STUDENT SOLUTIONS MANUAL FOR
CALCULUS: A COMPLETE COURSE – 10TH
EDITION BY ADAMS & ESSEX – FULL WORKED
SOLUTIONS
,
, INSTRUCTOR’S SOLUTIONS MANUAL SECTION P.1 (PAGE 10)
CHAPTER P. PRELIMINARIES 19. Given: 1/(2 — x) < 3.
CASE I. If x < 2, then 1 < 3(2 – x) = 6 − 3x , so 3x < 5
and x < 5/3. This case has solutions x < 5/3.
Section P.1 Real Numbers and the Real Line CASE II. If x > 2, then 1 > 3(2 x−) = 6 − 3x , so 3x > 5
(page 10) and x > 5/3. This case has solutions x > 2.
Solution: (−∞, 5/3) ∪ (2, ∞).
2
1. = 0.22222222 ·· · = 0.2 20. Given: (x + 1)/x ≥ 2.
9 CASE I. If x > 0, then x + 1 ≥ 2x , so x ≤ 1.
2. 1 CASE II. If x < 0, then x + 1 ≤ 2x , so x ≥ 1. (not
= 0.09090909 ··· = 0.09 possible)
11
3. If x = 0.121212 ·· ·, then 100x = 12.121212 ··· = 12 + x . Solution: (0, 1].
Thus 99x = 12 and x = 12/99 = 4/33. 21. Given: x 2 − 2x ≤ 0. Then x(x − 2) ≤ 0. This is only
4. If x = 3.277777 ·· ·, then 10x − 32 = 0.77777 ··· and possible if x ≥ 0 and x ≤ 2. Solution: [0, 2].
100x − 320 = 7 + (10x − 32), or 90x = 295. Thus 22. Given 6x 2 − 5x ≤ −1, then (2x − 1)(3x − 1) ≤ 0, so
x = 295/90 = 59/18. either x ≤ 1/ 2 and x ≥ 1/ 3, or x ≤ 1/ 3 and x ≥ .
The latter combination is not possible. The solution set is
5. 1/7 = 0.142857142857 ··· = 0.142857 [1/3, 1/2].
2/7 = 0.285714285714 ··· = 0.285714
3/7 = 0.428571428571 ··· = 0.428571 23. Given x 3 > 4x , we have x(x 2 − 4 ) > 0. This is possible
if x < 0 and x2 < 4, or if x > 0 and x2 > 4. The
4/7 = 0.571428571428 ··· = 0.571428 possibilities are, therefore, −2 < x < 0 or 2 < x < ∞.
note the same cyclic order of the repeating digits Solution: (−2, 0) ∪ (2, ∞).
5/7 = 0.714285714285 ··· = 0.714285 24. Given x 2−x ≤ 2, then x2− x −2 ≤ 0 so (x −2)(x +1) ≤ 0.
6/7 = 0.857142857142 ··· = 0.857142 This is possible if x ≤ 2 and x ≥ −1 or if x ≥ 2 and
x ≤ −1. The latter situation is not possible. The solution
6. Two different decimal expansions can represent the same
set is [−1, 2].
number. For instance, both 0.999999 ··· = 0.9 and
1.000000 ··· = 1.0 represent the number 1. x 4
25. Given: ≥ 1+ .
7. x ≥ 0 and x ≤ 5 define the interval [0, 5]. 2 x 2
CASE I. If x > 0, then x ≥ 2x + 8, so that
8. x < 2 and x ≥ −3 define the interval [−3, 2). x 2 − 2x − 8 ≥ 0, or (x − 4)(x + 2) ≥ 0. This is
possible for x > 0 only if x ≥ 4.
9. x > −5 or x < −6 defines the union CASE II. If x < 0, then we must have (x − 4)(x + 2) ≤ 0,
(−∞, −6) ∪ (−5, ∞). which is possible for x < 0 only if x ≥ −2.
10. x ≤ −1 defines the interval (−∞, −1]. Solution: [−2, 0) ∪ [4, ∞).
3 2
11. x > −2 defines the interval (−2, ∞). 26. Given: < .
x−1 x+1
12. x < 4 or x ≥ 2 defines the interval ( −∞ ,∞ ), that is, the CASE I. If x > 1 then (x − 1)(x + 1) > 0, so that
whole real line. 3(x +1) < 2(x −1 ). Thus x < 5 −. There are no solutions
in this case.
