AN24 Answers: Chapter 4 4.3 Assess Your Understanding (page 226) 5. a 6. ( ) f c 7. b 8. F 9. 0 10. T 11. ( ) = = R f 2 8; no 13. ( ) = = − R f 2 82; no 15. ( ) = − = R f 4 0; yes 17. ( ) = − = R f 4 1; no 19. ( ) = = R f 1 2 0; yes 21. 7; 3 or 1 positive; 2 or 0 negative 23. 6; 2 or 0 positive; 2 or 0 negative 25. 3; 2 or 0 positive; 1 negative 27. 4; 2 or 0 positive; 2 or 0 negative 29. 5; 0 positive; 3 or 1 negative 31. 6; 1 positive; 1 negative 33. ± ± 1, 1 3 35. ± ± 1, 5 37. ± ± ± ± 1, 3, 1 9 , 1 3 39. ± ± ± ± ± ± ± ± 1, 3, 9, 1 2 , 1 3 , 1 6 , 3 2 , 9 2 41. ± ± ± ± ± ± ± ± 1, 2, 3, 4, 6, 12, 1 2 , 3 2 43. ±±±±± ± ±±±±±±± ± ±± 1, 2, 4, 5, 10, 20, 1 2 , 5 2 , 1 3 , 2 3 , 4 3 , 5 3 , 10 3 , 20 3 , 1 6 , 5 6 45. ( ) ( )( )( ) − − = + + − f x x x x 3, 1, 2; 3 1 2 47. 1 2 , 3, 3; ( ) ( )( ) = − − f x x x 2 1 3 2 49. ( ) ( ) ( ) − = + + + f x x x x 1 3 ; 3 1 1 2 51. ( )( ) ( ) ( ) − + = − − + − − f x x x x 4, 3 5, 3 5; 4 3 5 3 5 53. ( ) ( )( )( ) − − = + + − f x x x x 2, 1, 1, 1; 2 1 1 2 55. ( ) ( )( )( )( ) − − = − + + − f x x x x x 7 3 , 1, 2 7 , 2; 2 1 3 7 7 2 57. ( ) ( ) + − = − − ⎛ + ⎝ ⎜⎜⎜ ⎞ ⎠ ⎟⎟ ⎟ − ⎛ − ⎝ ⎜⎜⎜ ⎞ ⎠ ⎟⎟ ⎟ f x x x x 3, 5 17 2 , 5 17 2 ; 3 5 17 2 5 17 2 59. ( ) ( ) ( )( ) − = + − + f x x x x 1 2 , 1 2 ; 2 1 2 1 2 2 61. − 2 2 , 2 2 , 2; ( )( )( ) ( ) ( ) = − − + + f x x x x x 2 2 1 2 1 2 1 2 63. − − 5.9, 0.3, 3 65. −3.8, 4.5 67. −43.5, 1, 23 69. { } −1, 2 71. { } − + − − 2 3 , 1 2, 1 2 73. { } − 1 3 , 5, 5 75. { } − − 3, 2 77. { } − 1 3 79. { } 1 2 , 2, 5 81. −1 and 1 2 21 22 1 83. −11 and 11 200 211 2200 11 85. −9 and 3 100 29 250 3 87. ( ) ( ) = − = f f 0 1; 1 10; Zero: 0.21 89. ( ) ( ) − = − − = − f f 5 58; 4 2; Zero: 4.04 91. ( ) ( ) = − = f f 1.4 0.17536; 1.5 1.40625; Zero: 1.41 101. ( ) ( ) ( )( ) = − + + f x x x x 1 1 2 2 y 16 x 5 (21, 0) (22, 0) (21.59, 21.62) (1, 0) (0, 2) (2, 12) (20.16, 2.08) 107. − − − 8, 4, 7 3 109. = k 5 111. −7 113. If ( ) = − f x x c , n n then ( ) = − = f c c c 0, n n so −x c is a factor of f . 115. 5 117. 7 in. 119. All the potential rational zeros are integers, so r either is an integer or is not a rational zero (and is therefore irrational). 121. b 123. Define ( ) ( ) ( ) = − − = + − f x x x x x 1 1. 3 2 3 2 Since ( ) = − f 0 1 and ( ) = f 1 1, there is at least one point in the interval where the two functions have the same y-value, so there is at least one intersection point in the interval. 125. No; by the Rational Zeros Theorem, 1 3 is not a potential rational zero. 127. No; by the Rational Zeros Theorem, 2 3 is not a potential rational zero. 128. ( ) ( ) = − − + f x x 3 5 71 2 129. [ ) 3, 8 130. = − y x 2 5 3 5 131. No real solutions. The complex solutions are ± i 3 3 2 132. [ ] −3, 2 and [ )∞ 5, 133. ( ] −∞ −, 3 and [ ] 2, 5 134. −5 and −1 135. ( ) ( ) ( ) − − 5, 0 , 1, 0 , 0, 3 136. ( ) ( ) − − 3, 2 , 2, 6 ,and ( ) 5, 1 137. Absolute minimum: ( ) − = − f 3 2; no absolute maximum 4.4 Assess Your Understanding (page 234) 5. one 6. − i 3 4 7. T 8. F 9. +i 4 11. − − i i , 3 13. − − i i , 5 15. −i 17. − − + i i 4 9 , 7 2 19. ( ) = − + − + = f x x x x x a 14 77 200208; 1 4 3 2 21. ( ) = − + − + − = f x x x x x x a 4 7 8 6 4; 1 5 4 3 2 23. ( ) = − + − + = f x x x x x a 6 10 6 9; 1 4 3 2 25. − i3, 5 27. − i4, 2, 1 4 29. + − i 2 5 , 3, 6 31. − − i2 , 7, 7, 2 3 33. ( ) ( ) − − − + = − + + ⎛ ⎝ ⎜⎜⎜ ⎞ ⎠ ⎟⎟ ⎟ + − ⎛ ⎝ ⎜⎜⎜ ⎞ ⎠ ⎟⎟ ⎟ i i f x x x i x i 1, 1 2 3 2 , 1 2 3 2 ; 1 3 1 2 3 2 1 2 3 2 103. ( ) ( ) ( ) = − − ⎛ ⎝ ⎜⎜⎜ ⎞ ⎠ ⎟⎟ ⎟ + ⎛ ⎝ ⎜⎜⎜ ⎞ ⎠ ⎟⎟ ⎟ + f x x x x x 2 2 2 2 2 2 2 1 2 y 4 x 5 (21, 29) (0, 2) (1.61, 210.09) (2, 0) , 0 2 2 Ï2 , 0 2 Ï2 (20.30, 2.23) 105. (a) 2350 26 6 400 (b) ( ) ( ) ( ) − − 2, 298 , 2, 214 , 4, 134 (c) The solutions are the same as the x-coordinates of the turning points. 93. ( ) ( )( )( ) = + − + f x x x x 1 2 3 y 30 x 5 (21, 0) (22.12, 4.06) (23, 0) (24, 218) (0.79, 28.21) (0, 26) (2, 0) (3, 24) 95. ( ) ( ) ( ) = − + f x x x 2 1 1 2 y 5 x 5 (0, 21) (1, 2) , 0 1 2 97. ( ) ( ) ( )( ) = − + + f x x x x 1 1 1 2 y 2.5 x 2.5 (21, 0) (0, 22) (1, 0) 99. ( ) ( ) ( )( ) = − + + f x x x x 2 1 2 1 2 2 y 2 x 2.5 (0, 22) , 0 1 2 , 0 2 1 2
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