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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 −
− 1
= for − 1 6 = 0, so and [where () = ] are not equal because (1) is undefined and
3. (a) The point (− 2 2) lies on the graph of , so (−2) = 2. Similarly, (0) = − 2 , (2) = 1, and (3) 2 5.
(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 0 5), the range of is approximately { | 0 5 ≤ ≤ 4 } = [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, the range is [− 85 115].
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 $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
240
pay
(^0) 0.25 0.50 0.75 29.50 29.75 30 hours
2
4
238 236
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 3
2 − 2 = 5 for : 3
2 − 2 = 5 ⇔ − 2 = − 3
2
3 2
2 −
5
- Since the equation determines
exactly one value of for each value of , the equation defines as a function of .
9. We solve
2
2 = 5 for :
2
2 = 5 ⇔ ( − 3)
2 = 5 −
2 ⇔ − 3 = ±
5 − ^2 ⇔
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(5)(−4)
2(5)
4 ^2 + 80
^2 + 20
(using the quadratic formula). Some input
values correspond to more than one output . (For instance, = 4 corresponds to = − 2 and to = 2 5 .) Thus, the
equation does not define as a function of .
11. We solve ( + 3)
3
3
- 1 = 2 ⇔ ( + 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 .
12 ¤^ CHAPTER 1 FUNCTIONS AND 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
before noon. These times are reasonable, considering the power consumption schedules of most individuals and
businesses.
24. 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.
25. Of course, this graph depends strongly on the
geographical location!
26. 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.)
27. As the price increases, the amount sold decreases. 28. The value of the car decreases fairly rapidly initially, then
somewhat less rapidly.
SECTION 1.1 FOUR WAYS TO REPRESENT A FUNCTION ¤^ 13
30. (a) (b)
(c) (d)
31. (a) (b) 9:00 AM corresponds to = 9. When = 9, the
temperature is about 74
◦ F.
32. (a) (b) The blood alcohol concentration rises rapidly, then slowly
decreases to near zero.
2 − + 2.
2 − 2 + 2 = 12 − 2 + 2 = 12.
2 − (−2) + 2 = 12 + 2 + 2 = 16.
2 − + 2.
2 − (−) + 2 = 3
2
( + 1) = 3( + 1)^2 − ( + 1) + 2 = 3(^2 + 2 + 1) − − 1 + 2 = 3^2 + 6 + 3 − + 1 = 3^2 + 5 + 4.
2 − + 2) = 6
2 − 2 + 4.
2 − (2) + 2 = 3(4
2 ) − 2 + 2 = 12
2 − 2 + 2.
[continued]
SECTION 1.1 FOUR WAYS TO REPRESENT A FUNCTION ¤^ 15
40. The function () =
2
^2 + 4 − 21
is defined for all values of except those for which
2
( + 7)( − 3) = 0 ⇔ = − 7 or = 3. Thus, the domain is { ∈ | 6 = − 7 3 } = (−∞ −7) ∪ (− 7 3) ∪ (3 ∞).
2 − 1 is defined for all real numbers. In fact
3
(), where () is a polynomial, is defined for all real numbers.
Thus, the domain is or (−∞ ∞).
2 + is defined when 3 − ≥ 0 ⇔ ≤ 3 and 2 + ≥ 0 ⇔ ≥ − 2. Thus, the domain is
− 2 ≤ ≤ 3 , or [− 2 3].
^2 − 5 is defined when ^2 − 5 0 ⇔ ( − 5) 0. Note that ^2 − 5 6 = 0 since that would result in
division by zero. The expression ( − 5) is positive if 0 or 5. (See Appendix A for methods for solving
inequalities.) Thus, the domain is (−∞ 0) ∪ (5 ∞).
is defined when + 1 6 = 0 [ 6 = − 1 ] and 1 +
= 0. Since 1 +
= − 1 ⇔ 1 = − − 1 ⇔ = − 2 , the domain is { | 6 = − 2 , 6 = − 1 } = (−∞ −2) ∪ (− 2 −1) ∪ (− 1 ∞).
is defined when ≥ 0 and 2 −
≥ 0. Since 2 −
0 ≤ ≤ 4 , the domain is [0 4].
