Solution manual for calculus and its applications 11th edition by bittinger

a First, we locate 1 on the horizontal axis and then we look vertically to find the point on the graph for which 1 is the first coordinate.. c First, we locate 2 on the vertical axis and

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Exercise Set R.1

1 Graph y= + x 4

We choose some x-values and calculate the corresponding y-values to find some ordered

pairs that are solutions of the equation Then

we plot the points and connect them with a smooth curve

2 Graph y= − x 1

3 Graph y= −3x

We choose some x-values and calculate the corresponding y-values to find some ordered

6 Graph 5 3

3

y= − x+

7 Graph x+ = y 5

We solve for y first

subtract from both sides commutative property

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9 Graph 8y−2x= 4

We solve for y first

add 2 to both sides divide both sides by 8

428

42 8

81124

x

x y x y x y x y

=

−+

12 5 6

526526

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20 Graph y= −5 x2

21 Graph y+ =1 x3

First we solve for y

3 3

11

=

According to this model, the world record for the mile in 1954 is approximately 3.97532 minutes To convert this to traditional minutes-seconds we multiply the decimal part by 60 seconds

( ) 58.51920.97532 60 =

Therefore the world record for the mile in 1954 was 3:58.5

Likewise, we substitute 2008 in for x to get

0.00582 2008 15.34763.66104

=According to the model, the world record for the mile in 2008 will be approximately 3.66104 minutes or 3:39.7

Finally, we substitute 2012 in for x to get

0.00582 2012 15.34763.63776

=According to the model, the world record for the mile in 2012 will be approximately 3.63776 minutes or 3:38.3

24 A=0.5t4+3.45t3−96.65t2+347.7 , 0t ≤ ≤ t 6

We substitute t= 2

( )4 ( )3 ( )2 ( )

0.5 2 3.45 2 96.65 2 347.7 2344.4

=Approximately 344.4 milligrams of ibuprofen will remain in the blood stream 2 hours after

400 mg have been swallowed

28

28 16281628161.3228

s t

t t t t

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27 a) Locate 20 on the horizontal axis and go

directly up to the graph Then move left to the vertical axis and read the value there

We estimate the number of hearing-impaired Americans of age 20 is about 1.8 million

Follow the same process for 40, 50, and 60

to determine the number of impaired Americans at each of those ages

hearing-We estimate the number of hearing-impaired Americans of age 40 is about 3.7 million

b) Locate 4 on the vertical axis and move horizontally across to the graph There are

two x-values that correspond to the y-value

of 4 They are 44 and 70, so there are approximately 4 million Americans age 44 who are hearing-impaired and

approximately 4 million Americans age 70 who are hearing-impaired

c) The highest point on the graph appears to

correspond to the x-value of 58 Therefore,

age 58 appears to be the age at which the greatest number of Americans are hearing-impaired

d) Visually, we cannot tell precisely which point is the highest point on the graph or

which x-value corresponds exactly to that

point The graph is not detailed enough to make that determination

28 a) We estimate the incidence of breast cancer

in 40-yr-old women is about 100 per 100,000 women

b) We estimate that at age 67 and at age 88 the incidence of breast cancer is about 400 per 100,000 women

c) The largest incidence of breast cancer occurs

in woman who are approximately 79 years

of age

d) Visually, we cannot tell precisely which point is the highest point on the graph or

which x-value corresponds exactly to that

point The graph is not detailed enough to make that determination

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e) There are 24 365⋅ =8760hours in one year

1

nt i

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d) 1

nt i

0.05

1000 1

4

1000 1.01251050.95

0.05

1000 1

365

1000 1.0001369861051.27

0.05

1000 1

8760

1000 1.0000057081051.27

(6.4%=0.064) for i, and 36 (3 12⋅ =36) for n

Then we use a calculator to perform the computation

36 36

0.0640.064

11212

18, 000

0.064

11

12550.86

34 30 years = ⋅30 12=360 months

100, 000; 0.048

360 360

0.0480.048

11212

100, 000

0.048

12524.67

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36 We substitute 50,000 for W, 0.0725

1

7 % 0.07254

⎞

⎟

⎝ in for i, and 20 for n Then we

proceed to solve for P

37 a) Locate 250,000 on the vertical axis and then

think of horizontal lines extending across the graph from this point The years for which the graph lies above this line are the years for which the deer population was at or above 250,000 Those time periods are

