Calculus: Early Transcendentals 9/e
Metric Version [Stewart], Chapter 1-16
SOLUTIONS
,TABLE OF CONTENTS
1. FUNCTIONS AND MODELS.
2. LIMITS AND DERIVATIVES.
3. DIFFERENTIATION RULES.
4. APPLICATIONS OF DIFFERENTIATION.
5. INTEGRALS.
6. APPLICATIONS OF INTEGRATION.
7. TECHNIQUES OF INTEGRATION.
8. FURTHER APPLICATIONS OF INTEGRATION.
9. DIFFERENTIAL EQUATIONS.
10. PARAMETRIC EQUATIONS AND POLAR COORDINATES.
11. SEQUENCES, SERIES, AND POWER SERIES.
12. VECTORS AND THE GEOMETRY OF SPACE.
13. VECTOR FUNCTIONS.
14. PARTIAL DERIVATIVES.
15. MULTIPLE INTEGRALS.
16. VECTOR CALCULUS.
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1 FUNCTIONS AND MODELS
1.1 Four Ways to Represent a Function
√ √
1. The functions () = + 2 − and () = + 2 − give exactly the same output values for every input value, so
and are equal.
2 − ( − 1)
2. () = = = for − 1 6= 0, so and [where () = ] are not equal because (1) is undefined and
−1 −1
(1) = 1.
3. (a) The point (−2 2) lies on the graph of , so (−2) = 2. Similarly, (0) = −2, (2) = 1, and (3) 25.
(b) Only the point (−4 3) on the graph has a value of 3, so the only value of for which () = 3 is −4.
(c) The function outputs () are never greater than 3, so () ≤ 3 for the entire domain of the function. Thus, () ≤ 3 for
−4 ≤ ≤ 4 (or, equivalently, on the interval [−4 4]).
(d) The domain consists of all values on the graph of : { | −4 ≤ ≤ 4} = [−4 4]. The range of consists of all the
values on the graph of : { | −2 ≤ ≤ 3} = [−2 3].
(e) For any 1 2 in the interval [0 2], we have (1 ) (2 ). [The graph rises from (0 −2) to (2 1).] Thus, () is
increasing on [0 2].
4. (a) From the graph, we have (−4) = −2 and (3) = 4.
(b) Since (−3) = −1 and (−3) = 2, or by observing that the graph of is above the graph of at = −3, (−3) is larger
than (−3).
(c) The graphs of and intersect at = −2 and = 2, so () = () at these two values of .
(d) The graph of lies below or on the graph of for −4 ≤ ≤ −2 and for 2 ≤ ≤ 3. Thus, the intervals on which
() ≤ () are [−4 −2] and [2 3].
(e) () = −1 is equivalent to = −1, and the points on the graph of with values of −1 are (−3 −1) and (4 −1), so
the solution of the equation () = −1 is = −3 or = 4.
(f) For any 1 2 in the interval [−4 0], we have (1 ) (2 ). Thus, () is decreasing on [−4 0].
(g) The domain of is { | −4 ≤ ≤ 4} = [−4 4]. The range of is { | −2 ≤ ≤ 3} = [−2 3].
(h) The domain of is { | −4 ≤ ≤ 3} = [−4 3]. Estimating the lowest point of the graph of as having coordinates
(0 05), the range of is approximately { | 05 ≤ ≤ 4} = [05 4].
5. From Figure 1 in the text, the lowest point occurs at about ( ) = (12 −85). The highest point occurs at about (17 115).
Thus, the range of the vertical ground acceleration is −85 ≤ ≤ 115. Written in interval notation, the range is [−85 115].
°
c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. 9
Telegram: @uni_k
, www.konkur.in
10 ¤ CHAPTER 1 FUNCTIONS AND MODELS
6. Example 1: A car is driven at 60 mih for 2 hours. The distance
traveled by the car is a function of the time . The domain of the
function is { | 0 ≤ ≤ 2}, where is measured in hours. The range
of the function is { | 0 ≤ ≤ 120}, where is measured in miles.
Example 2: At a certain university, the number of students on
campus at any time on a particular day is a function of the time after
midnight. The domain of the function is { | 0 ≤ ≤ 24}, where is
measured in hours. The range of the function is { | 0 ≤ ≤ },
where is an integer and is the largest number of students on
campus at once.
