Final Exam Review Problems – Math 123 – Intermediate Algebra Online – Fall 2013

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Problem 1: Your cell/mobile phone plan charges you $75 per month which includes 500 minutes of talk

time. Any additional minutes are charged at $0.40 per minute. Create a linear model for your phone bill

as a function of the talk time where x represents the number of minutes you talk over the regular 500

minutes included in the plan and y represents your monthly bill.

Linear model:

What is your bill for a month if in that month you talked for 700 minutes which included the regular 500

minutes of talk time that comes with the plan?

Problem 2: Find the equation of a line that passes through the point (5, 8) and is perpendicular to the line

3 4 1x y

Problem 3: A mother wants to invest $10,000 for her son’s future education. She invests a portion of the

money in the U.S. Government issued bond which earns 4% return per year and the remainder in a stable

stock which on the average returns 8% per year. If the total return on investment (ROI) for the year is

$620, how much money was invested in the stock?

Problem 4: You travel from San Francisco to Tokyo by air, a distance of about 8,400 kilometers. On

your way to Tokyo it took 16 hours because of the headwind and on your way back it took 14 hours

because of the tailwind. What are the speed of the wind and the speed of the plane?

Problem 5: In a right triangle one of the acute angles is 3 degrees more than 2 times the other acute

angle. Find the acute angles.

Problem 6: Find domain for this real function and label it on a number line or express it in set notation:

9 2( )

5 7

xf x

x

Problem 7: Find domain of the real function and label it on a number line or express it in set notation:

( ) 2 9f x x

Problem 7: Use vertical line test to determine whether the following form functions:

Problem 8: Do the following divisions: 5 4 23 4 3 4 4x x x x x

Problem 9: Factor the following completely:

a) 2 26 11 10x xy y

b) 210 21x x

c) 28 26 15x x

d) 221 56 28x x

e) 2 236 49m n

f) 4 216 1

81 256p q

g) 3 31

8x y

h) 3 31

12527

m n

i) 2

15 2 3 11 2 3 14x x

j) Solve the following equations by factoring: 210 19 6 0x x

Problem 10: The area of rectangular garden is 200 square meters. The length of the garden is 1 meter

more than 3 times its width; find the length and the width of the garden.

Problem 11:

(a) Simplify and express in terms of positive exponents only:

2 22 3 2

33 4

2 4

2

m p m p

m p

(b) Simplify

31

33

69

x y

x y

Problem 12:

a) Rationalize the denominator:

3 5

3 5

b) Simplify and express it in the form (a + ib):

6 1

3 2

i

i

c) Simplify: 2011i

Problem 13: Find the overlapping region for these inequalities by graphing the inequalities in the

rectangular coordinate system given below: 2 3 3

2 2

x y

x y

The first step is to graph the straight lines:

2 3 3

2 2

x y

x y

To graph a straight line we need two different points on the line.

In 2 3 3x y we solve for 2 3

3

xy

and pick any two convenient points for x and find the

corresponding y values:

The line is labeled L1

-10

-8

-6

-4

-2

0

2

4

6

8

10

-10 -8 -6 -4 -2 0 2 4 6 8 10

x y = (2x – 3)/3 (x, y)

0 –1 (0, –1)

3 1 (3, 1)

L1

L2 Overlapping region

To find the region of interest for L1 just pick a point that is not on the line. Since the line does not go

through the origin (0, 0) I pick this as my reference point and check it on the inequality

2 3 3 2 0 3 0 3 0 3x y which is a false statement. Hence for this line the region of

interest is the region below the line L1. Also we make the line solid because it includes the boundary.

Similarly in 2 2x y we solve for 2

2

xy

and pick any two convenient points for x and find the

corresponding y values:

The line is labeled L2. Once again we may use (0, 0) as our reference point since the point is not on the

line. We have 2 2 0 2 0 2 0 2x y which is a true statement; the region below L2 is the

region of interest. The line is made dotted indicating that the boundary points are not included.

Now if we compare the two regions of interest for 2 3 3 and 2 2x y x y as outlined by L1 and L2,

we obviously notice that the overlapping region is as labeled in the figure above.

Problem 14: Find the following for the parabola 24 16 15y x x

(a) Does the parabola open up or down? Why?