13. If −2x > 4, then x < −2. Solution: (−∞, −2) CASE II. If −1 < x < 1, then (x − 1)(x + 1 ) < 0, so
3(x + 1) > 2(x − 1). Thus x > −5. In this case all
14. If 3x + 5 ≤ 8, then 3x ≤ 8 − 5 − 3 and x ≤ 1. Solution:
(−∞, 1] numbers in (−1, 1) are solutions.
CASE III. If x < −1, then (x − 1)(x + 1) > 0, so that
15. If 5x − 3 ≤ 7 − 3x , then 8x ≤ 10 and x ≤ 5/4. Solution: 3(x + 1) < 2(x − 1). Thus x < −5. All numbers x < −5
(−∞, 5/4] are solutions.
6−x 3x − 4 Solutions : (−∞, −5) ∪ (−1, 1).
16. If 2 , then 6 − x ≥ 6x − 8. Thus 14 ≥ 7x
4 ≥ 27. If |x | = 3 then x = ±3.
and x ≤ 2. Solution: (−∞, 2]
28. If |x − 3| = 7, then x − 3 = ±7, so x = −4 or x = 10.
17. If 3(2 − x) < 2(3 + x), then 0 < 5x and x > 0. Solution:
(0, ∞) 29. If |2t + 5| = 4, then 2t + 5 = ±4, so t = −9/2 or
t = −1/2.
18. If x 2 < 9, then |x | < 3 and −3 < x < 3. Solution:
(−3, 3) 30. If|1 − t| = 1, then 1 − t = ±1, so t = 0 or t = 2.
1
, SECTION P.1 (PAGE 10) R. A. ADAMS: CALCULUS
31. If |8 − 3s| = 9, then 8 − 3s = ±9, so 3s = −1 or 17, and 2. From A(−1, 2) to B(4, −10), ω ,x = 4 − (−1) = 5 and
s = −1/3 or s = 17/3. ωy = −10 − 2 = −12. | AB| = 52 + (−12)2 = 13.
32. If s − 1 = 1, then s − 1 = ±1, so s = 0 or s = 4. 3. From A(3, 2) to B(−1, −2), ω, x = −1 − 3 = − 4 √an d
2 2 ωy = −2 − 2 = −4. | AB| = (−4)2 + (−4)2 = 4 2.
33. If |x| < 2, then x is in (−2, 2). 4. From A(0.5, 3) to B(2, 3), ωx = 2 − 0.5 = 1.5 and
34. If |x| ≤ 2, then x is in [−2, 2]. ωy = 3 − 3 = 0. | AB| = 1.5.
35. If |s − 1| ≤ 2, then 1 − 2 ≤ s ≤ 1 + 2, so s is in [−1, 3]. 5. Starting point: (−2, 3). Increments ωx = 4, ωy = −7.
New position is (−2 + 4, 3 + (−7)), that is, (2, −4).
36. If |t + 2| < 1, then −2 − 1 < t < −2 + 1, so t is in
(−3, −1). 6. Arrival point: (−2, −2). Increments ωx = −5, ωy = 1.
Starting point was (−2 − (−5), −2 − 1), that is, (3, −3).
37. If |3x − 7 | < 2, then 7 − 2 < 3x < 7 + 2, so x is in
(5/3, 3). 7. x 2 + y2 = 1 represents a circle of radius 1 centred at the
origin.
38. If |2x + 5| < 1, then −5 − 1 < 2x < −5 + 1, so x is in √
(−3, −2). 8. x 2 + y2 = 2 represents a circle of radius 2 centred at
the origin.
x x
39. If − 1 ≤ 1, then 1 − 1 ≤ ≤ 1 + 1, so x is in [0, 4]. 9. x 2 + y2 ≤ 1 represents points inside and on the circle of
2 2
x 1 radius 1 centred at the origin.
40. If 2 − < , then x/2 lies between 2 − (1/2) and
2 2 10. x 2 + y2 = 0 represents the origin.
2 + (1/2). Thus x is in (3, 5).
41. The inequality |x + 1| > |x − 3| says that the distance 11. y ≥ x 2 represents all points lying on or above the
parabola y = x2.
from x to −1 is greater than the distance from x to 3, so 12. 2
x must be to the right of the point half-way between 1 y < x represents all points lying below the parabola
and 3. Thus x > 1. y = x 2.
42. 13. The vertical line through (−2, 5/3) is x = −2; the hori-
|x − 3|2 < 2|x| ⇔ x 2 − 6x + 9 = (x − 3)2 < 4x 2
⇔ 3x + 6x − 9 > 0 ⇔ 3(x + 3)(x − 1) > 0. This zontal line through that point is y = 5/3.
