Student’s solutions manual : introductory mathematical analysis for business, economics, and the life and social sciences

719 Pages·2011·10.77 MB·English

by  Haeussler

Table of Contents
Chapter 0 1
Chapter 1 35
Chapter 2 54
Chapter 3 89
Chapter 4 132
Chapter 5 160
Chapter 6 177
Chapter 7 231
Chapter 8 295
Chapter 9 333
Chapter 10 357
Chapter 11 378
Chapter 12 423
Chapter 13 469
Chapter 14 539
Chapter 15 614
Chapter 16 658
Chapter 17 670
Chapter 0
Problems 0.1 x+2 x 2 x
7. True; = + = +1.
2 2 2 2
1. True; –13 is a negative integer.
⎛b⎞ ab
2. True, because −2 and 7 are integers and 7 ≠ 0. 8. True, because a⎜ ⎟= .
⎝c⎠ c
3. False, because the natural numbers are 1, 2, 3,
and so on. 9. False; the left side is 5xy, but the right side is
5x2y.
0
4. False, because 0= .
1 10. True; by the associative and commutative
properties, x(4y) = (x ⋅ 4)y = (4 ⋅ x)y = 4xy.
5
5. True, because 5= .
1 11. distributive
6. False, since a rational number cannot have 12. commutative
7
denominator of zero. In fact, is not a number 13. associative
0
at all because we cannot divide by 0. 14. definition of division
7. False, because 25=5, which is a positive 15. commutative and distributive
integer.
16. associative
8. True; 2 is an irrational real number.
17. definition of subtraction
9. False; we cannot divide by 0.
18. commutative
10. False, because the natural numbers are 1, 2, 3,
19. distributive
and so on, and 3 lies between 1 and 2.
20. distributive
11. True
21. 2x(y − 7) = (2x)y − (2x)7 = 2xy − (7)(2x)
12. False, since the integer 0 is neither positive nor
= 2xy − (7 · 2)x = 2xy − 14x
negative.
22. (a − b) + c = [a + (−b)] + c = a + (−b + c)
Problems 0.2
= a + [c + (−b)] = a + (c − b)
1. False, because 0 does not have a reciprocal.
23. (x + y)(2) = 2(x + y) = 2x + 2y
7 3 21
2. True, because ⋅ = =1. 24. 2[27 + (x + y)] = 2[27 + (y + x)] = 2[(27 + y) + x]
3 7 21 = 2[(y + 27) + x]
3. False; the negative of 7 is −7 because 25. x[(2y + 1) + 3] = x[2y + (1 + 3)] = x[2y + 4]
7 + (−7) = 0. = x(2y) + x(4) = (x · 2)y + 4x = (2x)y + 4x
= 2xy + 4x
4. False; 2(3 · 4) = 2(12) = 24, but
(2 · 3)(2 · 4) = 6 · 8 = 48. 26. (1 + a)(b + c) = 1(b + c) + a(b + c)
