x – 4| < 3-3 < x – 4 < 31 < x < 7

  32. Solving Absolute Value Inequality (Using Compound Inequalities) • Solve: |2x + 3| ≥ 52x + 3 ≥ 5 or 2x + 3 ≤ -5 2x ≥ 2 or 2x ≤ -8 x ≥ 1 or x ≤ -4 -4 1

  33. Your Turn • Solve inequalities using equivalent compound inequalities • |x – 2| < 5 • |2x – 5| ≥ 3

  34. 2x – 3y > 6 (3, 0) 2x – 3y < 6 (0, -2) 2x – 3y = 6 4.4 Linear Inequalities in 2 Variables • Solve: 2x – 3y ≥ 6 • Graph: 2x – 3y = 6To find y-intercept To find x-intercept y = 0 x = 0 2x = 6 -3y = 6 x = 3 y = -2

  35. Choose a test point in one half-plane and check with original inequality.2x – 3y ≥ 6Choose A (0, 0) as a test point0 – 0 ≥ 60 ≥ 6false—A is outside the solution set

  36. If A(0, 0) is not in solution set, the other half-plane is the solution set of 2x – 3y ≥ 6Because of ≥ , include the boundary line in the graph of the solution set. Graph of : {x | 2x – 3y ≥ 6} 2x – 3y = 6 2x – 3y < 6 A(0, 0) 2x – 3y > 6

  37. Your Turn • Graph the following inequality: • 4x – 2y ≥ 8 • x/4 + y/2 < 1

  38. Graphing System of Linear Inequalities • Graph solution set of:x – y < 12x + 3y ≥ 12 • Graph equations x – y = 1 2x + 3y = 12x-intercept: x-intercept: y = 0 y = 0 x – 0 = 1 2x + 0 = 12 x = 1 x = 6 (1, 0) (6, 0)

  39. Graphing System of Linear Inequalities • Graph equalities x – y = 1 2x + 3y = 12x-intercept: x-intercept: (1, 0) (6, 0)y-intercept: y-intercept: x = 0 x = 0 0 – y = 1 0 + 3y = 12 -y = 1 3y = 12 y = -1 y = 4 (0, -1) ( 0, 4) Points for: x – y = 1 Points for: 2x + 3y = 12 (1, 0) (0, -1) (6, 0) (0, 4)

  40. Graphing System of Linear Inequalities 2x + 3y = 12 x – y = 1 (0, 4) (1, 0) (6, 0) (0, 0) (1, 0)

  41. Graphing System of Linear Inequalities • Choose a point and check with original inequalities.Pick (0, 0) Pick (0, 0)Part of x – y < 1? Part of 2x + 3y ≥ 12? Check: Check: 0 – 0 < 1? 0 + 0 ≥ 12? 0 < 1? 0 ≥ 12? true false this half-plane other half-plane

  42. Graphing System of Linear Inequalities x – y = 1 2x + 3y = 12 (0, 0)

  43. Your Turn • Graph the solution set of the system:x – 3y < 62x + 3y ≥ -6

  44. 4.5 Linear Programming • Problem: • A division of a furniture company specializes inmanufacturing bookcases and computer desks. • The division makes $25 per bookcase and $55 per desk. • To maintain quality, the division can make a maximum of 80 bookcases and desks (total) per day

  45. 4.5 Linear Programming • Problem (cont.) • Because of customer demands, between 30 and 80 bookcases must be made daily. • Furthermore, at least 10 and not more than 30 desks must be made per day • How many bookcases and desks must be made each day to maximize profit?

  46. 4.5 Linear Programming • Solution • Use variables to represent quantitiesx = number of bookcases per monthy = desks per monthz = profit for month • Form objective functionz = 25x + 55y • Write constraints as inequalitiesx + y ≤ 8030 ≤ x ≤ 8010 ≤ y ≤ 30 • Graph the inequalities

  47. 4.5 Linear Programming • Graph the inequalities • x + y ≤ 80 x + y = 80line passes through (80, 0) and (0, 80) • 30 ≤ x ≤ 80y can be any value • 10 ≤ y ≤ 30x can be any value

  48. 30 ≤x ≤ 80 (0, 80) A B 10 ≤y ≤ 30 (80, 0) C D x + y ≤ 80 4.5 Linear Programming • Graph the inequalities

  49. x = 80 x = 30 x + y = 80 (0, 80) y = 30 A B y = 10 C D (80, 0) 4.5 Linear Programming • Determine the corners of the solution area To find A: x = 30 y = 30 (30, 30) To find B: y = 30 x + 30 = 80 x = 50 (50, 30) To find C: y = 10 x + 10 = 80 x = 70 (70, 10)To find D: (30, 10)

  50. 4.5 Linear Programming • Check the objective equation with the corner points Solution: 50 bookcases, 30 desks, resulting in $2900 profit

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