√ √
inequality holds if x < −3 or x > 1. 14. The vertical line through ( 2, −1.3) is x = 2; the
horizontal line through that point is y = −1.3.
43. |a| = a if and only if a ≥ 0. It is false if a < 0.
15. Line through (−1, 1) with slope m = 1 is
44. The equation |x − 1| = 1 − x holds if |x − 1| = −(x − 1),
y = 1 + 1(x + 1), or y = x + 2.
that is, if x − 1 < 0, or, equivalently, if x < 1.
16. Line through (−2, 2) with slope m = 1/2 is
45. The triangle inequality |x + y| ≤ |x | + |y| implies that
y = 2 + (1/2)(x + 2), or x − 2y = −6.
|x | ≥ |x + y| − |y|. 17. Line through (0, b) with slope m = 2 is y = b + 2x .
18. Line through (a, 0) with slope m = −2 is
Apply this inequality with x = a − b and y = b to get y = 0 − 2(x − a), or y = 2a − 2x .
|a − b| ≥ |a| − |b|. 19. At x = 2, the height of the line 2x + 3y = 6 is
y = (6 − 4)/3 = 2/3. Thus (2, 1) lies above the line.
Similarly, |a − b| = |b − a| ≥ |b| − |a|. Since |a| − |b| 20. At x = 3, the height of the line x − 4y = 7 is
y = (3 − 7)/4 = −1. Thus (3, −1) lies on the line.
is equal to either | a| − |b| or | b| − | a|, depending on the
sizes of a and b, we have 21. The line through (0, 0) and (2, 3) has slope
m = (3 − 0)/(2 − 0) = 3/2 and equation y = (3/2)x or
|a − b| ≥ |a| − |b| . 3x − 2y = 0.
22. The line through (−2, 1) and (2, −2) has slope
m = (−2 − 1)/(2 + 2) = −3/4 and equation
Section P.2 Cartesian Coordinates in the y = 1 − (3/4)(x + 2) or 3x + 4y = −2.
Plane (page 16) 23. The line through (4, 1) and (−2, 3) has slope
m = (3 − 1)/(−2 − 4) = −1/3 and equation
1. From A(0, 3) to B(4, 0), ωx, = 4 − 0 = 4 and 1
y = 1 − (x − 4) or x + 3y = 7.
ωy = 0 − 3 = −3. | AB| = 42 + (−3)2 = 5. 3
2
CALCULUS: A COMPLETE COURSE – 10TH
EDITION BY ADAMS & ESSEX – FULL WORKED
SOLUTIONS
,
, INSTRUCTOR’S SOLUTIONS MANUAL SECTION P.1 (PAGE 10)
CHAPTER P. PRELIMINARIES 19. Given: 1/(2 — x) < 3.
CASE I. If x < 2, then 1 < 3(2 – x) = 6 − 3x , so 3x < 5
and x < 5/3. This case has solutions x < 5/3.
Section P.1 Real Numbers and the Real Line CASE II. If x > 2, then 1 > 3(2 x−) = 6 − 3x , so 3x > 5
(page 10) and x > 5/3. This case has solutions x > 2.
Solution: (−∞, 5/3) ∪ (2, ∞).
2
1. = 0.22222222 ·· · = 0.2 20. Given: (x + 1)/x ≥ 2.
9 CASE I. If x > 0, then x + 1 ≥ 2x , so x ≤ 1.
2. 1 CASE II. If x < 0, then x + 1 ≤ 2x , so x ≥ 1. (not
= 0.09090909 ··· = 0.09 possible)
11
3. If x = 0.121212 ·· ·, then 100x = 12.121212 ··· = 12 + x . Solution: (0, 1].
Thus 99x = 12 and x = 12/99 = 4/33. 21. Given: x 2 − 2x ≤ 0. Then x(x − 2) ≤ 0. This is only
4. If x = 3.277777 ·· ·, then 10x − 32 = 0.77777 ··· and possible if x ≥ 0 and x ≤ 2. Solution: [0, 2].
100x − 320 = 7 + (10x − 32), or 90x = 295. Thus 22. Given 6x 2 − 5x ≤ −1, then (2x − 1)(3x − 1) ≤ 0, so
x = 295/90 = 59/18. either x ≤ 1/ 2 and x ≥ 1/ 3, or x ≤ 1/ 3 and x ≥ .
The latter combination is not possible. The solution set is
5. 1/7 = 0.142857142857 ··· = 0.142857 [1/3, 1/2].