46. The function () =
^2 − 4 − 5 is defined when
2 − 4 − 5 ≥ 0 ⇔ ( + 1)( − 5) ≥ 0. The polynomial
2 − 4 − 5 may change signs only at its zeros, so we test values of on the intervals separated by = − 1 and
= 5: (−2) = 7 0 , (0) = − 5 0 , and (6) = 7 0. Thus, the domain of , equivalent to the solution intervals
of () ≥ 0 , is { | ≤ − 1 or ≥ 5 } = (−∞ −1] ∪ [5 ∞).
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
2 ≥ 0
2 ^
= { | 2 ≥ ||} = [− 2 2]. From the graph,
the range is 0 ≤ ≤ 2 , or [0 2].
48. The function () =
2 − 4
− 2
is defined when − 2 6 = 0 ⇔ 6 = 2, so the
domain is { | 6 = 2} = (−∞ 2) ∪ (2 ∞). On its domain,
2 − 4
− 2
= + 2. Thus, the graph of is the
line = + 2 with a hole at (2 4).
16 ¤^ CHAPTER 1 FUNCTIONS AND MODELS
2
if ≥ 0
2
- 2 = 11, (0) = 0, and (2) = 2.
5 if 2
1 2 ^ −^3 if^ ^ ≥^2
(−3) = 5, (0) = 5, and (2) =
1 2 (2)^ −^ 3 =^ −^2.
+ 1 if ≤ − 1
2 if − 1
2 = 0, and (2) = 2
2 = 4.
− 1 if ≤ 1
7 − 2 if 1
(−3) = − 1 , (0) = − 1 , and (2) = 7 − 2(2) = 3.
if ≥ 0
− if 0
so () = + || =
2 if ≥ 0
0 if 0
Graph the line = 2 for ≥ 0 and graph = 0 (the axis) for 0
18 ¤^ CHAPTER 1 FUNCTIONS AND MODELS
59. Recall that the slope of a line between the two points ( 1 1 ) and ( 2 2 ) is =
and an equation of the line
connecting those two points is − 1 = ( − 1 ). The slope of the line segment joining the points (1 −3) and (5 7) is
, so an equation is − (−3) =
5 2 (^ −^ 1). The function is^ ^ () =^
5 2 ^ −^
11 2 ,^1 ≤^ ^ ≤^5.
60. The slope of the line segment joining the points (− 5 10) and (7 −10) is
, so an equation is
5 3 [^ −^ (−5)]. The function is^ ^ () =^ −^
5 3 ^ +^
5 3 ,^ −^5 ≤^ ^ ≤^7.
61. We need to solve the given equation for . + ( − 1)
2 = 0 ⇔ ( − 1)
2 = − ⇔ − 1 = ±
−. The expression with the positive radical represents the top half of the parabola, and the one with the negative
radical represents the bottom half. Hence, we want () = 1 −
−. Note that the domain is ≤ 0.
62. ^2 + ( − 2)^2 = 4 ⇔ ( − 2)^2 = 4 − ^2 ⇔ − 2 = ±
4 − ^2 ⇔ = 2 ±
4 − ^2. The top half is given by
the function () = 2 +
4 − ^2 , − 2 ≤ ≤ 2.
63. For 0 ≤ ≤ 3 , the graph is the line with slope − 1 and intercept 3 , that is, = − + 3. For 3 ≤ 5 , the graph is the line
with slope 2 passing through (3 0); that is, − 0 = 2( − 3), or = 2 − 6. So the function is
− + 3 if 0 ≤ ≤ 3
2 − 6 if 3 ≤ 5
64. For − 4 ≤ ≤ − 2 , the graph is the line with slope −
3 2 passing through^ (−^2 ^ 0);^ that is,^ ^ −^ 0 =^ −^
3 2 [^ −^ (−2)], or
3 2 ^ −^3. For^ −^2 ^2 , the graph is the top half of the circle with center^ (0^ 0)^ and radius^2. An equation of the circle
is
2
2 = 4, so an equation of the top half is =
4 − ^2. For 2 ≤ ≤ 4 , the graph is the line with slope
3 2 passing
through (2 0); that is, − 0 = 32 ( − 2), or = 32 − 3. So the function is
3 2 ^ −^3 if^ −^4 ≤^ ^ ≤ −^2 √ 4 − ^2 if − 2 2
3 2 ^ −^3 if^2 ≤^ ^ ≤^4
65. Let the length and width of the rectangle be and . Then the perimeter is 2 + 2 = 20 and the area is = .
Solving the first equation for in terms of gives =
= 10 − . Thus, () = (10 − ) = 10 −
2
. Since
lengths are positive, the domain of is 0 10. If we further restrict to be larger than , then 5 10 would be
the domain.
