Chapter Polynomial and Rational Functions Section.7 Nonlinear Inequalities You should be able to solve inequalities. Find the critical number.. Values that make the epression zero. Values that make the epression undefined Test one value in each test interval on the real number line resulting from the critical numbers. (c) Determine the solution intervals. Vocabulary Check. critical; test intervals. zeros; undefined values. P R C. <? <? (c) <? (d) 5 5 <? 6 < < No, is not a Yes, is a Yes, is a No, 5 is not a. 5 (c) (d) 5 5???? 8 Yes, 5 is a No, is not a 6? 8 Yes, is a Yes, 9? is a. 5 (c) 9 (d) 9 5 5? 7 Yes, 5 is a? 6 is undefined. No, is not a 9 9? 5 7 No, 9 is not a 9 9? Yes, 9 is a
Section.7 Nonlinear Inequalities. < (c) (d) <? <? <? <? 8 < No, is not a 5 < Yes, is a < Yes, is a 7 < No, is not a 5. 6 6., 9 5 7. 9 5 9 5 5 9 The critical numbers are and 5 9. 5 5 5 7 5 7 5 7 5 7, 5 8. The critical numbers are,,,. 9. 9 9 ±,,,,, Test: Is? Interval,,, -Value Value of 9 6 9 7 9 9 6 9 7 Conclusion Solution set:,. < 6 6 6 6 6, 6, 6 6 6 6, 6 6 6 6, 6 6 Solution interval: 6, 6 8 6 6 8
Chapter Polynomial and Rational Functions. < 5. < 5 6 8 7, 7,,, 7, 7,,,, Test: Is 7?, Interval -Value Value of Conclusion 7, 7 9 Solution intervals:,, 5 7, 7, 5 Solution set: 7, 7 8 6. 9. 5 5 5,, 5, 5,,, Test: Is 5? Interval -Value Value of 5 Conclusion, 5 5,, 6 7 7 5 5 7 7 Solution set:, 5, 7, 7 Solution interval:, 7 6 9 < 6 6 7 6, 7, 7 7 7, 7 7 8 6 5 5. < 6 6. 6,,,,,, Test: Is? Interval -Value Value of Conclusion,,, 6 6 6 6 6 Solution set:, >,,,, Solution intervals:,,
Section.7 Nonlinear Inequalities 5 7. 8.,,,,,, Test: Is? Interval -Value Value of Conclusion,,, 5 5 5 5 Solution set:, ± 6 Solution intervals: 5 + 5 6 8 ± 5 5, 5, 5 5, 5 5,, 5 5, 9. 8 5 8 5 Complete the square. 8 6 5 6 ± ± ±,,,,, Test: Is 8 5? Interval,,, -Value Value of 8 5 8 5 5 5 5 6 5 5 Conclusion + Solution set: <, 8 6. 6 5 6 5 6 ± 6 5, Solution interval:, 6 ± 56 9, 9 9 9, 6 ± 9 9 6 5 9, 9 6 5 9, 6 5 9 ± 9 9 + 5
6 Chapter Polynomial and Rational Functions. ±,,,,,,,, Test: Is? Interval -Value Value of Conclusion,,,., Solution set:,, 5 5 5 5 8,, 8, 8, 8 Solution interval:,. 9 9 8 9 9, ±,,,,,,, Test: Is? Interval,,,, -Value.5 Value of 67 8.55.5.5.75 7 Conclusion Solution set:,, 5. 8 6 6 8 5,,, 8 5, 8 5, 8 5 8 6, 8 5 Solution interval:,,,
Section.7 Nonlinear Inequalities 7 5. 6. Critical number: Test: Is? Interval -Value Value of Conclusion,,,,, 8 The critical numbers are imaginary: ± i So the set of real numbers is the solution set. Solution set: 7. 6 8.,,,,,, Test: Is? By testing an -value in each test interval in the inequality, we see that the solution set is:,,,,,, Solution interval:, 9.., ±,,,,,,, Test: Is? By testing an -value in each test interval in the inequality, we see that the solution set is:,,,,,, Solution intervals:,,..,,,,,, ) Test: Is? By testing an -value in each test interval in the inequality, we see that the solution set is:,,,,, Solution intervals:,, or,
