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BESTSOLUTION1
Solutions Manual Calculus Early Transcendentals
9th Edition James Stewart Daniel K Complete
Study Guide Derivatives Integrals And Practice
Problems Workbook 2026/ 2027
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BESTSOLUTION1
1 functions and models
1.1 four ways to represent a function
√ √
1. the functions ƒ(x) = x + 2 − x and g(u) = u + 2 − u give exactly the same output values for every input value, so ƒ
and g are equal.
x2 x x(x − 1)
− = x for x 1 = 0, so ƒ and g [where g(x) = x] are not equal because ƒ(1) is undefined and
2. ƒ(x) = = —
x−1 x −1
g(1) = 1.
3. (a) the point (−2 2) lies on the graph of g, so g(−2) = 2. similarly, g (0) = −2, g (2) = 1, and g (3) 2 5.
(b) only the point (−4 3) on the graph has a y-value of 3, so the only value of x for which g(x) = 3 is −4.
(c) the function outputs g(x) are never greater than 3, so g(x) ≤ 3 for the entire domain of the function. thus, g(x) ≤ 3 for
−4 ≤ x ≤ 4 (or, equivalently, on the interval [−4 4]).
(d) the domain consists of all x-values on the graph of g: {x | −4 ≤ x ≤ 4} = [−4 4]. the range of g consists of all the y-values
on the graph of g: {y | −2 ≤ y ≤ 3} = [−2 3].
(e) for any x1 x2 in the interval [0 2], we have g(x 1) g(x2). [the graph rises from (0 −2) to (2 1).] thus,
g(x) isincreasing on [0 2].
4. (a) from the graph, we have ƒ(−4) = −2 and g(3) = 4.
(b) since ƒ(−3) = −1 and g(−3) = 2, or by observing that the graph of g is above the graph of ƒ at x = −3, g(−3) is larger
than ƒ(−3).
(c) the graphs of ƒ and g intersect at x = −2 and x = 2, so ƒ(x) = g(x) at these two values of x.
(d) the graph of ƒ lies below or on the graph of g for −4 ≤ x ≤ −2 and for 2 ≤ x ≤ 3. thus, the intervals on which ƒ (x) ≤
g(x) are [−4 −2] and [2 3].
(e) ƒ (x) = −1 is equivalent to y = −1, and the points on the graph of ƒ with y-values of −1 are (−3 −1) and (4
−1), so the solution of the equation ƒ(x) = −1 is x = −3 or x = 4.
(f) for any x1 x2 in the interval [−4 0], we have g(x1) g(x2 ). thus, g(x) is decreasing on [−4 0].
(g) the domain of ƒ is {x | −4 ≤ x ≤ 4} = [−4 4]. the range of ƒ is {y | −2 ≤ y ≤ 3} = [−2 3].
(h) the domain of g is {x | −4 ≤ x ≤ 3} = [−4 3]. estimating the lowest point of the graph of g as having coordinates (0 0
5), the range of g is approximately {y | 0 5 ≤ y ≤ 4} = [0 5 4].
5. from figure 1 in the text, the lowest point occurs at about (t a) = (12 −85). the highest point occurs at about (17
115). thus, the range of the vertical ground acceleration is −85 ≤ a ≤ 115. written in interval notation, the range
is [−85 115].
c 2021 cengage learning. all rights re served. may not be scanne d, copied, or dupl icated, or posted to a publicly accessible website, in whole or in part.
° 9
2
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BESTSOLUTION1
10 ¤ chapter 1 functions and models
6. example 1:
distance d traveled by the car is a function of the time t. the
domain of the function is {t | 0 ≤ t ≤ 2}, where t is measured in
hours. the range
of the function is {d | 0 ≤ d ≤ 120}, where d is measured in miles.
example 2: at a certain university, the number of students n on
campus at any time on a particular day is a function of the time t
aftermidnight. the domain of the function is {t | 0 ≤ t ≤ 24}, where
t is
measured in hours. the range of the function is {n | 0 ≤ n ≤ k},
where n is an integer and k is the largest number of students on
campus at once.k
example 3: a certain employee is paid $8 00 per hour and works
a maximum of 30 hours per week. the number of hours worked
is rounded down to the nearest quarter of an hour. this
employee’s gross weekly pay p is a function of the number of
hours worked h. the domain of the function is [0 30] and the
range of the function is 0 0.25 0.50 0.75 29.50 29.75 30 hours
{0 2 00 4 00 238 00 240 00}.
