Practice Problems Section 2.1-2.7

8/10/2019 Practice Problems Section 2.1-2.7

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Practice problems (Sections 2.1-2.7)

1. Find the domain and range of the relation:

{(0, 9.1), (10, 6.7), (20, 10.7), (30, 13.2), (40, 21.2)}

The domain is the set of all first components.

Domain:

{0, 10, 20, 30, 40}.

The range is the set of all second components.

Range:

{9.1, 6.7, 10.7, 13.2, 21.2}

1. Find the domain and range of the relation:

{(3, 4), (3, 5), (4, 4), (4, 5)}

2a. Determine whether the relation is a function:

( )( )( )( ){ }1,2 , 3,4 , 5,6 , 5,8

5 corresponds to both 6 and 8. If any element in the domain

corresponds to more than one element in the range, the

relation is not a function.

Thus, the relation is not a function.

2a. Determine whether the relation is a function:

{(3, 4), (3, 5), (4, 4), (4, 5)}

2b. Determine whether the relation is a function:

( )( )( )( ){ }1,2 , 3,4 , 6,5 , 8,5

Every element in the domain corresponds to exactly one

element in the range. No two ordered pairs in the given

relation have the same first component and different second

components.

Thus, the relation is a function.

2b. Determine whether the relation is a function:

( )( )( )( ){ }3, 3 , 2, 2 , 1, 1 , 0,0- - - - - -

3. Solve each equation for yand then determine whether

the equation defines yas a function ofx.

3a. 2 6x y+ =

Subtract 2xfrom both sides to solve for y.

2 6

2 2 6 2

6 2

x y

x x y x

y x

+ =

- + = -

= -

For each value ofx, there is only one value of y, so the

equation defines yas a function ofx.

3. Solve each equation for yand then determine whether

the equation defines yas a function ofx.

3a.2 16x y+ =


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3b. 2 2 1x y+ =

Subtract 2x from both sides and then use the square root

property to solve for y.2 2

2 2 2 2

2 2

2

1

1

1

1

x y

x x y x

y x

y x

+ =

- + = -

= -

= -

For values ofxbetween 1 and 1, there are two values of y.

For example, ifx= 0, then y= 1. Thus, the equation does

not define yas a function ofx.

3b. 2x y=

4. If2( ) 2 7,f x x x= - + evaluate each of the following.

4a. f(x+ 4)

Substitutex+ 4 forx and then simplify. Place parentheses

aroundx+ 4 when making the substitution.

Use2 2 2( ) 2A B A AB B+ = + + to expand 2( 4)x + and the

distributive property to multiply 2(x+ 4). Then combine like

terms.

2

2

2

( 4) ( 4) 2( 4) 7

8 16 2 8 7

6 15

f x x x

x x x

x x

+ = + - + +

= + + - - +

= + +

4b. f(x)

Substitute xforx. Place parentheses around xwhen

making the substitution.

2

2

( ) ( ) 2( ) 7

2 7

f x x x

x x

- = - - - +

= + +

4. If2( ) 2 3,g x x x= + + evaluate each of the following.

4a. g(1)

4b. g(x+ 5)

4c. g(x)

5. Use the vertical line test to determine if the graph

represents y as a function ofx.

6. Use the vertical line test to determine if the graph

represents y as a function ofx.


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The graph passes the vertical line test and thus yis a function

ofx.

6a. The following is the graph of g.

Use the graph to find ( 20).g -

The graph indicates that to the left of 4,x= -

the graph is ata constant height of 2.

Thus, ( 20) 2.g - =

6a. The following is the graph of g.

Use the graph to find ( 4).g -

6b. Use the graph from Problem 7aabove to find the value

ofxfor which ( ) 1.g x = -

(1) 1g = -

The height of the graph is1 when 1.x =

6b. Use the graph from Problem 7aabove to find the value of

xfor which ( ) 1.g x =

7. Use the graph of the function to identify its domain andits range.

Inputs on thex-axis extend from2, excluding 2, to 1,

including 1.The domain is ( 2,1]- .

Outputs on the y-axis extend from1, including 1, to 2,

excluding 2.

The range is [ 1,2)- .

7. Use the graph of the function to identify its domain andits range.


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8. State the intervals on which the given function is

increasing, decreasing, or constant.

The intervals are stated in terms ofx-values.

When we start at the left and follow along the graph, at first

the graph is going up. This continues untilx= 1. The

function is increasing on the interval (, 1).

Atx= 1, the graph turns and moves downward until we get

tox= 1. The function is decreasing on the interval (1, 1).

Atx= 1, the graph turns again and continues in an upward

direction. The function is increasing on the interval (1, ).

8. State the intervals on which the given function is

increasing, decreasing, or constant.

9. Look at the graph in Solved Problem #1. Locate values at

which the functionfhas any relative maxima or minima.

What are these relative maxima or minima?

The graph has a turning point at x= 1. The value off(x) or y

atx= 1 is greater than the values of f(x) for values ofxnear1 (for values ofxbetween 2 and 0, for example). Thus,f

has a relative maximum atx= 1. The relative maximum is

the value off(x) or ycorresponding tox= 1. Using the

equation in the graph, we find that f(1) = (1)33(1) = 2.

