1. Sign-Charts and Inequalities

2. We use the factored polynomials or rational expressions to determine the signs of the outputs. Sign-Charts and Inequalities

3. We use the factored polynomials or rational expressions to determine the signs of the outputs. That is, given a formula f and a value for x, factor f to determine if the output is positive (f > 0) or negative (f < 0) when f is evaluated with x. Sign-Charts and Inequalities

4. We use the factored polynomials or rational expressions to determine the signs of the outputs. Example A: Determine the outcome is + or – for x 2 – 2x – 3 if x = -3/2, -1/2. That is, given a formula f and a value for x, factor f to determine if the output is positive (f > 0) or negative (f < 0) when f is evaluated with x. Sign-Charts and Inequalities

5. We use the factored polynomials or rational expressions to determine the signs of the outputs. Example A: Determine the outcome is + or – for x 2 – 2x – 3 if x = -3/2, -1/2. In factored form x 2 – 2x – 3 = (x – 3)(x + 1) That is, given a formula f and a value for x, factor f to determine if the output is positive (f > 0) or negative (f < 0) when f is evaluated with x. Sign-Charts and Inequalities

6. We use the factored polynomials or rational expressions to determine the signs of the outputs. Example A: Determine the outcome is + or – for x 2 – 2x – 3 if x = -3/2, -1/2. In factored form x 2 – 2x – 3 = (x – 3)(x + 1) Hence, for x = -3/2: (-3/2 – 3)(-3/2 + 1) That is, given a formula f and a value for x, factor f to determine if the output is positive (f > 0) or negative (f < 0) when f is evaluated with x. Sign-Charts and Inequalities

7. We use the factored polynomials or rational expressions to determine the signs of the outputs. Example A: Determine the outcome is + or – for x 2 – 2x – 3 if x = -3/2, -1/2. In factored form x 2 – 2x – 3 = (x – 3)(x + 1) Hence, for x = -3/2: (-3/2 – 3)(-3/2 + 1) is (–)(–) = + . That is, given a formula f and a value for x, factor f to determine if the output is positive (f > 0) or negative (f < 0) when f is evaluated with x. Sign-Charts and Inequalities

8. We use the factored polynomials or rational expressions to determine the signs of the outputs. Example A: Determine the outcome is + or – for x 2 – 2x – 3 if x = -3/2, -1/2. In factored form x 2 – 2x – 3 = (x – 3)(x + 1) Hence, for x = -3/2: (-3/2 – 3)(-3/2 + 1) is (–)(–) = + . And for x = -1/2: (-1/2 – 3)(-1/2 + 1) That is, given a formula f and a value for x, factor f to determine if the output is positive (f > 0) or negative (f < 0) when f is evaluated with x. Sign-Charts and Inequalities

9. We use the factored polynomials or rational expressions to determine the signs of the outputs. Example A: Determine the outcome is + or – for x 2 – 2x – 3 if x = -3/2, -1/2. In factored form x 2 – 2x – 3 = (x – 3)(x + 1) Hence, for x = -3/2: (-3/2 – 3)(-3/2 + 1) is (–)(–) = + . And for x = -1/2: (-1/2 – 3)(-1/2 + 1) is (–)(+) = – . That is, given a formula f and a value for x, factor f to determine if the output is positive (f > 0) or negative (f < 0) when f is evaluated with x. Sign-Charts and Inequalities

10. Example B. Determine the outcome is + or – for if x = -3/2, -1/2. x 2 – 2x – 3 x 2 + x – 2 Sign-Charts and Inequalities

11. Example B. Determine the outcome is + or – for if x = -3/2, -1/2. x 2 – 2x – 3 x 2 + x – 2 x 2 – 2x – 3 x 2 + x – 2 In factored form = (x – 3)(x + 1) (x – 1)(x + 2) Sign-Charts and Inequalities

12. Example B. Determine the outcome is + or – for if x = -3/2, -1/2. x 2 – 2x – 3 x 2 + x – 2 x 2 – 2x – 3 x 2 + x – 2 In factored form = (x – 3)(x + 1) (x – 1)(x + 2) Hence, for x = -3/2: (x – 3)(x + 1) (x – 1)(x + 2) = (–)(–) (–)(+) Sign-Charts and Inequalities

13. Example B. Determine the outcome is + or – for if x = -3/2, -1/2. x 2 – 2x – 3 x 2 + x – 2 x 2 – 2x – 3 x 2 + x – 2 In factored form = (x – 3)(x + 1) (x – 1)(x + 2) Hence, for x = -3/2: (x – 3)(x + 1) (x – 1)(x + 2) = (–)(–) (–)(+) < 0 Sign-Charts and Inequalities

14. Example B. Determine the outcome is + or – for if x = -3/2, -1/2. x 2 – 2x – 3 x 2 + x – 2 x 2 – 2x – 3 x 2 + x – 2 In factored form = (x – 3)(x + 1) (x – 1)(x + 2) Hence, for x = -3/2: (x – 3)(x + 1) (x – 1)(x + 2) = (–)(–) (–)(+) < 0 For x = -1/2: (x – 3)(x + 1) (x – 1)(x + 2) = (–)(+) (–)(+) Sign-Charts and Inequalities