1996 – 2000, and then a brief time between

2001 and 2002

b) Locate 200,000 on the vertical axis and then think of a horizontal line extending across the graph from this point The years for which the graph touches this line are the years for which the population was at 200,000 Those time periods are 1987 and

1990

c) Locate the highest point on the graph and extend a line vertically to the horizontal axis The year which the deer population was the highest was 1999

d) Locate the lowest point on the graph and extend a line vertically to the horizontal axis In this case there are two points that are exactly at 200,000 The years when the deer population was the lowest are 1987 and

38 Answers will vary, but might include

references to milder winters, increase food sources, reduced number of predators etc…

39 a) Using the formula ( )1 i n 1

in her account when she retires

b) Sally invested $1200 per year for 35 years Therefore, the total amount of her original payments is: $1200 35i =$42, 000 Since

the total amount in the account was

$206,780, the interest earned over the 35

$206, 780.16 $42, 000− =$164, 780.16 Therefore, $42,000 was the total amount of Sally’s payments and $164,780.16 was the total amount of her interest

40 a) Using the formula:

1

11

(8%=0.08) for i, and 15 (80 65− =15) for

n Then we use a calculator to perform the

computation

180 180

0.080.08

11212

206, 780.16

0.08

121976.10

b) Sally received 180 payments of $1976.10, therefore she received a total of

$1976.10 180i =$355, 698during the 15 years Of that 42,000 was what she originally contributed, leaving $313,698 in interest

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41 Graph y= −x 150

We use the following window

Next, we type the equation into the calculator

The resulting graph is:

42 Graph y=25− x

Using the following window:

The resulting graph is

43 Graph y=x3+2x2−4x−13

We use the following window:

Next, we type the equation in to the calculator

44 Graph y= 23 7− x Using a standard window

Results in the graph:

45 Graph 9.6x+4.2y= −100

First, we solve for y

subtract 9.6 from both sides

=

Next, we set the window to be:

Next, we type the equation into the calculator at the top of the next page

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46 Graph y= −2.3x2+4.8x− 9

We use the following window:

47 Graph x= +4 y2

First we solve for y

2 subtracting 4 2

from both sides taking the square root

8

y= ± − x

We use the standard window

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Exercise Set R.2

1 The correspondence is a function because each

member of the domain corresponds to only one member of the range

4 The correspondence is not a function because

one member of the domain, 6, corresponds to two members of the range, – 6 and – 7

member of the domain corresponds to only one member of the range, even though two members

of the domain, Quarter Pounder with Cheese ® and Big N’ Tasty with cheese ® correspond to

$3.20

iPod has exactly one amount of memory

iPod has exactly one owner If an iPod has more than one owner, this relationship would not be a function

iPod has exactly one number of songs at any given time

10 The correspondence is a function because each

iPod has exactly one number of Avril Lavigne songs at any given time

11 The correspondence is a function because any

number squared and then increased by 8, corresponds to exactly one number greater than

or equal to 8

12 The correspondence is a function because any

number raised to the fourth power corresponds

to exactly one nonnegative number

13 The correspondence is a function because every

female has exactly one biological mother

14 The correspondence is a function because every

male has exactly one biological father

15 This correspondence is not a function, because

it is reasonable to assume at least one avenue is intersected by more than one cross street

16 This correspondence is not a function, because

it is reasonable to assume that a textbook has more than one even-numbered page

shape has exactly one area

shape has exactly one perimeter

19 a) f x( )=4x− 3

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

133

4 44

93

4 33

113242

3434

34

3

4 11

34434

f f f

k k

=

+

=+

−

++ =

20 a) f x( )=3x+ 2

( )

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b) f x( )=3x+ 2

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

172

3 55

12131

2323

f f

k k

f k

t t t

x h

=+

2 2 2 2

231311

3303

2313

223253

32

3

g g g g

h ah a h

( ) ( ) ( ) ( )

( )

2 2

2

33

h

ah h h

=+

2 2 2 2

13433

44

5411

44

42

4

g g g g

v v

g v

ah h a

=+

=

+++

2

22

a h

+++

f x

x

=+

+

( ) ( )

Output is undefined.

11

( )2 ( )2

33

f a

a a

++

4

7

11

33

f t

t t

=+ =

+ ++ +

2 2 2

2 2

11

33

3362

33

f x h f x h

h

x h x

h

x h

h x h h hx x h

h x

h x h

x h

h x

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24 ( )

( )2

15

2 2

2 2 2

Output is undefined.

