Student Solutions Manual for Calculus: A Complete Course – 10th Edition by Adams & Essex – Full Worked Solutions

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
11 possible)
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 ≥ 1 /2.
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 x 2 − 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
2
8. x < 2 and x ≥ −3 define the interval [−3, 2). x − 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 , then 6 − x ≥ 6x − 8. Thus 14 ≥ 7x
4 ≥ 2 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
2 t = −1/2.
18. If x < 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 √ and
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. |x − 3| < 2|x | ⇔ x 2 − 6x + 9 = (x − 3)2 < 4x 2 13. The vertical line through (−2, 5/3) is x = −2; the hori-
zontal line through that point is y = 5/3.
⇔ 3x 2 + 6x − 9 > 0 ⇔ 3(x + 3)(x − 1) > 0. This √ √
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
is equal to either | a| − |b| or | b| − | a| , depending on the y = (3 − 7)/4 = −1. Thus (3, −1) lies on the line.
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

, INSTRUCTOR’S SOLUTIONS MANUAL SECTION P.2 (PAGE 16)



y
24. The line through (−2, 0) and (0, 2) has slope
m = (2 − 0)/(0 + 2) = 1 and equation y = 2 + x .
√
25. If m = −2 a√n d b = 2, then the line has equation
y = −2x + 2.

26. If m = −1/2 and b = −3, then the line has equation
y = −(1/2)x − 3, or x + 2y = −6. x

27. 3x + 4y = 12 has x -intercept a = 12/3 = 4 and y-
intercept b = 12/4 = 3. Its slope is −b/a = −3/4.
y Fig. P.2.30

31. line through (2, 1) parallel to y = x + 2 is y = x − 1; line
perpendicular to y = x + 2 is y = −x + 3.
32. line through (−2, 2) parallel to 2x + y = 4 is
3x + 4y = 12 2x + y = −2; line perpendicular to 2x + y = 4 is
x − 2y = −6.
33. We have
x 3x + 4y = −6 ⇐⇒ 6x + 8y = −12
2x − 3y = 13 6x − 9y = 39.
Fig. P.2.27
Subtracting these equations gives 17y = −51, so y = −3
28. x + 2y = −4 has x -intercept a = −4 and y-intercept and x = (13−9)/2 = 2. The intersection point is (2, −3).
b = −4/2 = −2. Its slope is −b/a = 2/(−4) = −1/2. 34. We have
y
2x + y = 8 ⇐⇒ 14x + 7y = 56
5x − 7y = 1 5x − 7y = 1.
x
Adding these equations gives 19x = 57, so x = 3 and
x + 2y = −4 y = 8 − 2x = 2. The intersection point is (3, 2).
35. If a ̸= 0 and b ̸= 0, then (x/a) + (y/b) = 1 represents
a straight line that is neither horizontal nor vertical, and
does not pass through the origin. Putting y = 0 we get
x/a = 1, so the x -intercept of this line is x = a; putting
Fig. P.2.28 x = 0 gives y/b = 1, so the y-intercept is y = b.

√ √ √ √ 36. The line (x/2) − (y/3) = 1 has x -intercept a = 2, and
29. 2x − 3y = 2 has x -inte√rcept a = 2/ 2 = 2 y-intercept b = −3.
and y-interc√ept b √= −2/ 3. Its slope is y
−b/a = 2/ 6
y 2 x


x y
x − =1
2 3
−3




Fig. P.2.36
Fig. P.2.29
37. The line through (2, 1) and (3, −1) has slope
30. 1.5x − 2y = −3 has x -intercept a = −3/1.5 = −2 and y- m = (−1 − 1)/(3 − 2) = −2 and equation
intercept b = −3/(−2) = 3/2. Its slope is −b/a = 3/4. y = 1 − 2(x − 2) = 5 − 2x . Its y-intercept is 5.




