SOLUTION MANUALS FOR
CALCULUS EARLY
TRANSCENDENTALS 9
th
EDITION BY JAMES
STEWART, DANIEL K.
CLEGG & SALEEM
WATSON 2023/2024 /
GRADED A+. 1 / 4
1 FUNCTIONS AND MODELS
1.1Four Ways to Represent a Function 1.The functions() = + √
- −
- −
and() = + √
give exactly the same output values for every input value, so andare equal.
2.() =
2 −
− 1
=
( − 1)
− 1
= for − 1 6= 0, soand[where() = ] are not equal because (1)is undefined and
(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 offor which() = 3is−4.(c) The function outputs()are never greater than 3, so() ≤ 3for the entire domain of the function. Thus,() ≤ 3for
−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 ofconsists of all the values on the graph of
:{ | −2 ≤
≤ 3} = [−2 3].
(e) For any1 2in 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) = −2and(3) = 4.(b) Since(−3) = −1and(−3) = 2, or by observing that the graph ofis above the graph ofat = −3,(−3)is larger than(−3).(c) The graphs ofandintersect at = −2and = 2, so () = ()at these two values of.(d) The graph oflies below or on the graph offor−4 ≤ ≤ −2and for2 ≤ ≤ 3. Thus, the intervals on which
() ≤
()are[−4 −2]and[2 3].(e) () = −1is equivalent to = −1, and the points on the graph ofwith values of −1are(−3 −1)and(4 −1), so the solution of the equation() = −1is = −3or = 4.(f) For any1
2in the interval[−4 0], we have(1)
(
2). Thus,()is decreasing on[−4 0].(g) The domain ofis{ | −4 ≤ ≤ 4} = [−4 4]. The range ofis{ | −2 ≤
≤ 3} = [−2 3].
(h) The domain ofis{ | −4 ≤ ≤ 3} = [−4 3]. Estimating the lowest point of the graph ofas having coordinates (0 05), the range ofis 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 2 / 4
10¤CHAPTER 1FUNCTIONS AND MODELS
6.Example 1: A car is driven at60mih for2hours. The distance
traveled by the car is a function of the time. The domain of the function is{ | 0 ≤ ≤ 2}, whereis measured in hours. The range of the function is{ | 0 ≤ ≤ 120}, whereis measured in miles.
Example 2: At a certain university, the number of studentson
campus at any time on a particular day is a function of the timeafter midnight. The domain of the function is{ | 0 ≤ ≤ 24}, whereis measured in hours. The range of the function is{ | 0 ≤ ≤ }, whereis an integer andis the largest number of students on campus at once.
Example 3: A certain employee is paid$800per hour and works a
maximum of30hours per week. The number of hours worked is rounded down to the nearest quarter of an hour. This employee’s gross weekly payis a function of the number of hours worked.The domain of the function is[0 30]and the range of the function is
{0 200 400
23800 24000}.
240 pay hours0.250.500.75029.50 29.7530 2 4 238 236
7.We solve3 − 5 = 7for:3 − 5 = 7⇔ −5 = −3 + 7⇔
= 3 5
−
7 5 . Since the equation determines exactly one value offor each value of, the equation definesas a function of.
8.We solve3 2
− 2 = 5for:3
2
− 2 = 5⇔ −2 = −3
2
+ 5⇔
= 3 2 2 − 5 2 . Since the equation determines exactly one value offor each value of, the equation definesas a function of.
9.We solve 2
+ ( − 3)
2
= 5for:
2
+ ( − 3)
2
= 5⇔ ( − 3)
2
= 5 −
2 ⇔
− 3 = ±
√
- −
2 ⇔
= 3 ±
√
- −
2 . Some input valuescorrespond to more than one output. (For instance, = 1corresponds to = 1and to = 5.) Thus, the equation does not defineas a function of.
10.We solve2 + 5 2
= 4for: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 valuescorrespond to more than one output. (For instance, = 4corresponds to = −2and to = 25.) Thus, the equation does not defineas a function of.
11.We solve( + 3) 3
+ 1 = 2for:( + 3)
3
+ 1 = 2⇔ ( + 3)
3
= 2 − 1⇔
- 3 =
3 √
2 − 1⇔
= −3 +
3 √ 2 − 1. Since the equation determines exactly one value offor each value of, the equation definesas 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. 3 / 4
SECTION 1.1FOUR WAYS TO REPRESENT A FUNCTION¤11
12.We solve2 − || = 0for:2 − || = 0⇔ || = 2⇔
= ±2. Some input valuescorrespond to more than one output. (For instance, = 1corresponds to = −2and to = 2.) Thus, the equation does not defineas a function of.
13.The height 60 in( = 60)corresponds to shoe sizes 7 and 8( = 7and = 8). Since an input valuecorresponds to more than output value, the table does not defineas a function of.
14.Each yearcorresponds to exactly one tuition cost. Thus, the table definesas 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 about138 ◦ 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 ◦
- Thus, the range ofis about[135,144].
- / 4
20.(a) The ring width varies from near0 mmto about16 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 mid 19th 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.