11

366

1 5

11

,5

0

5 515

f f

f k k

1

6

1 5

11

f t

t t

55

−+

b) The function squares the input, then subtracts 10 times the input, then adds 25, and then it takes the reciprocal of the result

25 Graph f x( )=2x− 5

First, we choose some values for x and compute

the values for f x( ), in order to form the ordered pairs that we will plot on the graph

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

26 Graph f x( )=3x− 1

27 Graph g x( )= − 4x

the values for g x( ), in order to form the ordered pairs that we will plot on the graph

( ) ( )

4141

g g

28 Graph g x( )= −2x

29 Graph f x( )=x2− 2

the values for f x( ), in order to form the ordered pairs that we will plot on the graph The calculations are shown at the top of the next page

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Choosing some values for x and evaluating the

function, we have:

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

2 2 2 2 2

2222

1211

22

12

22

f f f f f

Next we plot the input – output pairs from the

table and, in this case, draw the curve to complete the graph

30 Graph f x( )=x2+ 4

31 Graph ( ) 2

6

f x = −x

the values for f x , in order to form the ( )

ordered pairs that we will plot on the graph

( ) ( ) ( ) ( ) ( ) ( )

2 2 2 2 2

2262

5

66

5161

2262

f f f f f

the values for g x , in order to form the ( )

ordered pairs that we will plot on the graph

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

3 3 3 3 3

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Next we plot the input – output pairs from the table on the previous page and, in this case, draw the curve to complete the graph

34 Graph ( ) 1 3

2

g x = x

35 The graph is a function, it is impossible to draw

a vertical line that intersects the graph more than once

38 The graph is not that of a function A vertical

line can intersect the graph more than once

47 Graph x=y2− 2

a) First, we choose some values for y (since x

is expressed in terms of y) and compute the

values for x, in order to form the ordered pairs that we will plot on the graph

( ) ( )

2 2

12

222For 2;

b) The graph is not that of a function A vertical line can intersect the graph more than once

42. The graph is a function, it is impossible to draw

a vertical line that intersects the graph morethan once

43. The graph is a function, it is impossible to draw

a vertical line that intersects the graph morethan once

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48 a) Graph x=y2− 3

b) The graph is not that of a function A vertical line can intersect the graph more than once

h

xh h h h

h x h h

51 To find f( )− we need to locate which piece 1

defines the function on the domain that contains

First, we graph f x( )= for inputs less than 0 1

We note for any x-value less than 0, the graph is

the horizontal liney= Note that for1 f x( )= 1

( ) ( )

1211

f f

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The open circle indicates that ( )0,1 is not part of the graph

Next, we graph f x( )= − for inputs greater 1

than or equal to 0 We note for any x-value less

than 0, the graph is the horizontal line y= − 1Note that for f x( )= − 1

( ) ( ) ( )

f f f

First, we graph f x( )= for 6 x= −2

This graph consists of only one point, (−2, 6) The solid dot indicates that (−2, 6)is part of the graph

Next, we graph f x( )=x2for inputs x≠ −2 Note that for f x( )=x2

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

2 2 2 2 2

933

111

000

111

422

f f f f f

(−2, 4)is not part of the graph

58 Graph ( ) 3

5, for 1, for 1

Next, we graph g x( )= for 4 x=0 This part

of the graph consists of a single point The solid dot indicates that ( )0, 4 is part of the graph The solution is continued on the next page

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Next, we graph g x( )= + for inputs x 2 x>0

Creating the input – output table, we have:

graph

Next, we graph g x( )= − for inputs x 3 x>2

Creating the input – output table, we have:

3

x

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

2 2 2 2 2

6333

2311

3300

2311

6333

f f f f f

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Choosing some values for x and evaluating the

Since the input x=2is not defined on this part

of the graph, the point ( )2,1 is not part of the graph The open circle indicates that ( )2,1 is not part of the graph

0.06

800 1

4

800 1.015956.49

The patient’s approximate surface area is

2

1.818 m b) We substitute 170 for h and 100 for w

( )( )170 100

2.1733600

2

2.173m c) We substitute 170 for h and 50 for w

( )( )170 50

1.5373600

2

1.871m

69 a) Yes, the table represents a function Each

event is assigned exactly one scale of impact number

b) The inputs are the events; the outputs are the scale of impact numbers

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70 Solve the equation for y

− = −

=

We sketch a graph of the equation

No vertical line meets the graph more than once Thus, the equation represents a function