The parabola opens up because the coefficient of the square term 24x is 4 which is greater than zero. If

the coefficient of the square term is negative then the parabola opens down.

(b) Find the coordinates of the vertex: x-coordinate of the vertex is given by 2

bx

a

so the we have 16

22 2 4

bx

a

x y = (–x + 1)/2 (x, y)

2 0 (2, 0)

4 –1 (4, –1)

To find the y-coordinate of the vertex just plug in x = 2 in the equation of the parabola 24 16 15y x x and then solve for y.

2 24 16 15 4 2 16 2 15

16 32 15

1

y x x y

y

y

Hence the coordinates of the vertex are: ( , ) (2, 1)x y

(c) Find the x-intercepts if they exist: (Use

2 4

2

b b acx

a

if needed)

To find the x-intercept set the y = 0 and then solve for x in 24 16 15y x x

Thus we have 24 16 15 0x x

Using the quadratic equation solution formula we have

2 4

2

b b acx

a

and a = 4, b = –16, and c = 15 from

24 16 15 0x x

2( 16) ( 16) 4 4 15

2 4

16 256 240 16 16 16 4

2 4 8 8

16 4 20 5 16 4 12 3 or

8 8 4 8 8 2

x

x

x x

So the x-intercepts are at 5

,04

and 3

,02

Also note that if the solutions to the quadratic equation 24 16 15 0x x were to be complex

conjugates that would mean that the parabola did not have x-intercepts.

(d) Find the y-intercept:

To find the y-intercept just set x = 0 in the equation 24 16 15y x x and solve for y.

2 24 16 15 4 0 16 0 15 15y x x y y

So the y-intercept of the parabola is: (0, 15)

(e) Sketch the parabola:

The above is a sketch (approximation) of the parabola 24 16 15y x x

Problem 15: Solve for x using the quadratic formula:

2 4

2

b b acx

a

(a) 2 5 1 0x x

In this problem a = 1, b = 5, and c = 1. By putting these values in the formula

2 4

2

b b acx

a

we

get:

-10

-8

-6

-4

-2

0

2

4

6

8

10

-10 -8 -6 -4 -2 0 2 4 6 8 10

25 5 4 1 1 5 25 4 5 21

2 1 2 2x

Hence the solutions are 5 21

2x

and

5 21

2x

; note that the solutions are conjugates.

(b) 2 1 0x x

Here a = 1, b = 1, and c = 1. By putting these values in the formula

2 4

2

b b acx

a

we get:

21 1 4 1 1 1 1 4 1 3

2 1 2 2

1 1 3 1 3 since 1

2 2

x

ix i

Hence the solutions are 1 3

2

ix

and

1 3

2

ix

; note that the solutions are complex conjugates.

(c) 2 4 5 0x x

Here a = 1, b = –4, and c = 5. By putting these values in the formula

2 4

2

b b acx

a

we get:

2( 4) ( 4) 4 1 5 4 16 20 4 4

2 1 2 2

4 1 4 4 2 since 1

2 2

2(2 )2

2

x

ix i

ix i

Hence the solutions are 2x i and 2x i ; note that the solutions are complex conjugates.

Problem 16: Solve for x:

(a) 2 3 5x

Note that for example 1x means that the distance of the point x from the origin in the number line is

less than 1. This says that 1 1x ; also note that since absolute value of a number is its distance from

the origin it is always positive or zero, since a negative distance does not make sense. Graphically I mean

the following:

1 1 ( 1,1)x

Using this concept I can simplify 2 3 5 2 8x x

1 0

–1

i.e. 8 2 8x

Add 2 to all sides then you get 8 2 2 2 8 2x

So the solution set for x is: 6 10x

6 10 ( 6,10)x

In interval notation ( 6,10)

Problem 17:

Properties of logarithm:

log ( ) log log log log log log logy

a a a a a a a a

xxy x y x y x y x

y

(a) Simplify using the properties of logarithm:

23

2ln

xz

y

Using the property log log loga a a

xx y

y

we have

232 23

2ln ln( ) ln

xzxz y

y

Now using the property log ( ) log loga a axy x y we can further simplify:

1 1

2 2 2 2 2 23 3 3ln( ) ln ln ln ln ln ln ln (since )n nxz y x z y x z y a a

Now use this property log logy

a ax y x and simplify further:

