Download Review Exercises

1.1 Functions
1
Exercises 1.1
15. It appears that near x  1, the graph
goes vertical. A vertical line drawn
at that point would touch the graph
in multiple locations. However, if the
graph doesn’t actually go vertical
near x  1, then it is a function. One
drawback of reading a graph is that it
is sometimes difficult to tell if the
graph goes vertical or not.
1. At any given instant in time, you can
have only one weight. For example,
at age 19 years, 3 months, 2 days, 11
hours, 2 minutes, 27 seconds you
have only one weight. Thus your
weight is a function of your age.
3. Temperature is a function of the time
of day since at any time of the day
when the temperature is measured,
the measured temperature will be a
single value.
5. The input value 182 has two different
output values (32 and 47). Therefore,
the number of salmon in a catch is
not a function of the number of fish
caught.
17. Since any vertical line drawn will
touch the graph exactly once, the
graph is a function.
19. Since any vertical line drawn will
touch the graph exactly once, the
graph is a function.
21. y   x 2  4;  3  x  3,  10  y  1
7. C  4  39.95  4
 159.80
The total cost of four pairs of shoes
is $159.80.
9. H  2   16  2   120
2
 56
Two seconds after he jumped, the
cliff diver is 56 feet above the water.
23. y  2 x 2  1;  2  x  2,  3  y  2
11. As shown in the table, E  4  0.06 .
In the fourth quarter since December
1999, shares in the tortilla company
earned $0.06 per share.
13. As seen in the graph, P  4  18.72 .
On November 8, 2001, the closing
stock price of the computer company
was approximately $18.72. (One
drawback of reading a graph is that it
is difficult to be precise.)
Copyright © Houghton Mifflin Company. All rights reserved.
2
CHAPTER 1 Functions and Linear Models
25. y  5 x  x 2 ; x  1
From the graph, it appears that y  4
at x  1. We calculate the exact
value of y algebraically:
2
y  5 1  1
 5 1
4
27. y  
4  x2  4
; x  2
x 3  2 x 2  3x  6
From the graph, it appears that y  2
at x  2 . We calculate the exact
value of y algebraically:
y


4  2   4
 2   2  2 
4 4  4
3
2
2

 3  2   6
8  8  6  6
0
0
But division by zero is not a legal
operation. Therefore, the function is
not defined when x  2 .
Graphically speaking, there is a
“hole” in the graph at x  2 .

29. The function f ( p)  p 2  2 p  1 has
the domain of all real numbers.
r2  1
has a
r2  1
domain of all real numbers, since no
value of r will make the denominator
equal to zero.
31. The function h  r  
3t  3
is
4t  4
undefined when the denominator
equals zero.
4t  4  0
33. The function g  t  
4t  4
t 1
The domain of the function is all real
numbers except t  1 . That is,
t | t  1 .
35. The function f (a)  a 3  a  1 is
undefined when the radicand is
negative. So
a 1  0
a  1
The domain of the function is all real
numbers greater than or equal to –1.
That is, a | a  1 .
2x  6
is
x2  3
undefined when the radicand is
negative or the denominator is equal
to zero.
2x  6  0
37. The function f ( x) 
2 x  6
x  3
The denominator is always positive.
The domain of the function is the set
of all real numbers greater than or
equal to –3. That is, x | x  3 .
39. Since it doesn’t make sense to sell a
negative number of bags of candy, n
is nonnegative. The domain of the
candy profit function is the set of
whole numbers. That is,
n | n is a whole number .
Copyright © Houghton Mifflin Company. All rights reserved.
1.2 Linear Functions
41. Using our current calendaring
system, the domain of the function is
the set of whole numbers between 1
and the current year. For example, in
2006, the domain of the function is
n |1  n  2006 . (It is impossible
to calculate the average height of z
of a person yet to be born in future
years.)
x 1
is
x2  1
undefined whenever the
denominator is equal to zero.
x2  1  0
43. The function f ( x ) 
 x  1 x  1  0
x  1, x  1
The domain of the function is all real
numbers except x  1 and x  1.
45. Yes. Even though the domain value
of x  1 is listed twice in the table, it
is linked with the same range value,
y  6.
Exercises 1.2
3
5. The slope of the line passing through
(2, 2) and (5, 2) is
22
m
52
0