7. we solve 3x − 5y = 7 for y: 3x − 5y = 7 ⇔ −5y = −3x +7 ⇔ y = 3 x − 7 . since the equation determines exactly
5 5
one value of y for each value of x, the equation defines y as a function of x.
8. we solve 3x2 − 2y = 5 for y: 3x2 − 2y = 5 ⇔ −2y = −3x2 +5 ⇔ y = 3 x2 − 5 . since the equation determines
2 2
exactly one value of y for each value of x, the equation defines y as a function of x.
√
9. we solve x2 + (y − 3) 2 = 5 for y: x2 + (y − 3) 2 = 5 ⇔ (y − 3) 2 = 5 − x2 ⇔ Y − 3 = ± 5 − X2 ⇔
√
y = 3 ± 5 − x2. some input values x correspond to more than one output y. (for instance, x = 1 corresponds to y =
1 and to y = 5.) thus, the equation does not define y as a function of x.
10. we solve 2xy + 5y2 = 4 for y: 2xy + 5y2 = 4 ⇔ 5y2 + (2x) y − 4 = 0 ⇔
−2x ± √ √
−2x ± 4x2 + 8010 −x ± x2 + 20
y= 2(5) = = (using the quadratic formula). some input
5
values x correspond to more than one output y. (for instance, x = 4 corresponds to y = −2 and to y = thus,
theequation does not define y as a function of x.
√
11. we solve (y + 3) 3 + 1 = 2x for y: (y + 3) 3 + 1 = 2x ⇔ (y + 3) 3 = 2x − 1 ⇔ y+3= 2x − 1 ⇔
3√
y = −3 + 2x − 1. since the equation determines exactly one value of y for each value of x, the equation defines y as
a function of x.
3
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BESTSOLUTION1
section 1.1 four ways torepresent a function ¤ 11
12. we solve 2x − |y| = 0 for y: 2x − |y| = 0 ⇔ |y| = 2x ⇔ y = ±2x. some input values x correspond to more than one
output y. (for instance, x = 1 corresponds to y = −2 and to y = 2.) thus, the equation does not define y as a function of
x.
13. the height 60 in (x = 60) corresponds to shoe sizes 7 and 8 (y = 7 and y = 8). since an input value x corresponds to
more than output value y, the table does not define y as a function of x.
14. each year x corresponds to exactly one tuition cost y. thus, the table defines y as a function of x.
15. no, the curve is not the graph of a function because a vertical line intersects the curve more than once. hence, the
curve fails the vertical line test.
16. yes, the curve is the graph of a function because it passes the vertical line test. the domain is [−2 2] and the
range is [−1 2].
17. yes, the curve is the graph of a function because it passes the vertical line test. the domain is [−3 2] and the
range is [−3 −2) ∪ [−1 3].
18. no, the curve is not the graph of a function since for x = 0, ±1, and ±2, there are infinitely many points on the curve.
19. (a) when t = 1950, t ≈ 13 8 ◦c, so the global average temperature in 1950 was about 13 8 ◦ c.
(b) when t = 14 2 ◦c, t ≈ 1990.
(c) the global average temperature was smallest in 1910 (the year corresponding to the lowest point on the graph)
and largest in 2000 (the year corresponding to the highest point on the graph).
(d) when t = 1910, t ≈ 13 5 ◦c, and when t = 2000, t ≈ 14 4◦c. thus, the range of t is about [13 5, 14 4].
20. (a) the ring width varies from near 0 mm to about 1 6 mm, so the range of the ring width function is approximately [0 1
6].
(b) according to the graph, the earth gradually cooled from 1550 to 1700, warmed into the late 1700s, cooled again
into thelate 1800s, and has been steadily warming since then. in the mid-19th century, there was variation that
could have been associated with volcanic eruptions.
21. the water will cool down almost to freezing as the ice melts. then,
when the ice has melted, the water will slowly warm up to room
temperature.
22. the temperature of the pie would increase rapidly, level off to
oven temperature, decrease rapidly, and then level off to room
temperature.