= 1(b) + 1(c) + a(b) + a(c) = b + c + ab + ac
5. False; –x + y = y + (–x) = y – x.
6. True; (x + 2)(4) = (x)(4) + (2)(4) = 4x + 8.
1
Chapter 0: Review of Algebra ISM: Introductory Mathematical Analysis
27. x(y − z + w) = x[(y − z) + w] = x(y − z) + x(w) 51. X(1) = X
= x[y + (−z)] + xw = x(y) + x(−z) + xw
52. 3(x – 4) = 3(x) – 3(4) = 3x – 12
= xy − xz + xw
53. 4(5 + x) = 4(5) + 4(x) = 20 + 4x
28. –2 + (–4) = –6
54. –(x – 2) = –x + 2
29. –6 + 2 = –4
55. 0(–x) = 0
30. 6 + (–4) = 2
31. 7 – 2 = 5 ⎛ 1 ⎞ 8⋅1 8
56. 8⎜ ⎟= =
⎝11⎠ 11 11
32. 7 – (–4) = 7 + 4 = 11
5
33. −5 − (−13) = −5 + 13 = 8 57. =5
1
34. −a − (−b) = −a + b
14x 2⋅7⋅x 2x
58. = =
35. (–2)(9) = –(2 · 9) = –18 21y 3⋅7⋅y 3y
36. 7(–9) = –(7 · 9) = –63
3 3 3
59. = =−
37. (–2)(–12) = 2(12) = 24 −2x −(2x) 2x
38. 19(−1) = (−1)19 = −(1 · 19) = −19 2 1 2⋅1 2
60. ⋅ = =
3 x 3⋅x 3x
−1 ⎛ 9⎞
39. =−1⎜− ⎟=9
−1 ⎝ 1⎠ a a(3b) 3ab
9 61. (3b)= =
c c c
40. –(–6 + x) = –(–6) – x = 6 – x
⎛ 7 ⎞
41. –7(x) = –(7x) = –7x 62. (5a)⎜ ⎟=7
⎝5a⎠
42. –12(x – y) = (–12)x – (–12)(y) = –12x + 12y
−aby −a⋅by by
(or 12y – 12x)
63. = =
−ax −a⋅x x
43. –[–6 + (–y)] = –(–6) – (–y) = 6 + y
7 1 7⋅1 7
−3 3 1⋅3 1 64. ⋅ = =
44. −3÷15= =− =− =− y x y⋅x xy
15 15 5⋅3 5
2 5 2⋅5 10
−9 9 9⋅1 1 65. ⋅ = =
45. −9÷(−27)= = = = x y x⋅y xy
−27 27 9⋅3 3
1 1 3 2 3+2 5
−a a 66. + = + = =
46. (−a)÷(−b)= = 2 3 6 6 6 6
−b b
5 3 5 9 5+9 14 2⋅7 7
47. 2(–6 + 2) = 2(–4) = –8 67. + = + = = = =
12 4 12 12 12 12 2⋅6 6
48. 3[–2(3) + 6(2)] = 3[–6 + 12] = 3[6] = 18
3 7 9 14 9−14 −5 5⋅1 1
49. (–2)(–4)(–1) = 8(–1) = –8 68. − = − = = =− =−
10 15 30 30 30 30 5⋅6 6
50. (−12)(−12) = (12)(12) = 144
2
ISM: Introductory Mathematical Analysis Section 0.3
4 6 4+6 10 (a3)7 a3⋅7 a21
69. + = = =2 7. = =
5 5 5 5 (b4)5 b4⋅5 b20
X Y X −Y
70. − = ⎛x2 ⎞5 (x2)5 x2⋅5 x10
5 5 5 8. ⎜ ⎟ = = =
⎜⎝ y3⎟⎠ (y3)5 y3⋅5 y15
3 1 1 18 3 2 18−3+2 17
71. − + = − + = =
2 4 6 12 12 12 12 12 9. (2x2y3)3 =23(x2)3(y3)3 =8x2⋅3y3⋅3 =8x6y9
2 3 16 15 16−15 1
2
72. − = − = = ⎛w2s3⎞ (w2s3)2 (w2)2(s3)2 w2⋅2s3⋅2
5 8 40 40 40 40 10. ⎜ ⎟ = = =
⎜⎝ y2 ⎟⎠ (y2)2 y2⋅2 y4
73. 6 =6÷ x =6⋅ y = 6y w4s6
x y x x =
y y4
74. 3l = l ÷m = l ⋅ 1 = l 11. x9 =x9−5 =x4
m 3 1 3 m 3m x5
−x 6
75. y2 =− x ÷ z =− x ⋅xy =−x2 12. ⎛⎜2a4 ⎞⎟ = (2a4)6
xzy y2 xy y2 z yz ⎜⎝7b5 ⎟⎠ (7b5)6
26(a4)6
=
76. 7 is not defined (we cannot divide by 0). 76(b5)6
0 64a4⋅6
=
0 117,649b5⋅6
77. 7 =0 64a24
=
117,649b30
0
78. is not defined (we cannot divide by 0).
