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. () =
2 −
− 1
=
( − 1)
− 1
= for − 1 6= 0, so and [where () = ] are not equal because (1) is undefined and
(1) = 1.
3. (a) The point (−22) lies on the graph of , so (−2) = 2. Similarly, (0) = −2, (2) = 1, and (3) 25.
(b) Only the point (−43) 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 [−44]).
(d) The domain consists of all values on the graph of : { | −4 ≤ ≤ 4} = [−44]. The range of consists of all the
values on the graph of : { | −2 ≤ ≤ 3} = [−23].
(e) For any 1 2 in the interval [02], we have ( 1 ) ( 2 ). [The graph rises from (0−2) to (21).] Thus, () is
increasing on [02].
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 [23].
(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 [−40], we have ( 1 ) ( 2 ). Thus, () is decreasing on [−40].
(g) The domain of is { | −4 ≤ ≤ 4} = [−44]. The range of is { | −2 ≤ ≤ 3} = [−23].
(h) The domain of is { | −4 ≤ ≤ 3} = [−43]. Estimating the lowest point of the graph of as having coordinates
(005), the range of is approximately { | 05 ≤ ≤ 4} = [054].
5. From Figure 1 in the text, the lowest point occurs at about () = (12−85). The highest point occurs at about (17115).
Thus, the range of the vertical ground acceleration is −85 ≤ ≤ 115. Written in interval notation, the range is [−85115].
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
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CHAPTER 1 FUNCTIONSAND 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
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 is a function of the number of hours worked .
The domain of the function is [030] and the range of the function is
{02004002380024000}.
240
pay
hours 0.25 0.50 0.75
0
29.50 29.75 30
2
4
238
236
7. We solve 3−5 = 7 for : 3−5 = 7 ⇔ −5 = −3+7 ⇔ =
3
5 −
7
5 . Since the equation determines exactly
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 ⇔ =
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. () =
2 −
− 1
=
( − 1)
− 1
= for − 1 6= 0, so and [where () = ] are not equal because (1) is undefined and
(1) = 1.
3. (a) The point (−22) lies on the graph of , so (−2) = 2. Similarly, (0) = −2, (2) = 1, and (3) 25.
(b) Only the point (−43) 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 [−44]).
(d) The domain consists of all values on the graph of : { | −4 ≤ ≤ 4} = [−44]. The range of consists of all the
values on the graph of : { | −2 ≤ ≤ 3} = [−23].
(e) For any 1 2 in the interval [02], we have ( 1 ) ( 2 ). [The graph rises from (0−2) to (21).] Thus, () is
increasing on [02].
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 [23].
(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 [−40], we have ( 1 ) ( 2 ). Thus, () is decreasing on [−40].
(g) The domain of is { | −4 ≤ ≤ 4} = [−44]. The range of is { | −2 ≤ ≤ 3} = [−23].
(h) The domain of is { | −4 ≤ ≤ 3} = [−43]. Estimating the lowest point of the graph of as having coordinates
(005), the range of is approximately { | 05 ≤ ≤ 4} = [054].
5. From Figure 1 in the text, the lowest point occurs at about () = (12−85). The highest point occurs at about (17115).
Thus, the range of the vertical ground acceleration is −85 ≤ ≤ 115. Written in interval notation, the range is [−85115].
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 FUNCTIONSAND 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
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 is a function of the number of hours worked .
The domain of the function is [030] and the range of the function is
{02004002380024000}.
240
pay
hours 0.25 0.50 0.75
0
29.50 29.75 30
2
4
238
236
7. We solve 3−5 = 7 for : 3−5 = 7 ⇔ −5 = −3+7 ⇔ =
3
5 −
7
5 . Since the equation determines exactly
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 ⇔ =