3
2
2
−
5
2 . 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(5)
=
−2 ±
√ 4 2
+ 80
10
=
− ±
√ 2
+ 20
5
(using the quadratic formula). Some input
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 .
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SECTION 1.1 FOUR WAYS TO REPRESENTA FUNCTION
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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 [−22] and the range
is [−12].
17. Yes, the curve is the graph of a function because it passes the Vertical Line Test. The domain is [−32] and the range
is [−3−2) ∪ [−13].
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 [016].
(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.
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CHAPTER 1 FUNCTIONSAND MODELS
23. (a) The power consumption at 6 AM is 500 MW which is obtained by reading the value of power when = 6 from the
graph. At 6 PM we read the value of when = 18 obtaining approximately 730 MW
(b) The minimum power consumption is determined by finding the time for the lowest point on the graph, = 4 or 4 AM . The
maximum power consumption corresponds to the highest point on the graph, which occurs just before = 12 or right