2/7 = 0.285714285714 ··· = 0.285714
3/7 = 0.428571428571 ··· = 0.428571 23. Given x 3 > 4x , we have x(x 2 − 4 ) > 0. This is possible
if x < 0 and x2 < 4, or if x > 0 and x2 > 4. The
4/7 = 0.571428571428 ··· = 0.571428 possibilities are, therefore, −2 < x < 0 or 2 < x < ∞.
note the same cyclic order of the repeating digits Solution: (−2, 0) ∪ (2, ∞).
5/7 = 0.714285714285 ··· = 0.714285 24. Given x 2−x ≤ 2, then x2− x −2 ≤ 0 so (x −2)(x +1) ≤ 0.
6/7 = 0.857142857142 ··· = 0.857142 This is possible if x ≤ 2 and x ≥ −1 or if x ≥ 2 and
x ≤ −1. The latter situation is not possible. The solution
6. Two different decimal expansions can represent the same
set is [−1, 2].
number. For instance, both 0.999999 ··· = 0.9 and
1.000000 ··· = 1.0 represent the number 1. x 4
25. Given: ≥ 1+ .
7. x ≥ 0 and x ≤ 5 define the interval [0, 5]. 2 x 2
CASE I. If x > 0, then x ≥ 2x + 8, so that
8. x < 2 and x ≥ −3 define the interval [−3, 2). x 2 − 2x − 8 ≥ 0, or (x − 4)(x + 2) ≥ 0. This is
possible for x > 0 only if x ≥ 4.
9. x > −5 or x < −6 defines the union CASE II. If x < 0, then we must have (x − 4)(x + 2) ≤ 0,
(−∞, −6) ∪ (−5, ∞). which is possible for x < 0 only if x ≥ −2.
10. x ≤ −1 defines the interval (−∞, −1]. Solution: [−2, 0) ∪ [4, ∞).
3 2
11. x > −2 defines the interval (−2, ∞). 26. Given: < .
x−1 x+1
12. x < 4 or x ≥ 2 defines the interval ( −∞ ,∞ ), that is, the CASE I. If x > 1 then (x − 1)(x + 1) > 0, so that
whole real line. 3(x +1) < 2(x −1 ). Thus x < 5 −. There are no solutions
in this case.
13. If −2x > 4, then x < −2. Solution: (−∞, −2) CASE II. If −1 < x < 1, then (x − 1)(x + 1 ) < 0, so
3(x + 1) > 2(x − 1). Thus x > −5. In this case all
14. If 3x + 5 ≤ 8, then 3x ≤ 8 − 5 − 3 and x ≤ 1. Solution:
(−∞, 1] numbers in (−1, 1) are solutions.
CASE III. If x < −1, then (x − 1)(x + 1) > 0, so that
15. If 5x − 3 ≤ 7 − 3x , then 8x ≤ 10 and x ≤ 5/4. Solution: 3(x + 1) < 2(x − 1). Thus x < −5. All numbers x < −5
(−∞, 5/4] are solutions.
6−x 3x − 4 Solutions : (−∞, −5) ∪ (−1, 1).
16. If 2 , then 6 − x ≥ 6x − 8. Thus 14 ≥ 7x
4 ≥ 27. If |x | = 3 then x = ±3.
and x ≤ 2. Solution: (−∞, 2]
28. If |x − 3| = 7, then x − 3 = ±7, so x = −4 or x = 10.
17. If 3(2 − x) < 2(3 + x), then 0 < 5x and x > 0. Solution:
(0, ∞) 29. If |2t + 5| = 4, then 2t + 5 = ±4, so t = −9/2 or
t = −1/2.
18. If x 2 < 9, then |x | < 3 and −3 < x < 3. Solution:
(−3, 3) 30. If|1 − t| = 1, then 1 − t = ±1, so t = 0 or t = 2.
1
, SECTION P.1 (PAGE 10) R. A. ADAMS: CALCULUS
31. If |8 − 3s| = 9, then 8 − 3s = ±9, so 3s = −1 or 17, and 2. From A(−1, 2) to B(4, −10), ω ,x = 4 − (−1) = 5 and
s = −1/3 or s = 17/3. ωy = −10 − 2 = −12. | AB| = 52 + (−12)2 = 13.
32. If s − 1 = 1, then s − 1 = ±1, so s = 0 or s = 4. 3. From A(3, 2) to B(−1, −2), ω, x = −1 − 3 = − 4 √an d
2 2 ωy = −2 − 2 = −4. | AB| = (−4)2 + (−4)2 = 4 2.