3
0
7. y-intercept:  0,10
0  5 x  10
10  5 x
x  2
x-intercept:  2,0 
9. y-intercept:  0,11
0  2 x  11
11  2 x
x  5.5
x-intercept:  5.5,0 
11. 3  0  y  4
y  4
y  4
y-intercept:  0, 4
1. The slope of the line passing through
(2, 5) and (4, 3) is
53
m
24
2

2
 1
3x   0   4
3x  4
x
x-intercept:
4
3
 43 ,0
3. The slope of the line passing through
(1.2, 3.4) and (2.7, 3.1) is
3.4  3.1
m
1.2  2.7
0.3

1.5
 0.2
Copyright © Houghton Mifflin Company. All rights reserved.
4
CHAPTER 1 Functions and Linear Models
13. The slope of the line passing through
(2, 5) and (4, 3) is
53
m
24
2

2
 1
y  mx  b
y  1x  b
5  1 2   b plugging in  2,5 
b7
The slope-intercept form of the line
is y   x  7 . The standard form of
the line is x  y  7 . A point-slope
form of the line is y  5  1 x  2  .
17. The slope of the line passing through
 2, 2 and (5, 2) is
m
22
5   2 
0
7
0
Since the slope is equal to zero, this
line is a horizontal line. The slopeintercept form of the line is
y  0 x  2 and is commonly written
as y  2 . The standard form of the
line is 0 x  y  2 and is also often
written as y  2 . A point-slope form

of the line is y  2  0  x  2 .
19. y  4 x  2
15. The slope of the line passing through
(1.2, 3.4) and (2.7, 3.1) is
3.4  3.1
m
1.2  2.7
0.3

1.5
 0.2
y  mx  b
y  0.2 x  b
3.4  0.2 1.2   b plugging in 1.2,3.4 
3.4  0.24  b
b  3.64
The slope-intercept form of the line
is y  0.2 x  3.64 . The standard
form of the line is typically written
with integer coefficients. Therefore,
y  0.2 x  3.64
21. y  4  0.5  x  2
0.2 x  y  3.64
20 x  100 y  364
5 x  25 y  91 (dividing by 4)
is the standard form of the line.
A point-slope form of the line is
y  3.4  0.2  x  1.2  .
Copyright © Houghton Mifflin Company. All rights reserved.
1.2 Linear Functions
23. 2 x  3 y  5
5
31. A table containing exactly two points
each with different x-values will
always represent a linear function.
U.S. Average Personal
Year
Income (in terms of
year 2000 dollars)
1989
18,593
1999
28,525
Source: www.census.gov
25. y   23 x  43
The slope is given by
28525  18593 year 2000 dollars
m
1999  1989
years
 993.2 year 2000 dolllars per year.
Between 1989 and 1999, the U.S.
average personal income (in year
2000 dollars) increased by an
average of $993.20 per year.
33.
27.  0,3 and  3,5 are two points on
the line. The slope of the line is
53
m
3 0
2

3
The slope-intercept form of the line
2
is y  x  3 .
3
29. This is a vertical line with x-intercept
3,0 . The equation of the line is
x  3.
Months
Take Home Pay
(since
(dollars)
Sep 01)
0
3167.30
1
4350.31
Source: Employee pay stubs
The slope is given by
4350.31  3167.30 dollars
m
1 0
months
 1183.01 dollars per month.
Between September 2001 and
October 2001, the employee’s take
home pay increased at a rate of
$1183.01 per month.
Copyright © Houghton Mifflin Company. All rights reserved.
6
CHAPTER 1 Functions and Linear Models
35. If the table of data represents a linear
function then a linear function
passing through two of the points
will also pass through all other points
in the table.
Cost to Dispose of
Clean Wood at
Enumclaw Transfer
Station
500
$18.75
700
$26.25
900
$33.75
1000
$37.50
Source: www.dnr.metrokc.gov
Clean
Wood
(Pounds)
The slope of the line passing through
500,18.75 and  700, 26.25 is
26.25  18.75 dollars
700  500 pound
7.5