15. Example B. Determine the outcome is + or – for if x = -3/2, -1/2. x 2 – 2x – 3 x 2 + x – 2 x 2 – 2x – 3 x 2 + x – 2 In factored form = (x – 3)(x + 1) (x – 1)(x + 2) Hence, for x = -3/2: (x – 3)(x + 1) (x – 1)(x + 2) = (–)(–) (–)(+) < 0 For x = -1/2: (x – 3)(x + 1) (x – 1)(x + 2) = (–)(+) (–)(+) > 0 Sign-Charts and Inequalities

16. Example B. Determine the outcome is + or – for if x = -3/2, -1/2. x 2 – 2x – 3 x 2 + x – 2 x 2 – 2x – 3 x 2 + x – 2 In factored form = (x – 3)(x + 1) (x – 1)(x + 2) Hence, for x = -3/2: (x – 3)(x + 1) (x – 1)(x + 2) = (–)(–) (–)(+) < 0 For x = -1/2: (x – 3)(x + 1) (x – 1)(x + 2) = (–)(+) (–)(+) > 0 This leads to the sign charts of formulas. The sign- chart of a formula gives the signs of the outputs.

17. Here is an example, the sign chart of f = x – 1: Sign-Charts and Inequalities

18. Here is an example, the sign chart of f = x – 1: 1 f = 0 + + – – – – x – 1 Sign-Charts and Inequalities

19. Here is an example, the sign chart of f = x – 1: 1 f = 0 + + – – – – x – 1 The &quot;+&quot; indicates the region where the output is positive i.e. if 1 < x. Sign-Charts and Inequalities

20. Here is an example, the sign chart of f = x – 1: 1 f = 0 + + – – – – x – 1 The &quot;+&quot; indicates the region where the output is positive i.e. if 1 < x. Likewise, the &quot;–&quot; indicates the region where the output is negative, i.e. x < 1. Sign-Charts and Inequalities

21. Construction of the sign-chart of f. Here is an example, the sign chart of f = x – 1: 1 f = 0 + + – – – – x – 1 The &quot;+&quot; indicates the region where the output is positive i.e. if 1 < x. Likewise, the &quot;–&quot; indicates the region where the output is negative, i.e. x < 1. Sign-Charts and Inequalities

22. Construction of the sign-chart of f. I. Solve for f = 0 (and denominator = 0 if there is any denominator). Here is an example, the sign chart of f = x – 1: 1 f = 0 + + – – – – x – 1 The &quot;+&quot; indicates the region where the output is positive i.e. if 1 < x. Likewise, the &quot;–&quot; indicates the region where the output is negative, i.e. x < 1. Sign-Charts and Inequalities

23. Construction of the sign-chart of f. I. Solve for f = 0 (and denominator = 0 if there is any denominator). II. Draw the real line, mark off the answers from I. Here is an example, the sign chart of f = x – 1: 1 f = 0 + + – – – – x – 1 The &quot;+&quot; indicates the region where the output is positive i.e. if 1 < x. Likewise, the &quot;–&quot; indicates the region where the output is negative, i.e. x < 1. Sign-Charts and Inequalities

24. Construction of the sign-chart of f. I. Solve for f = 0 (and denominator = 0 if there is any denominator). II. Draw the real line, mark off the answers from I. III. Sample each segment for signs by testing a point in each segment. Here is an example, the sign chart of f = x – 1: 1 f = 0 + + – – – – x – 1 The &quot;+&quot; indicates the region where the output is positive i.e. if 1 < x. Likewise, the &quot;–&quot; indicates the region where the output is negative, i.e. x < 1. Sign-Charts and Inequalities

25. Construction of the sign-chart of f. I. Solve for f = 0 (and denominator = 0) if there is any denominator. II. Draw the real line, mark off the answers from I. III. Sample each segment for signs by testing a point in each segment. Here is an example, the sign chart of f = x – 1: 1 f = 0 + + – – – – x – 1 The &quot;+&quot; indicates the region where the output is positive i.e. if 1 < x. Likewise, the &quot;–&quot; indicates the region where the output is negative, i.e. x < 1. Fact : The sign stays the same for x's in between the values from step I (where f = 0 or f is undefined.) Sign-Charts and Inequalities

26. Example C. Let f = x 2 – 3x – 4 , use the sign - chart to indicate when is f = 0, f > 0, and f < 0 . Sign-Charts and Inequalities

27. Example C. Let f = x 2 – 3x – 4 , use the sign - chart to indicate when is f = 0, f > 0, and f < 0 . Solve x 2 – 3x – 4 = 0 (x – 4)(x + 1) = 0  x = 4 , -1 Sign-Charts and Inequalities

28. Example C. Let f = x 2 – 3x – 4 , use the sign - chart to indicate when is f = 0, f > 0, and f < 0 . Solve x 2 – 3x – 4 = 0 (x – 4)(x + 1) = 0  x = 4 , -1 Mark off these points on a line: (x-4)(x+1) 4 -1 Sign-Charts and Inequalities

29. Example C. Let f = x 2 – 3x – 4 , use the sign - chart to indicate when is f = 0, f > 0, and f < 0 . Solve x 2 – 3x – 4 = 0 (x – 4)(x + 1) = 0  x = 4 , -1 Mark off these points on a line: (x-4)(x+1) Select points to sample in each segment: 4 -1 Sign-Charts and Inequalities