3

, SECTION P.2 (PAGE 16) R. A. ADAMS: CALCULUS




38. The line through (—2, 5) and (k, 1) has x -intercept 3, so A = (2, −1), B,= (1, 3), C = (−3, 2)√
also passes through (3, 0). Its slope m satisfies 43. | AB| = (1 − 2)2 + (3 + 1)2 = 17
, √ √ √
1−0 0−5 | AC | = (−3 − 2)2 + (2 + 1)2 = 34 = 2 17
, √
=m = = −1.
k−3 3+2 | BC | = ( −3 − 1)2 + (2 − 3)2 = 17.
√
Since | AB | = |BC | and | AC | = 2 |AB |, triangle ABC
Thus k − 3 = −1, and so k = 2.
is an isosceles right-angled triangle with right angle at
B. Thus ABC D is a square if D is displaced from C
39. C = Ax + B. If C = 5, 000 when x = 10, 000 and by the same amount A is from B, that is, by increments
C = 6, 000 when x = 15, 000, then ωx = 2 − 1 = 1 and ωy = −1 − 3 = −4. Thus
D = (−3 + 1, 2 + (−4)) = (−2, −2).
10, 000 A + B = 5, 000
44. If M = (xm , ym ) is the midpoint of P1 P2 , then the dis-
15, 000 A + B = 6, 000 placement of M from P1 equals the displacement of P2
from M:
Subtracting these equations gives 5, 000 A = 1, 000, so
A = 1/5. From the first equation, 2, 000 + B = 5, 000, xm − x1 = x2 − xm , ym − y1 = y2 − ym .
so B = 3, 000. The cost of printing 100,000 pamphlets is
$100, 000/5 + 3, 000 = $23, 000. Thus xm = (x1 + x2 )/2 and ym = (y1 + y2 )/2.

◦ ◦
45. If Q = (xq , yq ) is the point on P1 P2 that is two thirds of
40. −40 and −40 is the same temperature on both the the way from P1 to P2, then the displacement of Q from
Fahrenheit and Celsius scales. P1 equals twice the displacement of P2 from Q:
C
40 xq − x1 = 2(x2 − xq ), yq − y1 = 2(y2 − yq ).
30
C =F Thus xq = (x1 + 2x2 )/3 and yq = (y1 + 2y2 )/3.
20
10 46. Let the coordinates of P be (x, 0) and those of Q be
(X, −2X). If the midpoint of PQ is (2, 1), then
-50 -40 -30 -20 -10 10 20 30 40 50 60 70 80 F
-10
5 (x + X)/2 = 2, (0 − 2X )/2 = 1.
-20 C = (F − 32)
9
-30 The second equation implies that X = −1, and the sec-
-40 ond then implies that x = 5. Thus P is (5, 0).
(− 40, − 40) ,
-50 47. (x − 2)2 + y2 = 4 says that the distance of (x, y) from
Fig. P.2.40 (2, 0) is 4, so the equation represents a circle of radius 4
centred at (2, 0).
, ,
A = (2, 1), B, = (6, 4), C = (5, −3)√ 48. (x − 2)2 + y2 = x 2 + (y − 2)2 says that (x, y) is
41. |AB | = (6 − 2)2 + (4 − 1)2 = 25 = 5 equidistant from (2, 0) and (0, 2). Thus (x, y) must
, √ lie on the line that is the right bisector of the line from
| AC | = , (5 − 2)2 + (−3 − 1)2 =√ 25 = √ 5 (2, 0) to (0, 2). A simpler equation for this line is x = y.
|BC | = (6 − 5)2 + (4 + 3)2 = 50 = 5 2.
Since | AB| = | AC |, triangle ABC is isosceles. 49. The line 2x + ky = 3 has slope m = −2/ k. This line
is perpendicular to 4x + y = 1, which has slope −4,
√ provided m = 1/4, that is, provided k = −8. The line is
42. A = (0, 0), B = (1, 3), C = (2, 0) parallel to 4x + y = 1 if m = −4, that is, if k = 1/2.
, √ √
| AB| = (1 − 0)2 + ( 3 − 0)2 = 4 = 2 50. For any value of k, the coordinates of the point of inter-
, √
| AC | = (2 − 0)2 + (0 − 0)2 = 4 = 2 section of x + 2y = 3 and 2x − 3y = −1 will also satisfy
, the equation
√ √
| BC| = (2 − 1)2 + (0 − 3)2 = 4 = 2.
Since | AB| = | AC | = |BC |, triangle ABC is equilateral. (x + 2y − 3) + k(2x − 3y + 1) = 0



4