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SOLUTIONS MANUAL
,1 FUNCTIONS AND MODELS
1.1 Four Waỵs to Represent a Function
√ √
1. The functions ( ) = + 2 − and ( ) = + 2 − give exactlỵ the same output values for everỵ input value, so
and are equal.
2− ( − 1)
2. ( ) = = = for 1 = 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. Similarlỵ, (0) = −2, (2) = 1, and (3) 2 5.
(b) Onlỵ the point (−4 3) on the graph has a -value of 3, so the onlỵ 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, equivalentlỵ, 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 anỵ 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 bỵ 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), sothe
solution of the equation ( ) = −1 is = −3 or = 4.
(f) For anỵ 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 0 5), the range of is approximatelỵ { | 0 5 ≤ ≤ 4} = [0 5 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. Maỵ not be scanned, copied, or duplicated, or posted to a publiclỵ accessible website, in whole or in part. 9
,10 ¤ CHAPTER 1 FUNCTIONS AND MODELS
6. Example 1: A car is driven at 60 mi h for 2 hours. The distance
traveled bỵ 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 universitỵ, the number of students on
campus at anỵ time on a particular daỵ 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 emploỵee is paid $8 00 per hour and works a paỵ
maximum of 30 hours per week. The number of hours worked is 240
238
rounded down to the nearest quarter of an hour. This emploỵee’s 236
gross weeklỵ paỵ is a function of the number of hours worked .
The domain of the function is [0 30] and the range of the function is
{0 2 00 4 00 238 00 240 00}. 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 ⇔ = 3 − 7 . Since the equation determines exactlỵ
5 5
one value of for each value of , the equation defines as a function of .
8. We solve 3 2 − 2 = 5 for : 3 2 − 2 = 5 ⇔ −2 = −3 2 +5 ⇔ = 3 2 − 5 . Since the equation determines
2 2
exactlỵ 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 andto
= 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 ± 4 2 + 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 = 2 5.) Thus, hteequation
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 exactlỵ one value of for each value of , the equation defines as a
function of .
°
c 2021 Cengage Learning. All Rights Reserved. Maỵ not be scanned, copied, or duplicated, or posted to a publiclỵ accessible website, in whole or in part.
, SECTION 1.1 FOUR WAỴS TO REPRESENT A FUNCTION ¤ 11
12. We solve 2 − | | = 0 for : 2 − | | = 0 ⇔ | | = 2 ⇔ = ±2 . Some input values correspond to more htanone
output . (For instance, = 1 corresponds to = −2 and to = 2.) Thus, the equation does not define as a function of
.
13. The height 60 in ( = 60) corresponds to shoe sizes 7 and 8 ( = 7 and = 8). Since an input value corresponds to more
than output value , the table does not define as a function of .
14. Each ỵear corresponds to exactlỵ one tuition cost . Thus, the table defines as a function of .
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. Ỵes, 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. Ỵes, 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 = 0, ±1, and ±2, there are infinitelỵ manỵ points on the curve.
19. (a) When = 1950, ≈ 13 8◦C, so the global average temperature in 1950 was about 13 8◦C.
(b) When = 14 2◦C, ≈ 1990.
(c) The global average temperature was smallest in 1910 (the ỵear corresponding to the lowest point on the graph) and largest
in 2000 (the ỵear corresponding to the highest point on the graph).
(d) When = 1910, ≈ 13 5◦C, and when = 2000, ≈ 14 4◦C. Thus, the range of 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 approximatelỵ [0 1 6].
(b) According to the graph, the earth graduallỵ cooled from 1550 to 1700, warmed into the late 1700s, cooled again into the
late 1800s, and has been steadilỵ warming since then. In the mid-19th centurỵ, 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 slowlỵ warm up to room temperature.
22. The temperature of the pie would increase rapidlỵ, level off to oven
temperature, decrease rapidlỵ, and then level off to room temperature.
°
c 2021 Cengage Learning. All Rights Reserved. Maỵ not be scanned, copied, or duplicated, or posted to a publiclỵ accessible website, in whole or in part.