66. Let the length and width of the rectangle be and . Then the area is = 16, so that = 16. The perimeter is
= 2 + 2 , so () = 2 + 2(16) = 2 + 32, and the domain of is 0 , since lengths must be positive
quantities. If we further restrict to be larger than , then 4 would be the domain.
SECTION 1.1 FOUR WAYS TO REPRESENT A FUNCTION ¤^ 19
67. Let the length of a side of the equilateral triangle be . Then by the Pythagorean Theorem, the height of the triangle satisfies
2
1 2
2 , so that
2 =
2 − 14
2 = 34
2 and =
√ 3 2 . Using the formula for the area^ ^ of a triangle,
1 2 (base)(height), we obtain^ () =^
1 2 ()
3 2
√ 3 4
2 , with domain 0.
68. Let the length, width, and height of the closed rectangular box be denoted by , , and , respectively. The length is twice
the width, so = 2. The volume of the box is given by = . Since = 8, we have 8 = (2 ) ⇒
2 ⇒ =
, and so = ( ) =
69. Let each side of the base of the box have length , and let the height of the box be . Since the volume is 2 , we know that
2 , so that = 2
2 , and the surface area is =
2
2
2 ) =
2
domain 0.
70. Let and denote the radius and the height of the right circular cylinder, respectively. Then the volume is given by
2 , and for this particular cylinder we have
2 = 25 ⇔
2
. Solving for and rejecting the negative
solution gives =
, so = () =
in.
71. The height of the box is and the length and width are = 20 − 2 , = 12 − 2 . Then = and so
2 ) = 4
3 − 64
2
The sides , , and must be positive. Thus, 0 ⇔ 20 − 2 0 ⇔ 10 ;
0 ⇔ 12 − 2 0 ⇔ 6 ; and 0. Combining these restrictions gives us the domain 0 6.
72. The area of the window is = +
1 2
1 2
2
, where is the height of the rectangular portion of the window.
The perimeter is = 2 + +
1 2 ^ = 30^ ⇔^2 ^ = 30^ −^ ^ −^
1 2 ^ ⇔^ ^ =^
1 4 (60^ −^2 ^ −^ ). Thus,
2
2 −
2
2 = 15 −
2 −
2 = 15 −
2
Since the lengths and must be positive quantities, we have 0 and 0. For 0 , we have 2 0 ⇔
1 2 ^0 ⇔^60 ^2 ^ +^ ^ ⇔^ ^
. Hence, the domain of is 0
73. 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
SECTION 1.1 FOUR WAYS TO REPRESENT A FUNCTION ¤^ 21
(b) For an odd function, (−) = −(). The graph of an odd function is
symmetric about the origin. We rotate the given portion of the graph of
through 180
◦ about the origin in order to complete it.
80. (a) The graph of an even function is symmetric about the axis. We reflect the
given portion of the graph of about the axis in order to complete it.
(b) The graph of an odd function is symmetric about the origin. We rotate the
given portion of the graph of through 180 ◦^ about the origin in order to
complete it.
^2 + 1
(−)^2 + 1
^2 + 1
^2 + 1
Since (−) = −(), is an odd function.
2
^4 + 1
2
(−)^4 + 1
2
^4 + 1
Since (−) = (), is an even function.
, so (−) =
Since this is neither () nor − (), the function is
neither even nor odd.
Since (−) = −(), is an odd function.
22 ¤^ CHAPTER 1 FUNCTIONS AND MODELS
2 −
4 .
2 −(−)
4 = 1+3
2 −
4 = ().
Since (−) = (), is an even function.
