Solution Manual For Single Variable Calculus, 8th Edition

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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 (1 3) is on the graph of , so  (1) = 3.
(b) When  = −1,  is about −02, so  (−1) ≈ −02.
(c)  () = 1 is equivalent to  = 1 When  = 1, we have  = 0 and  = 3.
(d) A reasonable estimate for  when  = 0 is  = −08.
(e) The domain of  consists of all -values on the graph of  . For this function, the domain is −2 ≤  ≤ 4, or [−2 4].
The range of  consists of all -values on the graph of  . For this function, the range is −1 ≤  ≤ 3, or [−1 3].
(f ) As  increases from −2 to 1,  increases from −1 to 3. Thus,  is increasing on the interval [−2 1].
4. (a) The point (−4 −2) is on the graph of  , so  (−4) = −2. The point (3 4) is on the graph of , so (3) = 4.
(b) We are looking for the values of  for which the -values are equal. The -values for  and  are equal at the points
(−2 1) and (2 2), so the desired values of  are −2 and 2.
(c)  () = −1 is equivalent to  = −1. When  = −1, we have  = −3 and  = 4.
(d) As  increases from 0 to 4,  decreases from 3 to −1. Thus,  is decreasing on the interval [0 4].
(e) The domain of  consists of all -values on the graph of  . For this function, the domain is −4 ≤  ≤ 4, or [−4 4].
The range of  consists of all -values on the graph of  . For this function, the range is −2 ≤  ≤ 3, or [−2 3].
(f ) The domain of  is [−4 3] and the range is [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, we get

10 ¤ CHAPTER 1 FUNCTIONS AND MODELS
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 $800 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 [0 30] and the range of the function is
{0 200 400     23800 24000}.
240
pay
hours0.25 0.50 0.750 29.50 29.75 30
2
4
238
236
7. 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.
8. 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].
9. 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].
10. No, the curve is not the graph of a function since for  = 0, ±1, and ±2, there are infinitely many points on the curve.
11. (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 year corresponding to the lowest point on the graph) and largest
in 2005 (the year corresponding to the highest point on the graph).
(d) When  = 1910,  ≈ 135◦C, and when  = 2005,  ≈ 145◦C. Thus, the range of  is about [135, 145].
12. (a) The ring width varies from near 0 mm to about 16 mm, so the range of the ring width function is approximately [0 16].
(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.
13. 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.
post , in

SECTION 1.1 FOUR WAYS TO REPRESENT A FUNCTION ¤ 11
14. Runner A won the race, reaching the finish line at 100 meters in about 15 seconds, followed by runner B with a time of about
19 seconds, and then by runner C who finished in around 23 seconds. B initially led the race, followed by C, and then A.
C then passed B to lead for a while. Then A passed first B, and then passed C to take the lead and finish first. Finally,
B passed C to finish in second place. All three runners completed the race.
15. (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
before noon. These times are reasonable, considering the power consumption schedules of most individuals and
businesses.
16. The summer solstice (the longest day of the year) is
around June 21, and the winter solstice (the shortest day)
is around December 22. (Exchange the dates for the
southern hemisphere.)
17. Of course, this graph depends strongly on the
geographical location!
18. The value of the car decreases fairly rapidly initially, then
somewhat less rapidly.
19. As the price increases, the amount sold decreases.
20. The temperature of the pie would increase rapidly, level
off to oven temperature, decrease rapidly, and then level
off to room temperature.
21.
or du cessi

12 ¤ CHAPTER 1 FUNCTIONS AND MODELS
22. (a) (b)
(c) (d)
23. (a) (b) 9:00 AM corresponds to  = 9. When  = 9, the
temperature  is about 74◦F.
24. (a) (b) The blood alcohol concentration rises rapidly, then slowly
decreases to near zero. Note that the BAC in this exercise is
measured in mgmL, not percent.
25.  () = 32 −  + 2
 (2) = 3(2)2 − 2 + 2 = 12 − 2 + 2 = 12
 (−2) = 3(−2)2 − (−2) + 2 = 12 + 2 + 2 = 16
 () = 32 −  + 2
 (−) = 3(−)2 − (−) + 2 = 32 +  + 2
 ( + 1) = 3( + 1)2 − ( + 1) + 2 = 3(2 + 2 + 1) −  − 1 + 2 = 32 + 6 + 3 −  + 1 = 32 + 5 + 4
2 () = 2 ·  () = 2(32 −  + 2) = 62 − 2 + 4
post , in