30. Example C. Let f = x 2 – 3x – 4 , use the sign - chart to indicate when is f = 0, f > 0, and f < 0 . Solve x 2 – 3x – 4 = 0 (x – 4)(x + 1) = 0  x = 4 , -1 Mark off these points on a line: (x-4)(x+1) 4 -1 Select points to sample in each segment: Test x = - 2, -2 Sign-Charts and Inequalities

31. Example C. Let f = x 2 – 3x – 4 , use the sign - chart to indicate when is f = 0, f > 0, and f < 0 . Solve x 2 – 3x – 4 = 0 (x – 4)(x + 1) = 0  x = 4 , -1 Mark off these points on a line: (x-4)(x+1) 4 -1 Select points to sample in each segment: Test x = - 2, get – * – = + . -2 Sign-Charts and Inequalities

32. Example C. Let f = x 2 – 3x – 4 , use the sign - chart to indicate when is f = 0, f > 0, and f < 0 . Solve x 2 – 3x – 4 = 0 (x – 4)(x + 1) = 0  x = 4 , -1 Mark off these points on a line: (x-4)(x+1) 4 -1 Select points to sample in each segment: Test x = - 2, get – * – = + . Hence the segment is positive. Draw + sign over it. -2 Sign-Charts and Inequalities

33. Example C. Let f = x 2 – 3x – 4 , use the sign - chart to indicate when is f = 0, f > 0, and f < 0 . Solve x 2 – 3x – 4 = 0 (x – 4)(x + 1) = 0  x = 4 , -1 Mark off these points on a line: (x-4)(x+1) + + + + + 4 -1 Select points to sample in each segment: Test x = - 2, get – * – = + . Hence the segment is positive. Draw + sign over it. -2 Sign-Charts and Inequalities

34. Example C. Let f = x 2 – 3x – 4 , use the sign - chart to indicate when is f = 0, f > 0, and f < 0 . Solve x 2 – 3x – 4 = 0 (x – 4)(x + 1) = 0  x = 4 , -1 Mark off these points on a line: (x-4)(x+1) + + + + + 0 4 -1 Select points to sample in each segment: Test x = - 2, get – * – = + . Hence the segment is positive. Draw + sign over it. -2 Test x = 0, get – * + = –. Sign-Charts and Inequalities

35. Example C. Let f = x 2 – 3x – 4 , use the sign - chart to indicate when is f = 0, f > 0, and f < 0 . Solve x 2 – 3x – 4 = 0 (x – 4)(x + 1) = 0  x = 4 , -1 Mark off these points on a line: (x-4)(x+1) + + + + + 0 4 -1 Select points to sample in each segment: Test x = - 2, get – * – = + . Hence the segment is positive. Draw + sign over it. -2 Test x = 0, get – * + = –. Hence this segment is negative. Put – over it. Sign-Charts and Inequalities

36. Example C. Let f = x 2 – 3x – 4 , use the sign - chart to indicate when is f = 0, f > 0, and f < 0 . Solve x 2 – 3x – 4 = 0 (x – 4)(x + 1) = 0  x = 4 , -1 Mark off these points on a line: (x-4)(x+1) + + + + + – – – – – 0 4 -1 Select points to sample in each segment: Test x = - 2, get – * – = + . Hence the segment is positive. Draw + sign over it. -2 Test x = 0, get – * + = –. Hence this segment is negative. Put – over it. Sign-Charts and Inequalities

37. Example C. Let f = x 2 – 3x – 4 , use the sign - chart to indicate when is f = 0, f > 0, and f < 0 . Solve x 2 – 3x – 4 = 0 (x – 4)(x + 1) = 0  x = 4 , -1 Mark off these points on a line: (x-4)(x+1) + + + + + – – – – – 0 4 -1 Select points to sample in each segment: Test x = - 2, get – * – = + . Hence the segment is positive. Draw + sign over it. -2 Test x = 0, get – * + = –. Hence this segment is negative. Put – over it. Test x = 5, get + * + = +. 5 Sign-Charts and Inequalities

38. Example C. Let f = x 2 – 3x – 4 , use the sign - chart to indicate when is f = 0, f > 0, and f < 0 . Solve x 2 – 3x – 4 = 0 (x – 4)(x + 1) = 0  x = 4 , -1 Mark off these points on a line: (x-4)(x+1) + + + + + – – – – – 0 4 -1 Select points to sample in each segment: Test x = - 2, get – * – = + . Hence the segment is positive. Draw + sign over it. -2 Test x = 0, get – * + = –. Hence this segment is negative. Put – over it. Test x = 5, get + * + = +. Hence this segment is positive. Put + over it. 5 Sign-Charts and Inequalities

39. Example C. Let f = x 2 – 3x – 4 , use the sign - chart to indicate when is f = 0, f > 0, and f < 0 . Solve x 2 – 3x – 4 = 0 (x – 4)(x + 1) = 0  x = 4 , -1 Mark off these points on a line: (x-4)(x+1) + + + + + – – – – – + + + + + 0 4 -1 Select points to sample in each segment: Test x = - 2, get – * – = + . Hence the segment is positive. Draw + sign over it. -2 Test x = 0, get – * + = –. Hence this segment is negative. Put – over it. Test x = 5, get + * + = +. Hence this segment is positive. Put + over it. 5 Sign-Charts and Inequalities

40. Example D. Make the sign chart of f = (x – 3) (x – 1)(x + 2) Sign-Charts and Inequalities

41. Example D. Make the sign chart of f = (x – 3) (x – 1)(x + 2) The root for f = 0 is from the zero of the numerator which is x = 3. Sign-Charts and Inequalities

