1. CHAPTER 1 PRELIMINARIES 1.1 REAL NUMBERS AND THE REAL LINE 1. Executing long division," 92. Executing long division," 11 0.1,2 9 0.2, 0.09,2 113 9 0.3, 0.18,3 118 9 0.8, 0.27,9 119 9 0.9 0.81,11 11 0.993. NT = necessarily true, NNT = Not necessarily true. Given: 2 < x < 6. a) NNT. 5 is a counter example. b) NT. 2 < x < 6 2 c 2 < x c 2 < 6 c 2 0 < x c 2 < 2. c) NT. 2 < x < 6 2/2 < x/2 < 6/2 1 < x < 3. d) NT. 2 < x < 6 1/2 > 1/x > 1/6 1/6 < 1/x < 1/2. e) NT. 2 < x < 6 1/2 > 1/x > 1/6 1/6 < 1/x < 1/2 6(1/6) < 6(1/x) < 6(1/2) 1 < 6/x < 3. f) NT. 2 < x < 6 x < 6 (x c 4) < 2 and 2 < x < 6 x > 2 cx < c2 cx + 4 < 2 c(x c 4) < 2. The pair of inequalities (x c 4) < 2 and c(x c 4) < 2 | x c 4 | < 2. g) NT. 2 < x < 6 c2 > cx > c6 c6 < cx < c2. But c2 < 2. So c6 < cx < c2 < 2 or c6 < cx < 2. h) NT. 2 < x < 6 c1(2) > c1(x) < c1(6) c6 < cx < c2 4. NT = necessarily true, NNT = Not necessarily true. Given: c1 < y c 5 < 1. a) NT. c1 < y c 5 < 1 c1 + 5 < y c 5 + 5 < 1 + 5 4 < y < 6. b) NNT. y = 5 is a counter example. (Actually, never true given that 4 y 6) c) NT. From a), c1 < y c 5 < 1, 4 < y < 6 y > 4. d) NT. From a), c1 < y c 5 < 1, 4 < y < 6 y < 6. e) NT. c1 < y c 5 < 1 c1 + 1 < y c 5 + 1 < 1 + 1 0 < y c 4 < 2. f) NT. c1 < y c 5 < 1 (1/2)(c1 + 5) < (1/2)(y c 5 + 5) < (1/2)(1 + 5) 2 < y/2 < 3. g) NT. From a), 4 < y < 6 1/4 > 1/y > 1/6 1/6 < 1/y < 1/4. h) NT. c1 < y c 5 < 1 y c 5 > c1 y > 4 cy < c4 cy + 5 < 1 c(y c 5) < 1. Also, c1 < y c 5 < 1 y c 5 < 1. The pair of inequalities c(y c 5) < 1 and (y c 5) < 1 | y c 5 | < 1. 5. c2x 4 x c2 6. 8 c 3x 5 c3x c3 x 1 7. 5x c $ ( c 3x 8x 10 x qqqqqqqqp x 1 5 48. 3(2 c x) 2(3 b x) 6 c 3x 6 b 2x 0 5x 0 x 9. 2x c" # 10." 56 cx 4 7x bc10 63xc4 27 6 c"c # x or c" 37 6qqqqqqqqp x 0 5xx 12 c 2x 12x c 16 28 14x 2 xqqqqqqqqq x 2Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

2. 2 11.Chapter 1 Preliminaries 4 5" 3(x c 2) (x c 6) 12(x c 2) 5(x c 6) 12x c 24 5x c 30 7x c6 or x c 6 7 12. c xb5 212b3x 4 c(4x b 20) 24 b 6x c44 10x c 22 x 5qqqqqqqqq x c22/513. y 3 or y c3 14. y c 3 7 or y c 3 c7 y 10 or y c4 15. 2t b 5 4 or 2t b & c4 2t c1 or 2t c9 t c " or t c 9 # # 16. 1 c t 1 or 1 c t c1 ct ! or ct c2 t 0 or t 2 17. 8 c 3s 18.s #9 29 or 8 c 3s c # c3s c 7 or c3s c 25 s # #c 1 1 ors #c 1 c1 s # 2 ors #7 6or s 25 6 ! s 4 or s 019. c2 x 2; solution interval (c2 2) 20. c2 x 2; solution interval [c2 2]qqqqqqqqp x c2 221. c3 t c 1 3 c2 t 4; solution interval [c2 4] 22. c1 t b 2 1 c3 t c1; solution interval (c3 c1)qqqqqqqqp t c3 c123. c% 3y c 7 4 3 3y 11 1 y solution interval 111 3;11 324. c1 2y b 5 " c6 2y c4 c3 y c2; solution interval (c3 c2) 25. c1 z 5c11 0z 5qqqqqqqqp y c3 c2 2 0 z 10;solution interval [0 10] 26. c2 c 1 2 c1 solution interval 0 and f(x) = 2x + 1. Suppose that | xc1 | < $ . Then | xc1 | < $ 2| xc1 | < 2$ | 2x c # | < 2$ | (2x + 1) c 3 | < 2$ | f(x) c f(1) | < 2$ 50. Let % > 0 be any positive number and f(x) = 2x + 3. Suppose that | x c 0 | < % /2. Then 2| x c 0 | < % and | 2x + 3 c3 | < %. But f(x) = 2x + 3 and f(0) = 3. Thus | f(x) c f(0) | < %. 51. Consider: i) a > 0; ii) a < 0; iii) a = 0. i) For a > 0, | a | a by definition. Now, a > 0 ca < 0. Let ca = b. By definition, | b | cb. Since b = ca, | ca | c(ca) a and | a | | ca | a. ii) For a < 0, | a | ca. Now, a < 0 ca > 0. Let ca b. By definition, | b | b and thus |ca| ca. So again | a | |ca|. iii) By definition | 0 | 0 and since c0 0, | c0 | 0. Thus, by i), ii), and iii) | a | | ca | for any real number. Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 5. Section 1.2 Lines, Circles and Parabolas Prove | x | > 0 x > a or x < ca for any positive number, a. For x 0, | x | x. | x | > a x > a. For x < 0, | x | cx. | x | > a cx > a x < ca. ii) Prove x > a or x < ca | x | > 0 for any positive number, a. a > 0 and x > a | x | x. So x > a | x | > a. For a > 0, ca < 0 and x < ca x < 0 | x | cx. So x < ca cx > a | x | > a.52. i)1=1 |1|=1 bb)lal lbl a" b a " b" bl bl lbl a b l bl "" b lbl lbl b b "53. a) b b b b" b " blal lbl54. Prove Sn kan k kakn for any real number a and any positive integer n. ka" k kak " a, so S" is true. Now, assume that Sk ak kak k is true form some positive integer 5 . Since ka" k kak " and ak kak k , we have akb" ak a" ak ka" k kak k kak " kak k+" . Thus, Skb" akb" kak k+" is also true. Thus by the Principle of Mathematical Induction, Sn l an l l a ln is true for all n positive integers. 1.2 LINES, CIRCLES, AND PARABOLAS 1. ?x c1 c (c3) 2, ?y c2 c 2 c4; d (?x)# b (?y)# 4 b 16 25 2. ?x c$ c (c1) c2, ?y 2 c (c2) 4; d (c2)# b 4# 25 3. ?x c8.1 c (c3.2) c4.9, ?y c2 c (c2) 0; d (c4.9)# b 0# 4.9 #4. ?x 0 c 2 c2, ?y 1.5 c 4 c2.5; d c2 b (c2.5)# 8.25 5. Circle with center (! !) and radius 1.6. Circle with center (! !) and radius 2.7. Disk (i.e., circle together with its interior points) with center (! !) and radius 3. 8. The origin (a single point). 9. m ?y ?xc1 c 2 c2 c (c1)3perpendicular slope c " 310. m ?y ?xc# c " 2 c (c2)c3 4perpendicular slope 4 3Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley5 6. 6Chapter 1 Preliminaries11. m ?y ?x3c3 c1 c 2012. m 14. (a) x 2(b) y c# c 0 c# c (c#); no slope15. (a) x 016. (a) x c1(b) y c2(b) y c1.34 3perpendicular slope 0perpendicular slope does not exist13. (a) x c1?y ?x(b) y 017. P(c1 1), m c1 y c 1 c1ax c (c1)b y cx 18. P(2 c3), m " # y c (c3) 19. P(3 4), Q(c2 5) m ?y ?x20. P(c8 0), Q(c1 3) m ?y ?x" #(x c 2) y 5c4 c2 c 3" #xc4 c " y c 4 c " (x c 3) y c " x b 5 5 53c0 c1 c (c8)3 7 yc03 7ax c (c8)b y 3 723 5xb21. m c 5 , b 6 y c 5 x b 6 4 422. m " , b c3 y #" #23. m 0, P(c12 c9) y c924. No slope, P " % x 324 7xc3 " 325. a c1, b 4 (0 4) and (c" 0) are on the line m ?y ?x0c4 c1 c 0 4 y 4x b 426. a 2, b c6 (2 0) and (! c6) are on the line m ?y ?xc6 c 0 0c2 3 y 3x c 627. P(5 c1), L: 2x b 5y 15 mL c 2 parallel line is y c (c1) c 2 (x c 5) y c 2 x b 1 5 5 5 28. P c2 2 , L: 2x b 5y 3 mL c 52 parallel line is y c 2 c 52 x c c2 y c 52 x b8 529. P(4 10), L: 6x c 3y 5 mL 2 m c " perpendicular line is y c 10 c " (x c 4) y c " x b 12 # # # 30. P(! 1), L: 8x c 13y 13 mL 8 13 m c 13 perpendicular line is y c 13 x b 1 8 8Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 7. Section 1.2 Lines, Circles and Parabolas 31. x-intercept 4, y-intercept 332. x-intercept c4, y-intercept c233. x-intercept 3, y-intercept c234. x-intercept c2, y-intercept 3C Band Bx c Ay C# y B AxcC A#"35. Ax b By C" y c A x b BB . Since c A A c1 is the Bproduct of the slopes, the lines are perpendicular. "slopecA, BC Band Ax b By C# y c A x b BC B#36. Ax b By C" y c A x b B. Since the lines have the samethey are parallel.37. New position axold b ?x yold b ?yb (c# b & 3 b (c6)) ($ c3). 38. New position axold b ?x yold b ?yb (6 b (c6) 0 b 0) (0 0). 39. ?x 5, ?y 6, B(3 c3). Let A (x y). Then ?x x# c x" 5 3 c x x c2 and ?y y# c y" 6 c3 c y y c9. Therefore, A (c# c9). 40. ?x " c " !, ?y ! c ! !Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley7 8. 8Chapter 1 Preliminaries41. C(! 2), a 2 x# b (y c 2)# 442. C(c$ 0), a 3 (x b 3)# b y# 943. C(c1 5), a 10 (x b 1)# b (y c 5)# 1044. C(" "), a 2 (x c 1)# b (y c 1)# 2 x 0 (0 c 1)# b (y c 1)# 2 (y c 1)# 1 y c 1 1 y 0 or y 2. Similarly, y 0 x 0 or x 2#45. C c3 c2 , a 2 x b 3 b (y b 2)# 4, #x 0 0 b 3 b (y b 2)# 4 (y b 2)# 1 y b 2 1 y c1 or y c3. Also, y 0 ## x b 3 b (0 b 2)# 4 x b 3 0 x c 3 # 46. C 3 " , a 5 (x c 3)# b y c " 25, so # # # x 0 (0 c 3)# b y c " 25 # # y c " 16 y c #" # 4 y9 ## or y c 7 . Also, y 0 (x c 3)# b 0 c " 25 # # (x c 3)# x399 4 311 # xc3 311 #Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 9. Section 1.2 Lines, Circles and Parabolas 47. x# b y# b 4x c 4y b % 0 x# b %B b y# c 4y c4 x# b 4x b 4 b y# c 4y b 4 4 (x b 2)# b (y c 2)# 4 C (c2 2), a 2.48. x# b y# c 8x b 4y b 16 0 x# c 8x b y# b 4y c16 x# c 8x b 16 b y# b 4y b 4 4 (x c 4)# b (y b 2)# 4 C (% c2), a 2.49. x# b y# c 3y c 4 0 x# b y# c 3y 4 x# b y# c 3y b 9 25 4 4 # x# b y c 3 #25 4 C 0 3 , #a 5. #50. x# b y# c 4x c #9 4 #0 x c 4x b y 9 4 # x# c 4x b 4 b y (x c 2)# b y# 25 425 4 C (2 0), a 5 . #51. x# b y# c 4x b 4y 0 x# c 4x b y# b 4y 0 x# c 4x b 4 b y# b 4y b 4 8 (x c 2)# b (y b 2)# 8 C(2 c2), a 8.Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley9 10. 10Chapter 1 Preliminaries52. x# b y# b 2x 3 x# b 2x b 1 b y# 4 (x b 1)# b y# 4 C (c1 0), a 2.c2 53. x c #ba c 2(1) 1 y (1)# c 2(1) c 3 c4 V (" c4). If x 0 then y c3. Also, y 0 x# c 2x c 3 0 (x c 3)(x b 1) 0 x 3 or x c1. Axis of parabola is x 1.4 54. x c #ba c 2(1) c2 y (c2)# b 4(c2) b 3 c1 V (c2 c1). If x 0 then y 3. Also, y 0 x# b 4x b 3 0 (x b 1)(x b 3) 0 x c1 or x c3. Axis of parabola is x c2.4 55. x c #ba c 2(c1) 2 y c(2)# b 4(2) 4 V (2 4). If x 0 then y 0. Also, y 0 cx# b 4x 0 cx(x c 4) 0 x 4 or x 0. Axis of parabola is x 2.4 56. x c #ba c 2(c1) 2 y c(2)# b 4(2) c 5 c1 V (2 c1). If x 0 then y c5. Also, y 0 cx# b 4x c 5 0 x# c 4x b 5 0 x 4 c 4 # no x intercepts. Axis of parabola is x 2.Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 11. Section 1.2 Lines, Circles and Parabolas 57. x c #ba c 2(c61) c3 c y c(c3)# c 6(c3) c 5 4 V (c3 %). If x 0 then y c5. Also, y 0 cx# c 6x c 5 0 (x b 5)(x b 1) 0 x c5 or x c1. Axis of parabola is x c3.c1 58. x c #ba c 2(2) " 4#" y 2 " c 4 b 3 23 4 8 V " 23 . If x 0 then y 3. 4 8Also, y 0 2x# c x b 3 0 x1c23 4 no x intercepts.Axis of parabola is x " . 41 59. x c #ba c 2(1/2) c1 " #(c1)# b (c1) b 4 7 2 V c" 7 . If x 0 then y 4. 2 yAlso, y 0 xc1 c 7 1" #x# b x b 4 0 no x intercepts.Axis of parabola is x c1. 60. x c #ba c 2(c21/4) 4 y c " (4)# b 2(4) b 4 8 4 V (4 8) . If x 0 then y 4. Also, y 0 c " x# b 2x b 4 0 4 xc2 8 c1/2 4 42.Axis of parabola is x 4.61. The points that lie outside the circle with center (! 0) and radius 7. 62. The points that lie inside the circle with center (! 0) and radius 5. 63. The points that lie on or inside the circle with center (" 0) and radius 2. 64. The points lying on or outside the circle with center (! 2) and radius 2. 65. The points lying outside the circle with center (! 0) and radius 1, but inside the circle with center (! 0), and radius 2 (i.e., a washer).Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley11 12. 12Chapter 1 Preliminaries66. The points on or inside the circle centered at (! !) with radius 2 and on or inside the circle centered at (c2 0) with radius 2.67. x# b y# b 6y 0 x# b (y b 3)# 9. The interior points of the circle centered at (! c3) with radius 3, but above the line y c3.68. x# b y# c 4x b 2y 4 (x c 2)# b (y b 1)# 9. The points exterior to the circle centered at (2 c1) with radius 3 and to the right of the line x 2.69. (x b 2)# b (y c 1)# 670. (x b 4)# b (y c 2)# 1671. x# b y# 2, x 172. x# b y# 4, (x c 1)# b (y c 3)# 1073. x# b y# 1 and y 2x 1 x# b 4x# 5x# x " 5and y 2 5 " 2 or x c 5 and y c 5 ." 