Chapter1 Assignments
P12 63, 64
63. AIR POLLUTION An environmental study of a certain suburban community
suggests that the average daily of carbon monoxide in the air will be c( p)  0.4 p  1
parts per million when the population is p thousand. It is estimated that t years from
now the population of the community will be p(t )  8  0.2t 2 thousand.
a. Express the level of carbon monoxide in the air as a function of time.
b. What will the level of carbon monoxide be 2 years from now?
c. When will the carbon monoxide level reach 6.2 parts per million?
Solution (Composition of Functions; P9 Example 1.1.11)
a. Since the level of carbon monoxide is related to the variable p by the equation
c( p)  0.4 p  1
and the variable p is related to the variable t by the equation
p(t )  8  0.2t 2
it follows that the composition function
c( p(t ))  c(8  0.2t 2 )  0.4  (8  0.2t 2 )  1  4.2  0.08t 2
expresses the level of carbon monoxide in the air as a function of the variable t .
b. When t  2 ,
c( p(2))  4.2  0.08  2 2  4.52
That is, the level of carbon monoxide in the air will be 4.52 parts per million 2
years from now.
c. Set c( p(t )) equal to 6.2 and solve for t to get
4.2  0.08t 2  6.2
0.08t 2  2
t2 
2
 25
0.08
t  25  5
That is, 5 years from now the level of carbon monoxide will be 6.2 parts per
million
64. MANUFACTURING COST At a certain factory, the total cost of
manufacturing q units during the daily production run is C (q)  q 2  q  900 dollars.
On a typical workday, q(t )  25t units are manufactured during the first t hours of a
production run.
a. Express the total manufacturing cost as a function of t .
b. How much will have been spent on production by the end of the third hour?
c. When will the total manufacturing cost reach $11,000?
Solution (Composition of Functions; P9 Example 1.1.11)
a. Since (the level of carbon monoxide is related to the variable p by the equation)
C (q)  q 2  q  900
and (the variable p is related to the variable t by the equation)
q (t )  25t
it follows that the composition function
C (q(t ))  (25t ) 2  25t  900  625t 2  25t  900
expresses total manufacturing cost as a function of t .
b. When t  3 ,
C (q(3))  625  32  25  3  900  6600
That is, $6600 will have been spent on production by the end of third hour.
c. Set C (q(t )) equal to 11,000 and solve for t to get
625t 2  25t  900  11000
t
 1  1  25 * 4 * 404  1  201

50
50
Since t  0 , we get t  4 .That is , the total manufacturing cost reach $11,000 by the
end of the forth hour.
P37 19 , 27 , 33
In Problems 19, 27 and 34, write an equation for the line with the given properties
19.Through (2, 0) with slope 1.
Solution (Linear Function; P32 Example 1.3.4)
Use the formula y  y0  m( x  x0 ) with ( x0 , y0 )  (2,0) and m  1 to get
y  0  1  ( x  2)
Rewrite it as
y  x2
1
5
2 1
3 4
27. Through ( ,1) and ( , ) .
Solution (Linear Function; P32 Example 1.3.2)
First comput the slope
1
1
45
m 4