3 −
5 , so
3 − (−)
5 = 1 + 3(−
3 ) − (−
5 )
3
5
Since this is neither () nor − (), the function is
neither even nor odd.
87. (i) If and are both even functions, then (−) = () and (−) = (). Now
( + )(−) = (−) + (−) = () + () = ( + )(), so + is an even function.
(ii) If and are both odd functions, then (−) = − () and (−) = −(). Now
( + )(−) = (−) + (−) = − () + [−()] = −[ () + ()] = −( + )(), so + is an odd function.
(iii) If is an even function and is an odd function, then ( + )(−) = (−) + (−) = () + [−()] = () − (),
which is not ( + )() nor −( + )(), so + is neither even nor odd. (Exception: if is the zero function, then
+ will be odd. If is the zero function, then + will be even .)
88. (i) If and are both even functions, then (−) = () and (−) = (). Now
( )(−) = (−)(−) = ()() = ( )(), so is an even function.
(ii) If and are both odd functions, then (−) = − () and (−) = −(). Now
( )(−) = (−)(−) = [− ()][−()] = ()() = ( )(), so is an even function.
(iii) If is an even function and is an odd function, then
( )(−) = (−)(−) = ()[−()] = −[ ()()] = −( )(), so is an odd function.
1.2 Mathematical Models: A Catalog of Essential Functions
1. (a) () =
3
2 is a polynomial function of degree 3. (This function is also an algebraic function.)
(b) () = cos
2 − sin is a trigonometric function.
(c) () =
√ 3 is a power function.
(d) () = 8
is an exponential function.
(e) =
^2 + 1
is an algebraic function. It is the quotient of a root of a polynomial and a polynomial of degree 2.
(f ) () = log 10 is a logarithmic function.
24 ¤^ CHAPTER 1 FUNCTIONS AND MODELS
(c) To belong to both families, an equation must have slope = 2, so the equation in part (b), = + (1 − 2 ),
becomes = 2 − 3. It is the only function that belongs to both families.
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. We graph () = ^3 − ^2 for = − 2 , 0 , 1 , and 3. For 6 = 0,
3 −
2 =
2 ( − ) has two intercepts, 0 and . The curve has
one decreasing portion that begins or ends at the origin and increases in
length as || increases; the decreasing portion is in quadrant II for 0 and
in quadrant IV for 0.
11. Because is a quadratic function, we know it is of the form () =
2
- + . The intercept is 18 , so (0) = 18 ⇒
= 18 and () =
2
- + 18. Since the points (3 0) and (4 2) lie on the graph of , we have
This is a system of two equations in the unknowns and , and subtracting (1) from (2) gives = 2. From (1) ,
3 (2) + = − 6 ⇒ = − 12 , so a formula for is () = 2^2 − 12 + 18.
12. is a quadratic function so () = ^2 + + . The y intercept is 1 , so (0) = 1 ⇒ = 1 and () = ^2 + + 1.
Since the points (− 2 2) and (1 − 2 5) lie on the graph of , we have
Then (1) + 2 · (2) gives us 6 = − 6 ⇒ = − 1 and from (2) , we have −1 + = − 3 5 ⇒ = − 2 5 , so a formula for
is () = −
2 − 2 5 + 1.
SECTION 1.2 MATHEMATICAL MODELS: A CATALOG OF ESSENTIAL FUNCTIONS ¤^ 25
13. Since (−1) = (0) = (2) = 0, has zeros of − 1 , 0 , and 2 , so an equation for is () = [ − (−1)]( − 0)( − 2),
or () = ( + 1)( − 2). Because (1) = 6, we’ll substitute 1 for and 6 for ().
6 = (1)(2)(−1) ⇒ − 2 = 6 ⇒ = − 3 , so an equation for is () = − 3 ( + 1)( − 2).
14. (a) For = 0 02 + 8 50 , the slope is 0 02 , which means that the average surface temperature of the world is increasing at
a rate of 0 02
◦ C per year. The intercept is 8 50 , which represents the average surface temperature in
◦ C in the
year 1900.
(b) = 2100 − 1900 = 200 ⇒ = 002(200) + 850 = 12 50 ◦C
15. (a) = 200, so = 0 0417 ( + 1) = 00417(200)( + 1) = 8 34 + 8 34. The slope is 8 34 , which represents the
change in mg of the dosage for a child for each change of 1 year in age.