2 " 2 Thus, A 5 5 , B c 5 c 5 are thepoints of intersection.Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 13. Section 1.2 Lines, Circles, and Parabolas 74. x b y 1 and (x c 1)# b y# 1 1 (cy)# b y# 2y# y " 2and x " c" y c 2 and x 1 bA " c" 2" 2 " 2 ." 2 and B 1 borThus," 2" c 2 are intersection points.75. y c x 1 and y x# x# c x 1 1 5 . # 1 b 5 3 b 5 If x # , then y x b 1 # . If x 1c# 5 , then y x b 1 3c# 5 . Thus, A 1b# 5 3b# 5 and B 1c# 5 3c# 5 x# c x c 1 0 x are the intersection points.76. y cx and C c(x c 1)# (x c 1)# x 3 5 . # 5 c3 3 c 5 x # , then y cx # . If x 3b# 5 , then y cx c 3b# 5 . Thus, A 3c# 5 5c3 and B 3b# 5 # x# c 3x b " 0 x If c 3b# 5 are the intersection points.77. y 2x# c 1 cx# 3x# 1 " " " " x 3 and y c 3 or x c 3 and y c 3 . " " " " Thus, A 3 c 3 and B c 3 c 3 are theintersection points.Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley13 14. 14Chapter 1 Preliminaries78. y x 4#3x 4# (x c 1)# 0 c 2x b 1# 0 3x c 8x b 4 (3x c 2)(x c 2) x 4#yx 4# x 2 and y 1, or x " 9 . Thus, A(2 1) and2 3 and 2 " B 3 9are the intersection points.79. x# b y# 1 (x c 1)# b y# x# (x c 1)# x# c 2x b 1 0 c2x b 1 x " . Hence # y# " c x # A " #3 # and3 4or y 3 # B " c #3 #. Thus,are theintersection points.80. x# b y# 1 x# b y y# y y(y c 1) 0 y 0 or y 1. If y 1, then x# " c y# 0 or x 0. If y 0, then x# 1 c y# 1 or x 1. Thus, A(0 1), B(" 0), and C(c1 0) are the intersection points.81. (a) A (69 0 in), B (68 .4 in) m (b) A (68 .4 in), B (10 4 in) m (c) A (10 4 in), B (5 4.6 in) m 82. The time rate of heat transfer across a material, to the temperature gradient across the material, k=cA? ? >? U?-kA ?X ?BB Xof the material.?U ?>68 c 69 .4 c 0 c2.5/in. 10 c 68 4 c .4 c16.1/in. 5 c 10 4.6 c 4 c8.3/in. ?U ?> , is directly ?X ?B (the slopes. Note that?U ?>proportional to the cross-sectional area, A, of the material, from the previous problem), and to a constant characteristicand?X ?Bare of opposite sign because heat flow is toward lowertemperature. So a small value of k corresponds to low heat flow through the material and thus the material is a good insulator.Since all three materials have the same cross section and the heat flow across each is the same (temperatures are not changing), we may define another constant, K, characteristics of the material: K c " Using the values of ?X from ?B B? X?the prevous problem, fiberglass has the smallest K at 0.06 and thus is the best insulator. Likewise, the wallboard is the poorest insulator, with K 0.4. 83. p kd b 1 and p 10.94 at d 100 k 10.94c" 100 0.0994. Then p 0.0994d b 1 is the diver'spressure equation so that d 50 p (0.0994)(50) b 1 5.97 atmospheres. 84. The line of incidence passes through (! 1) and (" 0) The line of reflection passes through (" 0) and (# ") 0 m 1c1 1 y c 0 1(x c 1) y x c 1 is the line of reflection. #cCopyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 15. Section 1.2 Lines, Circles, and Parabolas 85. C 5 9(F c 32) and C F F 86. m 37.1 10014 ?x ?x 14 .371 .5 9Fc160 94 915F c 160 or F c40 gives the same numerical reading. 9#14 Therefore, distance between first and last rows is (14)# b .371 40.25 ft.87. length AB (5 c 1)# b (5 c 2)# 16 b 9 5 length AC (4 c 1)# b (c# c #)# 9 b 16 5 length BC (4 c 5)# b (c# c 5)# 1 b 49 50 52 5 #88. length AB (1 c 0)# b 3 c 0 1 b 3 2 length AC (2 c 0)# b (0 c 0)# 4 b 0 2 #length BC (2 c 1)# b 0 c 3 1 b 3 2 89. Length AB (?x)# b (?y)# 1# b 4# 17 and length BC (?x)# b (?y)# 4# b 1# 17. 4 Also, slope AB c1 and slope BC " , so AB BC. Thus, the points are vertices of a square. The coordinate 4 increments from the fourth vertex D(x y) to A must equal the increments from C to B 2 c x ?x 4 and c1 c y ?y " x c2 and y c2. Thus D(c# c2) is the fourth vertex.90. Let A (x 2) and C (9 y) B (x y). Then 9 c x kADk and 2 c y kDCk 2(9 c x) b 2(2 c y) 56 and 9 c x 3(2 c y) 2(3(2 c y)) b 2(2 c y) 56 y c5 9 c x 3(2 c (c5)) x c12. Therefore, A (c12 2), C (9 c5), and B (c12 c5). 91. Let A(c" "), B(# $), and C(2 !) denote the points. Since BC is vertical and has length kBCk 3, let D" (c" 4) be located vertically upward from A and D# (c" c2) be located vertically downward from A so that kBCk kAD" k kAD# k 3. Denote the point D$ (x y). Since the slope of AB equals the slope of CD$ we have yc3 c " 3y c 9 cx b 2 or x c2 3x b 3y 11. Likewise, the slope of AC equals the slope 2 of BD$ so that y c 0 3 3y 2x c 4 or 2x c 3y 4. xc2Solving the system of equationsx b 3y "" we find x 5 and y 2 yielding the vertex D$ (5 #). 2x c 3y 4 I92. Let ax, yb, x ! and/or y ! be a point on the coordinate plane. The slope, m, of the segment a! !b to ax, yb is y . A 90 x" rotation gives a segment with slope mw c m c x . If this segment has length equal to the original segment, its endpoint ywill be acy, xb or ay, cxb, the first of these corresponds to a counter-clockwise rotation, the latter to a clockwise rotation. (a) (c" 4); (b) (3 c2); (c) (5 2); (d) (0 x);Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 16. 16Chapter 1 Preliminaries (e) (cy 0);(f) (cy x);(g) (3 c10)93. 2x b ky 3 has slope c 2 and 4x b y 1 has slope c4. The lines are perpendicular when c 2 (c4) c1 or k k k c8 and parallel when c 2 c4 or k " . k #94. At the point of intersection, 2x b 4y 6 and 2x c 3y c1. Subtracting these equations we find 7y 7 or y 1. Substitution into either equation gives x 1 (1 1) is the intersection point. The line through (1 1) and (" #) is vertical with equation x 1. 95. Let M(a b) be the midpoint. Since the two triangles shown in the figure are congruent, the value a must lie midway between x" and x# , so a x bx . # "#y by ##"Similarly, b .96. (a) L has slope 1 so M is the line through P(2 1) with slope c1; or the line y cx b 3. At the intersection point, Q, we have equal y-values, y x b 2 cx b 3. Thus, 2x 1 or x " . Hence Q has coordinates # 3 " 5 . The distance from P to L the distance from P to Q # # b c 3 # 18 # # # 4(b) L has slope c 4 so M has slope 33 43 2 # .and M has the equation 4y c 3x 12. We can rewrite the equations ofthe lines as L: x b y 3 and M: cB b 4 y 4. Adding these we get 25 y 7 so y 84 . Substitution 3 12 25 into either equation gives x 4 84 c 4 12 so that Q 12 84 is the point of intersection. The distance 3 25 25 25 25 3 4from P to L 4 c12 # 2584 # 25b 6 c22 5 .(c) M is a horizontal line with equation y b. The intersection point of L and M is Q(c" b). Thus, the distance from P to L is (a b 1)# b 0# ka b 1k . C (d) If B 0 and A 0, then the distance from P to L is A c x! as in (c). Similarly, if A 0 and B 0, the C B distance is B c y! . If both A and B are 0 then L has slope c A so M has slope A . Thus, BL: Ax b By C and M: cBx b Ay c Bx! b Ay! . Solving these equations simultaneously we find theB aAx bBy bCb aA bB b### # !# !#!#!##!##kAx bBy bCk A bB!#!#!###!## # !# !#b Thus, (?x)# b (?y)# aAx A ByBbCb b# The distance from!## !aA bB bcBCcA y bABx A bB#!, and (?y)# y#A aAx bBy bCb aA b B b# !P to Q equals (?x)# b (?y)# , where (?x)# BCbA aAy cBx b . A bB # b x aA bB bcACB ABy cB x A band y !ACcB aAy cBx b A bBpoint of intersection Q(x y) with x ..##!#!1.3 FUNCTIONS AND THEIR GRAPHS 1. domain (c_ _); range [1 _) 3. domain (! _); y in range y 2. domain [0 _); range (c_ 1] " t, t 0 y# " tand y ! y can be any positive real number range (! _).Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 17. Section 1.3 Functions and Their Graphs 4. domain [0 _); y in range y " 1 b t17, t 0. If t 0, then y 1 and as t increases, y becomes a smallerand smaller positive real number range (0 1]. 5. 4 c z# (2 c z)(2 b z) 0 z [c2 2] domain. Largest value is g(0) 4 2 and smallest value is g(c2) g(2) 0 0 range [0 2]. 6. domain (c2 2) from Exercise 5; smallest value is g(0) " and as 0 z increases to 2, g(z) gets larger and # larger (also true as z 0 decreases to c2) range < " _ . # 7. (a) Not the graph of a function of x since it fails the vertical line test. (b) Is the graph of a function of x since any vertical line intersects the graph at most once. 8. (a) Not the graph of a function of x since it fails the vertical line test. (b) Not the graph of a function of x since it fails the vertical line test. 9. y " c " x (a) No (x !; (c) No; if x "," x" xc " ! x 1 and x !. So,"" x(b) No; division by ! undefined; (d) ! "c " !;10. y # c x # c x ! x ! and x #. x ! x ! and x # x % So, ! x %. (a) No; (b) No; (c) ! % #x 11. base x; (height)# b # x# height 3 #x; area is a(x) " #(base)(height) " #(x) 3 # x3 4x# ;perimeter is p(x) x b x b x 3x. 12. s side length s# b s# d# s d 2; and area is a s# a " #d#13. Let D diagonal of a face of the cube and j the length of an edge. Then j# b D# d# and (by Exercise 10) 6d 3#. The surface area is 6j# 2d# and the volume is j$ d 314. The coordinates of P are x x so the slope of the line joining P to the origin is m " x, x m ,#d 3x x" x" m .#15. The domain is ac_ _b.16. The domain is ac_ _b.Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley$#d 3 3 $D# 2j# 3j# d# j (x 0). Thus,. 18. 18Chapter 1 Preliminaries17. The domain is ac_ _b.18. The domain is c_ !.19. The domain is ac_ !b r a! _b.20. The domain is ac_ !b r a! _b.21. Neither graph passes the vertical line test (a)(b)22. Neither graph passes the vertical line test (a)(b) xby" y1cx or or kx b yk 1 x b y c" y c" c x Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 19. Section 1.3 Functions and Their Graphs 23.x y0 025. y 1 12 024.x y0 11 02 0" , x0 26. y x x, 0 x3 c x, x 1 2x, 1 x27. (a) Line through a! !b and a" "b: y x Line through a" "b and a# !b: y cx b 2 x, 0 x 1 f(x) cx b 2, 1 x 2 2, ! x " ! " x # (b) f(x) 2 # x $ ! $ x % 28. (a) Line through a! 2b and a# !b: y cx b 2 Line through a2 "b and a& !b: m ! c " &c# cx b #, 0 x # f(x) " c$x b &, # x & $c" $c$ c ! ! c c" c" c $ c% #c! #(b) Line through ac" !b and a! c$b: m Line through a! $b and a# c"b: m f(x) c " , so y c " ax c 2b b " c " x b $ $ $& $c$, so y c$x c $ c#, so y c#x b $c$x c $, c" x ! c#x b $, ! x #29. (a) Line through ac" "b and a! !b: y cx Line through a! "b and a" "b: y " Line through a" "b and a$ !b: m !c" $c" cx c" x ! " !x" f(x) c"x b $ "x$ # #c" # c " , so y c " ax c "b b " c " x b # # #$ #(b) Line through ac# c"b and a! !b: y " x #Line through a! #b and a" !b: y c#x b # Line through a" c"b and a$ c"b: y c"Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley19 20. 