,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
,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.
°
, SECTION 1.1 FOUR WAYS TO REPRESENT A FUNCTION ¤ 11
12. We solve 2 − || = 0 for : 2 − || = 0 ⇔ || = 2 ⇔ = ±2. Some input values correspond to more than
one 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 year corresponds to exactly 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. 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 = 0, ±1, and ±2, there are infinitely many points on the curve.
19. (a) When = 1950, ≈ 138◦ C, so the global average temperature in 1950 was about 138◦ C.
(b) When = 142◦ C, ≈ 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 = 1910, ≈ 135◦ C, and when = 2000, ≈ 144◦ C. Thus, the range of is about [135, 144].
20. (a) The ring width varies from near 0 mm to about 16 mm, so the range of the ring width function is approximately [0 16].
(b) According to the graph, the earth gradually cooled from 1550 to 1700, warmed into the late 1700s, cooled again into the
late 1800s, and has been steadily warming since then. In the mid19th 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.
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.
°
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
,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.
°
, SECTION 1.1 FOUR WAYS TO REPRESENT A FUNCTION ¤ 11
12. We solve 2 − || = 0 for : 2 − || = 0 ⇔ || = 2 ⇔ = ±2. Some input values correspond to more than
one 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 year corresponds to exactly 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. 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 = 0, ±1, and ±2, there are infinitely many points on the curve.
19. (a) When = 1950, ≈ 138◦ C, so the global average temperature in 1950 was about 138◦ C.
(b) When = 142◦ C, ≈ 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 = 1910, ≈ 135◦ C, and when = 2000, ≈ 144◦ C. Thus, the range of is about [135, 144].
20. (a) The ring width varies from near 0 mm to about 16 mm, so the range of the ring width function is approximately [0 16].
(b) According to the graph, the earth gradually cooled from 1550 to 1700, warmed into the late 1700s, cooled again into the
late 1800s, and has been steadily warming since then. In the mid19th 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.
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.
°