8 Chapter Polynomial and Rational Functions. y y when or. 6 y when. 5 7. y y y 7 7 ± ± 8 ± 6 y 7 when, 6. y when. 5. y 8 8 y when, <. y 6 when. 8 6. y 6 6 y y 6 8 6 6 6 6 6 6 6 6 5 y when <,. y 6 when, 5 <. 7. 8., ±,,,,,,, Test: Is? By testing an -value in each test interval in the inequality, we see that the solution set is:,,, Solution interval:,,,,,
Section.7 Nonlinear Inequalities 9 9. 6. 6,,,,,, Test: Is? By testing an -value in each test interval in the inequality, we see that the solution set is:,, 5 6,,,, Solution interval:, 6 6 6. 5 5 >. 5 5 5 5 5 5 5 5, 5, 5, 5, 5, 5, 5 Test: Is 5? By testing an -value in each test interval in the inequality, we see that the solution set is: 5, 5 5 7 < 5 7,,,, 5 6 9 5 8 Solution intervals:,,
Chapter Polynomial and Rational Functions. 5 >. 5 6 > 5 5 6 6 5 5 8 6 7 Test: Is 5? By testing an -value in each test interval in the inequality, we see that the solution set is: 5,, 5 7 7 5, 5,, 5, 5,,,,, Solution intervals:, 6, 5 5 5 6,, 6,,, 6 6, 8 6 8 6 8 6 8 6 5. 9 9 5 9,, 6,,,,, 6, 6, 5 Test: Is? By testing an -value in each test interval in the inequality, we see that the solution set is:, 6, 6.,,, Solution intervals:,,, 6 8
Section.7 Nonlinear Inequalities 7. 9 8.,, ±,,,,,,,,, Test: Is? By testing an -value in each test interval in the inequality, we see that the solution set is:,, 6,,,,,, Solution intervals:,, 9. 5 Test: Is? By testing an -value in each test interval in the inequality, we see that the solution set is:,,, 5 5,,,,,,,,,,, ± 5 < 5. 5 7 6, Solution intervals:,,,, 6,,,, 6 6, 6 6 6 6 6,,, 6 6 6 5. y 8 y when <. y 6 when <. 6
Chapter Polynomial and Rational Functions 5. y y y 8 8 5 5 y when <. 8 6 6 6 y 8 when <. 5. y 6 6 y when or 6 This can also be epressed as y for all real numbers... This can also be epressed as < <. 5. y 5 5 y when. y y 5 5 6 6 y when <. 55. 56. Test: Is? ±,,,,, By testing an -value in each test interval in the inequality, we see that the domain set is:,,,,, Domain:,, 57. 7 58.,,,,,, Test: Is? By testing an -value in each test interval in the inequality, we see that the domain set is:,, 9 9,, 9, 9, 9 Domain:,
Section.7 Nonlinear Inequalities 59. 5 6. 5 7, 5, 7, 5, 5,,, 7, 7, Test: Is 5 7? By testing an -value in each test interval in the inequality, we see that the domain set is: 5, 7, 9,,,, Domain:,,,, 6.. 5.6. 6...9..5 ±.5,.5,.5,.5,.5, By testing an -value in each test interval in the inequality, we see that the solution set is:.5,.5..78 >...66 ±.,.,.,.,., Solution set:.,. 6..5.5.6 6. The zeros are.5 ±.5.5.6..5., 5.,.,., 5., 5., By testing an -value in each test interval in the inequality, we see that the solution set is:., 5...8. < 5...8..,. Solution set:.,.,.,.,.,., 65. >. 66.. 5... 5... 5.. 5. 7.8 8.68. 5..9,.6,.6,.6,.9,.9, By testing an -value in each test interval in the inequality, we see that the solution set is:.6,.9..7 > 5.8 5.8..7..7.6 7.98..7.9,.,.9.9,.., Solution interval:.9,..6 7.98..7.6 7.98..7.6 7.98..7