20Chapter 1 Preliminaries " #xf(x) c#x b # c"c# x ! !x" "x$30. (a) Line through T ! and aT "b: m # A, cA f(x) A cA!x T # T # x T T T x $# $T # x #T x #(b)1bx 0:x #x 0:x 24 xc1cx # 4 x1b4 x x (c2 0) r (% _)c1c 4 0 x 2x 0 x c2x c8 #31. (a) From the graph, x ## # # T , so y T x c T b 0 T x c " #!, 0 x T # # T T x c ", # x Tf(x) J(b)"c! TcaT#b0 (xc4)(xb2) #x0(xc4)(xb2) #x0 x 4 since x is positive; 4 x0 x c2xc8 2x #c1c0 x c2 since x is negative; sign of (x c 4)(x b 2) b b c c2 % Solution interval: (c# 0) r (% _)3 2 x c1 x b 1 3 2 x c1 x b 132. (a) From the graph, (b) Case x c1: x (c_ c5) r (c1 1) 3(xb1) x c12 3x b 3 2x c 2 x c5. Thus, x (c_ c5) solves the inequality. Case c1 x 1:3 x c12 x b13(xb1) x c12 3x b 3 2x c 2 x c5 which is true if x c1. Thus, x (c1 1) solves the inequality. 3 2 Case 1 x: xc1 xb1 3x b 3 2x c 2 x c5 which is never true if 1 x, so no solution here. In conclusion, x (c_ c5) r (c1 1). 33. (a) x 0 for x [0 1)(b) x 0 for x (c1 0]34. x x only when x is an integer. 35. For any real number x, n x n b ", where n is an integer. Now: n x n b " cn b " cx cn. By definition: cx cn and x n cx cn. So cx cx for all x d .Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 21. Section 1.3 Functions and Their Graphs2136. To find f(x) you delete the decimal or fractional portion of x, leaving only the integer part.37. v f(x) x"% c 2x22 c 2x %x$ c 72x# b $!)x; ! x 7 38. (a) Let h height of the triangle. Since the triangle is isosceles, AB # b AB # 2# AB 2 So, #h# b "# 2 h " B is at a! "b slope of AB c" The equation of AB is y f(x) cB b "; x ! ". (b) Ax 2x y 2xcx b " c2x# b #x; x ! ". 39. (a) Because the circumference of the original circle was )1 and a piece of length x was removed. (b) r )1#c x % c #x 1 1 x %1# # 4x c 1x %11 "'1 x c %x %1"'1xcx #1#a)1 c xb "'1x c x #%1b# ###4x 1##"'1x c x #1 "' c 16 c# ##(d) V " 1 r# h " 1 )1cx $ $ #1x # #1#(c) h "' c r# "' c % c40. (a) Note that 2 mi = 10,560 ft, so there are )!!# b x# feet of river cable at $180 per foot and a"! &'! c xb feet of land cable at $100 per foot. The cost is Caxb ")!)!!# b x# b "!!a"! &'! c xb. (b) Ca!b $" #!! !!! Ca&!!b $" "(& )"# Ca"!!!b $" ")' &"# Ca"&!!b $" #"# !!! Ca#!!!b $" #%$ ($# Ca#&!!b $" #() %(* Ca$!!!b $" $"% )(! Values beyond this are all larger. It would appear that the least expensive location is less than 2000 feet from the point P. 41. A curve symmetric about the x-axis will not pass the vertical line test because the points ax, yb and ax, cyb lie on the same vertical line. The graph of the function y faxb ! is the x-axis, a horizontal line for which there is a single y-value, !, for any x. 42. Pick 11, for example: "" b & "' # "' $# $# c ' #' faxb #axb&bc' ##' # "$ "$ c # "", the original number.c # x, the number you started with.Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 22. 22Chapter 1 Preliminaries1.4 IDENTIFYING FUNCTIONS; MATHEMATICAL MODELS 1. (a) linear, polynomial of degree 1, algebraic. (c) rational, algebraic.(b) power, algebraic. (d) exponential.2. (a) polynomial of degree 4, algebraic. (c) algebraic.(b) exponential. (d) power, algebraic.3. (a) rational, algebraic. (c) trigonometric.(b) algebraic. (d) logarithmic.4. (a) logarithmic. (c) exponential.(b) algebraic. (d) trigonometric.5. (a) Graph h because it is an even function and rises less rapidly than does Graph g. (b) Graph f because it is an odd function. (c) Graph g because it is an even function and rises more rapidly than does Graph h. 6. (a) Graph f because it is linear. (b) Graph g because it contains a! "b. (c) Graph h because it is a nonlinear odd function. 7. Symmetric about the origin Dec: c_ x _ Inc: nowhere8. Symmetric about the y-axis Dec: c_ x ! Inc: ! x _9. Symmetric about the origin Dec: nowhere Inc: c_ x ! !x_10. Symmetric about the y-axis Dec: ! x _ Inc: c_ x !Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 23. Section 1.4 Identifying Functions; Mathematical Models 11. Symmetric about the y-axis Dec: c_ x ! Inc: ! x _12. No symmetry Dec: c_ x ! Inc: nowhere13. Symmetric about the origin Dec: nowhere Inc: c_ x _14. No symmetry Dec: ! x _ Inc: nowhere15. No symmetry Dec: ! x _ Inc: nowhere16. No symmetry Dec: c_ x ! Inc: nowhereCopyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley23 24. 24Chapter 1 Preliminaries17. Symmetric about the y-axis Dec: c_ x ! Inc: ! x _18. Symmetric about the y-axis Dec: ! x _ Inc: c_ x !19. Since a horizontal line not through the origin is symmetric with respect to the y-axis, but not with respect to the origin, the function is even. and facxb acxbc& " ac x b&" x" c x cfaxb. Thus the function is odd. &&20. faxb xc& 21. Since faxb x# b " acxb# b " cfaxb. The function is even. 22. Since faxb x# b x facxb acxb# c x and faxb x# b x cfaxb caxb# c x the function is neither even nor odd. 23. Since gaxb x$ b x, gacxb cx$ c x cax$ b xb cgaxb. So the function is odd. 24. gaxb x% b $x# b " acxb% b $acxb# c " gacxb thus the function is even.26. gaxb x x c ";# #27. hatb " t c ";" acxb c" gacxb. Thus the function is even.gacxb c x x gacxb. So the function is odd. c" #" x c"#25. gaxb h a ct b " ct c " ;ch at b " " c t.Since hatb chatb and hatb hactb, the function is neither even nor odd.28. Since l t$ | l actb$ |, hatb hactb and the function is even. 29. hatb 2t b ", hactb c2t b ". So hatb hactb. chatb c2t c ", so hatb chatb. The function is neither even nor odd. 30. hatb 2l t l b " and hactb 2l ct l b " 2l t l b ". So hatb hactb and the function is even. 31. (a)The graph supports the assumption that y is proportional to x. The constant of proportionality is estimated from the slope of the regression line, which is 0.166.Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 25. Section 1.4 Identifying Functions; Mathematical Models (b)25The graph supports the assumption that y is proportional to x"# . The constant of proportionality is estimated from the slope of the regression line, which is 2.03.32. (a) Because of the wide range of values of the data, two graphs are needed to observe all of the points in relation to the regression line.The graphs support the assumption that y is proportional to $x . The constant of proportionality is estimated from the slope of the regression line, which is 5.00. (b) The graph supports the assumption that y is proportional to ln x. The constant of proportionality is extimated from the slope of the regression line, which is 2.99.33. (a) The scatterplot of y reaction distance versus x speed isAnswers for the constant of proportionality may vary. The constant of proportionality is the slope of the line, which is approximately 1.1.Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 26. 26Chapter 1 Preliminaries (b) Calculate x w speed squared. The scatterplot of x w versus y braking distance is:Answers for the constant of proportionality may vary. The constant of proportionality is the slope of the line, which is approximately 0.059. 34. Kepler's 3rd Law is Tadaysb !%"R$# , R in millions of miles. "Quaoar" is 4 "!* miles from Earth, or about 4 "!* b *$ "!' % "!* miles from the sun. Let R 4000 (millions of miles) and T a!%"ba%!!!b$# days "!$ (#$ days. 35. (a)The hypothesis is reasonable. (b) The constant of proportionality is the slope of the line (c) y(in.) a!)( in./unit massba"$ unit massb ""$" in. 36. (a))(%" c ! "! c !in./unit mass !)(% in./unit mass.(b)Graph (b) suggests that y k x$ is the better model. This graph is more linear than is graph (a). 1.5 COMBINING FUNCTIONS; SHIFTING AND SCALING GRAPHS g Dfg : x 1. Rf : c_ y _, Rg : y 0, Rf g : y 1, Rfg : y 0 b1. Df : c_ x _, Dg : x 1 Dfb2. Df : x b 1 0 x c1, Dg : x c 1 0 x 1. Therefore Df Rf Rg : y 0, Rf g : y 2, Rfg : y 0g Dfg : x 1.bbCopyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 27. Section 1.5 Combining Functions; Shifting and Scaling Graphs 3. Df : c_ x _, Dg : c_ x _ Df g : c_ x _ since g(x) 0 for any x; Dg f : c_ x _ since f(x) 0 for any x. Rf : y 2, Rg : y 1, Rf g : 0 y 2, Rgf : y " # 4. Df : c_ x _, Dg : x 0 Df g : x 0 since g(x) 0 for any x 0; Dg f : x 0 since f(x) 0 for any x 0. Rf : y 1, Rg : y 1, Rf g : 0 y 1, Rgf : y " 5. (a) (b) (c) (d) (e) (f) (g) (h)f(g(0)) f(c3) 2 g(f(0)) g(5) 22 f(g(x)) f(x# c 3) x# c 3 b 5 x# b 2 g(f(x)) g(x b 5) (x b 5)# c 3 x# b 10x b 22 f(f(c5)) f(0) 5 g(g(2)) g(1) c2 f(f(x)) f(x b 5) (x b 5) b 5 x b 10 g(g(x)) g(x# c 3) (x# c 3)# c 3 x% c 6x# b 6" 6. (a) f g " f 2 c 3 # 3 (b) g f " g c " 2 # # " (c) f(g(x)) f x b 1 " x b1c1(d) g(f(x)) g(x c 1) " (xc1) b 1(e) f(f(2)) f(1) 0 (f) g(g(2)) g " 3" 4 3cx xb1 " x3 4(g) f(f(x)) f(x c 1) (x c 1) c 1 x c 2 " xb" (h) g(g(x)) g x b 1 "b 1 x b # (x c1 and x c2) b "x1" " # 4 7. (a) u(v(f(x))) u v " u x 4 x c 5 x c 5 x " " 4 (b) u(f(v(x))) u af ax# bb u x 4 x c 5 x c 5###### " (c) v(u(f(x))) v u " v 4 x c 5 4 c 5 x x(d) v(f(u(x))) v(f(4x c 5)) v 4x " 5 4x " 5 c c " 4x c 58. (a) h(g(f(x))) h g x h x 4 4" (4x c 5)x 4 #(f) f(v(u(x))) f(v(4x c 5)) f a(4x c 5)# b #(e) f(u(v(x))) f au ax# bb f a4 ax# b c 5b #c 8 x c 8(b) h(f(g(x))) h f x h x 4 x c 8 2x c 8 4 4 4 4 x c 8 x c 2 4 4x c 8 4 x#c 2(c) g(h(f(x))) g h x g 4x c 8 (d) g(f(h(x))) g(f(4x c 8)) g 4x c 8c (e) f(g(h(x))) f(g(4x c 8)) f 4x 4 8 f(x c 2) x c 2 (f) f(h(g(x))) f h x f 4 x c 8 f(x c 8) x c 8 449. (a) y f(g(x)) (c) y g(g(x)) (e) y g(h(f(x)))(b) y j(g(x)) (d) y j(j(x)) (f) y h(j(f(x)))10. (a) y f(j(x)) (c) y h(h(x)) (e) y j(g(f(x)))(b) y h(g(x)) g(h(x)) (d) y f(f(x)) (f) y g(f(h(x))) Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley27 28. 28Chapter 1 Preliminaries g(x)f(x)(f g)(x)(a)xc7xx c 7(b)xb23x3(x b 2) 3x b 6(c)x#x c 5x# c 5(d)x xc1x xc1" xc1 " x1b(f)c(e)x x 1 x x 1 c1" x(b) afgbaxb gaxbc" g ax bx x c (xc1)xx" x12. (a) afgbaxb lgaxbl c11.x" lx c "l . x xb""c" g ax bx xb""cx xb"" g ax b" xb"" gaxb sogaxb x b ".(c) Since afgbaxb gaxb lxl, gaxb x . (d) Since afgbaxb fx l x l, faxb x# . (Note that the domain of the composite is ! _.) #The completed table is shown. Note that the absolute value sign in part (d) is optional. gaxb faxb afgbaxb " " lxl xc" lx c "l xc" xxb"xx# xx#x xb"lxl lxl13. (a) fagaxbb 1 b 1 1bx x x gafaxbb 1 x b 1(b) Domain afgb: 0, _, domain agf b: c1, _ (c) Range afgb: 1, _, range agf b: 0, _ 14. (a) fagaxbb 1 c 2x b x gafaxbb 1 c kxk (b) Domain afgb: 0, _, domain agf b: 0, _ (c) Range afgb: 0, _, range agf b: c_, 1 15. (a) y c(x b 7)#(b) y c(x c 4)#16. (a) y x# b 3(b) y x# c 517. (a) Position 4(b) Position 1(c) Position 2(d) Position 318. (a) y c(x c 1)# b 4(b) y c(x b 2)# b 3(c) y c(x b 4)# c 1(d) y c(x c 2)#Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 29. Section 1.5 Combining Functions; Shifting and Scaling Graphs 19.20.21.22.23.24.25.26.Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley29 30. 30Chapter 1 Preliminaries27.28.29.30.31.32.33.34.35.36.Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 31. Section 1.5 Combining Functions; Shifting and Scaling Graphs 37.38.39.40.41.42.43.44.Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley31 32. 32Chapter 1 Preliminaries45.46.47.48.49. (a) domain: [0 2]; range: [# $](b) domain: [0 2]; range: [c1 0](c) domain: [0 2]; range: [0 2](d) domain: [0 2]; range: [c1 0](e) domain: [c2 0]; range: [! 1](f) domain: [1 3]; range: [! "]Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 33. Section 1.5 Combining Functions; Shifting and Scaling Graphs (g) domain: [c2 0]; range: [! "](h) domain: [c1 1]; range: [! "]50. (a) domain: [0 4]; range: [c3 0](b) domain: [c4 0]; range: [! $](c) domain: [c4 0]; range: [! $](d) domain: [c4 0]; range: [" %](e) domain: [# 4]; range: [c3 0](f) domain: [c2 2]; range: [c3 0]Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley33 34. 34Chapter 1 Preliminaries (g) domain: [" 5]; range: [c3 0](h) domain: [0 4]; range: [0 3]51. y 3x# c 3 52. y a2xb# c 1 %x# c 1#" ax$b" #" #xb1b* x##54. y 1 b" x#53. y " " b #55. y %x b 1 56. y 3x b 1 # 57. y % c x " 16 c x# # #58. y " % c x# $ 59. y " c a3xb$ " c 27x$ x )$$60. y " c x " c #"# "# 61. Let y c#x b " faxb and let gaxb x"# , haxb x b " , iaxb #x b " , and # # "# jaxb c#x b " faBb. The graph of haxb is the graph of gaxb shifted left #" #unit; the graph of iaxb is the graphof haxb stretched vertically by a factor of #; and the graph of jaxb faxb is the graph of iaxb reflected across the x-axis.Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 35. Section 1.5 Combining Functions; Shifting and Scaling Graphs 62. Let y " cx # faxb Let gaxb acxb"# , haxb acx b #b"# , and iaxb " # ac xb #b"# " cx #35 faxbThe graph of gaxb is the graph of y x reflected across the x-axis. The graph of haxb is the graph of gaxb shifted right two units. And the graph of iaxb is the graph of haxb compressed vertically by a factor of #.63. y faxb x$ . Shift faxb one unit right followed by a shift two units up to get gaxb ax c "b3 b #.64. y a" c Bb$ b # cax c "b$ b ac#b faxb. Let gaxb x$ , haxb ax c "b$ , iaxb ax c "b$ b ac#b, and jaxb cax c "b$ b ac#b. The graph of haxb is the graph of gaxb shifted right one unit; the graph of iaxb is the graph of haxb shifted down two units; and the graph of faxb is the graph of iaxb reflected across the x-axis.Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 36. Chapter 1 Preliminaries65. Compress the graph of faxb and gaxb # x#" x" #x .Then shift gaxb vertically down 1 unit toc ".b"" # #B#66. Let faxb horizontally by a factor of 2 to get gaxb b"" x##get haxb " #x" xb"" "#B#36b " Since # "%, we see that the graph offaxb stretched horizontally by a factor of 1.4 and shifted up 1 unit is the graph of gaxb.67. Reflect the graph of y faxb x across the x-axis to get gaxb cx. $$68. y faxb ac#xb#$ ac"ba#bx#$ ac"b#$ a#xb#$ a#xb#$ . So the graph of faxb is the graph of gaxb x#$ compressed horizontally by a factor of 2.Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 37. Section 1.5 Combining Functions; Shifting and Scaling Graphs 70. A*sin((2*Pi/B)*(x-C))+D1; A:=3; C:=0; D1:=0; f_list := [seq( f(x), B=[1,3,2*Pi,5*Pi] )]; plot( f_list, x=-4*Pi..4*Pi, scaling=constrained, color=[red,blue,green,cyan], linestyle=[1,3,4,7], legend=["B=1","B=3","B=2*Pi","B=3*Pi"], title="#67 (Section 1.6)" ); Mathematica Clear[a, b, c, d, f, x] f[x_]:=a Sin[21/b (x c c)] + d Plot[f[x]/.{a 3, b 1, c 0, d 0}, {x, c41, 41 }] 67. (a) The graph stretches horizontally.Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley45 46. 46Chapter 1 Preliminaries (b) The period remains the same: period l B l. The graph has a horizontal shift of" #period.68. (a) The graph is shifted right C units.(b) The graph is shifted left C units. (c) A shift of one period will produce no apparent shift. l C l ' 69. The graph shifts upwards l D lunits for D ! and down l D lunits for D !70. (a) The graph stretches l A l units.(b) For A !, the graph is inverted. 1.7 GRAPHING WITH CALCULATORS AND COMPUTERS 1-4.The most appropriate viewing window displays the maxima, minima, intercepts, and end behavior of the graphs and has little unused space.Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 47. Section 1.7 Graphing with Calculators and Computers 1. d.2. c.3. d.4. b.5-30.For any display there are many appropriate display widows. The graphs given as answers in Exercises 5c30 are not unique in appearance.5. c2 5 by c15 406. c4 4 by c4 47. c2 6 by c250 508. c1 5 by c5 30Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley47 48. 48Chapter 1 Preliminaries9. c4 4 by c5 510. c2 2 by c2 811. c2 6 by c5 412. c4 4 by c8 813. c" ' by c" %14. c" ' by c" &15. c3 3 by ! "!16. c" # by ! "Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 49. Section 1.7 Graphing with Calculators and Computers 17. c& " by c& &18. c& " by c# %19. c% % by ! $20. c& & by c# #21. c"! "! by c' '22. c& & by c# #23. c' "! by c' '24. c$ & by c# "!Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley49 50. 50Chapter 1 Preliminaries25. c!!$ !!$ by c"#& "#&26. c!" !" by c$ $27. c$!! $!! by c"#& "#&28. c&! &! by c!" !"29. c!#& !#& by c!$ !$30. c!"& !"& by c!!# !!&31. x# b #x % b %y c y# y # cx# c #x b ). The lower half is produced by graphing y # c cx# c #x b ).32. y# c "'x# " y " b "'x# . The upper branch is produced by graphing y " b "'x# .Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 51. Section 1.7 Graphing with Calculators and Computers 33.34.35.36.37.3839.40.41. (a) y "!&*"%x c #!(%*(# (b) m "!&*"% dollars/year, which is the yearly increase in compensation.Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley51 52. 52Chapter 1 Preliminaries (c)(d) Answers may vary slightly. y a"!&*14ba#!"!b c #!(%*(# $&$ 899 42. (a) Let C cost and x year. C a(*'!("bx c "' "!( (b) Slope represents increase in cost per year (c) C a#'$("%bx c "!' (d) The median price is rising faster in the northeast (the slope is larger). 43. (a) Let x represent the speed in miles per hour and d the stopping distance in feet. The quadratic regression function is d !!)''x# c "*(x b &!". (b)(c) From the graph in part (b), the stopping distance is about $(! feet when the vehicle is (# mph and it is about & feet when the speed is )& mph. Algebraically: dquadratic a(#b !!)''a(#b# c "*(a(#b b &!" $'(' ft. dquadratic a)&b !!)''a)&b# c "*(a)&b b &!" #) ft. (d) The linear regression function is d ')*x c "%!% dlinear a(#b ')*a(#b c "%!% $&&( ft and dlinear a)&b ')*a)&b c "%!% %% ft. The linear regression line is shown on the graph in part (b). The quadratic regression curve clearly gives the better fit.44. (a) The power regression function is y %%%'%(x!&""%"% .Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 53. Chapter 1 Practice Exercises (b)(c) 152 km/h (d) The linear regression function is y !*"$'(&x b %")**(' and it is shown on the graph in part (b). The linear regession function gives a speed of "%# km/h when y "" m. The power regression curve in part (a) better fits the data. CHAPTER 1 PRACTICE EXERCISES 1. ( b 2x $ #x c% x c# 2.c 3x "! x c "! $3." & axc "b " ax c #b %ax c "b &ax c #b %xc$ # c %bx $ax c $b c#a% b xb $qqqqqqqq x c "! $ %x c % &x c "! ' x4. $x c * c) c #x &x " x 5.qqqqqqqq x " &" &lx b " l ( x b " ( or cax b "b ( x ' or x c)6. ly c $ l % c% y c $ % c" y ( 7. " c x #$ # "cx # c $ or " c #x #$ # c x c & or c x # # #" # cx c& or cx " x & or x c" b( 8. #x$ & c& #xb( $ & c1& #x b ( 1& c22 #x 8 c"" x %9. Since the particle moved to the y-axis, c# b ?x ! ?x 2. Since ?y 3?x 6, the new coordinates are (x b ?x y b ?y) (c# b # & b ') (0 11). 10. (a)Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley53 54. 54Chapter 1 Preliminaries (b)line ABslope 10 c 1 9 3 2 c 8 c6 c # 10 c 6 4 2 2 c (c4) 6 3 6 c (c3) 9 3 c% c 2 c6 c # 1 c (c3) 4 2 8c2 6 3 6c6 0 c% c 14 3BC CD DA CEBD is vertical and has no slope (c) Yes; A, B, C and D form a parallelogram. (d) Yes. The line AB has equation y c 1 c 3 (x c 8). Replacing x by 14 gives y c 3 14 c 8 b " # 3 # 3 c 3 c 10 b 1 5 b 1 6. Thus, E 14 6 lies on the line AB and the points A, B and E are collinear. # 3 3 (e) The line CD has equation y b 3 c 3 (x c 2) or y c 3 x. Thus the line passes through the origin. # # 11. The triangle ABC is neither an isosceles triangle nor is it a right triangle. The lengths of AB, BC and AC are 53, 72 and 65, respectively. The slopes of AB, BC and AC are 7 , c1 and " , respectively. #812. P(x 3x b 1) is a point on the line y 3x b 1. If the distance from P to (! 0) equals the distance from P to (c$ %), then x# b (3x b 1)# (x b 3)# b (3 c 3x)# x# b 9x# b 6x b 1 x# b 6x b 9 b 9 c 18x b 9x# 18x 17 or x 17 y 3x b 1 3 17 b 1 23 . Thus the point is P 17 23 . 18 18 6 18 6 13. y $ax c "b b ac'b y $x c * 14. y c " ax b "b b # y c " x b # #$ #15. x ! 16. m c# c ' " c ac$bc) % c# y c#ax b $b b ' y c#x17. y # 18. m &c$ c# c $# c # # c & y c & ax c $ b b $ y c & x b#" &19. y c$x b $ 20. Since #x c y c# is equivalent to y #x b #, the slope of the given line (and hence the slope of the desired line) is 2. y #a x c "b b " y # x c & 21. Since %x b $y "# is equivalent to y c % x b %, the slope of the given line (and hence the slope of the desired line) is $ c % . y c % ax c 4b c "2 y c % x c $ $ $#! $22. Since $x c &y " is equivalent to y $ x c " , the slope of the given line is & & c5. $yc 5 ax $b #b c $ y c5x $c"* $$ &and the slope of the perpendicular line is23. Since " x b " y " is equivalent to y c $ x b $, the slope of the given line is c $ and the slope of the perpendicular line # $ # # is # . y # ax b "b b # y # x b $ $ $) $24. The line passes through a! c&b and a$ !b. m ! c ac&b $c!& $& y $x c &Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 55. Chapter 1 Practice Exercisessurface area gives S %1 r# %1 $V %1"##$#C A 1 #1 C %1 .. The volume is V % 1 r$ r $V . Substitution into the formula for $ %1 $26. The surface area is S %1 r# r %S 1C #1#25. The area is A 1 r# and the circumference is C #1 r. Thus, r .27. The coordinates of a point on the parabola are ax x# b. The angle of inclination ) joining this point to the origin satisfies28. tan ) rise runh &!!x x#the equation tan ) x. Thus the point has coordinates ax x# b atan ) tan# )b. h &!! tan ) ft.29.30.Symmetric about the origin.Symmetric about the y-axis.31.32.NeitherSymmetric about the y-axis.33. yacxb acxb# b " x# b " yaxb. Even. 34. yacxb acxb& c acxb$ c acxb cx& b x$ b x cyaxb. Odd. 35. yacxb " c cosacxb " c cos x yaxb. Even.x b" cx b#x csec x tan x cyaxb. Odd.b" c xx c#x cyaxb. Odd. $%$csin x cos x%acxb b" acxb c#acxb#%$37. yacxb sinacxb cos acxb#36. yacxb secacxb tanacxb 38. yacxb " c sinacxb " b sin x. Neither even nor odd. 39. yacxb cx b cosacxb cx b cos x. Neither even nor odd. 40. yacxb acxb% c " x% c " yaxb. Even.Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley55 56. 56Chapter 1 Preliminaries41. (a) The function is defined for all values of x, so the domain is ac_ _b. (b) Since l x l attains all nonnegative values, the range is c# _. 42. (a) Since the square root requires " c x !, the domain is c_ ". (b) Since " c x attains all nonnegative values, the range is c# _. 43. (a) Since the square root requires "' c x# !, the domain is c% %. (b) For values of x in the domain, ! "' c x# "', so ! "' c x# %. The range is ! %. 44. (a) The function is defined for all values of x, so the domain is ac_ _b. (b) Since $#cx attains all positive values, the range is a" _b. 45. (a) The function is defined for all values of x, so the domain is ac_ _b. (b) Since #ecx attains all positive values, the range is ac$ _b. 46. (a) The function is equivalent to y tan #x, so we require #x k1 #for odd integers k. The domain is given by x k1 %forodd integers k. (b) Since the tangent function attains all values, the range is ac_ _b. 47. (a) The function is defined for all values of x, so the domain is ac_ _b. (b) The sine function attains values from c" to ", so c# #sina$x b 1b # and hence c$ #sina$x b 1b c " ". The range is c3 1. 48. (a) The function is defined for all values of x, so the domain is ac_ _b. (b) The function is equivalent to y x# , which attains all nonnegative values. The range is ! _. &49. (a) The logarithm requires x c $ !, so the domain is a$ _b. (b) The logarithm attains all real values, so the range is ac_ _b. 50. (a) The function is defined for all values of x, so the domain is ac_ _b. (b) The cube root attains all real values, so the range is ac_ _b. 51. (a) The function is defined for c% x %, so the domain is c% %. (b) The function is equivalent to y l x l, c% x %, which attains values from ! to # for x in the domain. The range is ! #. 52. (a) The function is defined for c# x #, so the domain is c# #. (b) The range is c" ". 53. First piece: Line through a! "b and a" !b. m Second piece: Line through a" "b and a# !b. m faxb " c x, ! x " # c x, " x #54. First piece: Line through a! !b and a2 5b. m Second piece: Line through a2 5b and a4 !b. m faxb J10 c5 2 x, 5x 2 ,!c" c" "c! " ! c " c" #c" "c" y cx b " " c x c" y cax c "b b " cx b # # c x5c! 5 5 2c! 2 y 2x c ! c 5 c5 c 5 4 2 2 2y c 5 ax c 2b b 5 c 5 x b 10 10 c 2 2!x2 (Note: x 2 can be included on either piece.) 2x4 Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley5x 2 57. Chapter 1 Practice Exercises " 55. (a) afgbac"b fagac"bb f c" b # fa"b " # or & x, x !(d) aggbaxb gagaxbb g x"b # " #b "" "x"xb#x b #%(c) aff baxb fafaxbb f " x" b# # "(b) agf ba#b gafa#bb g " 2" " " b # x b #56. (a) afgbac"b fagac"bb fc" b " fa!b # c ! # (b) agf ba#b faga#bb ga# c #b ga!b ! b " " $$(c) aff baxb fafaxbb fa# c xb # c a# c xb x (d) aggbaxb gagaxbb gx b " x b " b " $$$#57. (a) afgbaxb fagaxbb fx b # # c x b # cx, x c#. agf baxb fagaxbb ga# c x# b a# c x# b b # % c x# (b) Domain of fg: c# _ Domain of gf: c# # (c) Range of fg: c_ # Range of gf: ! # 58. (a) afgbaxb fagaxbb f" c x " c x " c x. %agf baxb fagaxbb gx " c x (b) Domain of fg: c_ " Domain of gf: ! " (c) Range of fg: ! _ Range of gf: ! " 59.60.The graph of f# (x) f" akxkb is the same as the graph of f" (x) to the right of the y-axis. The graph of f# (x) to the left of the y-axis is the reflection of y f" (x), x 0 across the y-axis.The graph of f# (x) f" akxkb is the same as the graph of f" (x) to the right of the y-axis. The graph of f# (x) to the left of the y-axis is the reflection of y f" (x), x 0 across the y-axis.Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley57 58. 58Chapter 1 Preliminaries61.62.The graph of f# (x) f" akxkb is the same as the graph of f" (x) to the right of the y-axis. The graph of f# (x) to the left of the y-axis is the reflection of y f" (x), x 0 across the y-axis.It does not change the graph.63.64.The graph of f# (x) f" akxkb is the same as the graph of f" (x) to the right of the y-axis. The graph of f# (x) to the left of the y-axis is the reflection of y f" (x), x 0 across the y-axis. 65.66.Whenever g" (x) is positive, the graph of y g# (x) kg" (x)k is the same as the graph of y g" (x). When g" (x) is negative, the graph of y g# (x) is the reflection of the graph of y g" (x) across the x-axis. 67.The graph of f# (x) f" akxkb is the same as the graph of f" (x) to the right of the y-axis. The graph of f# (x) to the left of the y-axis is the reflection of y f" (x), x 0 across the y-axis.It does not change the graph.Whenever g" (x) is positive, the graph of y g# (x) kg" (x)k is the same as the graph of y g" (x). When g" (x) is negative, the graph of y g# (x) is the reflection of the graph of y g" (x) across the x-axis.Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 59. Chapter 1 Practice Exercises5968.Whenever g" (x) is positive, the graph of y g# (x) kg" (x)k is the same as the graph of y g" (x). When g" (x) is negative, the graph of y g# (x) is the reflection of the graph of y g" (x) across the x-axis.69.70.period 1period 4171.72.period 2period 473.74.period 21period 2175. (a) sin B sin1 3b cb # b 2 sin1 3 23 # 3. By the theorem of Pythagoras,a# b b# c# a c# c b# 4 c 3 1. b c2 c c2 sin 312 1 3#(b) sin B sin34 3#4 . Thus, a c# c b# 3 c (2)# 4 376. (a) sin A a c a c sin A(b) tan A a b a b tan A77. (a) tan B b a a(b) sin A a c cb tan Ba sin ACopyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley2 3. 60. Chapter 1 Preliminaries (c) sin A a ca c c cb c #78. (a) sin A #6079. Let h height of vertical pole, and let b and c denote the distances of points B and C from the base of the pole, measured along the flatground, respectively. Then, tan 50 h , tan 35 h , and b c c 10. c b Thus, h c tan 50 and h b tan 35 (c b 10) tan 35 c tan 50 (c b 10) tan 35 c (tan 50 c tan 35) 10 tan 35 35 c tan10 tantan 35 h c tan 50 50c 10 tan 35 tan 50 tan 50ctan 35 16.98 m.80. Let h height of balloon above ground. From the figure at the right, tan 40 h , tan 70 h , and a b b 2. Thus, a b h b tan 70 h (2 c a) tan 70 and h a tan 40 (2 c a) tan 70 a tan 40 a(tan 40 b tan 70) 70 2 tan 70 a tan 2 tan tan 70 h a tan 40 40b 2 tan 70 tan 40 tan 40btan 70 1.3 km.81. (a)(b) The period appears to be 41. (c) f(x b 41) sin (x b 41) b cos xb41 sin (x b 21) b cos x b 21 sin x b cos # # since the period of sine and cosine is 21. Thus, f(x) has period 41.x #82. (a)(b) D (c_ 0) r (! _); R [c1 1] (c) f is not periodic. For suppose f has period p. Then f #" b kp f #" sin 21 0 for all 1 1 integers k. Choose k so large that" #1b kp " 1 0" (1/21)bkp 1. But thenf #" b kp sin (1/#1")bkp 0 which is a contradiction. Thus f has no period, as claimed. 1Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 61. Chapter 1 Additional and Advanced Exercises CHAPTER 1 ADDITIONAL AND ADVANCED EXERCISES 1. (a) The given graph is reflected about the y-axis.(b) The given graph is reflected about the x-axis.(c) The given graph is shifted left 1 unit, stretched vertically by a factor of 2, reflected about the x-axis, and then shifted upward 1 unit.2. (a)(d) The given graph is shifted right 2 units, stretched vertically by a factor of 3, and then shifted downward 2 units.(b)3. There are (infinitely) many such function pairs. For example, f(x) 3x and g(x) 4x satisfy f(g(x)) f(4x) 3(4x) 12x 4(3x) g(3x) g(f(x)). 4. Yes, there are many such function pairs. For example, if g(x) (2x b 3)$ and f(x) x"$ , then (f g)(x) f(g(x)) f a(2x b 3)$ b a(2x b 3)$ b"$ 2x b 3.5. If f is odd and defined at x, then f(cx) cf(x). Thus g(cx) f(cx) c 2 cf(x) c 2 whereas cg(x) c(f(x) c 2) cf(x) b 2. Then g cannot be odd because g(cx) cg(x) cf(x) c 2 cf(x) b 2 4 0, which is a contradiction. Also, g(x) is not even unless f(x) 0 for all x. On the other hand, if f is even, then g(x) f(x) c 2 is also even: g(cx) f(cx) c 2 f(x) c 2 g(x). 6. If g is odd and g(0) is defined, then g(0) g(c0) cg(0). Therefore, 2g(0) 0 g(0) 0.Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley61 62. 62Chapter 1 Preliminaries7. For (x y) in the 1st quadrant, kxk b kyk 1 b x x b y 1 b x y 1. For (x y) in the 2nd quadrant, kxk b kyk x b 1 cx b y x b 1 y 2x b 1. In the 3rd quadrant, kxk b kyk x b 1 cx c y x b 1 y c2x c 1. In the 4th quadrant, kxk b kyk x b 1 x b (cy) x b 1 y c1. The graph is given at the right. 8. We use reasoning similar to Exercise 7. (1) 1st quadrant: y b kyk x b kxk 2y 2x y x. (2) 2nd quadrant: y b kyk x b kxk 2y x b (cx) 0 y 0. (3) 3rd quadrant: y b kyk x b kxk y b (cy) x b (cx) 0 0 all points in the 3rd quadrant satisfy the equation. (4) 4th quadrant: y b kyk x b kxk y b (cy) 2x 0 x. Combining these results we have the graph given at the right:sin 4 4sin A asin B bsin B 3 sin B 3 411 111. By the law of cosines, a# b# b c# c 2bc cos A cos A 1 4sin3 b bc ca 2bc #10. By the law of sines,# b#sin 4 b33 42 2.2 # 2 b3 c2 2(2)(3)#12. By the law of cosines, c# a# b b# c 2ab cos C 2# b 3# c (2)(2)(3) cos1 4#sin B b3 2 8##sin A asin 3 3#3 sin (1/4) sin (1/3)9. By the law of sines,. 3. 4 4 b 9 c 12 2 # 13 c 62 c 13 c 62 , since c 0. 4b16c9 16b4 c5 a bb c c 2(2)(2)(4) 2ab 231 25 256 16 .4b16c25 16#######5 c 16 . Since 0 C 1, sin C 1 c cos# C 1 c#14. By the law of cosines, c# a# b b# c 2ab cos C cos C #Since 0 B 1, sin B 1 c cos# B 1 c#11 16 .#15. (a) sin# x b cos# x 1 sin# x 1 c cos# x (1 c cos x)(1 b cos x) (1 c cos x) 1ccos x sin xsin x 1bcos xsin x 1bcos x(b) Using the definition of the tangent function and the double angle formulas, we have cos 2 x#c"cos 2 x1ccos x 1bcos x.#b" # sin x cos x # ####tan# x #Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley#2 b4 c3 (2)(2)(4) 315 16 .