42. Example D. Make the sign chart of f = (x – 3) (x – 1)(x + 2) The root for f = 0 is from the zero of the numerator which is x = 3. The zeroes of the denominator x = 1, -2 are the values where f is undefined (UDF). Sign-Charts and Inequalities

43. Example D. Make the sign chart of f = (x – 3) (x – 1)(x + 2) The root for f = 0 is from the zero of the numerator which is x = 3. The zeroes of the denominator x = 1, -2 are the values where f is undefined (UDF). Mark these values on a real line. Sign-Charts and Inequalities

44. Example D. Make the sign chart of f = (x – 3) (x – 1)(x + 2) The root for f = 0 is from the zero of the numerator which is x = 3. The zeroes of the denominator x = 1, -2 are the values where f is undefined (UDF). Mark these values on a real line. (x – 3) (x – 1)(x + 2) -2 1 3 UDF UDF f=0 Sign-Charts and Inequalities

45. Example D. Make the sign chart of f = Select a point to sample in each segment: (x – 3) (x – 1)(x + 2) The root for f = 0 is from the zero of the numerator which is x = 3. The zeroes of the denominator x = 1, -2 are the values where f is undefined (UDF). Mark these values on a real line. (x – 3) (x – 1)(x + 2) -2 1 3 UDF UDF f=0 -3 0 2 4 Sign-Charts and Inequalities

46. Example D. Make the sign chart of f = Select a point to sample in each segment: Test x = -3, we get a (x – 3) (x – 1)(x + 2) The root for f = 0 is from the zero of the numerator which is x = 3. The zeroes of the denominator x = 1, -2 are the values where f is undefined (UDF). Mark these values on a real line. (x – 3) (x – 1)(x + 2) -2 1 3 UDF UDF f=0 -3 ( – ) ( – )( – ) = – segment. 0 2 4 Sign-Charts and Inequalities

47. Example D. Make the sign chart of f = Select a point to sample in each segment: Test x = -3, we get a (x – 3) (x – 1)(x + 2) The root for f = 0 is from the zero of the numerator which is x = 3. The zeroes of the denominator x = 1, -2 are the values where f is undefined (UDF). Mark these values on a real line. (x – 3) (x – 1)(x + 2) -2 1 3 UDF UDF f=0 -3 ( – ) ( – )( – ) = – segment. 0 2 4 Test x = 0, we get a ( – ) ( – )( + ) = + segment. Sign-Charts and Inequalities

48. Example D. Make the sign chart of f = Select a point to sample in each segment: Test x = -3, we get a (x – 3) (x – 1)(x + 2) The root for f = 0 is from the zero of the numerator which is x = 3. The zeroes of the denominator x = 1, -2 are the values where f is undefined (UDF). Mark these values on a real line. (x – 3) (x – 1)(x + 2) -2 1 3 UDF UDF f=0 -3 ( – ) ( – )( – ) = – segment. 0 2 4 Test x = 0, we get a ( – ) ( – )( + ) = + segment. Test x = 2, we get a ( – ) ( + )( + ) segment. = – Sign-Charts and Inequalities

49. Example D. Make the sign chart of f = Select a point to sample in each segment: Test x = -3, we get a (x – 3) (x – 1)(x + 2) The root for f = 0 is from the zero of the numerator which is x = 3. The zeroes of the denominator x = 1, -2 are the values where f is undefined (UDF). Mark these values on a real line. (x – 3) (x – 1)(x + 2) -2 1 3 UDF UDF f=0 -3 ( – ) ( – )( – ) = – segment. 0 2 4 Test x = 0, we get a ( – ) ( – )( + ) = + segment. Test x = 2, we get a ( – ) ( + )( + ) segment. = – Test x = 4, we get a ( + ) ( + )( + ) segment. = + – – – – + + + – – – + + + + Sign-Charts and Inequalities

50. The easiest way to solve a polynomial or rational inequality is use the sign-chart. Sign-Charts and Inequalities

51. The easiest way to solve a polynomial or rational inequality is use the sign-chart. To do this, I. set one side of the inequality to 0, Sign-Charts and Inequalities

52. The easiest way to solve a polynomial or rational inequality is use the sign-chart. To do this, I. set one side of the inequality to 0, II. factor the expression and draw the sign-chart, Sign-Charts and Inequalities

53. The easiest way to solve a polynomial or rational inequality is use the sign-chart. To do this, I. set one side of the inequality to 0, II. factor the expression and draw the sign-chart, III. read off the answer from the sign chart. Sign-Charts and Inequalities

54. Example E. Solve x 2 – 3x > 4 The easiest way to solve a polynomial or rational inequality is use the sign-chart. To do this, I. set one side of the inequality to 0, II. factor the expression and draw the sign-chart, III. read off the answer from the sign chart. Sign-Charts and Inequalities

55. Example E. Solve x 2 – 3x > 4 The easiest way to solve a polynomial or rational inequality is use the sign-chart. To do this, I. set one side of the inequality to 0, II. factor the expression and draw the sign-chart, III. read off the answer from the sign chart. Set one side to 0, we get x 2 – 3x – 4 > 0; Sign-Charts and Inequalities