5
BESTSOLUTION1
Solutions Manual Calculus Early Transcendentals
9th Edition James Stewart Daniel K Complete
Study Guide Derivatives Integrals And Practice
Problems Workbook 2026/ 2027
, Stuvia.com - The Marketplace to Buy and Sell your Study Material
BESTSOLUTION1
1 functions and models
1.1 four ways to represent a function
√ √
1. the functions ƒ(x) = x + 2 − x and g(u) = u + 2 − u give exactly the same output values for every input value, so ƒ
and g are equal.
x2 x x(x − 1)
− = x for x 1 = 0, so ƒ and g [where g(x) = x] are not equal because ƒ(1) is undefined and
2. ƒ(x) = = —
x−1 x −1
g(1) = 1.
3. (a) the point (−2 2) lies on the graph of g, so g(−2) = 2. similarly, g (0) = −2, g (2) = 1, and g (3) 2 5.
(b) only the point (−4 3) on the graph has a y-value of 3, so the only value of x for which g(x) = 3 is −4.
(c) the function outputs g(x) are never greater than 3, so g(x) ≤ 3 for the entire domain of the function. thus, g(x) ≤ 3 for
−4 ≤ x ≤ 4 (or, equivalently, on the interval [−4 4]).
(d) the domain consists of all x-values on the graph of g: {x | −4 ≤ x ≤ 4} = [−4 4]. the range of g consists of all the y-values
on the graph of g: {y | −2 ≤ y ≤ 3} = [−2 3].
(e) for any x1 x2 in the interval [0 2], we have g(x 1) g(x2). [the graph rises from (0 −2) to (2 1).] thus,
g(x) isincreasing on [0 2].
4. (a) from the graph, we have ƒ(−4) = −2 and g(3) = 4.
(b) since ƒ(−3) = −1 and g(−3) = 2, or by observing that the graph of g is above the graph of ƒ at x = −3, g(−3) is larger
than ƒ(−3).
(c) the graphs of ƒ and g intersect at x = −2 and x = 2, so ƒ(x) = g(x) at these two values of x.
(d) the graph of ƒ lies below or on the graph of g for −4 ≤ x ≤ −2 and for 2 ≤ x ≤ 3. thus, the intervals on which ƒ (x) ≤
g(x) are [−4 −2] and [2 3].
(e) ƒ (x) = −1 is equivalent to y = −1, and the points on the graph of ƒ with y-values of −1 are (−3 −1) and (4
−1), so the solution of the equation ƒ(x) = −1 is x = −3 or x = 4.
(f) for any x1 x2 in the interval [−4 0], we have g(x1) g(x2 ). thus, g(x) is decreasing on [−4 0].
(g) the domain of ƒ is {x | −4 ≤ x ≤ 4} = [−4 4]. the range of ƒ is {y | −2 ≤ y ≤ 3} = [−2 3].
(h) the domain of g is {x | −4 ≤ x ≤ 3} = [−4 3]. estimating the lowest point of the graph of g as having coordinates (0 0
5), the range of g is approximately {y | 0 5 ≤ y ≤ 4} = [0 5 4].
5. from figure 1 in the text, the lowest point occurs at about (t a) = (12 −85). the highest point occurs at about (17
115). thus, the range of the vertical ground acceleration is −85 ≤ a ≤ 115. written in interval notation, the range
is [−85 115].
c 2021 cengage learning. all rights re served. may not be scanne d, copied, or dupl icated, or posted to a publicly accessible website, in whole or in part.
° 9
2
, Stuvia.com - The Marketplace to Buy and Sell your Study Material
BESTSOLUTION1
10 ¤ chapter 1 functions and models
6. example 1:
distance d traveled by the car is a function of the time t. the
domain of the function is {t | 0 ≤ t ≤ 2}, where t is measured in
hours. the range
of the function is {d | 0 ≤ d ≤ 120}, where d is measured in miles.
example 2: at a certain university, the number of students n on
campus at any time on a particular day is a function of the time t
aftermidnight. the domain of the function is {t | 0 ≤ t ≤ 24}, where
t is
measured in hours. the range of the function is {n | 0 ≤ n ≤ k},
where n is an integer and k is the largest number of students on
campus at once.k
example 3: a certain employee is paid $8 00 per hour and works
a maximum of 30 hours per week. the number of hours worked
is rounded down to the nearest quarter of an hour. this
employee’s gross weekly pay p is a function of the number of
hours worked h. the domain of the function is [0 30] and the
range of the function is 0 0.25 0.50 0.75 29.50 29.75 30 hours
{0 2 00 4 00 238 00 240 00}.