#a bc c b #ac 135 121 256 1613. By the law of cosines, b# a# b c# c 2ac cos B cos B 63. Chapter 1 Additional and Advanced Exercises 16. The angles labeled # in the accompanying figure are equal since both angles subtend arc CD. Similarly, the two angles labeled ! are equal since they both subtend arc AB. Thus, triangles AED and BEC are similar which ) implies acc 2a cos c cb b ab (a c c)(a b c) b(2a cos ) c b) a# c c# 2ab cos ) c b# c# a# b b# c 2ab cos ).17. As in the proof of the law of sines of Section P.5, Exercise 57, ah bc sin A ab sin C ac sin B the area of ABC " (base)(height) " ah " bc sin A " ab sin C " ac sin B. # # # # # a# b# sin# Cbb cc #ab###" 16a# b# " c a 9a b 4# ##4a# b# c aa# b b# c c# b " 4" c.aa b b c c b 4a b #a# b# a" c cos# Cb #" 16" 4#a bb cc 2ab# #" 4# #a# h # a# b# a" c cos# Cb . By the law of cosines, c# a# b b# c 2ab cos C cos C Thus, (area of ABC)# " 4#" 4(base)# (height)# #" 4#18. As in Section P.5, Exercise 57, (Area of ABC)# ca2ab b aa# b b# c c# bb a2ab c aa# b b# c c# bbd" ca(a b b)# c c# b ac# c (a c b)# bd 16 c((a b b) b c)((a b b) c c)(c b (a c b))(c c (a c b))d b b b b < a b # b c ca b#b b c a c # b c a b # c c s(s c a)(s c b)(s c c), where s ab#bc ." 16Therefore, the area of ABC equals s(s c a)(s c b)(s c c) . 19. 1. 2. 3. 4. 5. 6.b b c c (a b c) b c a, which is positive since a b. Thus, a b c b b c. b c c c (a c c) b c a, which is positive since a b. Thus, a c c b c c. c 0 and a b c c 0 c and b c a are positive (b c a)c bc c ac is positive ac bc. a b and c 0 b c a and cc are positive (b c a)(cc) ac c bc is positive bc ac. Since a 0, a and " are positive " 0. a a " Since 0 a b, both " and b are positive. By (3), a b and " 0 a " b " or 1 b a a a a a 1 " b7.b ab a" a" b0 " b" a . " a " " and b are both negative, i.e., 0 and b 0. By (4), a b and " " " " 1 1 b b b by (4) since b 0 b " . a aab0 " b by (3) since" a 0 b " a " a a20. (a) If a 0, then 0 kak kbk b 0 0 kak# kbk# . Since kak# kak kak ka# k a# and kbk# b# we obtain a# b# . If a 0 then kak 0 and kak kbk a# b# . On the other hand, if a# b# then a# kak# kbk# b# 0 kbk# c kak# akbk c kakb akbk b kakb . Since akbk b kakb 0 and the product akbk c kakb akbk b kakb is positive, we must have akbk c kakb 0 kbk kak . Thus kak kbk a# b# . (b) ab kabk cab c2 kabk by Exercise 19(4) above a# c 2ab b b# kak# c 2 kak kbk b kbk# , since kak# a# and kbk# b# . Factoring both sides, (a c b)# akak c kbkb# ka c bk kkak c kbkk , by part (a). 21. The fact that ka" b a# b b an k ka" k b ka# k b b kan k holds for n 1 is obvious. It also holds for n 2 by the triangle inequality. We now show it holds for all positive integers n, by induction. Suppose it holds for n k 1: ka" b a# b b ak k ka" k b ka# k b b kak k (this is the induction hypothesis). Then ka" b a# b b ak b ak 1 k kaa" b a# b b ak b b ak 1 k ka" b a# b b ak k b kak 1 k (by the triangle inequality) ka" k b ka# k b b kak k b kak 1 k (by the induction hypothesis) and the inequality holds for n k b 1. Hence it holds for all n by induction. bbbbCopyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley63 64. 64Chapter 1 Preliminaries22. The fact that ka" b a# b b an k ka" k c ka# k c c kan k holds for n 1 is obvious. It holds for n 2 by Exercise 21(b), since ka" b a# k ka" c (ca# )k kka" k c kca# kk kka" k c ka# kk ka" k c ka# k . We now show it holds for all positive integers n by induction. Suppose the inequality holds for n k 1. Then ka" b a# b b ak k ka" k c ka# k c c kak k (this is the induction hypothesis). Thus ka" b b ak b ak 1 k kaa" b b ak b c acak 1 bk kkaa" b b ak bk c kcak 1 kk (by Exercise 21(b)) kka" b b ak k c kak 1 kk ka" b b ak k c kak 1 k ka" k c ka# k c c kak k c kak 1 k (by the induction hypothesis). Hence the inequality holds for all n by induction. bbbbbb23. If f is even and odd, then f(cx) cf(x) and f(cx) f(x) f(x) cf(x) for all x in the domain of f. Thus 2f(x) 0 f(x) 0. f(cx) b f(c(cx)) f(x) b#f(cx) E(x) E # even function. Define O(x) f(x) c E(x) f(x) c f(x) b#f(cx) f(x) c#f(cx) . Then f( O(cx) f(cx) c # c(cx)) f(cx)#c f(x) c f(x) c#f(cx) cO(x) O is an odd function24. (a) As suggested, let E(x) f(x) b f(cx) # E(cx) is an f(x) E(x) b O(x) is the sum of an even and an odd function. (b) Part (a) shows that f(x) E(x) b O(x) is the sum of an even and an odd function. If also f(x) E" (x) b O" (x), where E" is even and O" is odd, then f(x) c f(x) 0 aE" (x) b O" (x)b c (E(x) b O(x)). Thus, E(x) c E" (x) O" (x) c O(x) for all x in the domain of f (which is the same as the domain of E c E" and O c O" ). Now (E c E" )(cx) E(cx) c E" (cx) E(x) c E" (x) (since E and E" are even) (E c E" )(x) E c E" is even. Likewise, (O" c O)(cx) O" (cx) c O(cx) cO" (x) c (cO(x)) (since O and O" are odd) c(O" (x) c O(x)) c(O" c O)(x) O" c O is odd. Therefore, E c E" and O" c O are both even and odd so they must be zero at each x in the domain of f by Exercise 23. That is, E" E and O" O, so the decomposition of f found in part (a) is unique. cb 4a #b 4ab c a x bb # 2acb 4a ## #25. y ax# b bx b c a x# b b x b abc(a) If a 0 the graph is a parabola that opens upward. Increasing a causes a vertical stretching and a shift of the vertex toward the y-axis and upward. If a 0 the graph is a parabola that opens downward. Decreasing a causes a vertical stretching and a shift of the vertex toward the y-axis and downward. (b) If a 0 the graph is a parabola that opens upward. If also b 0, then increasing b causes a shift of the graph downward to the left; if b 0, then decreasing b causes a shift of the graph downward and to the right. If a 0 the graph is a parabola that opens downward. If b 0, increasing b shifts the graph upward to the right. If b 0, decreasing b shifts the graph upward to the left. (c) Changing c (for fixed a and b) by ?c shifts the graph upward ?c units if ?c 0, and downward c?c units if ?c 0. 26. (a) If a 0, the graph rises to the right of the vertical line x cb and falls to the left. If a 0, the graph falls to the right of the line x cb and rises to the left. If a 0, the graph reduces to the horizontal line y c. As kak increases, the slope at any given point x x! increases in magnitude and the graph becomes steeper. As kak decreases, the slope at x! decreases in magnitude and the graph rises or falls more gradually. (b) Increasing b shifts the graph to the left; decreasing b shifts it to the right. (c) Increasing c shifts the graph upward; decreasing c shifts it downward. 27. If m 0, the x-intercept of y mx b 2 must be negative. If m 0, then the x-intercept exceeds 0 mx b 2 and x " #2 xcm " # 0 m c4.Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley" # 65. Chapter 1 Additional and Advanced Exercises 28. Each of the triangles pictured has the same base b v?t v(1 sec). Moreover, the height of each triangle is the same value h. Thus " (base)(height) #" #bh A" A# A$ . In conclusion, the object sweeps out equal areas in each one second interval.a 29. (a) By Exercise #95 of Section 1.2, the coordinates of P are ab0 bb0 # b . Thus the slope # # #of OP b/2 a/2b a . b c0 The slope of AB 0ca c b . The line a of their slopes is c" b c b c b a a a#is perpendicular to OP when a b.#(b)?y ?xsegments AB and OP are perpendicular when the product . Thus, b# a# a b (since both are positive). Therefore, ABCopyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley65 66. 66Chapter 1 PreliminariesNOTES:Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 67. CHAPTER 2 LIMITS AND CONTINUITY 2.1 RATES OF CHANGE AND LIMITS 1. (a) Does not exist. As x approaches 1 from the right, g(x) approaches 0. As x approaches 1 from the left, g(x) approaches 1. There is no single number L that all the values g(x) get arbitrarily close to as x 1. (b) 1 (c) 0 2. (a) 0 (b) c1 (c) Does not exist. As t approaches 0 from the left, f(t) approaches c1. As t approaches 0 from the right, f(t) approaches 1. There is no single number L that f(t) gets arbitrarily close to as t 0. 3. (a) True (d) False(b) True (e) False(c) False (f) True4. (a) False (d) True(b) False (e) True(c) True5.xlim x 0 kx k x kx kdoes not exist becausex kx kx x 1 if x 0 andapproaches c1. As x approaches 0 from the right,x kx kx kxkx cx c1 if x 0. As x approaches 0 from the left,approaches 1. There is no single number L that allthe function values get arbitrarily close to as x 0. 6. As x approaches 1 from the left, the values of" x c1become increasingly large and negative. As x approaches 1from the right, the values become increasingly large and positive. There is no one number L that all the " function values get arbitrarily close to as x 1, so lim xc1 does not exist. x17. Nothing can be said about f(x) because the existence of a limit as x x! does not depend on how the function is defined at x! . In order for a limit to exist, f(x) must be arbitrarily close to a single real number L when x is close enough to x! . That is, the existence of a limit depends on the values of f(x) for x near x! , not on the definition of f(x) at x! itself. 8. Nothing can be said. In order for lim f(x) to exist, f(x) must close to a single value for x near 0 regardless of x0the value f(0) itself. 9. No, the definition does not require that f be defined at x 1 in order for a limiting value to exist there. If f(1) is defined, it can be any real number, so we can conclude nothing about f(1) from lim f(x) 5. x110. No, because the existence of a limit depends on the values of f(x) when x is near 1, not on f(1) itself. If lim f(x) exists, its value may be some number other than f(1) 5. We can conclude nothing about lim f(x), x1x1whether it exists or what its value is if it does exist, from knowing the value of f(1) alone.Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 68. 68Chapter 2 Limits and Continuity11. (a) f(x) ax# c *b/(x b 3) x c3.1 f(x) c6.1c3.001 c6.001c3.0001 c6.0001c3.00001 c6.00001c3.000001 c6.000001c2.99 c5.99c2.9 c5.9x f(x)c3.01 c6.01c2.999 c5.999c2.9999 c5.9999c2.99999 c5.99999c2.999999 c5.999999The estimate is lim f(x) c6. x c$(b)x c9 xb3 #(c) f(x) (x b 3)(x c 3) xb3 x c 3 if x c3, and lim (x c 3) c3 c 3 c6. x c$12. (a) g(x) ax# c #b/ x c 2 x g(x)1.4 2.814211.41 2.824211.414 2.828211.4142 2.8284131.41421 2.8284231.414213 2.828426(b)x c2 x c 2 #(c) g(x) x b 2 x c 2 x c 2 x b 2 if x 2, and13. (a) G(x) (x b 6)/ ax# b 4x c 12b x c5.9 c5.99 G(x) c.126582 c.1251564 x G(x)c6.1 c.123456c6.01 c.124843c5.999 c.1250156 c6.001 c.124984limx #c5.9999 c.1250015 c6.0001 c.124998x b 2 2 b 2 22.c5.99999 c.1250001 c6.00001 c.124999c5.999999 c.1250000 c6.000001 c.124999Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 69. Section 2.1 Rates of Change and Limits (b)xb6 ax b 4x c 12b #(c) G(x) xb6 (x b 6)(x c 2)" xc#14. (a) h(x) ax# c 2x c 3b / ax# c 4x b 3b x 2.9 2.99 h(x) 2.052631 2.005025 x h(x)3.1 1.9523803.01 1.995024"if x c6, and limx c' x c 2" c' c 2 c " c0.125. 