200
 $0.0375 per pound
m
The equation of the line passing
through these points is given by
y  0.0375 x  b
18.75  0.0375  500   b
18.75  18.75  b
b0
y  0.0375 x.
We evaluate the linear equation at
x  900 and x  1000 .
y  0.0375  900   33.75
y  0.0375 1000   37.50
These results match the table data.
The data table does represent a linear
function. It costs an average of
$0.0375 per pound to dispose of
clean wood.
37. Let x be the number of servings of
WheatiesTM and y be the grams of
fiber consumed. We have
y  2.1x  3.3
since each serving of cereal contains
2.1 grams of fiber and the banana
contains 3.3 grams of fiber. We must
solve
8  2.1x  3.3
4.7  2.1x
x  2.238
3
In order to consume 8-grams of
fiber, you would need to eat 3
servings ( 2 14 cups) of Wheaties
along with the large banana.
39. The slope of the line is
7   1
m
44
8

0
 undefined.
Therefore, the line is a vertical line.
Although the y-values change, every
point on a vertical line has the same
x-value . The x-value of each of the
points is x  4 . The equation of the
vertical line is x  4 .
41. The slope-intercept form is
y  mx  b and the point-slope form
is y  y1  m  x  x1  . For a vertical
line, the slope m is undefined and
thus may not be substituted into
either of the two forms.
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1.2 Linear Functions
7
43. Vertical lines are not functions since
they fail the Vertical Line Test.
However, all non-vertical lines are
functions.
The slope of the line is
y  y1
m 2
x2  x1
c
0
 b
c
0
a
c
 b
c

a
c a
  
b c
a
 .
b
45. The slope of the line y  4 is 0 since
y  4 may be written as y  0 x  4 .
The slope of a line perpendicular to
this line has slope
1
m
0
 undefined.
Therefore, the perpendicular line is a
vertical line. Since all points on a
vertical line have the same x-values,
the equation of the line is x  3 .
47. We can find the x-intercept by
plugging in y  0 and the y-intercept
by plugging in x  0 .
ax  b  0   c
ax  c
x
c
a
c 
The x-intercept is  , 0  .
a 
a  0   by  c
by  c
c
b
 c
The y-intercept is  0,  .
 b
y
49. We will first find the equation of the
line passing through  2, 7  and
 5,13 .
13  7
52
6