We say thatfhas a relative maximum of 2 atx= 1.

The graph has a second turning point at x= 1. The value of

f(x) or yatx= 1 is less than the values off(x) for values ofx

near 1 (for values ofxbetween 0 and 2, for example). Thus,f

has a relative minimum at

x= 1. The relative minimum is the value off(x) or y

corresponding tox= 1. Using the equation in the graph, we

find thatf(1) = (1)33(1) = 2. We say thatfhas a relative

minimum of 2 atx= 1.

Note that the relative maximum occurs where the functions

changes from increasing to decreasing and the relative

minimum occurs where the graph changes from decreasing

to increasing.

9. The graph of a functionfis given below. Locate values at

which the functionfhas any relative maxima or minima.

What are these relative maxima or minima? Read y-values

from the graph, as needed, since the equation is not

given.


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2 2

2 2 2

2 2 2

2

( ) ( )

[ 2( ) ( ) 5] ( 2 5)

2( 2 ) 5 2 5

2 4 2 5 2 5

4 2

( 4 2 1)

4 2 1, 0

f x h f x

h

x h x h x x

h

x xh h x h x x

hx xh h x h x x

h

xh h h

h

h x h

h

x h h

+ -

- + + + + - - + +=

- + + + + + + - -=

- - - + + + + - -=

- - +=

- - +=

= - - +

12. Write the point-slope form of the equation of the line

with slope 6 that passes through the point (2, 5).- Then

solve the equation for y.

Begin by finding the point-slope equation of a line.

( )

( )

( )

1 1

( 5) 6 2

5 6 2

y y m x x

y x

y x

- = -

- - = -

+ = -

Now solve this equation for y.

( )5 6 2

5 6 12

6 17

y x

y x

y x

+ = -

+ = -

= -

12. Write the point-slope form of the equation of the line

with slope 3- that passes through the point ( 2, 3).- -

Then solve the equation for y.

13. A line passes through the points ( 2, 1)- - and ( 1, 6).- -

Find the equation of the line in point-slope form and

then solve the equation for y.

Begin by finding theslope:6 ( 1) 5

51 ( 2) 1

m - - - -

= = = -

- - -

Using the slope and either point, find the point-slope

equation of a line.

( )

( )

( )

( )

( )

( )

1 1 1 1

or

( 1) 5 ( 2) ( 6) 5 ( 1)

1 5 2 6 5 1

y y m x x y y m x x

y x y x

y x y x

- = - - = -

- - = - - - - - = - - -

+ = - + + = - +

To obtain slope-intercept form, solve the above equation for

y:

13. A line passes through the points ( 3, 1)- - and (2,4). Find

the equation of the line in point-slope form and then

solve the equation for y.


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( ) ( )1 5 2 or 6 5 1

1 5 10 6 5 5

5 11 5 11

y x y x

y x y x

y x y x

+ = - + + = - +

+ = - - + = - -

= - - = - -

14. Graph:3

( ) 15

f x x= +

The y-intercept is 1, so plot the point (0,1).

The slope is3

.5

m =

Find another point by going up 3 units and to the right 5

units.

Use a straightedge to draw a line through the two points.

14. Graph:3

( ) 24

f x x= -

15. Find the slope and y-intercept of the line whose

equation is 3 6 12 0.x y+ - =

Solve for y.3 6 12 0

6 3 12

6 3 12

6 6

3 12

6 6

12

2

x y

y x

y x

y x

y x

+ - =

= - +

- +=

-= +

= - +

The coefficient ofx,1

,2

- is the slope, and the constant term,

2, is the y-intercept.

15. Find the slope and y-intercept of the line whose equation

is 2 3 18 0.x y+ - =

16. Graph: 3 2 6 0x y- - =

Find thexintercept by setting y= 0.

16. Graph: 6 2 12 0x y- - =


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3 2 6 0

3 2(0) 6 0

3 6

2

x y

x

x

x

- - =

- - =

=

=

Find the yintercept by settingx= 0.3 2 6 0

3(0) 2 6 0

2 6

3

x y

y

y

y

- - =

- - =

- =

= -

Plot the points and draw the line that passes through them.

17. The amount of carbon dioxide in the atmosphere,

measured in parts per million, has been increasing as a

result of the burning of oil and coal. The buildup of gases

and particles is believed to trap heat and raise the

planets temperature. When the atmospheric

concentration of carbon dioxide is 317 parts per million,

the average global temperature is 57.04F. When the

atmospheric concentration of carbon dioxide is 354 parts

per million, the average global temperature is 57.64F.

Write a linear function that models average global

temperature,f(x), for an atmospheric concentration of

carbon dioxide ofxparts per million. Use the function to

project the average global temperature when the

atmospheric concentration of carbon dioxide is 600 parts per

million.

Write the equation of the line through the points

(317, 57.04) and (354, 57.64). First find the slope.

2 1

2 1

57.64 57.04 0.60.016

354 317 37

y ym

x x

- -= = =

- -

Use this slope and the point (317, 57.04) in the point-slope

form.