0 (x3)6 x3⋅6 x18
13. = = =x18−4 = x14
x(x3) x1+3 x4
79. 0 · 0 = 0
Problems 0.3 (x2)3(x3)2 x2⋅3x3⋅2 x6x6 x12
14. = = =
(x3)4 x3⋅4 x12 x12
1. (23)(22)=23+2 =25(=32)
x12−12 = x0 =1
2. x6x9 =x6+9 =x15
15. 25=5
3. w4w8 =w4+8 =w12 16. 481=3
4. z3zz2 =z3+1+2 =z6 17. 7−128 =−2
x3x5 x3+5 x8
5. = = 18. 0.04 =0.2
y9y5 y9+5 y14
1 41 1
19. 4 = =
6. (x12)4 =x12⋅4 =x48 16 416 2
3
Chapter 0: Review of Algebra ISM: Introductory Mathematical Analysis
–8 3–8 −2 2 3 3 13 39 39 39
20. 3 = = =− 36. = ⋅ = = =
27 327 3 3 13 13 13 132 132 13
21. (49)1/2 = 49 =7 37. (9z4)1/2 = 9z4 = 32(z2)2 = 32 (z2)2
=3z2
22. (64)1/3 =364 =4
3 3
23. 93/2 =( 9)3 =(3)3 =27 38. (16y8)3/4 =⎡⎢⎣416y8⎤⎥⎦ =⎡⎢⎣4(2y2)4⎤⎥⎦ =(2y2)3
=8y6
1 1 1 1
24. (9)−5/2 = = = =
(9)5/2 ( 9)5 35 243 ⎛27t3⎞2/3 ⎛⎡3t⎤3⎞2/3 ⎡3t⎤2 9t2
39. ⎜ ⎟ =⎜ ⎟ = =
⎜⎝ 8 ⎟⎠ ⎜⎝⎢⎣2⎥⎦ ⎟⎠ ⎢⎣2⎥⎦ 4
1 1 1 1
25. (32)–2/5 = = = =
(32)2/5 (532)2 (2)2 4 ⎛256⎞−3/4 ⎛⎡ 4 ⎤4⎞−3/4 ⎡ 4 ⎤−3 4−3
40. ⎜ ⎟ =⎜⎢ ⎥ ⎟ =⎢ ⎥ =
⎝ x12 ⎠ ⎜⎣x3⎦ ⎟ ⎣x3⎦ (x3)−3
1 1 1 ⎝ ⎠
26. (0.09)–1/2 = (0.09)1/2 = 0.09 = 0.3 = 4−3 = x9 = x9
x−9 43 64
1 10
= =
3 3
10 a5b−3 1 1 1 a5
41. =a5⋅b−3⋅ =a5⋅ ⋅ =
c2 c2 b3 c2 b3c2
⎛ 1 ⎞4/5 ⎛ 1 ⎞4 ⎛1⎞4 1
27. ⎜⎝32⎟⎠ =⎜⎜⎝5 32⎟⎟⎠ =⎜⎝2⎟⎠ =16 42. 5 x2y3z–10 =x2/5y3/5z–10/5 = x2/5y3/5
z2
⎛ 64⎞2/3 ⎛ 64⎞2 ⎛ 4⎞2 16
28. ⎜⎝−27⎟⎠ =⎜⎜⎝3−27⎟⎟⎠ =⎜⎝−3⎟⎠ = 9 43. 5m−2m−7 =5m−2+(−7) =5m−9 = m59
29. 50 = 25⋅2 = 25⋅ 2 =5 2 1
44. x+y–1= x+
y
30. 354 =327⋅2 =327 32 =332
1 1
45. (3t)–2 = =
31. 32x3 =32 3x3 =x 32 (3t)2 9t2
32. 4x = 4 x =2 x 1
46. (3−z)–4 =
(3−z)4
33. 16x4 = 16 x4 =4x2
47. 55x2 =(5x2)1/5 =51/5(x2)1/5 =51/5x2/5
x 4x 4x
34. 4 = =
16 416 2 48. (X3Y−3)−3 =(X3)−3(Y−3)−3
= X−9Y9
35. 2 8−5 27+3128=2 4⋅2−5 9⋅3+364⋅2
Y9
=2⋅2 2−5⋅3 3+432 =
X9
=4 2−15 3+432
4
ISM: Introductory Mathematical Analysis Section 0.3
49. x− y =x1/2−y1/2 4 4 4(2x)1/2 4 2x