82.999 2.0005002.9999 2.0000502.99999 2.0000052.999999 2.00000053.001 1.9995003.0001 1.9999503.00001 1.9999953.000001 1.999999(b)x c 2x c3 x c 4x b 3 ##(c) h(x) (x c 3)(x b 1) (x c 3)(x c 1)xb1 xc115. (a) f(x) ax# c 1b / akxk c 1b x c1.1 c1.01 f(x) 2.1 2.01 x f(x)c.9 1.9c.99 1.99if x 3, and limxb1x $ xc13b1 3c14 # 2.c1.001 2.001c1.0001 2.0001c1.00001 2.00001c1.000001 2.000001c.999 1.999c.9999 1.9999c.99999 1.99999c.999999 1.999999(b)Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley69 70. Chapter 2 Limits and Continuity (c) f(x) x c" kx k c 1 #70(x b 1)(x c 1) J (x b x c 1c 1) 1)(x c(x b 1) x b 1, x 0 and x 1 , and lim (1 c x) 1 c (c1) 2. x c1 1 c x, x 0 and x c116. (a) F(x) ax# b 3x b 2b / a2 c kxkb x c2.1 c2.01 F(x) c1.1 c1.01 c1.9 c.9x F(x)c1.99 c.99c2.001 c1.001c2.0001 c1.0001c2.00001 c1.00001c2.000001 c1.000001c1.999 c.999c1.9999 c.9999c1.99999 c.99999c1.999999 c.999999(b)x b 3x b 2 2 c kx k #(c) F(x) (x b 2)(x b 1)# J (x b 2)(xc x ) b" 2bx17. (a) g()) (sin ))/) ) .1 g()) .998334, x0 , and lim (x b 1) c2 b 1 c1. x c# x b 1, x 0 and x c2.01 .999983.001 .999999.0001 .999999.00001 .999999.000001 .999999c.1 .998334c.01 .999983c.001 .999999c.0001 .999999c.00001 .999999c.000001 .99999918. (a) G(t) (1 c cos t)/t# t .1 G(t) .499583.01 .499995.001 .499999.0001 .5.00001 .5.000001 .5c.1 .499583c.01 .499995c.001 .499999c.0001 .5c.00001 .5c.000001 .5) g()) lim g()) 1)!(b)t G(t)lim G(t) 0.5t!Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 71. Section 2.1 Rates of Change and Limits (b)Graph is NOT TO SCALE 19. (a) f(x) x""cx x .9 f(x) .348678 x f(x)1.1 .385543.99 .366032.999 .367695.9999 .367861.99999 .367877.999999 .3678791.01 .3697111.001 .3680631.0001 .3678971.00001 .3678811.000001 .367878lim f(x) 0.36788x1(b)Graph is NOT TO SCALE. Also the intersection of the axes is not the origin: the axes intersect at the point (1 2.71820). 20. (a) f(x) a3x c 1b /x x .1 f(x) 1.161231.01 1.104669.001 1.099215.0001 1.098672.00001 1.098618.000001 1.098612c.1 1.040415c.01 1.092599c.001 1.098009c.0001 1.098551c.00001 1.098606c.000001 1.098611x f(x)lim f(x) 1.0986x!(b)21. lim 2x 2(2) 422. lim 2x 2(0) 023. lim (3x c 1) 3 " c 1 0 324. limx#$ "xx!c1x 1 3xc1c" 3(1) c 1" c#Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley71 72. lim 3x(2x c 1) 3(c1)(2(c1) c 1) 926.x c"1 #27. lim x sin x 11 #sin1 #28. xlim1 cos x 1 c13(c1) 2(c1)c13 c3 c1cos 1 1 c1c" 1 c1" 1 c1g(1) c g(c1) 1 c (c1)31. (a)?h ?th 34 c h 4 3 4 c 4?g ?tg(1) c g(0) 1c0(2 c 1) c (2 b 1) 1c011R(2) c R(0) 2c034.?P ?)P(2) c P(1) 2c135. (a) f(1) c f(c") 1 c (c1)2c0 #10(b)?g ?xg(0)cg(c2) 0c(c2)0c4 # c2(b)?h ?th ch6 c6?g ?tg(1) c g(c1) 1 c (c1)c1 c 1 #11?R ?)133.?f ?x 8 b1 c 1 #4 c12 c13c" #650 c 225 20 c 10 650 c 375 20 c 14 650 c 475 20 c 16.5 650 c 550 20 c 18Q# (14 375) Q$ (16.5 475) Q% (18 550)3c3 3 1(2 c 1) c (2 c ") #102c20Slope of PQ Q" (10 225)0 c 31(8 c 16 b 10)c(" c % b &) 1Q(b)#1 c1 2(b)#?g ?x 19130. (a)28 c 9 11f(3) c f(2) 3c#11?f ?x129. (a)32. (a)lim x c1 2xc1#x3x#25.Chapter 2 Limits and Continuity #72?p ?t 42.5 m/sec 45.83 m/sec 50.00 m/sec 50.00 m/sec(b) At t 20, the Cobra was traveling approximately 50 m/sec or 180 km/h. 36. (a)Slope of PQ Q Q" (5 20) Q# (7 39) Q$ (8.5 58) Q% (9.5 72)80 c 20 10 c 5 80 c 39 10 c 7 80 c 58 10 c 8.5 80 c 72 10 c 9.5?p ?t 12 m/sec 13.7 m/sec 14.7 m/sec 16 m/sec(b) Approximately 16 m/sec 37. (a)(b)?p ?t174 c 62 1994 c 1992112 # 56 thousand dollars per year(c) The average rate of change from 1991 to 1992 is ?p ?t The average rate of change from 1992 to 1993is ?p ?t62 c 27 1992 c 1991 111 c 62 1993 c 1992 35 thousand dollars per year. 49 thousand dollars per year.So, the rate at which profits were changing in 1992 is approximatley " a35 b 49b 42 thousand dollars per year. #Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 73. Section 2.1 Rates of Change and Limits 38. (a) F(x) (x b 2)/(x c 2) x 1.2 F(x) c4.0 ?F ?x ?F ?x ?F ?x?g ?x ?g ?x 1.1 c3.41.01 c3.041.001 c3.0041.0001 c3.00041 c3c4.0 c (c3) c5.0; 1.2 c 1 c3.04 c (c3) c4.04; 1.01 c 1 c3.!!!% c (c3) c4.!!!%; 1.0001 c 1?F ?x ?F ?x c3.4 c (c3) c4.4; 1.1 c 1 c3.004 c (c3) c4.!!%; 1.001 c 1 g(2) c g(1) #2 c " 0.414213 2c1 c1 1 b hc" g(1 b h) c g(1) (1 b h) c 1 h?g ?xg(1.5) c g(1) 1.5 c 1(b) The rate of change of F(x) at x 1 is c4. 39. (a)1.5 c " 0.5 0.449489(b) g(x) x 1bh 1 b h1.1 1.048801.01 1.0049871.001 1.00049981.0001 1.00004991.00001 1.0000051.000001 1.00000051 b h c 1 /h0.48800.49870.49980.4990.50.5(c) The rate of change of g(x) at x 1 is 0.5. (d) The calculator gives lim h!# ""T f(T) af(T) c f(2)b/aT c 2b 2 T61#3 ". #" c6c T T Tc##c 1 T c Tc#"(b)f(T) c f(2) Tc## "ii)f(3) c f(2) 3c2"c40. (a) i)1 b hc" h2.1 0.476190 c0.23812cT #T(T c 2)2cT c#T(2 c T)2.01 0.497512 c0.2488 c #" , T 2 T2.001 0.499750 c0.25002.0001 0.4999750 c0.25002.00001 0.499997 c0.25002.000001 0.499999 c0.2500(c) The table indicates the rate of change is c0.25 at t 2. " " (d) lim c#T c 4 T#41-46. Example CAS commands: Maple: f := x -> (x^4 c 16)/(x c 2); x0 := 2; plot( f(x), x x0-1..x0+1, color black, title "Section 2.1, #41(a)" ); limit( f(x), x x0 ); In Exercise 43, note that the standard cube root, x^(1/3), is not defined for x (surd(x+1, 3) c 1)/x. Mathematica: (assigned function and values for x0 and h may vary) Clear[f, x] f[x_]:=(x3 c x2 c 5x c 3)/(x b 1)2 x0= c1; h = 0.1; Plot[f[x],{x, x0 c h, x0 b h}] Limit[f[x], x x0]Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley73 74. 74Chapter 2 Limits and Continuity2.2 CALCULATING LIMITS USING THE LIMIT LAWS5.lim ax$ c 2x# b 4x b 8b (c2)$ c 2(c2)# b 4(c2) b 8 c8 c 8 c 8 b 8 c16x c#lim 8(t c 5)(t c 7) 8(6 c 5)(6 c 7) c8 xb3lim x # xb69.lim y c& 5 c y#y2b3 2b6y # y b 5y b 6 #13.5 8(c5) 5 c (c5)yb210. lim12.6.t'7.11.lim (10 c 3x) 10 c 3(12) 10 c 36 c26x 1#lim acx# b 5x c 2b c(2)# b 5(2) c 2 c4 b 10 c 2 48. 25 102b2 (2) b 5(#) b 6 #4.2.x##3.lim (2x b 5) 2(c7) b 5 c14 b 5 c9x c(lim 3s(2s c 1) 3 2 (x^4-81)/(x-3);x0 := 3; plot( f(x), x=x0-1..x0+1, color=black, # (a) title="Section 2.3, #61(a)" ); L := limit( f(x), x=x0 ); # (b) epsilon := 0.2; # (c) plot( [f(x),L-epsilon,L+epsilon], x=x0-0.01..x0+0.01, color=black, linestyle=[1,3,3], title="Section 2.3, #61(c)" ); q := fsolve( abs( f(x)-L ) = epsilon, x=x0-1..x0+1 ); # (d) delta := abs(x0-q); plot( [f(x),L-epsilon,L+epsilon], x=x0-delta..x0+delta, color=black, title="Section 2.3, #61(d)" ); for eps in [0.1, 0.005, 0.001 ] do # (e)Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 86. 86Chapter 2 Limits and Continuity q := fsolve( abs( f(x)-L ) = eps, x=x0-1..x0+1 ); delta := abs(x0-q); head := sprintf("Section 2.3, #61(e)n epsilon = %5f, delta = %5fn", eps, delta ); print(plot( [f(x),L-eps,L+eps], x=x0-delta..x0+delta, color=black, linestyle=[1,3,3], title=head )); end do: Mathematica (assigned function and values for x0, eps and del may vary): Clear[f, x] y1: L c eps; y2: L b eps; x0 1; f[x_]: (3x2 c (7x b 1)Sqrt[x] b 5)/(x c 1) Plot[f[x], {x, x0 c 0.2, x0 b 0.2}] L: Limit[f[x], x x0] eps 0.1; del 0.2; Plot[{f[x], y1, y2},{x, x0 c del, x0 b del}, PlotRange {L c 2eps, L b 2eps}]2.4 ONE-SIDED LIMITS AND LIMITS AT INFINITY 1. (a) True (e) True (i) False(b) True (f) True (j) False(c) False (g) False (k) True(d) True (h) False (l) False2. (a) True (e) True (i) True(b) False (f) True (j) False(c) False (g) True (k) True(d) True (h) Truelim f(x) bx#2 #b " #, lim f(x) $ c # " x#c3. (a)(b) No, lim f(x) does not exist because lim f(x) lim f(x) b 1 3, lim f(x) x%4 #bcx%x#4 #x#blim f(x) b"$cx#(c)(d) Yes, lim f(x) 3 because 3 lim f(x) lim f(x) 2 # 1, lim f(x) $ c # ", f(2) 2 x#clim f(x) bx#x%c4. (a)x%bx%(b) Yes, lim f(x) 1 because " lim f(x) lim f(x) cx c"limx c"cx c"(d) Yes, lim f(x) 4 because 4 x#f(x) 3 c (c1) 4 f(x) limx c"bx#limbx c"f(x) 3 c (c1) 4,blimcx#(c)f(x)5. (a) No, lim f(x) does not exist since sin " does not approach any single value as x approaches 0 x lim f(x) lim 0 0 x!cx!clim f(x) does not exist because lim f(x) does not existx!x!b(c)bx!(b)6. (a) Yes, lim g(x) 0 by the sandwich theorem since cx g(x) x when x 0 b bx!(b) No, lim g(x) does not exist since x is not defined for x 0 cx!(c) No, lim g(x) does not exist since lim g(x) does not exist x!cx!Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 87. Section 2.4 One-Sided Limits and Limits at Infinity lim f(x) " lim f(x)x1x1c(b)b7. (a(c) Yes, lim f(x) 1 since the right-hand and left-hand x1limits exist and equal 1lim f(x) 0 lim f(x)x1x1b(b)c8. (a)(c) Yes, lim f(x) 0 since the right-hand and left-hand x1limits exist and equal 09. (a) domain: 0 x 2 range: 0 y 1 and y 2 (b) xlimc f(x) exists for c belonging to (0 1) r (" #) (c) x 2 (d) x 010. (a) domain: c_ x _ range: c" y 1 (b) xlimc f(x) exists for c belonging to (c_ c1) r (c" ") r (" _) (c) none (d) nonelimx1c" 1 1 x b # " c # 0 ! x b" " xb1 xb6 3cx 1b1 1b6 3c1 " 7 2 1 x 7 1 7 # 1 7 h b 4h b 5 c 5 h b 4h b 5 b 5 h h b 4h b 5 b 5h!b limh(h b 4) h h b 4h b 5 b 5 #b###bbh!h!ah b 4h b 5b c 5 h h b 4h b 5 b 5# lim lim#h b 4h b 5 c 5 h#ch!blim12.2( 5 x b5 c2 " x b 1 2x b x c# b " (c#c2) bc2) (2) # 1 x ) b( #15.limx1 x b 2 c0.5 b 2 3/2 3 xc1 c0.5 b 1 1/2#14.limx c#c13.limx c!&b11.0b4 5 b 52 5Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley87 88. limx c#limx c# limx1b2x (x c 1) kx c 1 klimbakx c 1k x c 1 for x 1b lim 2x 2 x1 limx1c2x (x c 1) kx c 1 k2x (x c 1) c(x c 1)akx c 1k c(x c 1) for x 1bc lim c2x c2 (b)bclimh!tan 2x x26. lim2tt ! tan t27. limx!h!x! 2 lim" 3tsin t t ! cos t x csc 2x cos 5x)!sin cos 2x 2x x lim lim 2 lim" cos 5xx b x cos x2 36x cos x lim sin xxcos x b x!" sin 3h 3h" 3(where ) kt) (where ) 3y)3 4"sinlim9" lim x ! cos 2x x!" 312 sin 2x #x " 3(where ) 3h)122 2 lim cos t lim" sin t 9 2 " " 2 t!tt" lim cos 5x x ! sin 2x x!x ! sin x sin 2xx ! sin x cos xk1k3 sin ) ) 4 )lim! " # lim#29. limt cos t sin tt! limsin 2xx ! x cos 2x28. lim 6x# (cot x)(csc 2x) lim x!)!" 3 h lim !sin ) ) k limsin 3y 3 4 ylim! 3y 3h sin 3h lim sinx2x x!k sin ) ) lim3 sin 3y " 4 ylim! 3y limh sin 3h25. limx!k sin kt ktsin 3y 4yy!t!!23. lim limcsin kt tt!(where x 2))1sin x xx!c22. lim limc!)sin 2) 2))!t%)21. lim) )lim at c tb 4 c 3 1(b))bt%lim)$c13 3cc) )lim at c tb 4 c 4 020. (a)24.(x b 3)(c1) c(c2 b 3) c1blim)$akx b 2k c(x b 2) for x c2b2x (x c 1) (x c 1)x119. (a)akx b 2k x b 2 for x c2bb (x b 3) c(xb #2) (x )x c# c 2116 (x b 3) (c2) b 3 1climx1kx b 2 k xb2(x b 3)c(b)(xb2) (xb#)(x b 3)climx1limx c#c(0 b 11) 6 b 6h 6 b 5h b 11h b 6b18. (a)blimx c#bc(b)kx b2 k x b2(x b 3)ch(5h b 11)h!#clim6 c 5h b 11h b 6 6 5h 11h 6 6 b 5h b 11h b 6 h b b b limh 6 b 5h b 11h b 6x c#c6 c a5h b 11h b 6b #h!h!## lim17. (a) limc6 c 5h b 11h b 6 h#limh!