7. we solve 3x − 5y = 7 for y: 3x − 5y = 7 ⇔ −5y = −3x +7 ⇔ y = 3 x − 7 . since the equation determines exactly
5 5
one value of y for each value of x, the equation defines y as a function of x.
8. we solve 3x2 − 2y = 5 for y: 3x2 − 2y = 5 ⇔ −2y = −3x2 +5 ⇔ y = 3 x2 − 5 . since the equation determines
2 2
exactly one value of y for each value of x, the equation defines y as a function of x.
√
9. we solve x2 + (y − 3) 2 = 5 for y: x2 + (y − 3) 2 = 5 ⇔ (y − 3) 2 = 5 − x2 ⇔ Y − 3 = ± 5 − X2 ⇔
√
y = 3 ± 5 − x2. some input values x correspond to more than one output y. (for instance, x = 1 corresponds to y =
1 and to y = 5.) thus, the equation does not define y as a function of x.
10. we solve 2xy + 5y2 = 4 for y: 2xy + 5y2 = 4 ⇔ 5y2 + (2x) y − 4 = 0 ⇔
−2x ± √ √
−2x ± 4x2 + 8010 −x ± x2 + 20
y= 2(5) = = (using the quadratic formula). some input
5
values x correspond to more than one output y. (for instance, x = 4 corresponds to y = −2 and to y = thus,
theequation does not define y as a function of x.
√
11. we solve (y + 3) 3 + 1 = 2x for y: (y + 3) 3 + 1 = 2x ⇔ (y + 3) 3 = 2x − 1 ⇔ y+3= 2x − 1 ⇔
3√
y = −3 + 2x − 1. since the equation determines exactly one value of y for each value of x, the equation defines y as
a function of x.
3
, Stuvia.com - The Marketplace to Buy and Sell your Study Material
BESTSOLUTION1
section 1.1 four ways torepresent a function ¤ 11
12. we solve 2x − |y| = 0 for y: 2x − |y| = 0 ⇔ |y| = 2x ⇔ y = ±2x. some input values x correspond to more than one
output y. (for instance, x = 1 corresponds to y = −2 and to y = 2.) thus, the equation does not define y as a function of
x.
13. the height 60 in (x = 60) corresponds to shoe sizes 7 and 8 (y = 7 and y = 8). since an input value x corresponds to
more than output value y, the table does not define y as a function of x.
14. each year x corresponds to exactly one tuition cost y. thus, the table defines y as a function of x.
15. no, the curve is not the graph of a function because a vertical line intersects the curve more than once. hence, the
curve fails the vertical line test.
16. yes, the curve is the graph of a function because it passes the vertical line test. the domain is [−2 2] and the
range is [−1 2].
17. yes, the curve is the graph of a function because it passes the vertical line test. the domain is [−3 2] and the
range is [−3 −2) ∪ [−1 3].
18. no, the curve is not the graph of a function since for x = 0, ±1, and ±2, there are infinitely many points on the curve.
19. (a) when t = 1950, t ≈ 13 8 ◦c, so the global average temperature in 1950 was about 13 8 ◦ c.
(b) when t = 14 2 ◦c, t ≈ 1990.
(c) the global average temperature was smallest in 1910 (the year corresponding to the lowest point on the graph)
and largest in 2000 (the year corresponding to the highest point on the graph).
(d) when t = 1910, t ≈ 13 5 ◦c, and when t = 2000, t ≈ 14 4◦c. thus, the range of t is about [13 5, 14 4].
20. (a) the ring width varies from near 0 mm to about 1 6 mm, so the range of the ring width function is approximately [0 1
6].
(b) according to the graph, the earth gradually cooled from 1550 to 1700, warmed into the late 1700s, cooled again
into thelate 1800s, and has been steadily warming since then. in the mid-19th century, there was variation that
could have been associated with volcanic eruptions.
21. the water will cool down almost to freezing as the ice melts. then,
when the ice has melted, the water will slowly warm up to room
temperature.
22. the temperature of the pie would increase rapidly, level off to
oven temperature, decrease rapidly, and then level off to room
temperature.
5