61. = = =
2x (2x)1/2 (2x)1/2(2x)1/2 2x
u−2v−6w3 w3−(−5) w8
50. = = 2 2x
vw−5 u2v1−(−6) u2v7 =
x
51. x24xy–2z3 =x2(xy–2z3)1/4 =x2x1/4y–2/4z3/4 y y y(2y)1/2 y 2y
62. = = =
x9/4z3/4 2y (2y)1/2 (2y)1/2(2y)1/2 2y
=
y1/2 2y
=
2
52. 4a−3b−2a5b−4 =(a−3b−2)1/4a5b−4
=a−3/4b−1/2a5b−4 1 1 1(3x)2/3 3(3x)2
63. = = =
=a17/4b−9/2 33x (3x)1/3 (3x)1/3(3x)2/3 3x
a17/4 39x2
=
=
b9/2 3x
53. (2a−b+c)2/3 =3(2a−b+c)2 2 2 2⋅y1/3 2y1/3 23 y
64. = = = =
33 y2 3y2/3 3y2/3⋅y1/3 3y 3y
54. (ab2c3)3/4 =4(ab2c3)3 =4a3b6c9
12 12
65. = = 4 =2
1 1
55. x–4/5 = = 3 3
x4/5 5 x4
18
66. = 9 =3
56. 2x1/2−(2y)1/2 =2 x− 2y 2
57. 3w−3/5−(3w)−3/5 = 3 − 1 67. 52 = 52 = 52⋅a1/2b3/4
w3/5 (3w)3/5 4a2b a2/4b1/4 a1/2b1/4⋅a1/2b3/4
= 3 − 1 = 3 − 1 21/5a1/2b3/4 24/20a10/20b15/20
= =
5w3 5(3w)3 5w3 527w3 ab ab
(24a10b15)1/20 2016a10b15
= =
58. [(x–4)1/5]1/6 =[x−4/5]1/6 =x–4/30 = x–2/15 ab ab
1 1
= x2/15 =15x2 68. 2 = 2 = 21/2⋅32/3 = 23/634/6
33 31/3 31/3⋅32/3 3
6 6 6⋅51/2 6 5 (2334)1/6 6648
59. = = = = =
5 51/2 51/2⋅51/2 5 3 3
60. 3 = 3 = 3⋅21/4 =342 = 342 69. 2x2y–3x4 =2x6y–3 = 2x6
48 81/4 81/4⋅21/4 416 2 y3
3 3⋅u1/2v1/2 3u1/2v1/2
70. = =
u5/2v1/2 u5/2v1/2⋅u1/2v1/2 u3v
5
Chapter 0: Review of Algebra ISM: Introductory Mathematical Analysis
243 243 82. 75k4 =(75k4)1/2 =[(25k4)(3)]1/2
71. = = 81=9
3 3 =[(5k2)23]1/2 =5k231/2
72. {[(3a3)2]−5}−2 ={[32a6]−5}−2
(ab−3c)8 a8b−24c8 a5c14
={3−10a−30}−2 83. = =
(a−1c2)−3 a3c−6 b24
=320a60
20 1 26y6 84. 37(49) =37⋅72 =373 =7
73. = =
(2–2x1/2y–2)3 2–6x3/2y–6 x3/2
2
(x2)3 ⎡ x3 ⎤ x6 (x3)2
64y6⋅x1/2 64y6x1/2 85. ÷⎢ ⎥ = ÷
= x3/2⋅x1/2 = x2 x4 ⎢⎣(x3)2⎥⎦ x4 (x6)2
x6
=x2÷ = x2÷x6−12 = x2÷x−6
s5 s5/2 s15/6 x12
74. = = =s11/6
3s2 s2/3 s4/6 =x2÷ 1 =x2⋅x6 = x8
x6
75. 3 x2yz33 xy2 =3(x2yz3)(xy2) =3 x3y3z3
86. (–6)(–6) = 36 =6
=xyz
Note that (–6)2 ≠−6 since –6 < 0.