1

2 231

ln ln ln ln 2ln 2ln3

x z y x z y

10 0

–6

(b) Simplify using the properties of logarithm:

3 5

2 2log

x y

r

Using the property log log loga a a

xx y

y

we have

3 523 5

2 2 22log log ( ) log

x yx y r

r

Now using the property log ( ) log loga a axy x y we can further simplify:

2 23 35 52 2 2 2 2

1 1 1

23 52 2 2

log ( ) log log log log

log log log (since )n n

x y r x y r

x y r a a

Now use this property log logy

a ax y x and simplify further:

1 1

23 52 2 2 2 2 2

1 1log log log log log 2log

3 5x y r x y r

(c) Solve for t using the properties of logarithm: 10 10log ( 15) log 2t t

We use the property log log log ( )a a ax y xy to simplify 10 10log ( 15) log 2t t

2

10 10log ( 15) 2 log ( 15 ) 2t t t t

Now use the definition of logarithm log y

a t y a t to simplify 2

10log ( 15 ) 2t t

2 2 2

2

15 10 15 100 0

20 5 100 0 ( 20) 5( 20) 0

( 20)( 5) 0

20 0 or 5 0

20 or 5

t t t t

t t t t t t

t t

t t

t t

But negative value of t is not acceptable since log is not defined for negative values, i.e. the argument of

logarithm may not be negative. Hence t = 20.

(d) Solve for x: 2 2log ( 4) log ( 2) 4x x

Using the property log log loga a a

xx y

y

we have

2 2

2

log ( 4) log ( 2) 4

4log 4

2

x x

x

x

Now use the definition of logarithm log y

a t y a t to simplify 2

4log 4

2

x

x

4

2

4 4 4log 4 2 16

2 2 2

4( 2) 16 ( 2)

2

4 16 32 15 32 4 15 28

15

28

x x x

x x x

xx x

x

x x x x

x

(e) Solve for t:

2 2 2log 3 1 log 5 log 11t

Problem 18: Find the amount of money in an account after 12 years if $5000 is deposited at 7% annual

interest compounded as follows.

(a) Quarterly (b) Daily (c) Continuously

( ) 1

( )

nt

rt

rA t P

n

A t Pe

(a) Quarterly: You would need to find all the values in the formula ( ) 1

ntr

A t Pn

As given in the problem t = 12 years, r = 0.07, P = $5,000 and n is the number of periods in a year

which in this case is 4, since there are 4 quarters in a year, so n = 4. Note that n = 4 means that interest

is collected 4 times a year.

Put the values in the formula:

4 12

48

( ) 1

0.07(12) $5,000 1

4

0.07(12) $5,000 1

4

(12) $11,497.99

ntr

A t Pn

A

A

A

(b) Daily: You would need to find all the values in the formula ( ) 1

ntr

A t Pn

As given in the problem t = 12 years, r = 0.07, P = $5,000 and n is the number of periods in a year

which in this case is 365, since there are 365 days in a year, so n = 365. Note that n = 365 means that

interest is collected 365 times a year.

Put the values in the formula:

365 12

4380

( ) 1

0.07(12) $5,000 1

365

0.07(12) $5,000 1

365

(12) $11,580.90

ntr

A t Pn

A

A

A

(c) Continuously: If interest is collected continuously, i.e. at every instant then you would use this

formula ( ) rtA t Pe

As given in the problem t = 12 years, r = 0.07 and P = $5,000.

Put the values in the formula:

0.07 12

( )

(12) $5,000

(12) $11,581.83

rtA t Pe

A e

A

Note the difference in the amount of interest paid for different compounding periods as indicated by the

total amount received after 12 years:

Principal Amount Compounding Period Interest Charged Total after 12 years

$5,000 Quarterly $6,498 $11,498

$5,000 Daily $6,581 $11,581

$5,000 Continuously $6,582 $11,582

Problem 19: A sample of 400g of lead 210 decays to polonium 210 according to the function defined by 0.032( ) 400 tA t e where t is in years.

(a) How much lead will be left in the sample after 25 years?

Substitute t = 25 years and simplify.

0.032 25

0.032 25

( ) 400

( ) 400

( ) 179.73

A t e

A t e

A t g

(b) How long will it take the initial sample to decay to half of its original amount?