71 First we solve the equation for y

2 2 2

subtract 3 from both sides divide both sides by 2 take the square root of both sides

5432

52

5 2

52

x

x x y x y x y x y

+

=+

= ++

=+

= ±

We sketch a graph of the equation

We can see that a vertical line will intersect the graph more than once; therefore, this is not a function

72 First we solve the equation for y

( ) ( )

Next, we sketch a graph of the equation

We can see that a vertical line will intersect the graph more than once; therefore, this is not a function

73 First, we solve the equation for y

( )

3 2 3 3 3 3

88

Note: since y must be non-negative to satisfy the

original equation, we only graph the points for

which y is non-negative

Next, we sketch a graph of the equation:

No vertical line meets the graph more than once Thus, the equation represents a function

74 Answers will vary The vertical line test

works, because you are locating the values

to which each input corresponds For a relation to be a function, each input must correspond with exactly one output

Therefore, if a vertical line intersects the graph in more than one point, that particular input corresponds to more than one output and the graph is not a function

75 Yes, x=4is in the domain of f in

Exercises 51-54 The function is defined for all

=

−

We begin by setting up the table:

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Next, we will type in the equation into the graphing editor

Now, we are able to look at the table:

Now, we are able to look at the table We will

have to enter the appropriate values of x

From the table we conclude:

( ) ( ) ( ) ( )

23022024

f f f f

After entering the function into the graphing editor, we get:

In order to graph ( ) 2

34

f x x

After entering the function into the graphing editor, we get:

80 Answers will vary Some ordered pairs are

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21 a) First, we locate 1 on the horizontal axis and

then we look vertically to find the point on the graph for which 1 is the first coordinate From that point, we look to the vertical axis

to find the corresponding y-coordinate, 3

Thus, f ( )1 = 3

b) The domain is the set of all x-values of the

points on the graph The domain is

{− −3, 1,1, 3,5} c) First, we locate 2 on the vertical axis and then we look horizontally to find any points

on the graph for which 2 is the second coordinate One such point exists, ( )3, 2

Thus the x-value for which f x( )= is 23

x=

d) The range is the set of all y-values of the

points on the graph The range is

{−2, 0, 2, 3, 4}

22 a) f( )1 = − 1b) The domain is {− − − −4, 3, 2, 1, 0,1, 2} c) The point on the graph with the second coordinate 2 is (−2, 2) Thus the x-value for

which f x( )= is2 x= − 2d) The range is {− −2, 1, 0,1, 2, 3, 4}

Thus, f( )1 = 4

points on the graph The domain is

{− −5, 3,1, 2, 3, 4, 5} c) First, we locate 2 on the vertical axis and then we look horizontally to find any points

on the graph for which 2 is the second coordinate Three such point exists,

(−5, 2 ;) (−3, 2 ; and 4, 2) ( ) Thus the

x-values for which f x( )= are 2 {− −5, 3, 4}

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{−3, 2, 4,5}

24 a) f( )1 = 2

b) The domain is {− − −6, 4, 2, 0,1, 3, 4} c) The points on the graph with the second coordinate 2 are ( )1, 2 and 3, 2 Thus the ( )

x-values for which f x( )= are2 { }1, 3 d) The range is {− −5, 2, 0, 2,5}

then we look vertically to find the point on the graph for which 1 is the first coordinate

From that point, we look to the vertical axis

to find the corresponding y-coordinate, -1

Thus, f( )1 = − 1

points on the graph These extend from -2 to

4 Thus, the domain is {x| 2− ≤ ≤x 4}, or

in interval notation [−2, 4] c) First, we locate 2 on the vertical axis and then we look horizontally to find any points

on the graph for which 2 is the second coordinate One such point exists, ( )3, 2

Thus the x-value for which f x( )= is 23

x=

3 Thus, the range is {y| 3− ≤ ≤y 3}, or, in interval notation [−3, 3]

26 a) f( )1 ≈2.5

b) The domain is {x| 3− ≤ ≤x 5 or } [ ]−3,5 c) The point on the graph with the second coordinate 2 appears to be (2.25, 2 Thus )

the x-value for which f x( )= is2 x=2.25

d) The range is {y|1≤ ≤y 4 , or, 1, 4} [ ]

to find the corresponding y-coordinate, -2

Thus, f ( )1 = − 2

2 Thus, the domain is {x| 4− ≤ ≤x 2}, or,

on the graph for which 2 is the second coordinate One such point exists, (−2, 2)