2
1
52
 ( )
3
5
1
5
45
1
45
43
y  1   ( x  ) or y   x 
52
5
52
52
33. Through (3,5) and perpendicular to the line x  y  4
Then use the point-slope formula with ( ,1) as given point ( x0 , y0 ) to get
Solution (Linear Function; P35 Example 1.3.5(b))
By rewriting the equation x  y  4 in the slope-intercept form y   x  4 , we see
that this line has the slope ml  1 . A line perpendicular to line x  y  4 must have
slope m  
P56
1
 1 . Since the required line contains point (3,5), we have
ml
y 5  x3
y  x2
45 , 48
45. SUPPLY AND DEMAND Producers will supply x units of a certain
commodity to the market when the price is p  S (x) dollars per unit, and consumers
will demand (buy) x units when the price is p  D(x) dollars per unit, where
S ( x)  2 x  15
and
D( x) 
385
x 1
a. Find the equilibrium production level xe and the equilibrium price p e .
b. Draw the supply and demand curves on the same graph.
c. Where does the supply curve cross the y axis? Describe the economic
significance of this point.
Solution ( P49 Example 1.4.6)
a. Maket equilibrium occurs when
S ( x)  D( x)
385
x 1
2
2 x  17 x  15  385
2 x 2  17 x  370  0
( x  10)( 2 x  37)  0
2 x  15 
x  10 or x  18.5
Since only positive values of the production level x are meaningful, we reject
x  18.5 and conclude that equilibrium occurs when xe  10 . The corresponding
equilibrium price can be substiduting x  10 into either the supply function or the
demand function. Thus,
pe  S (10)  2(10)  15  35
b.
Equilibrium
point
(10, 35)
c. when the supply curve crosses the y axis,
S (0)  15
thus the supply curve crosses the y axis at (0,15) . This point mean that no units
will be produced until the price is at least $15.
48. BREAK-EVEN ANALYSIS A furniture manufacturer can selll dining room
tables for $70 apiece. The manufacturer’s total cost consists of a fixed overhead of
$8,000 plus production costs of 30 per table.
a. How many tables must the manufacturer sell to break even?
b. How many tables must the manufacturer sell to make a profit of $6,000?
c. What will be the manufacturer’s profit or loss if 150 tables are sold?
d. On the same set of axes, graph the manufacturer’s total revenue and total cost
fuctions. Explain how the overhead can be read from the graph.
Solution (P51 Example 1.4.7)
If x is the number of units the manufactured and sold, the total revenue is given by
R( x)  70 x and the total cost by C ( x)  8000  30 x .
a. To find the break-even point, set R ( x ) equal to C ( x) and solve
70x  8000  30 x
40x  8000
so that x  200
It follows that the manufacturer will have to sell 200 units to break even.
b. The profit P ( x ) is revenue minus cost. Hence
P( x)  R( x)  C ( x)  70 x  (8000  30 x)  40 x  8000
To determine the number of units that must be sold to make a profit of $6,000,
set the formula for profit P ( x ) equal to 6,000 and solve
40x  8000  6000
x  350
so that
It follows that the manufacturer will have to sell 350 units to make a profit of
$6,000.
c. The profit from the sale of 150 tables is
P(150)  40(150)  8000  2000
The minus sign indicates a loss, and it follows that the manufacturer will loss
$2,000 if 150 tables are sold.
d.
R(x)
y
Break-even
point
8000
0
C(x)
(200, 14000)
x
From the graph, we can find that the overhead is the y intercept of C ( x) .
P69 11
In Problem 11, find the indicated limit if it exists.
2
11. lim ( x  1) ( x  1)
x3
Solution (P63 Example 1.5.2)
Apply the properties of limits to obtain
lim( x 1) 2 ( x 1)  (lim( x 1)) 2  lim( x 1)  (3 1) 2(3 1)  16
x 3
x 3
x 3
P70 15 , 19 , 23 , 25 , 27 , 31 , 35
In Problems 15, 19, 23 and 25 , find the indicated limit if it exists
15. lim
x 5
x3
5 x
Solution (P64 Example 1.5.4)
The quotient rule for limits does not apply in this case since the limit of the
denominator is
lim(5  x)  0
x 5
x  3)  8 , which is not equal to zero, you can
Since the limit of the numerator is lim(
x 5
conclude that the limit of the quotient does not exist.
19. lim
x 5
x 2  3x  10
x5
0
0
Solution ( ; P64 Example 1.5.5)
Both the numerator and the denominator approach 0 as x approach 5. Simplify the
quotient to obtain:
x 2  3x  10 ( x  5)( x  2)

 x2
x 5
x 5
and then take the limit to get
23.
x 2  3x  10
lim
 lim( x  2)  7
x 5
x 5
x 5
x2  x  6
lim 2
x 2 x  3 x  2
0
0
Solution ( ; P64 Example 1.5.5)
Both the numerator and the denominator approach 0 as x approach -2. Simplify the
quotient to obtain:
x 2  x  6 ( x  2)( x  3) x  3


x 2  3x  2 ( x  2)( x  1) x  1
and then take the limit to get
( x  3)
x2  x  6
x  3 xlim
2