3
2
m
y  2x  b
7  2  2  b
7  4b
b3
So y  2 x  3 . We’ll now plug in the
point 10,b  and solve for b.
b  2 10   3
b  23
In order for the table of data to
represent a linear function, b must
equal 23.
Copyright © Houghton Mifflin Company. All rights reserved.
8
CHAPTER 1 Functions and Linear Models
51. The point of intersection of the lines
is  a, b  . This point is the only point
that satisfies both of the linear
equations.
Exercises 1.3
1.
a. The scatter plot shows that the
data is near linear.
b. Using linear regression on the
TI-83 Plus, we determine
The equation of the line of best
fit is y  6.290 x  110.9 .
c. m  6.290 means that the Harbor
Capital Appreciation Fund share
price is dropping at a rate of
$6.29 per month.
The y-intercept means that in
month 0 of 2000 the fund price
was $110.94. Since the months
of 2000 begin with 1 not 0, the yintercept does not represent the
price at the end of January. It
could, however, be interpreted as
being the price at the end of
December 1999.
d. This model is a useful tool to
show the trend in the stock price
between October and
December 2000. Since stock
prices tend to be volatile, we are
somewhat skeptical of the
accuracy of data values outside
of that domain.
3. a. The scatter plot shows that the data
is near linear.
b. Using linear regression on the
TI-83 Plus, we determine
The equation of the line of best fit
is y  1165.9 x  80,887 .
c. m  1165.9 means that
Washington State public university
enrollment is increasing by about
1166 students per year.
b  80,887 means that, according
to the model, Washington State
public university enrollment was
80,887 in 1990.
Copyright © Houghton Mifflin Company. All rights reserved.
1.3 Linear Models
d. The model fits the data extremely
well as shown by the graph of the
line of best fit.
This model could be used by
Washington State legislators and
university administrators in
budgeting and strategic planning.
9
The y-intercept means that in
1995 (year 0), North Carolina
was ranked 40th out of the 50
states. (Only positive whole
number rankings make sense.)
d. This model is not a highly
accurate representation of the
data as shown by the graph of the
line of best fit so it should be
used with caution.
5. a. The scatter plot shows that
although the data are not linear,
they are nonincreasing.
b. Using linear regression on the
TI-83 Plus, we determine
The equation of the line of best
fit is y  3.4 x  39.6 .
c. m  3.4 means that the per
capita income ranking of North
Carolina is changing at a rate of
3.4 places per year. That is, the
state is moving up in the rankings
by about 3 places per year.
However, an incumbent
government official could use the
model in a 2000 reelection
campaign as evidence that the
state’s economy had improved
during his or her tenure in office.
The official might also use the
model to claim that the trend of
improvement will continue if he
or she is reelected.
7. a. The minimum fees and the
rounding of the weight of the trash
make this particular function
somewhat complicated. We will
first find how much trash may be
disposed of for the minimum fees.
The amount of trash that can be
disposed of for the minimum
disposal fee is
$13.72
 0.1663 ton
$82.50 per ton
2000 lbs
0.1663 ton 
 332.6 lbs
1 ton
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10
CHAPTER 1 Functions and Linear Models
The largest multiple of twenty that
is less than or equal to 332.6 is
320. Therefore, a rounded weight
of 320 pounds of trash may be
disposed of for the minimum
disposal fee.
The amount of trash that can be
disposed of for the minimum
moderate risk waste fee is
$1.00
 0.3831 ton
$2.61 per ton
2000 lbs
0.3831 ton 
 766.3 lbs
1 ton
The largest multiple of twenty that
is less than or equal to 766.3 is
760.
For each of the following
functions, we let x represent the
weight of the trash rounded to the
nearest 20 pounds.
The cost of disposing 320 pounds
of trash or less (including tax) is
C ( x )  13.72  11.036 
 $15.25
The cost of disposing between 340
and 760 pounds of trash (including
tax) is
 x
C  x  
82.50   1 1.036 
 2000

  0.04125 x  11.036 
 x
C  x  
82.50  2.61  1.036 
 2000

  0.04256 x 1.036 
 0.04409 x
Combining the individual functions
into a single piecewise cost
function we have
0  x  320
15.25