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14 ¤ CHAPTER 1 FUNCTIONS AND MODELS
36.  () =  + 1
1 + 1
 + 1
is defined when  + 1 6 = 0 [ 6 = −1] and 1 + 1
 + 1 6 = 0. Since 1 + 1
 + 1 = 0 ⇔
1
 + 1 = −1 ⇔ 1 = − − 1 ⇔  = −2, the domain is { |  6 = −2,  6 = −1} = (−∞ −2) ∪ (−2 −1) ∪ (−1 ∞).
37.  () = 2 − √ is defined when  ≥ 0 and 2 − √ ≥ 0. Since 2 − √ ≥ 0 ⇔ 2 ≥ √ ⇔ √ ≤ 2 ⇔
0 ≤  ≤ 4, the domain is [0 4].
38. () = √4 − 2. Now  = √4 − 2 ⇒ 2 = 4 − 2 ⇔ 2 + 2 = 4, so
the graph is the top half of a circle of radius 2 with center at the origin. The domain
is  | 4 − 2 ≥ 0

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SECTION 1.1 FOUR WAYS TO REPRESENT A FUNCTION ¤ 15
44.  () =
 −1 if  ≤ 1
7 − 2 if   1
 (−3) = −1,  (0) = −1, and  (2) = 7 − 2(2) = 3.
45. || =
  if  ≥ 0
− if   0
so () =  + || =
2 if  ≥ 0
0 if   0
Graph the line  = 2 for  ≥ 0 and graph  = 0 (the -axis) for   0
46. () = | + 2| =
 + 2 if  + 2 ≥ 0
−( + 2) if  + 2  0
=
 + 2 if  ≥ −2
− − 2 if   −2
47. () = |1 − 3| =
 1 − 3 if 1 − 3 ≥ 0
−(1 − 3) if 1 − 3  0
=
 1 − 3 if  ≤ 1
3
3 − 1 if   1
3
48. || =
 if  ≥ 0
− if   0 and
| + 1| =
 + 1 if  + 1 ≥ 0
−( + 1) if  + 1  0 =
 + 1 if  ≥ −1
− − 1 if   −1
so () = || + | + 1| =



 + ( + 1) if  ≥ 0
− + ( + 1) if −1 ≤   0
− + (− − 1) if   −1
=



2 + 1

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18 ¤ CHAPTER 1 FUNCTIONS AND MODELS
64. We can summarize the monthly cost with a piecewise
defined function.
() =
35 if 0 ≤  ≤ 400
35 + 010( − 400) if   400
65. We can summarize the amount of the fine with a
piecewise defined function.
 () =



15(40 − ) if 0 ≤   40
0 if 40 ≤  ≤ 65
15( − 65) if   65
66. For the first 1200 kWh, () = 10 + 006.
For usage over 1200 kWh, the cost is
() = 10 + 006(1200) + 007( − 1200) = 82 + 007( − 1200).
Thus,
() =
10 + 006 if 0 ≤  ≤ 1200
82 + 007( − 1200) if   1200
67. (a) (b) On $14,000, tax is assessed on $4000, and 10%($4000) = $400.
On $26,000, tax is assessed on $16,000, and
10%($10,000) + 15%($6000) = $1000 + $900 = $1900.
(c) As in part (b), there is $1000 tax assessed on $20,000 of income, so
the graph of  is a line segment from (10,000 0) to (20,000 1000).
The tax on $30,000 is $2500, so the graph of  for   20,000 is
the ray with initial point (20,000 1000) that passes through
(30,000 2500).
68. One example is the amount paid for cable or telephone system repair in the home, usually measured to the nearest quarter hour.
Another example is the amount paid by a student in tuition fees, if the fees vary according to the number of credits for which
the student has registered.
69.  is an odd function because its graph is symmetric about the origin.  is an even function because its graph is symmetric with
respect to the -axis.
post , in

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SECTION 1.1 FOUR WAYS TO REPRESENT A FUNCTION ¤ 19
70.  is not an even function since it is not symmetric with respect to the -axis.  is not an odd function since it is not symmetric
about the origin. Hence,  is neither even nor odd.  is an even function because its graph is symmetric with respect to the
-axis.
71. (a) Because an even function is symmetric with respect to the -axis, and the point (5 3) is on the graph of this even function,
the point (−5 3) must also be on its graph.
(b) Because an odd function is symmetric with respect to the origin, and the point (5 3) is on the graph of this odd function,
the point (−5 −3) must also be on its graph.
72. (a) If  is even, we get the rest of the graph by reflecting
about the -axis.
(b) If  is odd, we get the rest of the graph by rotating
180◦ about the origin.
73.  () = 
2 + 1 .
 (−) = −
(−)2 + 1 = −
2 + 1 = − 
2 + 1 = −().
Since  (−) = −(),  is an odd function.
74.  () = 2
4 + 1 .
 (−) = (−)2
(−)4 + 1 = 2
4 + 1 =  ().
Since  (−) =  (),  is an even function.
75.  () = 
 + 1 , so (−) = −
− + 1 = 
 − 1 .
Since this is neither  () nor − (), the function  is
neither even nor odd.
76.  () =  ||.
 (−) = (−) |−| = (−) || = −( ||)
= − ()
Since  (−) = −(),  is an odd function.
or du cessi