lim

5
x 2 x 2  3 x  2
x 2 x  1
lim ( x  1)
lim
x 2
25. lim
x4
x 2
x4
0
0
Solution ( ; P64 Example 1.5.6)
Both the numerator and the denominator approach 0 as x approach 5. to simplify the
quotient, we rationalize the numerator:
x  2 ( x  2)( x  2)
x4



x4
( x  4)( x  2)
( x  4)( x  2)
and then take the limit to get
1
x 2
lim
x 4
x 2
1
1
 lim

x 4
x4
x 2 4
f ( x) and lim f ( x) . If the limiting value is
For problems 27, 31 and 35, find xlim
 
x  
infinite, indicate whether it is   or   .
27. f ( x)  x 3  4 x 2  4
Solution
Since the values of this function increase without bound as x increases, we get
lim ( x3  4 x2  4)  
x 
Similarly, we can obtain
lim ( x3  4 x2  4)  
x 
31.
f ( x) 
x 2  2x  3
2 x 2  5x  1


Solution ( ; P67 Example 1.5.8)
The highest power in the denominator is x 2 . Divide the numerator and the
denominator by x 2 to get
x2  2x  3
1  2 / x  3/ x 2 1  0  0 1

lim


x  2 x 2  5 x  1
x  2  5 / x  1/ x 2
200 2
x2  2x  3
1  2 / x  3/ x 2 1  0  0 1
lim
 lim


x  2 x 2  5 x  1
x  2  5 / x  1/ x 2
200 2
3x 2  6 x  2
f ( x) 
2x  9
lim
35.


Solution ( ; P68 Example 1.5.10)
The highest power in the denominator is x . Divide the numerator and the
denominator by x to get
3x 2  6 x  2
3x  6  2 / x
 lim
x 
x 
2x  9
29/ x
lim
since
2
lim (3 x  6  )   and
x
x 
9
lim (2  )  2
x
x 
it follows that
3x 2  6 x  2
 
x 
2x  9
lim
Similarly, we can get
3x 2  6 x  2
3x  6  2 / x
 lim
 
x 
x 
2x  9
29/ x
lim
P71 51
51. PER CAPITA EARNINGS Studies indicate that t years form now, the
population of a certiain country will be p  0.2t  1,500 thousand people, and that
gross earnings of the country will be E million dollars, where
E (t )  9t 2  0.5t  179
a. Express the per capita earnings of the country P  E / p as a function of time t .
(take care with the units.)
b. What happens to the per capita earnings in the long run (as t   )
Solution (P68 Example 1.5.9)
a. Both the gross earnings of the country E and the population p are functions of
time t . Hence the per capita earnings of the country is
P(t )  E (t ) / p(t ) 
9t 2  0.5t  179
0.2t  1500
thousand dollars per person .
b. Since
lim P(t )  lim
t 
t 
9t 2  0.5t  179
9  0.5 / t  179 / t 2
 lim
 15 ,
t 
0.2t  1500
0.2  1500 / t
the per capita earnings will be 15 thousand dollar per person ( $1,500 per person)
in the long run (as t   ).
P81 11 , 13
In Problems 11 and 13, find the indicated one-sided limit. If the limiting value is
infinite, indicate whether it is   or   .
11. lim
x 3

x 1  2
x3
Solution
lim
x 3
x 1 2
( x  1  2)( x  1  2)
 lim
 lim
x

3
x3
x3
( x  3)( x  1  2)
2 x 2  x
lim
f
(
x
)
lim
f
(
x
)
13. 
and 
, where f ( x)  
x 3
x 3
3  x
Solution (P74 Example 1.6.1)
Since f ( x )  2 x 2  x for x  3 , we have
lim f ( x )  lim (2 x 2  x )  15
x 3
x 3
1
x 1  2

if x  3
if x  3
1
4
Similarly, f ( x )  3  x for x  3 , so
lim f ( x )  lim ( 3  x )  0
x  3
x3
P82 19 , 23 , 26 , 33 , 39
In problems 19, 23 and 26, decide if the given function is continuous at the specified
value of x.
19. f ( x) 
x 1
x 1
at x  1 .
Solution (Verify the three criteria for continuity in P77)
Since f (1) is undifined, the function f ( x) 
x  1
2
23. f ( x)  
x 1
is not continuous at x  1.
x 1
if x  2
at x  2
if x  2
Solution (P78 Example 1.6.6)
Since the function f ( x ) approaches 3 as x approaches 2 from the left side and
f ( x ) does not exist. Thus,
approuches 2 as x approacher 2 from the right side, lim
x 2
this function is not continuous at x  2
 x2 1
26. f ( x)   x  1
x 2 - 3