56. Example E. Solve x 2 – 3x > 4 The easiest way to solve a polynomial or rational inequality is use the sign-chart. To do this, I. set one side of the inequality to 0, II. factor the expression and draw the sign-chart, III. read off the answer from the sign chart. Set one side to 0, we get x 2 – 3x – 4 > 0; factor, we've (x – 4)(x + 1) > 0. Sign-Charts and Inequalities

57. Example E. Solve x 2 – 3x > 4 The easiest way to solve a polynomial or rational inequality is use the sign-chart. To do this, I. set one side of the inequality to 0, II. factor the expression and draw the sign-chart, III. read off the answer from the sign chart. Set one side to 0, we get x 2 – 3x – 4 > 0; factor, we've (x – 4)(x + 1) > 0. The roots are x = -1, 4. Sign-Charts and Inequalities

58. Example E. Solve x 2 – 3x > 4 4 -1 The easiest way to solve a polynomial or rational inequality is use the sign-chart. To do this, I. set one side of the inequality to 0, II. factor the expression and draw the sign-chart, III. read off the answer from the sign chart. Draw the sign-chart, (x – 4)(x + 1) Set one side to 0, we get x 2 – 3x – 4 > 0; factor, we've (x – 4)(x + 1) > 0. The roots are x = -1, 4. Sign-Charts and Inequalities

59. Example E. Solve x 2 – 3x > 4 0 4 -1 -2 5 The easiest way to solve a polynomial or rational inequality is use the sign-chart. To do this, I. set one side of the inequality to 0, II. factor the expression and draw the sign-chart, III. read off the answer from the sign chart. Draw the sign-chart, sample the points x = -2, 0, 5 (x – 4)(x + 1) Set one side to 0, we get x 2 – 3x – 4 > 0; factor, we've (x – 4)(x + 1) > 0. The roots are x = -1, 4. Sign-Charts and Inequalities

60. Example E. Solve x 2 – 3x > 4 0 4 -1 -2 5 The easiest way to solve a polynomial or rational inequality is use the sign-chart. To do this, I. set one side of the inequality to 0, II. factor the expression and draw the sign-chart, III. read off the answer from the sign chart. Draw the sign-chart, sample the points x = -2, 0, 5 (x – 4)(x + 1) + + + Set one side to 0, we get x 2 – 3x – 4 > 0; factor, we've (x – 4)(x + 1) > 0. The roots are x = -1, 4. Sign-Charts and Inequalities

61. Example E. Solve x 2 – 3x > 4 0 4 -1 -2 5 The easiest way to solve a polynomial or rational inequality is use the sign-chart. To do this, I. set one side of the inequality to 0, II. factor the expression and draw the sign-chart, III. read off the answer from the sign chart. Draw the sign-chart, sample the points x = -2, 0, 5 (x – 4)(x + 1) + + + – – – – – – Set one side to 0, we get x 2 – 3x – 4 > 0; factor, we've (x – 4)(x + 1) > 0. The roots are x = -1, 4. Sign-Charts and Inequalities

62. Example E. Solve x 2 – 3x > 4 0 4 -1 -2 5 The easiest way to solve a polynomial or rational inequality is use the sign-chart. To do this, I. set one side of the inequality to 0, II. factor the expression and draw the sign-chart, III. read off the answer from the sign chart. Draw the sign-chart, sample the points x = -2, 0, 5 (x – 4)(x + 1) + + + – – – – – – + + + + Set one side to 0, we get x 2 – 3x – 4 > 0; factor, we've (x – 4)(x + 1) > 0. The roots are x = -1, 4. Sign-Charts and Inequalities

63. Example E. Solve x 2 – 3x > 4 0 4 -1 The solutions the positive region, {x < -1} U {4 < x} -2 5 4 -1 The easiest way to solve a polynomial or rational inequality is use the sign-chart. To do this, I. set one side of the inequality to 0, II. factor the expression and draw the sign-chart, III. read off the answer from the sign chart. Draw the sign-chart, sample the points x = -2, 0, 5 (x – 4)(x + 1) + + + – – – – – – + + + + Set one side to 0, we get x 2 – 3x – 4 > 0; factor, we've (x – 4)(x + 1) > 0. The roots are x = -1, 4. Sign-Charts and Inequalities

64. Example E. Solve x 2 – 3x > 4 0 4 -1 The solutions the positive region, {x < -1} U {4 < x} -2 5 4 -1 Note: The empty dot means those numbers are excluded. The easiest way to solve a polynomial or rational inequality is use the sign-chart. To do this, I. set one side of the inequality to 0, II. factor the expression and draw the sign-chart, III. read off the answer from the sign chart. Draw the sign-chart, sample the points x = -2, 0, 5 (x – 4)(x + 1) + + + – – – – – – + + + + Set one side to 0, we get x 2 – 3x – 4 > 0; factor, we've (x – 4)(x + 1) > 0. The roots are x = -1, 4. Sign-Charts and Inequalities

65. Sign-Charts and Inequalities For a fractional inequality, we may not clear the denominator as in the case for equation.

66. Sign-Charts and Inequalities For a fractional inequality, we may not clear the denominator as in the case for equation. Instead we set one side of the inequality to be 0 and put the expression in factored form,

67. Sign-Charts and Inequalities For a fractional inequality, we may not clear the denominator as in the case for equation. Instead we set one side of the inequality to be 0 and put the expression in factored form, i.e. combine into one fraction,

68. Sign-Charts and Inequalities For a fractional inequality, we may not clear the denominator as in the case for equation. Instead we set one side of the inequality to be 0 and put the expression in factored form, i.e. combine into one fraction, then do the sign-chart of the expression.