This amount of time needed for the initial amount of 400g to reduce to 200g is called half-life, a very

important concept in radioactive decay.

Use the formula 0.032( ) 400 tA t e and set A(t) = 200g

0.032

0.032

0.032

200 400

200 400

400 400

1

2

t

t

t

e

e

e

Now to solve for t use the definition of ln (natural log) or equivalently use the inverse function of e which

is ln.

0.032

0.032

1

2

1ln ln (Apply ln to both sides)

2

1ln(1) ln 2 0.032 (Since ln ln(1) ln 2)

2

ln 2 0.032

ln 2

0.032

21.66 years

t

t

e

e

t

t

t

t

Problem 20: Carbon-14 has a half-life of about 5700 years. An ancient artifact found has about 0.87g of

Carbon-14 left whereas it is estimated that at the time the artifact was built had 5 grams of Carbon-14 in

it. How old is the artifact?

This problem is a very important one. For solving a problem similar to this a Nobel Prize in chemistry

was awarded and this discovery revolutionized archeology and our understanding of the history of the

planet.

The key to solving this problem is to find the decay rate. This could be done by using the fact that half-

life of Carbon-14 is 5700 years. The general formula for exponential growth or decay is0( ) ktA t A e

where A0 is the initial amount and k is the exponential growth or decay. Positive value of k means

exponential growth and negative value of k is exponential decay.

In this problem A0 = 5 grams plug this value in 0( ) ktA t A e and then solve for k using the half-life

concept.

0

5700

( )

55

2

kt

k

A t A e

e

We use the concept that the initial amount of 5 grams will become 5/2 grams in 5700 years and plugged

these values in the general formula 0( ) ktA t A e

Hence we have the following:

5700

5700

5700

55

2

1 (by dividing both sides by 5)

2

1ln ln (Applied natural log (ln) to both sides)

2

1ln(1) ln 2 5700 (Since ln ln(1) ln 2)

2

ln 2

5700

k

k

k

e

e

e

k

k

So the formula for exponential decay becomes:

ln 2

5700( ) 5t

A t e

Now the amount left is 0.87g hence we have

ln 2

5700

ln 2

5700

ln 2

5700

ln 2

5700

ln 2

5700

( ) 5

0.87 5

0.87 5

5 5

0.87

5

0.87ln ln

5

0.87 ln 2ln

5 5700

0.87ln

5

ln 2

5700

14,300 years

t

t

t

t

t

A t e

e

e

e

e

t

t

t

Hence the artifact is about 14,300 years.

Problem 21: Sketch the following exponential function and label the y-intercept.

1( ) 4

3

x

f x

Simplify the given function:

1 1 1 1( ) 4 ( ) 4 ( ) 4 ( ) 4

13 11

333

( ) 4 1 3

( ) 4 3

x

x x

xx

x

x

f x f x f x f x

f x

f x

So we can graph the simplified function: ( ) 4 3xf x as is done below:

0

50

100

150

200

250

300

350

-6 -4 -2 0 2 4 6

Series1

Problem 22: Consider the following functions: 3

( ) 5 3 ( )5

xf x x g x

(a) Find ( )f g x

By definition

( ) ( ( ))

3

5

35 3

5

3 3

f g x f g x

xf

x

x

x

Hence ( ) ( ( ))f g x f g x x

(b) Find ( )g f x

-10

-8

-6

-4

-2

0

2

4

6

8

10

-10 -8 -6 -4 -2 0 2 4 6 8 10

By definition:

( ) ( ( ))

5 3

( 5 3) 3

5

5 3 3

5

5

5

g f x g f x

g x

x

x

x

x

Hence ( ) ( ( ))g f x g f x x

(c) Compare the results you found in parts (a) and (b). Can you draw a conclusion about the functions?

Parts (a) and (b) tell me that ( ) ( ( ))f g x f g x x and ( ) ( ( ))g f x g f x x

i.e. ( ) ( )f g x g f x x

So the functions are inverses of each other. Also note that if two functions are inverses of each other,

then their graphs are reflections of each other about the line y = x.

Problem 23: Two cars are 300 miles apart and are traveling toward each other on the same road. They

meet each other after driving 3 hours. One car is traveling 20 mph slower than the other. Find the speeds

of the cars.