76. (43)8 =(31/4)8 =38/4 =32 =9
8s–2 4 4
87. − =− =−
77. 32(32)−2/5 =32(25)−2/5 2s3 s3s2 s5
=32(2−2) 88. (a5b−3 c)3 =(a5)3(b−3)3(c1/2)3
1
=32⋅
22 =a15b−9c3/2
= 9 a15c3/2
=
4
b9
2/5
78. ⎛⎜5 x2y⎞⎟ =[(x2y)1/5]2/5 =(x2y)2/25 ⎛ 3x3y2 ⎞4
⎝ ⎠ 89. (3x3y2÷2y2z−3)4 =⎜ ⎟
=x4/25y2/25 ⎜⎝2y2z−3⎟⎠
4
⎛3x3z3⎞
79. (2x–1y2)2 =22x–2y4 = 4y4 =⎜⎜⎝ 2 ⎟⎟⎠
x2 (3x3z3)4
=
3 3 3⋅y2/3x3/4 (2)4
80. 3 y4 x = y1/3x1/4 = y1/3x1/4⋅y2/3x3/4 = 34x12z12
24
3x3/4y2/3
= 81x12z12
xy =
16
81. x x2y3 xy2 =x1/2(x2y3)1/2(xy2)1/2 90. 1 = 1 = 1 = 1 =8x10
=x1/2(xy3/2)(x1/2y)=x2y5/2 ( 126xx–23)2 ((1261/12/2)2)2((xx–32))22 126xx–64 8x110
6
ISM: Introductory Mathematical Analysis Section 0.4
Problems 0.4 18. −{−6a − 6b + 6 + 10a + 15b − a[2b + 10]}
= −{4a + 9b + 6 − 2ab − 10a}
1. 8x – 4y + 2 + 3x + 2y – 5 = 11x – 2y – 3
= −{−6a + 9b + 6 − 2ab}
= 6a − 9b − 6 + 2ab
2. 6x2−10xy+2+2z−xy+4
=6x2−11xy+2z+6 19. x2+(4+5)x+4(5)=x2+9x+20
3. 8t2−6s2+4s2−2t2+6=6t2−2s2+6 20. u2+(5+2)u+2(5)=u2+7u+10

4. x+2 x+ x+3 x =7 x 21. (w+2)(w−5)=w2+(−5+2)x+2(−5)
=w2−3w−10
5. a+2 3b− c+3 3b
= a+5 3b− c 22. z2+(–7−3)z+(–7)(–3)= z2−10z+21
6. 3a + 7b − 9 − 5a − 9b − 21 = −2a − 2b − 30
23. (2x)(5x)+[(2)(2)+(3)(5)]x+3(2)
7. 6x2 –10xy+ 2−2z+xy−4 =10x2+19x+6
=6x2−9xy−2z+ 2−4 24. (t)(2t) + [(1)(7) + (−5)(2)]t + (−5)(7)
=2t2−3t−35
8. x+2 x− x−3 x =− x
25. X2+2(X)(2Y)+(2Y)2 = X2+4XY+4Y2
9. x+ 2y− x− 3z = 2y− 3z
26. (2x)2−2(2x)(1)+12 =4x2−4x+1
10. 8z – 4w – 3w + 6z = 14z – 7w
11. 9x + 9y – 21 – 24x + 6y – 6 = –15x + 15y – 27 27. x2−2(5)x+52 =x2−10x+25
12. u − 3v − 5u − 4v + u − 3 = −3u − 7v − 3
( )2
28. (1⋅2) x +[(1)(5)+(–1)(2)] x+(–1)(5)
13. 5x2−5y2+xy−3x2−8xy−28y2
=2x+3 x−5
=2x2−33y2−7xy
29. ( 3x)2+2( 3x)(5)+(5)2
14. 2 – [3 + 4s – 12] = 2 – [4s – 9] = 2 – 4s + 9
= 11 – 4s =3x+10 3x+25
15. 2{3[3x2+6−2x2+10]}=2{3[x2+16]} 30. ( y)2−32 = y−9
=2{3x2+48}=6x2+96
31. (2s)2−12 =4s2−1
16. 4{3t + 15 – t[1 – t – 1]} = 4{3t + 15 – t[–t]}
=4{3t+15+t2}=4t2+12t+60
32. (z2)2−(3w)2 =z4−9w2
17. −5(8x3+8x2−2(x2−5+2x))
33. x2(x+4)−3(x+4)
=−5(8x3+8x2−2x2+10−4x)
=x3+4x2−3x−12
=−5(8x3+6x2−4x+10)
=−40x3−30x2+20x−50 34. x(x2+x+3)+1(x2+x+3)
=x3+x2+3x+x2+x+3
=x3+2x2+4x+3
7
Chapter 0: Review of Algebra ISM: Introductory Mathematical Analysis
35. x2(3x2+2x−1)−4(3x2+2x−1) 2x3 7x 4 4
46. − + =2x2−7+
=3x4+2x3−x2−12x2−8x+4 x x x x
=3x4+2x3−13x2−8x+4
6x5 4x3 1 1
47. + − =3x3+2x−
36. 3y(4y3+2y2−3y)−2(4y3+2y2−3y) 2x2 2x2 2x2 2x2
=12y4+6y3−9y2−8y3−4y2+6y
3y−4−9y−5
=12y4−2y3−13y2+6y 48.