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SECTION 1.2 MATHEMATICAL MODELS: A CATALOG OF ESSENTIAL FUNCTIONS ¤ 21
2. (a)  =  is an exponential function (notice that  is the exponent).
(b)  =  is a power function (notice that  is the base).
(c)  = 2(2 − 3) = 22 − 5 is a polynomial of degree 5.
(d)  = tan  − cos  is a trigonometric function.
(e)  = (1 + ) is a rational function because it is a ratio of polynomials.
(f )  = √3 − 1(1 + 3
√) is an algebraic function because it involves polynomials and roots of polynomials.
3. We notice from the figure that  and  are even functions (symmetric with respect to the -axis) and that  is an odd function
(symmetric with respect to the origin). So (b)  = 5 must be  . Since  is flatter than  near the origin, we must have
(c)  = 8 matched with  and (a)  = 2 matched with .
4. (a) The graph of  = 3 is a line (choice ).
(b)  = 3 is an exponential function (choice  ).
(c)  = 3 is an odd polynomial function or power function (choice  ).
(d)  = 3
√ = 13 is a root function (choice ).
5. The denominator cannot equal 0, so 1 − sin  6 = 0 ⇔ sin  6 = 1 ⇔  6 = 
2 + 2. Thus, the domain of
 () = cos 
1 − sin  is  |  6 = 
2 + 2,  an integer.
6. The denominator cannot equal 0, so 1 − tan  6 = 0 ⇔ tan  6 = 1 ⇔  6 = 
4 + . The tangent function is not defined
if  6 = 
2 + . Thus, the domain of (

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22 ¤ CHAPTER 1 FUNCTIONS AND MODELS
8. All members of the family of linear functions () = 1 + ( + 3) have
graphs that are lines passing through the point (−3 1).
9. All members of the family of linear functions () =  −  have graphs
that are lines with slope −1. The -intercept is .
10. The vertex of the parabola on the left is (3 0), so an equation is  = ( − 3)2 + 0. Since the point (4 2) is on the
parabola, we’ll substitute 4 for  and 2 for  to find . 2 = (4 − 3)2 ⇒  = 2, so an equation is  () = 2( − 3)2.
The -intercept of the parabola on the right is (0 1), so an equation is  = 2 +  + 1. Since the points (−2 2) and
(1 −25) are on the parabola, we’ll substitute −2 for  and 2 for  as well as 1 for  and −25 for  to obtain two equations
with the unknowns  and .
(−2 2): 2 = 4 − 2 + 1 ⇒ 4 − 2 = 1 (1)
(1 −25): −25 =  +  + 1 ⇒  +  = −35 (2)
2

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SECTION 1.2 MATHEMATICAL MODELS: A CATALOG OF ESSENTIAL FUNCTIONS ¤ 23
14. (a) (b) The slope of −4 means that for each increase of 1 dollar for a
rental space, the number of spaces rented decreases by 4. The
-intercept of 200 is the number of spaces that would be occupied
if there were no charge for each space. The -intercept of 50 is the
smallest rental fee that results in no spaces rented.
15. (a) (b) The slope of 9
5 means that  increases 9
5 degrees for each increase
of 1◦C. (Equivalently,  increases by 9 when  increases by 5
and  decreases by 9 when  decreases by 5.) The  -intercept of
32 is the Fahrenheit temperature corresponding to a Celsius
temperature of 0.
16. (a) Let  = distance traveled (in miles) and  = time elapsed (in hours). At
 = 0,  = 0 and at  = 50 minutes = 50 · 1
60 = 5
6 h,  = 40. Thus we
have two points: (0 0) and  5
6  40, so  = 40 − 0
5
6 − 0 = 48 and so  = 48.
(b)
(c) The slope is 48 and represents the car’s speed in mih.
17. (a) Using  in place of  and  in place of , we find the slope to be 2 − 1
2 − 1
= 80 − 70
173 − 113 = 10
60 = 1
6 . So a linear
equation is  − 80 = 1
6 ( − 173) ⇔  − 80 = 1
6  − 173
6 ⇔  = 1
6  + 307
6
 307
6 = 5116 .
(b) The slope of 1
6 means that the temperature in Fahrenheit degrees increases one-sixth as rapidly as the number of cricket
chirps per minute. Said differently, each increase of 6 cricket chirps per minute corresponds to an increase of 1◦F.
(c) When  = 150, the temperature is given approximately by  = 1
6 (150) + 307
6 = 7616 ◦F ≈ 76 ◦F.
18. (a) Let  denote the number of chairs produced in one day and  the associated
cost. Using the points (100 2200) and (300 4800), we get the slope
4800−2200
300−100 = 2600
200 = 13. So  − 2200 = 13( − 100) ⇔
 = 13 + 900.
(b) The slope of the line in part (a) is 13 and it represents the cost (in dollars)
of producing each additional chair.
(c) The -intercept is 900 and it represents the fixed daily costs of operating
the factory.
19. (a) We are given change in pressure
10 feet change in depth = 434
10 = 0434. Using  for pressure and  for depth with the point
(  ) = (0 15), we have the slope-intercept form of the line,  = 0434 + 15.
or du cessi

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