Example 3: A certain employee is paid $800 per hour and works a pay
maximum of 30 hours per week. The number of hours worked is 240
238
rounded down to the nearest quarter of an hour. This employee’s 236
gross weekly pay is a function of the number of hours worked .
4
The domain of the function is [0 30] and the range of the function is 2
{0 200 400 23800 24000}. 0 0.25 0.50 0.75 29.50 29.75 30 hours
7. We solve 3 − 5 = 7 for : 3 − 5 = 7 ⇔ −5 = −3 + 7 ⇔ = 35 − 75 . Since the equation determines exactly
one value of for each value of , the equation defines as a function of .
8. We solve 32 − 2 = 5 for : 32 − 2 = 5 ⇔ −2 = −32 + 5 ⇔ = 32 2 − 52 . Since the equation determines
exactly one value of for each value of , the equation defines as a function of .
√
9. We solve 2 + ( − 3)2 = 5 for : 2 + ( − 3)2 = 5 ⇔ ( − 3)2 = 5 − 2 ⇔ ⇔ − 3 = ± 5 − 2
√
= 3 ± 5 − 2 . Some input values correspond to more than one output . (For instance, = 1 corresponds to = 1 and
to = 5.) Thus, the equation does not define as a function of .
10. We solve 2 + 5 2 = 4 for : 2 + 5 2 = 4 ⇔ 5 2 + (2) − 4 = 0 ⇔
√ √
−2 ± (2)2 − 4(5)(−4) −2 ± 42 + 80 − ± 2 + 20
= = = (using the quadratic formula). Some input
2(5) 10 5
values correspond to more than one output . (For instance, = 4 corresponds to = −2 and to = 25.) Thus, the
equation does not define as a function of .
√
11. We solve ( + 3)3 + 1 = 2 for : ( + 3)3 + 1 = 2 ⇔ ( + 3)3 = 2 − 1 ⇔ + 3 = 3
2 − 1 ⇔
√
3
= −3 + 2 − 1. Since the equation determines exactly one value of for each value of , the equation defines as a
function of .
°
c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
Telegram: @uni_k
Metric Version [Stewart], Chapter 1-16
SOLUTIONS
,TABLE OF CONTENTS
1. FUNCTIONS AND MODELS.
2. LIMITS AND DERIVATIVES.
3. DIFFERENTIATION RULES.
4. APPLICATIONS OF DIFFERENTIATION.
5. INTEGRALS.
6. APPLICATIONS OF INTEGRATION.
7. TECHNIQUES OF INTEGRATION.
8. FURTHER APPLICATIONS OF INTEGRATION.
9. DIFFERENTIAL EQUATIONS.
10. PARAMETRIC EQUATIONS AND POLAR COORDINATES.
11. SEQUENCES, SERIES, AND POWER SERIES.
12. VECTORS AND THE GEOMETRY OF SPACE.
13. VECTOR FUNCTIONS.
14. PARTIAL DERIVATIVES.
15. MULTIPLE INTEGRALS.
16. VECTOR CALCULUS.
,www.konkur.in
1 FUNCTIONS AND MODELS
1.1 Four Ways to Represent a Function
√ √
1. The functions () = + 2 − and () = + 2 − give exactly the same output values for every input value, so
and are equal.
2 − ( − 1)
2. () = = = for − 1 6= 0, so and [where () = ] are not equal because (1) is undefined and
−1 −1
(1) = 1.
3. (a) The point (−2 2) lies on the graph of , so (−2) = 2. Similarly, (0) = −2, (2) = 1, and (3) 25.
(b) Only the point (−4 3) on the graph has a value of 3, so the only value of for which () = 3 is −4.
(c) The function outputs () are never greater than 3, so () ≤ 3 for the entire domain of the function. Thus, () ≤ 3 for
−4 ≤ ≤ 4 (or, equivalently, on the interval [−4 4]).
(d) The domain consists of all values on the graph of : { | −4 ≤ ≤ 4} = [−4 4]. The range of consists of all the
values on the graph of : { | −2 ≤ ≤ 3} = [−2 3].
(e) For any 1 2 in the interval [0 2], we have (1 ) (2 ). [The graph rises from (0 −2) to (2 1).] Thus, () is
increasing on [0 2].