69. Example E. Use the sign-chart to solve x – 2 4 < 2 Sign-Charts and Inequalities For a fractional inequality, we may not clear the denominator as in the case for equation. Instead we set one side of the inequality to be 0 and put the expression in factored form, i.e. combine into one fraction, then do the sign-chart of the expression.

70. Example E. Use the sign-chart to solve Set one side to 0, put the expression in the factored form x – 2 4 < 2 Sign-Charts and Inequalities x – 2 4 < – 2 0 For a fractional inequality, we may not clear the denominator as in the case for equation. Instead we set one side of the inequality to be 0 and put the expression in factored form, i.e. combine into one fraction, then do the sign-chart of the expression.

71. Example E. Use the sign-chart to solve Set one side to 0, put the expression in the factored form x – 2 4 < 2 Sign-Charts and Inequalities x – 2 4 < – 2 0 x – 2 4 < – 0 x – 2 2(x – 2) For a fractional inequality, we may not clear the denominator as in the case for equation. Instead we set one side of the inequality to be 0 and put the expression in factored form, i.e. combine into one fraction, then do the sign-chart of the expression.

72. Example E. Use the sign-chart to solve Set one side to 0, put the expression in the factored form x – 2 4 < 2 Sign-Charts and Inequalities x – 2 4 < – 2 0 x – 2 4 < – 0 x – 2 2(x – 2) x – 2 4 – 2(x – 2) < 0 For a fractional inequality, we may not clear the denominator as in the case for equation. Instead we set one side of the inequality to be 0 and put the expression in factored form, i.e. combine into one fraction, then do the sign-chart of the expression.

73. Example E. Use the sign-chart to solve Set one side to 0, put the expression in the factored form x – 2 4 < 2 Sign-Charts and Inequalities x – 2 4 < – 2 0 x – 2 4 < – 0 x – 2 2(x – 2) x – 2 4 – 2(x – 2) < 0 x – 2 8 – 2x < 0 For a fractional inequality, we may not clear the denominator as in the case for equation. Instead we set one side of the inequality to be 0 and put the expression in factored form, i.e. combine into one fraction, then do the sign-chart of the expression.

74. Hence the problem is transformed to Sign-Charts and Inequalities x – 2 8 – 2x < 0

75. Hence the problem is transformed to Sign-Charts and Inequalities x – 2 8 – 2x < 0 To solve it we do the sign-chart of . x – 2 8 – 2x

76. Hence the problem is transformed to Sign-Charts and Inequalities x – 2 8 – 2x < 0 To solve it we do the sign-chart of . x – 2 8 – 2x The roots from the numerator is x = 4 and the root from the denominator is 2,

77. Hence the problem is transformed to 4 Sign-Charts and Inequalities x – 2 8 – 2x < 0 To solve it we do the sign-chart of . x – 2 8 – 2x The roots from the numerator is x = 4 and the root from the denominator is 2, hence x – 2 8 – 2x 2 UDF

78. Hence the problem is transformed to 0 4 Sign-Charts and Inequalities x – 2 8 – 2x < 0 To solve it we do the sign-chart of . x – 2 8 – 2x The roots from the numerator is x = 4 and the root from the denominator is 2, hence Test x = 0, we get a ( + ) ( – ) segment. = – x – 2 8 – 2x 2 UDF

79. Hence the problem is transformed to 0 4 Sign-Charts and Inequalities x – 2 8 – 2x < 0 To solve it we do the sign-chart of . x – 2 8 – 2x The roots from the numerator is x = 4 and the root from the denominator is 2, hence Test x = 0, we get a ( + ) ( – ) segment. = – – – – – – – x – 2 8 – 2x 2 UDF

80. Hence the problem is transformed to 0 4 Sign-Charts and Inequalities x – 2 8 – 2x < 0 To solve it we do the sign-chart of . x – 2 8 – 2x The roots from the numerator is x = 4 and the root from the denominator is 2, hence Test x = 3, we get a ( + ) ( + ) = + segment. Test x = 0, we get a ( + ) ( – ) segment. = – 3 – – – – – – x – 2 8 – 2x 2 UDF

81. Hence the problem is transformed to 0 4 Sign-Charts and Inequalities x – 2 8 – 2x < 0 To solve it we do the sign-chart of . x – 2 8 – 2x The roots from the numerator is x = 4 and the root from the denominator is 2, hence Test x = 3, we get a ( + ) ( + ) = + segment. Test x = 0, we get a ( + ) ( – ) segment. = – 3 + + + – – – – – – x – 2 8 – 2x 2 UDF

82. Hence the problem is transformed to 0 4 Sign-Charts and Inequalities x – 2 8 – 2x < 0 To solve it we do the sign-chart of . x – 2 8 – 2x The roots from the numerator is x = 4 and the root from the denominator is 2, hence Test x = 3, we get a ( + ) ( + ) = + segment. Test x = 0, we get a ( + ) ( – ) segment. = – 3 5 Test x = 5, we get a ( + ) segment. ( – ) = – + + + – – – – – – x – 2 8 – 2x 2 UDF