if x  1
at x  1
if x  1
Solution (P78 Example 1.6.6)
f (1) is defined and f ( 1)  2 . Since
x2  1
 lim ( x  1)  2
x  1
x  1
x  1 x  1 
and
lim f ( x )  lim ( x 2  3)  2
lim f ( x )  lim
x 1
x 1
lim f ( x ) exists and lim f ( x )  2  f ( 1) .
x  1
x  1
Thus the funcion is continuous at x  1
In Problems 33 and 39, list all the value of x for which the given function is not
continuous.
33. f ( x) 
3x  2
( x  3)( x  6)
Solution (P78 Example 1.6.6)
This functin is a rational function, it follows that f ( x ) will be continuous
everywhere except x  3 and x  6 .
3x  2
if x  0
x  x
if x  0
39. f ( x)  
2
Solution (P78 Example 1.6.6)
Since 3 x  2 and x 2  x are both polynomials, it follows that f ( x ) will be
f ( x ) does not exist,
continuous everywhere except possibly at x  0 . We find that lim
x 0
since f ( x ) approaches -2 from the left and 0 from the right. It follows that f ( x ) is
coninuous at every real number except 0.
P83 44
44. WATER POLLUTION A ruptured pipe in a North Sea oil rig produces a
circular oil slick that is y meters thick at a distance x meters from the rupture.
Turbulence makes it diffucult to directly measure the thickness of the slick at the
source (where x  0 ), but for x  0 , it is found that
y
0.5( x 2  3x)
x3  x 2  4x
Assuming the oil slick is continuously distributed, how thick would you expect it to
be at the source?
Solution
Clealy the function y 
0.5( x 2  3x)
is not defined at x  0 . Since the oil thick is
x3  x 2  4x
continuously distributed, we can expect the oil thich will be
0.5( x 2  3 x )
0.5( x  3)
lim f ( x )  lim 3
 lim 2
 0.425
x 0
x 0 x  x 2  4 x
x 0 x  x  4
meters at the source.
P84 51 , 55
In Problem 51, find the values of the constant A such that the function f(x) will be
continuous for all x.
 Ax  3
51. f ( x)  
3  x  2 x
if x  2
2
if x  2
Solution (P79 Example 1.6.7)
Since Ax  3 and 3  x  2 x 2 are both polynomials, it follows that f ( x ) will be
continuous everywhere excpet possibly at x  2 . Moreover, f ( x ) approaches
2 A  3 as x approaches 2 from the left side and approuches 9 as x approacher 2 from
f ( x ) exists, we must have 2 A  3  9 or A  6 , in which
the right side. Thus, for lim
x 2
case
lim f ( x )  9  f ( 2)
x2
This means that f ( x ) is continouous for all x when A  6 .
55. Show that the equation
interval 0  x  8
3
x  8  9 x 2 / 3  29 has at least one solution for the
Solution (P80 Example 1.6.9)
Let f ( x )  3 x  8  9 x 2 / 3  29 . Then f (0)  31 and f (8)  7 .Since
lim f ( x)  f (c)
x c
c  (0,8)
lim f ( x)  f (0)
x 0 
and
lim f ( x)  f (8)
x 8 
f ( x ) is continuous for 0  x  8 . And  31  0  7 ,the function can attain 0. So the
equation 3 x  8  9 x 2 / 3  29 has at least one solution for the interval 0  x  8 .
OR
Let f ( x)  3 x  8  9 x 2 / 3 . Then f (0)  2 and f (8)  36 . Since
lim f ( x)  f (c)
x c
c  (0,8)
lim f ( x)  f (0)
x 0 
and
lim f ( x)  f (8)
x 8 
f ( x ) is continuous for 0  x  8 . And  2  29  36 , the function can attain 29. So the
equation 3 x  8  9 x 2 / 3  29 has at least one solution for the interval 0  x  8 .