#16.Chapter 2 Limits and Continuity #882x lim 3 cos x x!x cos x sin x cos xx lim sin x x!x sin x" cos x" # 1 (1) 2x sin 2x" #3"13b limxx ! sin x" " " lim sin x lim cos x b lim sin x (1)(1) b 1 2x!x!x c x b sin x #x #30. limxx!x lim # cx!x!" #xb " sin x 0 c # x" #b " (1) 0 #Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 89. Section 2.4 One-Sided Limits and Limits at Infinity2) #)4x 5xsin lim cos 3x 3x" sin 8xtan 3xx ! sin 4xx ! sin 8x 3 8 xlim! 36. limy!x!x!" # )lim! 5 4 limy!2) sin 2) sin 5x 5x5 4 xlim! sin lim cos 3x 3x 3 81114x sin 4x8x 3x11 5 4" #115 43 83 8sin 5y 4 5y lim siny3y cos 4y cos 5y 345y 4y sin 3 sin 3y sin 4y cos 5y y cos 4y sin 5yy!5y cos 5y lim sin 3y sin 4y sin 5y cos 4y 354 3y 4y y!Note: In these exercises we use the result" #" sin 8xx!" 8x cos 3x sin 3x sin 8x 3xsin 3y cot 5y y cot 4y sin ) )1111 "lim m x _ xn 1 since ) sin h 0 as h 0)!sin 5x35. lim)! lim sin 5x sin 4x) ! sin 2)34. limsin ) ) lim 1 since ) 1 c cos t 0 as t 0sin lim sin 2)) sin )33. limsin ) )Example 6 and the power rule in Theorem 8:limx _12 512 5 0 wheneverm n 0. This result follows immediately from" xm n sin (sin h) sin hh!)!limx _37. (a) c3"lim x _ xm n(b) c338. (a) 1" x m n (b) 139. (a)" #(b)" #40. (a)" 8(b)" 841. (a) c 5 3 3 4(b)45.limt_46. r _ lim" xcos ) 3)" 3)2 c t b sin t t b cos t x lim _ 47. (a) x lim _lim) c_ lim2 tt_r b sin r 2r b 7 c 5 sin r2x b 3 5x b 7sin 2x xr_ lim x lim _ 0 by the Sandwich Theoremcos ) 3) 0 by the Sandwich Theoremc 1 b sin t t 1 b cos t t1 b sin r r 2 b 7 c 5 sin r r r 2b 3 x 5b 7 xb7 48. (a) x lim x c2x b x b 7 x lim x _ _ (b) 2 (same process as part (a))0c1b0 1b0 c1r_ lim2 51b0 2b0c0(b)2 b x7 1 c x b x b x7 "44. c 3") # "sin 2x x$43. c " x3 4" #2 5(same process as part (a))2$42. (a)(b) c 5 3Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 0m n 0. 32. lim limsin(1 c cos t) 1ccos tt!31. lim89$#$ 90. Chapter 2 Limits and Continuity bx# ""49. (a) x lim _xb1 x b3 x lim _1 b x350. (a) x lim _3x b 7 x c2 x lim _1 c x251. (a) x lim _7x x c 3x b 6x52. (a) x lim _" x c 4x b 1x0(b) 0 (same process as part (a))0(b) 0 (same process as part (a))#b x7#3 x# x lim _7 1 c 3 b x6 x$ "#31 bx bx 110 x# "2 b x5 c x b x69 ##$$%%#%(same process as part (a)) c2 c x2 b x3c" 1 c 7 b x7 b x9 x%1cx 1bx x c_ lim2xc1b$# " x lim _58. x lim _ x c_ lim1c 1bx x 2 b x 2 c x x lim _ 2 x2xb" c1 c11_x2b3xxbb7xx_xc5b 3 x $" "$$$c %c#c "c$"&) $&x c 5x b 3 2x b x c 41c x0!""" &$ " &) &"*" &"" "& &2x c x b 7 x b 3x b xxb x" # ""c61. x lim _7 x$"c &" $"c &"#"% x lim _x x b x 3c x c_ lim60. x lim _$$#62. x c_ lim2 x# "$bx cx#x c x x b x59. x c_ lim# x lim _ c1&"# " &"# "$x 56. (a) x lim x c 7xcb 7x b 9 x lim _ _ (b) c1 (same process as part (a))#"$#(b) c 2 (same process as part (a)) 32 x b x 3x c 7c2 3#"#3 b 3 c x5 x% x lim _#c2x c 2x b 3 3x b 3x c 5x$55. (a) x lim _57. x lim _%$'9b x x lim _$ " $ "#9 #(b) 0 (same process as part (a))0'#(b)9x b x 2x b 5x c x b 6(b) 7 (same process as part (a))!x1 c x4 b x53. (a) x lim 10x bxx b 31 x lim _ _ (b) 0 (same process as part (a)) 54. (a) x lim _(# x lim _$ "90c4 x c5 #63. Yes. If lim f(x) L lim f(x), then xlima f(x) L. If lim f(x) lim f(x), then xlima f(x) does not exist. cxabxaxacbxa64. Since xlimc f(x) L if and only if lim f(x) L and lim f(x) L, then xlimc f(x) can be found by calculating xc xc lim f(x). cbbxc65. If f is an odd function of x, then f(cx) cf(x). Given lim f(x) 3, then lim f(x) c$. x!cbx!Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 91. Section 2.4 One-Sided Limits and Limits at Infinityf(x) g(x)67. Yes. If x lim _climx c#limx c#f(x) because we don't know lim f(x). x#f(x) 7. However, nothingbcan be said aboutcx#b66. If f is an even function of x, then f(cx) f(x). Given lim f(x) 7 then91 2 then the ratio of the polynomials' leading coefficients is 2, so x c_ limf(x) g(x) 2 as well.68. Yes, it can have a horizontal or oblique asymptote. f(x) g(x)69. At most 1 horizontal asymptote: If x lim _ L, then the ratio of the polynomials' leading coefficients is L, so L as well. ax b x b c a x c x b x b x b x c x ## ## #x cx x lim _#1 b x b 1 c x "#"x bxbb x b x c x b x b x c x# 70. x lim x# b x c x# c x x lim x# b x c x# c x x _ _ x 2x 2 2 x lim x lim 1 b1 1 _ _#f(x) lim x c_ g(x)#71. For any % 0, take N 1. Then for all x N we have that kf(x) c kk kk c kk 0 %. 72. For any % 0, take N 1. Then for all y cN we have that kf(x) c kk kk c kk 0 %. 73. I (5 5 b $ ) 5 x & b $ . Also, x c 5 % x c 5 %# x & b %# . Choose $ %# lim x c 5 0. bx&74. I (% c $ %) % c $ x 4. Also, % c x % % c x %# x % c %# . Choose $ %# lim % c x 0. cx%x 75. As x 0c the number x is always negative. Thus, kxk c (c1) % cx b 1 % 0 % which is always xlimx!ctrue independent of the value of x. Hence we can choose any $ 0 with c$ x ! x kx k c1.76. Since x #b we have x 2 and kx c 2k x c 2. Then, kxc2k c " xc2 c " % 0 % x c2 x c2 which is always true so long as x #. Hence we can choose any $ !, and thus # x # b $limx %!!bx %!!x c2 k x c2 k 1.x 400. Just observe that if 400 x 401, then x 400. Thus if we choose $ ", we have for anyc77. (a)limx c#b kxc2k c " %. Thus, x c2number % ! that 400 x 400 b $ lx c 400l l400 c 400l ! %. (b) lim x 399. Just observe that if 399 x 400 then x 399. Thus if we choose $ ", we have for any number % ! that 400 c $ x 400 lx c 399l l399 c 399l ! %. (c) Since lim x lim x we conclude that lim x does not exist. x %!!lim f(x) lim x 0 0; x c 0 % c% x % ! x %# for x positive. Choose $ %# x!bblim f(x) 0.x!bx!" " lim f(x) lim x# sin x 0 by the sandwich theorem since cx# x# sin x x# for all x 0.cx!x!c(b)b78. (a)x %!!cx %!!Since kx# c 0k kcx# c 0k x# % whenever kxk %, we choose $ % and obtain x# sin " c 0 % xif c$ x 0.Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 92. 92Chapter 2 Limits and Continuity (c) The function f has limit 0 at x! 0 since both the right-hand and left-hand limits exist and equal 0. )0 )" sin ) 1, ) x 3b 4 x 5 x _ 2c xlim3 b 4t limt 0 2 c 5t80. 3 #coslimxx c_ 1 b x lim)!cos ) 1b)" 1" 1, ) x , t " x"x " lim " lim zz 1, z x xx_z!" 3 b 2 cos " lim (3 b 2))(cos )) (3)(1) 3, ) x x xlimx _)03 " " " lim x c cos x 1 b sin x lim a3)# c cos )b (1 b sin )) (0 c 1)(1 b 0) c1, ) x x_)!b#84." limb83.3x b 4 x _ 2x c 5lim" x"82.x sin"81.limx _c79.2.5 INFINITE LIMITS AND VERTICAL ASYMPTOTES_4bx ! ax blimx!2 x12. limx! x# &"tan x _14.limx"# b 1c&" lim(b)2 3x$#_limx! positive positive_ negative negative c_negative positivepositive c_ c_ lim"x ! ax b# $"(b)lim x ! x (xb1)cc1 #_sec x _lim (1 b csc )) c_)!clim (2 c cot )) c_ and lim (2 c cot )) _, so the limit does not exist" x c4" x c4_ positive"positive " (xb2)(xc2) c_" positivenegative limx c#limx c#cx#" (xb2)(xc2)b## #c b climx c# limx#c(d)limx c# lim" x c4b(c)limx#" x c4 #(b)limx#b17. (a))!b)!c16._c"positive negative _3x 2xb10&"15.blimx $"13.x8.limx c&cx! x2 3x2411. limpositive positive " x c3$"limx!#10. (a)limx!6.limx$ c_5 2x#b9. (a) negative positive_4.limx!c42.positive negative c_2x x b8lim x ( (xc7) positive positive c_3 x c2c7.limx c)_b5.limx#" 3xb3.limx!c1." (xb2)(xc2) c_" positivenegative " (xb2)(xc2)_ negative"negative Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 93. Section 2.5 Infinite Limits and Vertical Asymptotescb$#c#$##$#$b#$b#$#$(x c 2)(x c ") x(x c #)(x b 2)positive negative ,x2" 4,x2xc1 x" 4,x2 negative negativenegative positive lim(x c 1) x(x b #)x#(x c 2)(x c ") x(x c #)(x b 2)limx c#(x c 2)(x c ") x(x c #)(x b 2) lim(x c 2)(x c ") x(x c #)(x b 2) lim c_ xc" x(x b #)cc b blimx!x"x#x! x"b# limcb#x c 3x b 2 x c 4xx!" 4xc1 xclim lim c_x c#x c 3x b 2 x c 4xxc" x(x b #)(x c 2)(x c 1) x (x c 2)b limxc1 xb$x c 3x b 2 x c 4xx#(x c 2)(x c 1) x (x c 2)bx# limx# c_negative negativenegative positiveb#and lim(x c 2)(x c 1) x (x c 2)#b#limx!x c 3x b 2 x c 4x lim#$limx"x!(x c 2)(x c 1) x (x c 2)x c1 2x b 4 c_###limx#cx#(x c 2)(x c 1) x (x c 2)# limlimx c#0cx c 3x b 2 x c 2xx#20 #b4(b)# limbcx c 3x b 2 x c 2xx!#b# lim limx03 #bc" 4bx c 3x b 2 x c 2xlim 2c"$ c 2c"$ 0(x b 1)(x c 1) 2x b 4x c 3x b 2 x c 2xx c#" positive #bx c1 2x b 4x"b limlimlim_ positive positive limx#" cx_x c 3x b 2 x c 2xx!"" negative " c c1 limx## c_#" ##x c1 2x b 4 #(e)x c1 2x b 4 #(d)#(c)2" cxc$(b)" xc$"c22. (a)x!" xc#(e)limx#x!$#b(d)limx#negative negativenegative 0 b lim#(c)lim c_" x#(b)limx!x (xb1)(xc1)#c21. (a)negative positivenegative cxx c#lim_x #xx!limx c"x (xb1)(xc1)c#(d)limx c" 0 b limlim x c1 #limpositive positivenegative " x(d)x" c_clim # x 2(c)x (xb1)(xc1)x #(c)20. (a)positive positivepositive b_climx!x x c1blim#limx c"x x c1x (xb1)(xc1)bc(b)limx c"x!x"#19. (a) limx"c(d) limx x c1b(c)limx"x x c1 #(b)limx"b18. (a)" #(4)(x c 1) x(x b #)" 8_(x c 1) x(x b #)_(x c 1) x(x b #)negative negativepositive negative negativepositive 0 (1)(3)0negative positivepositive _negative negativepositive so the function has no limit as x 0. x" b2 (x c 1)_(d)limx!limx"c(b)" t_b 7 c_ x" b2 (x c 1)_ x" b2 (x c 1)_$#_ x-1; plot( [f(x),g(x)], x=-5..5, legend=["y = f(x)","y = g(x)"], title="#91(a) (Section 5.6)" ); q1 := [ -5, -2, 1, 4 ]; # (b) q2 := [seq( fsolve( f(x)=g(x), x=q1[i]..q1[i+1] ), i=1..nops(q1)-1 )]; for i from 1 to nops(q2)-1 do # (c) area[i] := int( abs(f(x)-g(x)),x=q2[i]..q2[i+1] ); end do; add( area[i], i=1..nops(q2)-1 ); # (d) Mathematica: (assigned functions may vary) Clear[x, f, g] f[x_] = x2 Cos[x] g[x_] = x3 c x Plot[{f[x], g[x]}, {x, c2, 2}] After examining the plots, the initial guesses for FindRoot can be determined. pts = x/.Map[FindRoot[f[x]==g[x],{x, #}]&, {c1, 0, 1}] i1=NIntegrate[f[x] c g[x], {x, pts[[1]], pts[[2]]}] i2=NIntegrate[f[x] c g[x], {x, pts[[2]], pts[[3]]}] i1 b i2Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley343 344. 344Chapter 5 IntegrationCHAPTER 5 PRACTICE EXERCISES 1. (a) Each time subinterval is of length ?t 0.4 sec. The distance traveled over each subinterval, using the midpoint rule, is ?h " avi b vi 1 b ?t, where vi is the velocity at the left endpoint and vi 1 the velocity at # bbthe right endpoint of the subinterval. We then add ?h to the height attained so far at the left endpoint vi to arrive at the height associated with velocity vi 1 at the right endpoint. Using this methodology we build the following table based on the figure in the text: t (sec) 0 0.4 0.8 1.2 1.6 2.0 2.4 2.8 3.2 3.6 4.0 4.4 4.8 5.2 5.6 6.0 v (fps) 0 10 25 55 100 190 180 165 150 140 130 115 105 90 76 65 h (ft) 0 2 9 25 56 114 188 257 320 378 432 481 525 564 592 620.2 bt (sec) v (fps) h (ft)6.4 50 643.26.8 37 660.67.2 25 6727.6 12 679.48.0 0 681.8NOTE: Your table values may vary slightly from ours depending on the v-values you read from the graph. Remember that some shifting of the graph occurs in the printing process. The total height a