83. Hence the problem is transformed to 0 4 Sign-Charts and Inequalities x – 2 8 – 2x < 0 To solve it we do the sign-chart of . x – 2 8 – 2x The roots from the numerator is x = 4 and the root from the denominator is 2, hence Test x = 3, we get a ( + ) ( + ) = + segment. Test x = 0, we get a ( + ) ( – ) segment. = – 3 5 Test x = 5, we get a ( + ) segment. ( – ) = – + + + – – – – – – – – – – – x – 2 8 – 2x 2 UDF

84. Hence the problem is transformed to 0 4 Sign-Charts and Inequalities x – 2 8 – 2x < 0 To solve it we do the sign-chart of . x – 2 8 – 2x The roots from the numerator is x = 4 and the root from the denominator is 2, hence Test x = 3, we get a ( + ) ( + ) = + segment. Test x = 0, we get a ( + ) ( – ) segment. = – 3 5 Test x = 5, we get a ( + ) segment. ( – ) = – + + + – – – – – – – – – – – x – 2 8 – 2x 2 UDF The answers are the negative portions as shown,

85. Hence the problem is transformed to 0 4 Sign-Charts and Inequalities x – 2 8 – 2x < 0 To solve it we do the sign-chart of . x – 2 8 – 2x The roots from the numerator is x = 4 and the root from the denominator is 2, hence Test x = 3, we get a ( + ) ( + ) = + segment. Test x = 0, we get a ( + ) ( – ) segment. = – 3 5 Test x = 5, we get a ( + ) segment. ( – ) = – + + + – – – – – – – – – – – x – 2 8 – 2x 2 UDF The answers are the negative portions as shown, or that {x < –2} U {4 ≤ x }.

86. Example F. Solve x – 2 2 < x – 1 3 Sign-Charts and Inequalities

87. Example F. Solve x – 2 2 < x – 1 3 Set the inequality to 0, x – 2 2 x – 1 3 < 0 Sign-Charts and Inequalities

88. Example F. Solve x – 2 2 < x – 1 3 Set the inequality to 0, x – 2 2 x – 1 3 < 0 Put the expression into factored form, Sign-Charts and Inequalities

89. Example F. Solve x – 2 2 < x – 1 3 Set the inequality to 0, x – 2 2 x – 1 3 < 0 Put the expression into factored form, x – 2 2 x – 1 3 = (x – 2)(x – 1) 2(x – 1) – 3(x – 2) Sign-Charts and Inequalities

90. Example F. Solve x – 2 2 < x – 1 3 Set the inequality to 0, x – 2 2 x – 1 3 < 0 Put the expression into factored form, x – 2 2 x – 1 3 = (x – 2)(x – 1) 2(x – 1) – 3(x – 2) = (x – 2)(x – 1) – x + 4 Sign-Charts and Inequalities

91. Example F. Solve x – 2 2 < x – 1 3 Set the inequality to 0, x – 2 2 x – 1 3 < 0 Put the expression into factored form, x – 2 2 x – 1 3 = (x – 2)(x – 1) 2(x – 1) – 3(x – 2) = (x – 2)(x – 1) – x + 4 Hence the inequality is (x – 2)(x – 1) – x + 4 < 0 Sign-Charts and Inequalities

92. Example F. Solve x – 2 2 < x – 1 3 Set the inequality to 0, x – 2 2 x – 1 3 < 0 Put the expression into factored form, x – 2 2 x – 1 3 = (x – 2)(x – 1) 2(x – 1) – 3(x – 2) = (x – 2)(x – 1) – x + 4 Hence the inequality is (x – 2)(x – 1) – x + 4 < 0 It has root at x = 4, and its undefined at x = 1, 2. Sign-Charts and Inequalities

93. Example F. Solve x – 2 2 < x – 1 3 Set the inequality to 0, x – 2 2 x – 1 3 < 0 Put the expression into factored form, x – 2 2 x – 1 3 = (x – 2)(x – 1) 2(x – 1) – 3(x – 2) = (x – 2)(x – 1) – x + 4 Hence the inequality is (x – 2)(x – 1) – x + 4 < 0 Draw the sign chart by sampling x = 0, 3/2, 3, 5 It has root at x = 4, and its undefined at x = 1, 2. Sign-Charts and Inequalities

94. Example F. Solve x – 2 2 < x – 1 3 Set the inequality to 0, x – 2 2 x – 1 3 < 0 Put the expression into factored form, x – 2 2 x – 1 3 = (x – 2)(x – 1) 2(x – 1) – 3(x – 2) = (x – 2)(x – 1) – x + 4 Hence the inequality is (x – 2)(x – 1) – x + 4 < 0 Draw the sign chart by sampling x = 0, 3/2, 3, 5 It has root at x = 4, and its undefined at x = 1, 2. 4 1 2 UDF UDF (x – 2)(x – 1) – x + 4 Sign-Charts and Inequalities

95. Example F. Solve x – 2 2 < x – 1 3 Set the inequality to 0, x – 2 2 x – 1 3 < 0 Put the expression into factored form, x – 2 2 x – 1 3 = (x – 2)(x – 1) 2(x – 1) – 3(x – 2) = (x – 2)(x – 1) – x + 4 Hence the inequality is (x – 2)(x – 1) – x + 4 < 0 Draw the sign chart by sampling x = 0, 3/2, 3, 5 It has root at x = 4, and its undefined at x = 1, 2. 4 1 0 5 2 3/2 3 UDF UDF (x – 2)(x – 1) – x + 4 Sign-Charts and Inequalities