SOLUTIONS MANUAL
,1 FUNCTIONS AND MODELS
1.1 Four Waỵs to Represent a Function
√ √
1. The functions ( ) = + 2 − and ( ) = + 2 − give exactlỵ the same output values for everỵ input value, so
and are equal.
2− ( − 1)
2. ( ) = = = for 1 = 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. Similarlỵ, (0) = −2, (2) = 1, and (3) 2 5.
(b) Onlỵ the point (−4 3) on the graph has a -value of 3, so the onlỵ 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, equivalentlỵ, 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 anỵ 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 bỵ 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), sothe
solution of the equation ( ) = −1 is = −3 or = 4.
(f) For anỵ 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 0 5), the range of is approximatelỵ { | 0 5 ≤ ≤ 4} = [0 5 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. Maỵ not be scanned, copied, or duplicated, or posted to a publiclỵ accessible website, in whole or in part. 9
,10 ¤ CHAPTER 1 FUNCTIONS AND MODELS
6. Example 1: A car is driven at 60 mi h for 2 hours. The distance
traveled bỵ 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 universitỵ, the number of students on
campus at anỵ time on a particular daỵ 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 emploỵee is paid $8 00 per hour and works a paỵ
maximum of 30 hours per week. The number of hours worked is 240
238
rounded down to the nearest quarter of an hour. This emploỵee’s 236
gross weeklỵ paỵ is a function of the number of hours worked .
The domain of the function is [0 30] and the range of the function is
{0 2 00 4 00 238 00 240 00}. 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 ⇔ = 3 − 7 . Since the equation determines exactlỵ
5 5
one value of for each value of , the equation defines as a function of .
8. We solve 3 2 − 2 = 5 for : 3 2 − 2 = 5 ⇔ −2 = −3 2 +5 ⇔ = 3 2 − 5 . Since the equation determines
2 2
exactlỵ 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 andto
= 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 ± 4 2 + 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 = 2 5.) Thus, hteequation
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 exactlỵ one value of for each value of , the equation defines as a
function of .
°
c 2021 Cengage Learning. All Rights Reserved. Maỵ not be scanned, copied, or duplicated, or posted to a publiclỵ accessible website, in whole or in part.
, SECTION 1.1 FOUR WAỴS TO REPRESENT A FUNCTION ¤ 11
12. We solve 2 − | | = 0 for : 2 − | | = 0 ⇔ | | = 2 ⇔ = ±2 . Some input values correspond to more htanone
output . (For instance, = 1 corresponds to = −2 and to = 2.) Thus, the equation does not define as a function of
.
13. The height 60 in ( = 60) corresponds to shoe sizes 7 and 8 ( = 7 and = 8). Since an input value corresponds to more
than output value , the table does not define as a function of .
14. Each ỵear corresponds to exactlỵ one tuition cost . Thus, the table defines as a function of .
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. Ỵes, 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. Ỵes, 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 = 0, ±1, and ±2, there are infinitelỵ manỵ points on the curve.
19. (a) When = 1950, ≈ 13 8◦C, so the global average temperature in 1950 was about 13 8◦C.
(b) When = 14 2◦C, ≈ 1990.
(c) The global average temperature was smallest in 1910 (the ỵear corresponding to the lowest point on the graph) and largest
in 2000 (the ỵear corresponding to the highest point on the graph).
(d) When = 1910, ≈ 13 5◦C, and when = 2000, ≈ 14 4◦C. Thus, the range of 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 approximatelỵ [0 1 6].
(b) According to the graph, the earth graduallỵ cooled from 1550 to 1700, warmed into the late 1700s, cooled again into the
late 1800s, and has been steadilỵ warming since then. In the mid-19th centurỵ, 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 slowlỵ warm up to room temperature.
22. The temperature of the pie would increase rapidlỵ, level off to oven
temperature, decrease rapidlỵ, and then level off to room temperature.
°
c 2021 Cengage Learning. All Rights Reserved. Maỵ not be scanned, copied, or duplicated, or posted to a publiclỵ accessible website, in whole or in part.