17. When the literacy rate for adult females in a country is

0%, the infant mortality rate is 254 (per thousand). When

the literacy rate for adult females is 60%, the infant

mortality rate is 110. Write a linear function that models

child mortality,f (x), per thousand, for children under five

in a country wherex% of adult women are literate. Use

the function to predict the child mortality rate in a

country where 80% of adult females are literate.


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1 1( )

57.04 0.016( 317)

57.04 0.016 5.072

0.016 51.968

y y m x x

y x

y x

y x

- = -

- = -

- = -

= +

Using function notation and rounding the constant, we have

( ) 0.016 52.0f x x= +

To predict the temperature when the atmospheric

concentration of carbon dioxide is 600 parts per million, find

f(600).

(600) 0.016(600) 52.0 61.6f = + =

The model predicts an average global temperature of 61.6F

when the atmospheric concentration of carbon dioxide is 600

parts per million.

18. Write an equation of the line passing through ( 2,5)-

and parallel to the line whose equation is 3 1.y x= +

Express the equation in point-slope form and slope-

intercept form.

Since the line is parallel to 3 1,y x= + we know it will have

slope 3.m =

We are given that it passes through ( 2,5).- We use the

slope and point to write the equation in point-slope form.( )

( )( )

( )

1 1

5 3 2

5 3 2

y y m x x

y x

y x

- = -

- = - -

- = +

Point-Slope form: ( )5 3 2y x- = +

Solve for yto obtain slope-intercept form.

( )5 3 2

5 3 6

3 11( ) 3 11

y x

y x

y xf x x

- = +

- = +

= +

= +

Slope-Intercept form: 3 11y x= +

18. Write an equation of the line passing through ( 8, 10)- -

and parallel to the line whose equation

is 4 3.y x= - + Express the equation in point-slope form

and slope-intercept form.

19. Write an equation of the line passing through ( 2, 6)- -

and perpendicular to the line whose equation is

3 12 0.x y+ - = Express the equation in point-slope form

and general form.

19. Write an equation of the line passing through (4, 7)- and

perpendicular to the line whose equation is 2 3 0x y- - =

.Express the equation in point-slope form and general

form.


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First, find the slope of the line 3 12 0.x y+ - =

Solve the given equation for yto obtain slope-intercept form.

3 12 0

3 12

14

3

x y

y x

y x

+ - =

= - +

= - +

Since the slope of the given line is1

,3

- the slope of any line

perpendicular to the given line is 3.

We use the slope of 3 and the point ( 2, 6)- - to write the

equation in point-slope form. Then gather the variable and

constant terms on one side with zero on the other side.

( )

( ) ( )( )( )

1 1

6 3 26 3 2

y y m x x

y xy x

- = -

- - = - -

+ = +

6 3 6

0 3 or 3 0

y x

x y x y

+ = +

= - - =

20. In 1990, there 9.0 million men living alone and in

2008, there were 14.7 million men living alone. Use the

ordered pairs (1990, 9.0) and (2008, 14.7) to find the

slope of the line through the points. Express the slope

correct to two decimal places and describe what it

represents.

Change in 14.7 9.0

Change in 2008 1990

5.70.32

18

ym

x

-= =

-

=

The number of men living alone increased at an average rate

of approximately 0.32 million men per year.

20. In 1994, 617 active-duty servicemembers were

discharged under the dont ask, dont tell policy. In

1998, 1163 were discharged under the policy. Use the

ordered pairs (1994, 617) and (1998, 1163) to find the

slope of the line through the points. Express the slope

correct to the nearest whole number and describe what it

represents.

21. Find the average rate of change of the function fromx1

tox2.

21a.3( )f x x= fromx1= 0 tox2= 1

2 1

2 1

3 3

( ) ( ) (1) (0)

1 0

1 0

1

1

f x f x f f

x x

- -=

- -

-=

=

The average rate of change is 1.

21. Find the average rate of change of the function fromx1to

x2.

21a. ( ) 3f x x= fromx1= 0 tox2= 5


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33. Use the graph of2( )f x x= to graph

2( ) 2( 1) 3.g x x= - +

The graph of gis the graph offhorizontally shifted to the

right 1 unit, vertically stretched by a factor of 2, and vertically

shifted up 3 units. Beginning with a point on the graph off,

add 1 to eachx-coordinate, then multiply each y-coordinate

by 2, and finally add 3 to each

y-coordinate.

(1, 1)(0, 1) (0, 2) (0, 5)

(0, 0) (1, 0) (1, 0) (1, 3)

(1, 1) (2, 1) (2, 2) (2, 5)

33. Use the graph of3( )f x x= to graph

21( ) ( 3) 2.2

h x x= - - (You can follow the example

problem or do as I taught you by doing it in steps and

describing each transformation!!)

34. Find the domain of5

( ) .24 3

xj x

x=

-

The function contains both an even root and division. The

expression under the radical must be nonnegative and the

denominator cannot equal 0. Thus, 24 3xmust be greater

than 0.

24 3 0

24 3

8 or 8

x

x

x x

- >

>

>

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Practice Problems Section 2.1-2.7