Thus the x-value for which f x( )= are 22

x= −

3 Thus, the range is {y| 3− ≤ ≤y 3}, or in interval notation [−3, 3]

28 a) f( )1 ≈2.25 b) The domain is {x| 4− ≤ ≤x 3 or } [−4, 3] c) The point on the graph with the second coordinate 2 appears to be ( )0, 2 Thus the

x-value for which f x( )= is2 x= 0.d) The range is {y| 5− ≤ ≤y 4 , or, } [−5, 4]

Thus, f ( )1 = 3

3 Thus, the domain is {x| 3− ≤ ≤x 3}, or,

on the graph for which 2 is the second coordinate Two such point exists,

(−1.4, 2 and 1.4, 2) ( ) Thus the x-values for

which f x( )= are 2 {−1.4,1.4}

4 Thus, the range is {y| 5− ≤ ≤y 4}, or in interval notation [−5, 4]

30 a) f( )1 ≈ 2b) The domain is {x| 5− ≤ ≤x 4 or } [−5, 4]

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c) The points on the graph with the second

coordinate 2 are all the points with the

x-value in the set {x|1≤ ≤x 4} Thus the

x-values for which f x( )= are 2

{x|1≤ ≤x 4}, or [ ]1, 4 d) The range is {y| 3− ≤ ≤y 2 , or } [−3, 2]

Thus, f( )1 = 1

5 However, the open circle at the point

( )5, 2 indicates that 5 is not in the domain

Thus, the domain is {x| 5− ≤ <x 5}, or in interval notation [−5,5)

c) First, we locate 2 on the vertical axis and then we look horizontally to find any points

on the graph for which 2 is the second

coordinate We notice all the points with

x-values in the set {x| 3≤ < Thus the x- x 5}

values for which f x( )= are 2

{x| 3≤ <x 5 , or 3,5} [ )

{− −2, 1, 0,1, 2}

32 a) f( )1 = 2

b) The domain is {x| 4− ≤ ≤x 4 or } [−4, 4] c) The points on the graph with the second

coordinate 2 are all the points with the

x-value in the set {x| 0< ≤x 2} Thus the

x-values for which f x( )= are 2

=+Set the denominator equal to 0 and solve

3 03

x x

+ =

= −The domain is

{x x| is a real number and x≠ − , or, in 3}

interval notation, (− ∞ − ∪ − ∞ , 3) ( 3, )

35 f x( )= 2x

Since the function value cannot be calculated when the radicand is negative, the domain is all real numbers for which 2x≥ We find them 0

by solving the inequality

0 setting the radicand dividing both sides by 2

02

0

x x

We can calculate the function value for all

values of x, so the domain is the set of all real

numbers \

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38 ( ) 2

3

f x =x +

of f

setting the denominator equal to 0 adding 12 to both sides dividing both sides by 6

6

6 12 2

x x x

f x

x

=

− Solve 3x− = 6 0

numbers \

42 f x( )= − x 4

of f

setting the denominator equal to 0 adding 2 to both sides dividing both sides by 2

027

2772

x

x x x

2 is not in the domain of f, while all

other real numbers are The domain of f is

9

2

x=

Thus, 9

2 is not in the domain of f, while all

other real numbers are The domain of f is

setting the radicand 0 subtracting 4 from both sides dividing both sides by 5

4 5 0

5 4 4 5

x x x

x x

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47 ( ) 2

g x =x − x+

numbers \

48 g x( )=4x3+5x2−2x

=

−

Since the function value cannot be calculated when the denominator is equal to 0, we solve the following equation to find those real numbers that must be excluded from the domain

of g

2 2

setting the denominator equal to 0 adding 25 to both sides taking the square root or both sides

25 0

25 255

x x x x

=

= ±

= ±Thus, −5 and 5 are not in the domain of g,

while all other real numbers are The domain of

g is {x x| is a real number and x≠ −5, x≠ 5};

or, in interval notation,

(− ∞ − ∪ −, 5) ( 5,5) ( )∪ 5,∞

50 ( ) 2

136

x

g x x

x x

numbers \

52 g x( )= + x 7

of f

( )( )