3y
−6y−9
37. x{2(x2−2x−35)+4[2x2−12x]} =
3y
=x{2x2−4x−70+8x2−48x} −6y 9
= −
=x{10x2−52x−70} 3y 3y
=10x3−52x2−70x =−2−3
y
38. [(2z)2−12](4z2+1)=[4z2−1](4z2+1)
x
=(4z2)2−12 =16z4−1
49. x+5 x2+5x−3
39. x(3x + 2y – 4) + y(3x + 2y – 4) + 2(3x + 2y – 4) x2+5x
=3x2+2xy−4x+3xy+2y2−4y+6x+4y−8 −3
=3x2+2y2+5xy+2x−8 Answer: x+ −3
x+5
40. [x2+(x+1)]2
x−1
=(x2)2+2x2(x+1)+(x+1)2 50. x−4 x2−5x+4
=x4+2x3+2x2+x2+2x+1 x2−4x
=x4+2x3+3x2+2x+1 –x+4
–x+4
41. (2a)3+3(2a)2(3)+3(2a)(3)2+(3)3 0
Answer: x – 1
=8a3+36a2+54a+27
3x2−8x+17
42. (3y)3−3(3y)2(2)+3(3y)(2)2−(2)3
51. x+2 3x3−2x2+ x− 3
=27y3−54y2+36y−8
3x3+6x2
43. (2x)3−3(2x)2(3)+3(2x)(3)2−33 –8x2+ x
–8x2−16x
=8x3−36x2+54x−27
17x− 3

44. x3+3x2(2y)+3x(2y)2+(2y)3 17x+34
–37
=x3+6x2y+12xy2+8y3
–37
Answer: 3x2−8x+17+
x+2
z2 18z
45. − =z−18
z z
8
ISM: Introductory Mathematical Analysis Section 0.5
x3+x2+3x+3 z+2
52. x−1 x4+0x3+2x2+0x+1 56. z2−z+1 z3+z2+ z
x4− x3 z3−z2+ z
x3+2x2 2z2

x3– x2 2z2−2z+2
3x2+0x 2z−2

3x2−3x Answer: z+2+ 2z−2
z2−z+1
3x+1

3x−3
Problems 0.5
4
1. 2(ax + b)
4
Answer: x3+x2+3x+3+
x−1
2. 2y(3y – 2)
x2−2x+4 3. 5x(2y + z)
53. x+2 x3+0x2+0x+0
4. 3x2y(1−3xy2)
x3+2x2
−2x2+0 5. 4bc(2a3−3ab2d+b3cd2)
−2x2−4x
4x+0 6. 6u2v(uv2+3w4−12v2)

4x+8
−8 7. z2−72 =(z+7)(z−7)
8
Answer: x2−2x+4−
x+2 8. (x + 2)(x − 3)
3x−1 9. (p+3)(p+1)
2
54. 2x+3 6x2+8x+1 10. (s – 4)(s – 2)
6x2+9x
−x+1 11. (4x)2−32 =(4x+3)(4x−3)

−x−3
2 12. (x + 6)(x – 4)
5
2
13. (a + 7)(a + 5)
5
1
Answer: 3x− + 2
2 2x+3 14. (2t)2−(3s)2 =(2t+3s)(2t−3s)
x−2 15. x2+2(3)(x)+32 =(x+3)2
55. 3x+2 3x2−4x+3
3x2+2x 16. (y – 10)(y – 5)
–6x+3
–6x−4 17. 5(x2+5x+6)
=5(x+3)(x+2)
7
7
Answer: x−2+ 18. 3(t2+4t−5)
3x+2
=3(t−1)(t+5)
9