C  x   0.04274 x  1.036 , 340  x  760
0.04409 x
780  x

For values of x  780 , C  x  is
directly proportional to x.
b. We must first round the weight of
the trash to the nearest 20 pounds.
One way to do this is to divide the
weight of the trash by 20, round
the number to the nearest whole
number, and multiply the result by
20. That is,
230
 11.5
20
 12
12  20  240
Since this weight is below 320
pounds, the minimum $15.25 fee
will be charged.
c. How much will it cost to drop off
513 pounds of trash?
 0.04274 x  1.036
The cost of disposing of 780
pounds of trash or more is
513
 25.65
20
 26
12  26  520
C  520   0.04274  520   1.036
 $23.26
Copyright © Houghton Mifflin Company. All rights reserved.
1.3 Linear Models
d. The cost per pound is lowest when
780 or more pounds of trash are
disposed of. As a construction
company, we would try to keep the
weight of our trash deliveries at or
above 780 pounds.
9. a. Let C be the total cost (in dollars)
of using t anytime minutes of
phone time during a given month.
(If a fraction of a minute is used, t
is the next whole number value.
For example, if 3.2 anytime
minutes are used then t  4 .) The
$29.99 fee covers the first 200
minutes. The $0.40 per minute fee
is only charged on the minutes
used after the first 200 minutes.
Thus we have
29.99, 0  t  200
C t   
29.99  0.40  t  200  , t  200
which may be rewritten as
29.99, 0  t  200
C t   
0.4t  50.01, t  200
b. For the Qwest plan we have
C  300   0.35  300   22.51
 $82.49
For the Sprint plan we have
C  300   0.4  300   50.01
 $69.99
11
11. a. Let t be the production year of a
Toyota Land Cruiser 4-Wheel
DriveTM and V be the value of the
vehicle in 2001. We have
1995, 21125 and
 2000, 43650 . We must find the
linear function passing through
these points. We have
43650  21125
m
2000  1995
 4505
so V  4505t  b . We solve for b
using the point  2000, 43650 .
43650  4505  2000   b
b  8,966,350.
The linear model is
V  4505t  8966350 .
b. We have
V 1992   4505 1992   8966350
 7610
According to the model, a 1992
Toyota Land Cruiser was valued
at $7610 in 2001.
c. The model substantially
underestimated the value of a
1992 Land Cruiser. If we draw a
scatter plot of the value of a
1992, 1995, and 2000 Land
Cruisers in 2001, we see that the
vehicle value function is not
linear.
The Sprint plan is the best deal for
a customer who uses 300 anytime
minutes. If a plan was selected that
offered 300 anytime minutes for a
higher basic fee, it would likely
cost the customer even less.
Copyright © Houghton Mifflin Company. All rights reserved.
12
CHAPTER 1 Functions and Linear Models
13. a. Let F be the number of grams of
fat in x chef salads. We have
F  8x .
The number of fat grams is
directly proportional to the
number of chef salads.
b. We can consume up to 80 grams
of fat.
80  8x
x  10
Thus a person on a 2500-calorie
diet can consume up less than 10
salads.
c. One package of ranch salad
dressing contains 18 grams of fat.
F  8 x  18
80  8 x  18
62  8 x
x  7.75
A person on a 2500-calorie diet
can consume up to 7.75 salads
before exceeding the fat
requirement.
d.
F  8 x  18
65  8 x  18
15. Fees imposed on enplaning
passengers include the flight segment
tax ($3), the PFC (up to $4.50), and a
security fee ($2.50). That is, up to
$10 in fees is charged for each
segment.
Let s be the number of segments and
C be the maximum total cost of the
ticket (in dollars) from Seattle to
Phoenix. We have
C  10s  222 .
Note that s  1 .
17. A business determines that the
revenue generated by selling x cups
of coffee, y bagels, and z muffins is
given by
R  ax  by  cz .
a is the price of a cup of coffee, b is
the price of a bagel, and c is the price
of a muffin.
19. The equation of the line of best fit
for the data in the table is y  0 .
x
y
0
–1
1
1
2
0
3
0
4
1
5
–1
6
0
47  8 x
x  5.875
A person on a 2000-calorie diet
can consume up to 5.875 salads
before exceeding the fat
requirement.
The correlation coefficient r  0
indicates that the model doesn’t fit
the data well.
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1.3 Linear Models
13
23.The line of best fit is y  1 . This
model fits the table of data perfectly.
In addition, we can see from the
scatter plot that the model doesn’t
fit well.
21. The slope-intercept form of a line
passing through the origin is
y  mx  0
 mx
This implies that y is directly
proportional to x.
If the linear model y  mx fits the
original data well, it is likely that the
dependent variable of the original
data is directly proportional to the
independent variable. However, if
the linear model y  mx does not fit
the data well, the dependent and
independent variables of the original
data are not directly proportional.
For example, consider the table of
data
x
0
1
2
3
4
5
y
0
1
-1
-1
1
0
The line of best fit is y  0 x which
indicates that x and y are directly
proportional. However, we can see
from the table that y is not directly
proportional to x.
However, from the calculator display
we see that r is undefined.
The correlation coefficient is formally
defined as
n   xy     x   y 
r
2
2
n   x2     x  n   y 2     y 
The  symbol tells us to add each
of the terms. Since we have four data
points, n  4 for this data set.
We will calculate each term
individually and then substitute the
result into the correlation coefficient
formula.
n   xy   4  0  1  1  1  2  1  3  1
 4  6
 24
  x   y    0  1  2  31  1  1  1
 6  4
 24
n   x 2   4  02  12  22  32 
 4 14 
 56
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14
CHAPTER 1 Functions and Linear Models
  x    0  1  2  3
2
2
25.
 36
n   y 2   4 12  12  12  12 
 4  4
 16
  y   1  1  1  1
2
2
 42
 16
Thus we have
24  24
r
56  36 16  16
0