(b) For a newborn, = 0, so = 8 34 mg.
16. (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.
17. (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.
18. (a) Jari is traveling faster since the line representing her distance versus time is steeper than the corresponding line for Jade.
(b) At = 0, Jade has traveled 10 miles. At = 6, Jade has traveled 16 miles. Thus, Jade’s speed is
16 miles − 10 miles
6 minutes − 0 minutes
= 1 mimin. This is
1 mile
1 minute
×
60 minutes
1 hour
= 60 mih
At = 0, Jari has traveled 0 miles. At = 6, Jari has traveled 7 miles. Thus, Jari’s speed is
7 miles − 0 miles
6 minutes − 0 minutes
mimin or
7 miles
6 minutes
×
60 minutes
1 hour
= 70 mih
(c) From part (b), we have a slope of 1 (mileminute) for the linear function modeling the distance traveled by Jade and
from the graph the intercept is 10. Thus, () = 1 + 10 = + 10. Similarly, we have a slope of
7 6 milesminute^ for
SECTION 1.2 MATHEMATICAL MODELS: A CATALOG OF ESSENTIAL FUNCTIONS ¤^ 27
24. (a) = and = 39 kPa when = 0671 m
3 , so 39 = 0 671 ⇔ = 39(0671) = 26 169.
(b) When = 0 916 , = 26 169 = 26 169 0 916 ≈ 28 6 , so the pressure is reduced to approximately 28 6 kPa.
25. (a) =
3 so doubling the windspeed gives = (2)
3 = 8(
3 ). Thus, the power output is increased by a factor
of eight.
(b) The area swept out by the blades is given by =
2 , where is the blade length, so the power output is
3 =
2
3
. Doubling the blade length gives = (2)
2
3 = 4(
2
3 ). Thus, the power output is increased
by a factor of four.
(c) From part (b) we have =
2
3 , and = 0214 kgm
3 , = 30 m gives
kg
m^3
· 900 m
2 ·
3 = 192 6
3 kg m
For = 10 ms, we have
m
s
kg
m
m
2 · kg
s^3
≈ 605 ,000 W
Similarly, = 15 ms gives = 650, 025 ≈ 2 , 042 ,000 W and = 25 ms gives = 3, 009 , 375 ≈ 9 , 454 ,000 W.
26. (a) We graph ( ) = (5 67 × 10
− 8 )
4 for 100 ≤ ≤ 300 :
(b) From the graph, we see that as temperature increases, energy increases—slowly at first, but then at an increasing rate.
27. (a) The data appear to be periodic and a sine or cosine function would make the best model. A model of the form
() = cos() + seems appropriate.
(b) The data appear to be decreasing in a linear fashion. A model of the form () = + seems appropriate.
28. (a) The data appear to be increasing exponentially. A model of the form () = · ^ or () = · ^ + seems
appropriate.
(b) The data appear to be decreasing similarly to the values of the reciprocal function. A model of the form () =
seems appropriate.
28 ¤^ CHAPTER 1 FUNCTIONS AND MODELS
Exercises 29– 33: Some values are given to many decimal places. The results may depend on the technology used—rounding is left to the reader.
29. (a)
A linear model does seem appropriate.
(b) Using the points (4000 14 1) and (60, 000 8 2), we obtain
( − 4000) or, equivalently,
(c) Using a computing device, we obtain the regression line = − 0 0000997855 + 13 950764.
The following commands and screens illustrate how to find the regression line on a TI 84 Plus calculator.
Enter the data into list one (L1) and list two (L2). Press to enter the editor.
Find the regession line and store it in Y 1. Press.
Note from the last figure that the regression line has been stored in Y 1 and that Plot1 has been turned on (Plot1 is
highlighted). You can turn on Plot1 from the Y= menu by placing the cursor on Plot1 and pressing or by
pressing.
Now press to produce a graph of the data and the regression
line. Note that choice 9 of the ZOOM menu automatically selects a window
that displays all of the data.
(d) When = 25, 000 , ≈ 11 456 ; or about 11 5 per 100 population.