Chapter Polynomial and Rational Functions 67. s 6t v t s 6t 6t 68. 6t 6t 6tt t, t It will be back on the ground in seconds. 6t 6t > 8 6t 6t 8 6t t t t t t 6 < t < 6 seconds s 6t v t s 6t 8t 6t 8t 6tt 8 It will be back on the ground in 8 seconds. 6t 8t 8,,,,,, Solution set: 6t t t 8 t 8 6t 8t < 8 seconds t < seconds and seconds < t 8 seconds 69. L W W 5 L 7. LW 5 L5 L 5 L 5L 5 By the Quadratic Formula we have: L 5 ± 55 Test: Is L 5L 5? Solution set: 5 55 L 5 55.8 meters L 6. meters L W W L LW 8 L L 8 L L 8 By the Quadratic Formula we have: L ± Test: Is L L 8? Solution set: L 5.97 feet L 7. feet 7. R 75.5 and C 5, P R C 75.5 5,.5 5 5, P 75,.5 5 5, 75,.5 5,,,, 5, (These were obtained by using the Quadratic Formula.),,,,, 5,, 5,, By testing -values in each test interval in the inequality, we see that the solution set is,, 5, or, 5,. The price per unit is 7. What is the price per unit? When 9,: R $,88,,88, 9, When,: R $,,,,, $ per unit $ per unit Solution interval: $. p $. p R 75.5. For,, p $55. For 5,, p $5. Therefore, for, 5,, $5. p $55..
Section.7 Nonlinear Inequalities 5 7. C.t.6t 5.5t 9., t 8 t C 7.5 6 7.6 8 7.9 7.6 76.8 79.6 (c) C 75 when t.. C will be greater than 75% when t, which corresponds to. (d) C will be between 85% and % t C when t is between 7 and. These 6 8. values correspond to the years 7 to. 7 85. 8 87.8 9 9.5 9.5 96.8.. (e) 85 C when 6.8 t.89 or 7 t. (f) The model is a third-degree polynomial and as t, C. 7. Maimum safe load d 6 8 Load.9 559.9, 6,78,79 L 5,, 5,, 5, d 6 8 Depth of the beam 68.5d 7. 7. 68.5d.67 d.8 d The minimum depth is.8 inches. 75. R R R R RR R R R R R R Since R, we have R R R R R R. Since R, the only critical number is R. The inequality is satisfied when R ohms. 76. (c) N.t 9.6t 7 t 5 So the number of master s degrees earned by women eceeded, in 995. N.t 9.6t 7 t 6. So the number of master s degrees earned by women will eceed, in 6. and (d) Master's degrees earned (in thousands) 8 6 N 6 8 Year ( 99) N = N = t
6 Chapter Polynomial and Rational Functions 77. True The test intervals are,,,,,, and,. 78. True The y-values are greater than zero for all values of. 79. b 8. To have at least one real solution, b 6. This occurs when b or b. This can be written as,,. b To have at least one real solution, b b 6. This inequality is true for all real values of b. Thus, the interval for b such that the equation has at least one real solution is,. 8. b 8. b 5 To have at least one real solution, b. To have at least one real solution, b b b b ± ±,,,,, Test: Is b? Solution set:,, b 5 b. This occurs when b or b. Thus, the interval for b such that the equation has at least one real solution is,,. 8. If a and c, then b can be any real number. If a and c, then for b ac to be greater than or equal to zero, b is restricted to b < ac or b > ac. The center of the interval for b in Eercises 79 8 is. 8. a, b + + + + + a b (c) The real zeros of the polynomial 85. 5 5 86. 6 7 87. ) 88. 5 7 9 89. Area lengthwidth 9. Area baseheight bb b b