33. If |x| < 2, then x is in (−2, 2). 4. From A(0.5, 3) to B(2, 3), ωx = 2 − 0.5 = 1.5 and
34. If |x| ≤ 2, then x is in [−2, 2]. ωy = 3 − 3 = 0. | AB| = 1.5.
35. If |s − 1| ≤ 2, then 1 − 2 ≤ s ≤ 1 + 2, so s is in [−1, 3]. 5. Starting point: (−2, 3). Increments ωx = 4, ωy = −7.
New position is (−2 + 4, 3 + (−7)), that is, (2, −4).
36. If |t + 2| < 1, then −2 − 1 < t < −2 + 1, so t is in
(−3, −1). 6. Arrival point: (−2, −2). Increments ωx = −5, ωy = 1.
Starting point was (−2 − (−5), −2 − 1), that is, (3, −3).
37. If |3x − 7 | < 2, then 7 − 2 < 3x < 7 + 2, so x is in
(5/3, 3). 7. x 2 + y2 = 1 represents a circle of radius 1 centred at the
origin.
38. If |2x + 5| < 1, then −5 − 1 < 2x < −5 + 1, so x is in √
(−3, −2). 8. x 2 + y2 = 2 represents a circle of radius 2 centred at
the origin.
x x
39. If − 1 ≤ 1, then 1 − 1 ≤ ≤ 1 + 1, so x is in [0, 4]. 9. x 2 + y2 ≤ 1 represents points inside and on the circle of
2 2
x 1 radius 1 centred at the origin.
40. If 2 − < , then x/2 lies between 2 − (1/2) and
2 2 10. x 2 + y2 = 0 represents the origin.
2 + (1/2). Thus x is in (3, 5).
41. The inequality |x + 1| > |x − 3| says that the distance 11. y ≥ x 2 represents all points lying on or above the
parabola y = x2.
from x to −1 is greater than the distance from x to 3, so 12. 2
x must be to the right of the point half-way between 1 y < x represents all points lying below the parabola
and 3. Thus x > 1. y = x 2.
42. 13. The vertical line through (−2, 5/3) is x = −2; the hori-
|x − 3|2 < 2|x| ⇔ x 2 − 6x + 9 = (x − 3)2 < 4x 2
⇔ 3x + 6x − 9 > 0 ⇔ 3(x + 3)(x − 1) > 0. This zontal line through that point is y = 5/3.
√ √
inequality holds if x < −3 or x > 1. 14. The vertical line through ( 2, −1.3) is x = 2; the
horizontal line through that point is y = −1.3.
43. |a| = a if and only if a ≥ 0. It is false if a < 0.
15. Line through (−1, 1) with slope m = 1 is
44. The equation |x − 1| = 1 − x holds if |x − 1| = −(x − 1),
y = 1 + 1(x + 1), or y = x + 2.
that is, if x − 1 < 0, or, equivalently, if x < 1.
16. Line through (−2, 2) with slope m = 1/2 is
45. The triangle inequality |x + y| ≤ |x | + |y| implies that
y = 2 + (1/2)(x + 2), or x − 2y = −6.
|x | ≥ |x + y| − |y|. 17. Line through (0, b) with slope m = 2 is y = b + 2x .
18. Line through (a, 0) with slope m = −2 is
Apply this inequality with x = a − b and y = b to get y = 0 − 2(x − a), or y = 2a − 2x .
|a − b| ≥ |a| − |b|. 19. At x = 2, the height of the line 2x + 3y = 6 is
y = (6 − 4)/3 = 2/3. Thus (2, 1) lies above the line.
Similarly, |a − b| = |b − a| ≥ |b| − |a|. Since |a| − |b| 20. At x = 3, the height of the line x − 4y = 7 is
y = (3 − 7)/4 = −1. Thus (3, −1) lies on the line.
is equal to either | a| − |b| or | b| − | a|, depending on the
sizes of a and b, we have 21. The line through (0, 0) and (2, 3) has slope
m = (3 − 0)/(2 − 0) = 3/2 and equation y = (3/2)x or
|a − b| ≥ |a| − |b| . 3x − 2y = 0.
22. The line through (−2, 1) and (2, −2) has slope
m = (−2 − 1)/(2 + 2) = −3/4 and equation
Section P.2 Cartesian Coordinates in the y = 1 − (3/4)(x + 2) or 3x + 4y = −2.
Plane (page 16) 23. The line through (4, 1) and (−2, 3) has slope
m = (3 − 1)/(−2 − 4) = −1/3 and equation
1. From A(0, 3) to B(4, 0), ωx, = 4 − 0 = 4 and 1
y = 1 − (x − 4) or x + 3y = 7.
ωy = 0 − 3 = −3. | AB| = 42 + (−3)2 = 5. 3
2