0
Because we can’t divide by zero and
there is a zero in the denominator of
the expression, the correlation
coefficient r is undefined.
From the scatter plot, it appears that
the first four points are in a line and
the last three points are in a line. The
slope of the first line is m  2 . The
slope of the second line is m  0.5 .
Both lines pass through  4,2  .
The equation of the first line is
y  2 x  b
2  2  4   b
2  8  b
b  10
So y  2 x  10.
The equation of the second line is
y  0.5 x  b
2  0.5  4   b
2  2b
b0
So y  0.5 x
The piecewise function is
 2 x  10  2  x  4
y
4  x  10
0.5x
Copyright © Houghton Mifflin Company. All rights reserved.
Chapter 1 Review Exercises
Review Exercises
Section 1.1
1.
C  2   49.95  2 
 99.90
The cost to buy two pairs of shoes is
$99.90.
3.
H  2   16  2   100
2
 64  100
 36
The cliff diver is 36 feet above the
water 2 seconds after he jumps from a
100-foot cliff.
5. f ( p)  p 2  9 p  15
The domain of the function is all real
numbers.
r 1
r2  1
Since r  1 and r  1 make the
denominator equal zero, the domain of
the function is all real numbers except
r  1 and r  1 .
7. C  r  
2
9. At t  2 , P  21.03 . The stock price
of the computer company at the end
of the day two days after Dec 16,
2001, was about $21.03.
11. The graph passes the Vertical Line
Test, since no vertical line drawn
will cross the graph more than once.
Thus, the graph represents a
function.
15
Section 1.2
13. (–3, –4) and (0, –2)
2    4 
m
0   3 

2
3
15. (7, 11) and (8, 0)
0  11
m
87
11

1
 11
17. y  3x  18
The y-intercept is  0,18 .
0  3x  18
3x  18
x6
The x-intercept is  6,0 .
19. (2, 5) and (4, 3)
35
42
 1
y  1x  b
m
5  1 2   b
b7
So y  1x  7
In standard form, x  y  7 .
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16
CHAPTER 1 Functions and Linear Models
Section 1.3
21. a. Let x be the number of large
orders of french fries consumed.
The amount of fat consumed is
given by F  x   29 x grams.
b. Let y be the number of Big N’
TastyTM sandwiches consumed.
The amount of fat consumed is
given by G  y   32 y grams.
c. The total amount of fat consumed
by eating z combination meals is
H  z    32  29 z .
b. We have
v t   7312.5t  14,558,975 .
v 1999   7312.5 1999   14,558,975
 $58,712.50
c. The linear model was extremely
effective at accurately predicting
the value of the 1999 MercedesBenz Roadster 2-door SL500.
The $212.50 difference between
the predicted value of the model
and the NADA guide average
value was negligible.
H 1   32  29 1
 61
Only one combination meal may
be eaten. Eating more than one
will exceed the RDA for fat (at
most 65 grams).
23. a. Let t be the production year of the
Mercedes-Benz Roadster 2-door
SL500TM and let v  t  be the value
of the car in 2001. We have
v t 
t
1998 $51,400
2000 $66,025
The slope of the car value
function is
66025  51400
m
2000  1998
 $7312.5 per year
We have
v  t   7312.5t  b
51400  7312.5 1998   b
b  14,558,975
So v  t   7312.5t  14,558,975
is the linear model for the data.
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