4. (a) From the graph, we have (−4) = −2 and (3) = 4.
(b) Since (−3) = −1 and (−3) = 2, or by observing that the graph of is above the graph of at = −3, (−3) is larger
than (−3).
(c) The graphs of and intersect at = −2 and = 2, so () = () at these two values of .
(d) The graph of lies below or on the graph of for −4 ≤ ≤ −2 and for 2 ≤ ≤ 3. Thus, the intervals on which
() ≤ () are [−4 −2] and [2 3].
(e) () = −1 is equivalent to = −1, and the points on the graph of with values of −1 are (−3 −1) and (4 −1), so
the solution of the equation () = −1 is = −3 or = 4.
(f) For any 1 2 in the interval [−4 0], we have (1 ) (2 ). Thus, () is decreasing on [−4 0].
(g) The domain of is { | −4 ≤ ≤ 4} = [−4 4]. The range of is { | −2 ≤ ≤ 3} = [−2 3].
(h) The domain of is { | −4 ≤ ≤ 3} = [−4 3]. Estimating the lowest point of the graph of as having coordinates
(0 05), the range of is approximately { | 05 ≤ ≤ 4} = [05 4].
5. From Figure 1 in the text, the lowest point occurs at about ( ) = (12 −85). The highest point occurs at about (17 115).
Thus, the range of the vertical ground acceleration is −85 ≤ ≤ 115. Written in interval notation, the range is [−85 115].
°
c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. 9
Telegram: @uni_k
, www.konkur.in
10 ¤ CHAPTER 1 FUNCTIONS AND MODELS
6. Example 1: A car is driven at 60 mih for 2 hours. The distance
traveled by the car is a function of the time . The domain of the
function is { | 0 ≤ ≤ 2}, where is measured in hours. The range
of the function is { | 0 ≤ ≤ 120}, where is measured in miles.
Example 2: At a certain university, the number of students on
campus at any time on a particular day is a function of the time after
midnight. The domain of the function is { | 0 ≤ ≤ 24}, where is
measured in hours. The range of the function is { | 0 ≤ ≤ },
where is an integer and is the largest number of students on
campus at once.
Example 3: A certain employee is paid $800 per hour and works a pay
maximum of 30 hours per week. The number of hours worked is 240
238
rounded down to the nearest quarter of an hour. This employee’s 236
gross weekly pay is a function of the number of hours worked .
4
The domain of the function is [0 30] and the range of the function is 2
{0 200 400 23800 24000}. 0 0.25 0.50 0.75 29.50 29.75 30 hours
7. We solve 3 − 5 = 7 for : 3 − 5 = 7 ⇔ −5 = −3 + 7 ⇔ = 35 − 75 . Since the equation determines exactly
one value of for each value of , the equation defines as a function of .
8. We solve 32 − 2 = 5 for : 32 − 2 = 5 ⇔ −2 = −32 + 5 ⇔ = 32 2 − 52 . Since the equation determines
exactly one value of for each value of , the equation defines as a function of .
√
9. We solve 2 + ( − 3)2 = 5 for : 2 + ( − 3)2 = 5 ⇔ ( − 3)2 = 5 − 2 ⇔ ⇔ − 3 = ± 5 − 2
√
= 3 ± 5 − 2 . Some input values correspond to more than one output . (For instance, = 1 corresponds to = 1 and
to = 5.) Thus, the equation does not define as a function of .
10. We solve 2 + 5 2 = 4 for : 2 + 5 2 = 4 ⇔ 5 2 + (2) − 4 = 0 ⇔
√ √
−2 ± (2)2 − 4(5)(−4) −2 ± 42 + 80 − ± 2 + 20
= = = (using the quadratic formula). Some input
2(5) 10 5
values correspond to more than one output . (For instance, = 4 corresponds to = −2 and to = 25.) Thus, the
equation does not define as a function of .
√
11. We solve ( + 3)3 + 1 = 2 for : ( + 3)3 + 1 = 2 ⇔ ( + 3)3 = 2 − 1 ⇔ + 3 = 3
2 − 1 ⇔
√
3
= −3 + 2 − 1. Since the equation determines exactly one value of for each value of , the equation defines as a
function of .
°
c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
Telegram: @uni_k