2 setting the denominator equal to 0

factoring the quadratic equation

x x

=

−+ =

f x ≤ , so we scan the graph from left to

right looking for the values of x for which the graph lies on or below the x axis Those values extend from -1 to 2 So the set of x-values for

which f x( )≤ is 0 {x| 1− ≤ ≤x 2}, or, in interval notation, [−1, 2]

56 The graph crosses the line y= at every integer 1

value of x Therefore, the set of x-values for

which g x( )= is 1 {x x| is an integer}

57 a) We use the compound interest formula from

Theorem 2 in section R.1 and substitute

5000 for P, 2 for n and 0.08 (8%) for i The

equation for this function is derived on the next page

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Using the information from the previous page, we derive the function

( )

2 2

10.08

b) The independent variable t is the time in

years the principal has been invested in the account It would not make sense to have time be a negative number in this case

Therefore, the domain is the set of all negative real numbers {t| 0≤ < ∞ t }

non-58 a) We use the compound interest formula from

Theorem 2 in section R.1 and substitute

3000 for P, 365 for n and 0.05 (5%) for i

The equation for this function is:

365 365

b) The independent variable t is the time in

years the principal has been invested in the account It would not make sense to have time be a negative number in this case

Therefore, the domain is the set of all negative real numbers {t| 0≤ < ∞ t }

non-59 a) The graph extends from x= to 0 x=84.7,

so the domain, in interval notation, of the

function N is [0,84.7 ]

b) The graph extends from N x( )= to 0

( ) 4.6

N x = million Therefore, the range,

in interval notation, of the function N is

[0, 4, 600, 000 ]

c) Answers will vary We would target the 50 year old to 60 year old age group, because that is the age group that has the most number of hearing-impaired Americans

60 a) The domain in interval notation is [25,102 ]

b) The range in interval notation is [0, 455 ]

c) Answers will vary Since we are looking for the greatest increase in the incidence of breast cancer, we are looking for the steepest portion of the graph It appears that the greatest increase occurs from age 50 to age

60

61 a) The graph extends from t= to 0 t=70, so

the domain, in interval notation, of the

function L is [0, 70 ]

b) The graph extends from L t( )= to 8

L t( )=75, so the range, in interval notation,

of the function L is [ ]8, 75

62 a) The x-value where the cancer rate is 50 per

100,000 is approximately t=26or the year

1966

b) There are two t-values where the cancer rate

is 70 per 100,000 The first t-value is t=41

or the year 1981, and the second t-value is

function The value x= is the solution to the 2equation f x( )= − 5

64 Answers may vary Given a domain each value in the domain corresponds to exactly one value in the range It is certainly possible for each member of the domain to correspond with

itself Consider the function: y= Here the x

domain and the range are the same set

65 Answers may vary Consider the function

( ) 1

3

f x x

=

− Here the number 3 is not in the

domain of f because replacing x with 3 results in

division by zero

Trang 28

66 The range in interval notation for each function

Trang 29

line whose x-intercept is the point (−4.5, 0)

6 Graph x= −1.5

7 Graph 3.75y= The graph consists of all ordered pairs whose second coordinate is 3.75 This results in a

horizontal line whose y-intercept is the point

(0, 3.75 )

8 Graph 2.25y=

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9 Graph y= − 2x

Using Theorem 4, The graph of y is the straight

line through the origin ( )0, 0 and the point

(1, 2− We plot these two points and connect )

them with a straight line

The function y = − has slope –2, and y-2x

connect them with a straight line

The function f x( )=0.5x has slope 0.5, and

any number for x and then determine y by

substitution

When x= −1,y= − − = − 3( )1 4 7 When x=0,y=3 0( )− = − 4 4 When x=2,y=3 2( )− = 4 2

We organize these values into an input – output table

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15 Graph g x( )= − + x 3

First, we make a table of values We choose

any number for x and then determine y by

adding 3 to both sides

adding 3 to both sides subtracting 2 from both sides

= − +The slope is –2

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The y-intercept is ( )0, 2

commutative property of equality subtracting 7

dividing by 3 simplifying

3 7

7 3

1

6 2

y= x+

35 y−y1=m x( −x1)

( ) ( ( ) ) Substituting

Simplifying Adding 3

y y

=

− −+

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