96. Example F. Solve x – 2 2 < x – 1 3 Set the inequality to 0, x – 2 2 x – 1 3 < 0 Put the expression into factored form, x – 2 2 x – 1 3 = (x – 2)(x – 1) 2(x – 1) – 3(x – 2) = (x – 2)(x – 1) – x + 4 Hence the inequality is (x – 2)(x – 1) – x + 4 < 0 Draw the sign chart by sampling x = 0, 3/2, 3, 5 It has root at x = 4, and its undefined at x = 1, 2. 4 1 0 5 + + + 2 3/2 3 UDF UDF (x – 2)(x – 1) – x + 4 Sign-Charts and Inequalities

97. Example F. Solve x – 2 2 < x – 1 3 Set the inequality to 0, x – 2 2 x – 1 3 < 0 Put the expression into factored form, x – 2 2 x – 1 3 = (x – 2)(x – 1) 2(x – 1) – 3(x – 2) = (x – 2)(x – 1) – x + 4 Hence the inequality is (x – 2)(x – 1) – x + 4 < 0 Draw the sign chart by sampling x = 0, 3/2, 3, 5 It has root at x = 4, and its undefined at x = 1, 2. 4 1 0 5 + + + – – 2 3/2 3 UDF UDF (x – 2)(x – 1) – x + 4 Sign-Charts and Inequalities

98. Example F. Solve x – 2 2 < x – 1 3 Set the inequality to 0, x – 2 2 x – 1 3 < 0 Put the expression into factored form, x – 2 2 x – 1 3 = (x – 2)(x – 1) 2(x – 1) – 3(x – 2) = (x – 2)(x – 1) – x + 4 Hence the inequality is (x – 2)(x – 1) – x + 4 < 0 Draw the sign chart by sampling x = 0, 3/2, 3, 5 It has root at x = 4, and its undefined at x = 1, 2. 4 1 0 5 + + + – – + + + + 2 3/2 3 UDF UDF (x – 2)(x – 1) – x + 4 Sign-Charts and Inequalities

99. Example F. Solve x – 2 2 < x – 1 3 Set the inequality to 0, x – 2 2 x – 1 3 < 0 Put the expression into factored form, x – 2 2 x – 1 3 = (x – 2)(x – 1) 2(x – 1) – 3(x – 2) = (x – 2)(x – 1) – x + 4 Hence the inequality is (x – 2)(x – 1) – x + 4 < 0 Draw the sign chart by sampling x = 0, 3/2, 3, 5 It has root at x = 4, and its undefined at x = 1, 2. 4 1 0 5 + + + – – + + + + – – – – 2 3/2 3 UDF UDF (x – 2)(x – 1) – x + 4 Sign-Charts and Inequalities

100. Example F. Solve x – 2 2 < x – 1 3 Set the inequality to 0, x – 2 2 x – 1 3 < 0 Put the expression into factored form, x – 2 2 x – 1 3 = (x – 2)(x – 1) 2(x – 1) – 3(x – 2) = (x – 2)(x – 1) – x + 4 Hence the inequality is (x – 2)(x – 1) – x + 4 < 0 Draw the sign chart by sampling x = 0, 3/2, 3, 5 It has root at x = 4, and its undefined at x = 1, 2. 4 1 0 5 + + + – – + + + + – – – – 2 3/2 3 UDF UDF (x – 2)(x – 1) – x + 4 We want the shaded negative region, i.e. {1 < x < 2} U {4 < x}. Sign-Charts and Inequalities

101. Sign-Charts and Inequalities Exercise A. Draw the sign–charts of the following formulas. 1. (x – 2)(x + 3) 4. (2 – x)(x + 3) 5. –x(x + 3) 7. (x + 3) 2 9. x(2x – 1)(3 – x) 12. x 2 (2x – 1) 2 (3 – x) 13. x 2 (2x – 1) 2 (3 – x) 2 14. x 2 – 2x – 3 16. 1 – 15. x 4 – 2x 3 – 3x 2 (x – 2) (x + 3) 2. (2 – x) (x + 3) 3. – x (x + 3) 6. 8. –4(x + 3) 4 x (3 – x)(2x – 1) 10. 11. x 2 (2x – 1)(3 – x) 1 x + 3 17. 2 – 2 x – 2 18. 1 2x + 1 19. – 1 x + 3 – 1 2 x – 2 20. – 2 x – 4 1 x + 2

102. Sign-Charts and Inequalities Exercise B. Use the sign–charts method to solve the following inequalities. 21. (x – 2)(x + 3) > 0 23. (2 – x)(x + 3) ≥ 0 28. x 2 (2x – 1) 2 (3 – x) ≤ 0 29. x 2 – 2x < 3 33. 1 < 32. x 4 > 4x 2 (2 – x) (x + 3) 22. – x (x + 3) 24. 27. x 2 (2x – 1)(3 – x) ≥ 0 1 x 34. 2 2 x – 2 35. 1 x + 3 2 x – 2 36. > 2 x – 4 1 x + 2 25. x(x – 2)(x + 3) x (x – 2)(x + 3) 26. ≥ 0 30. x 2 + 2x > 8 30. x 3 – 2x 2 < 3x 31. 2x 3 < x 2 + 6x ≥ ≥ 0 ≤ 0 0 ≤ 37. 1 < 1 x 2