Advanced Algebra Topics COMPASS Review – revised Summer 2013

You will be allowed to use a calculator on the COMPASS test. Acceptable calculators are basic calculators,

scientific calculators, and approved models of graphing calculators. For more information, see the JCCC

Testing Services website at http://www.jccc.edu/testing/ or call 913-469-4439.

1. If 3 6( ) 2 2x

f x x x and ( ) ( 1)( 3)g x x x , find (2) ( 3)f g .

a. 12 b. 63 c. 27 d. 0

2. If 2( ) 3f x x x , ( ) 2 1g x x , and 2( ) 2 5h x x , find ( ) ( ) ( )f x g x h x .

a. 2 5 4x x b. 3 24 3 13 5x x x c. 3 24 7 5x x x d. 4 22 5 4x x

3. If ( ) 3 2f x x and 3 2( ) 12 8 9 6g x x x x , find ( )

( )

g x

f x.

a. 3 2

3 2

12 8 9 6

x

x x x

b. 2 12

3 24 3

xx

c. 24 9 3x x d. 24 3x

4. If ( ) 3 2f x x and 2( ) 4 1g x x x , find ))(( xfg .

a. 2 7 3x x b. 2 4 1x x c. 29 24 13x x d. 23 12 1x x

5. If 2( ) 4 2 1f x x x , find ( 3)f x .

a. 24 22 31x x b. 24 2 4x x c. 24 2 31x x d. 24 10 7x x

6. If ( ) (3 2)f x x x , find ( ) ( )f x h f x .

a. 26 3 2xh h h b. h c. 2 2 23 3 2x h h h d. 6 2x

7. If ( ) 3 2f x x and 2( ) 4 1g x x x , find [ (2)]f g .

a. 12 b. 1 c. 1 d. 11

8. If 3( ) 4 1f x x , find 1( )f x .

a.3

1 1( )

4

xf x

b. 31( ) 4 1f x x c. 3

11

4( ) 1f x x d.

3

1 1( )

4 1f x

x

9. If ( ) 6 3f x x , and g is a function such that ( ( )) ( ( ))f g x g f x x for all values of x , find ( )g x .

a. ( ) 3 6g x x b. 1 16 2

( )g x x c. ( ) 36 21g x x d. ( ) 6 3g x x

10. If ( )f x contains the point (4,1), then )(1 xf must contain what point?

a. ( 4,1) b. (1,4) c. (4, 1) d. 14

(1, )

11. The value V in dollars of a piece of equipment t months after its purchase is given by the function

( ) 7250 125V t t . Find 1(50)V .

a. 7247.5 b. 0.001 c. 1000 d. 57.6

2

12. Find the domain of 2

2

1

4( )

x

xf x

.

a. ( ,0) (0, ) c. ( , 2) ( 2, 1) (1,2) (2, )

b. ( , 1) ( 1,1) (1, ) d. ( , 2) ( 2,2) (2, )

13. Which describes all real values of x for which ( ) 3 9 1f x x is a real number?

a. | 3x x b. | 3x x c. | 0x x d. | 0 3x x

14. If the minimum value of ( )f x occurs at 3x , at which value of x does the minimum value of ( 2) 1f x

occur?

a. 2x b. 5x c. 4x d. 6x

15. Which describes the set of all real values of y for which ( )y f x if 2

2

1

4( )

x

xf x

?

a. | 2 and 2y y y b. | 1 y y c. 14

| 1 or y y y d. all real numbers

16. Find the range of ( ) 3 9 1f x x .

a. 3, b. 1, c. 0, d. [0,3]

17. If 3( ) 2f x kx x and ( 2) 6f , find (4)f .

a. 168 b. 24 c. 216 d. 56

18. If 4 1x y and 3 5z x , find an expression for z in terms of y .

a. z = 4y + 1 b. z = 12y – 2 c. z = 12y – 19 d. z = 3y – 5

19. The demand d for items priced at p dollars per item is given by 9000

( ) 31

d pp

, and the revenue R of

selling d items at price p is given by R pd . Find R as a function of p.

a. 9000

( ) 31

R pp

b. 2

3 8997( )

pR p

p p

c.

23 8997( )

1

p pR p

p

d.

9000( ) 3

1R p p

p

20. Find the vertex of the parabola 2( ) 3 6 5f x x x .

a. 3, 5 b. (1, 8) c. ( 1, 2) d. 1, 8

21. Find the solutions to the equation 28 2 3x x

a. 2 2 10x b. 10

28x c. 8 2 10x d.

10

22x

22. Which equation has 0,2, 4 as its solution set?

a. 2 4 0x b. ( 2)( 4) 0x x x c. ( 2)( 4) 0x x x d. (2 4) 0x x

3

23. What are the zeros of the function 3 2( ) 2 9 18f x x x x ?

a. x = 3, x = –3, x = 2 c. x = 3, x = –3, x = –2

b. x = 0, x = –3, x = 3, x = –2, x = 2 d. x = 0, x = 9, x = 2, x = –2

24. Which cubic function has ,2,0 xx and 5x as zeros?

a. 3 2( ) 7 10f x x x x b. 3 2( ) 7 10f x x x x c. 3 2( ) 3 10f x x x x d. 3 2( ) 3 10f x x x x

25. If the domain of 2 2( ) 9 9f x x x is all real numbers, what is the solution set for ( ) 0f x ?

a. 9, 9 b. 3, 3 c. 3, 3, 9 d. 3, 3,3 , 3i i

26. Factor 4( ) 81f x x completely over the complex numbers.

a. ( ) ( 3)( 3)( 3)( 3)f x x x x x c. ( ) ( 3)( 3)( 3 )( 3 )f x x x x i x i

b. 2( ) ( 3)( 3)( 9)f x x x x d. 2( ) ( 3 )( 3 )( 9)f x x i x i x

27. Which quadratic function has a corresponding graph with a vertex at (2, 3) and contains the point ( 2,5) ?

a. 2( ) ( 2) 3f x x b. 21

2( ) 2 1f x x x c. 2( ) ( 2) 3f x x d. 2( ) 4 1f x x x

28. Simplify 2

32( )a

a. 8

3a b. 4

3a c. 2

3a d. 1

3a

29. Simplify 2

5

2

3

2

3

2

1

23 baba

a. 3 15

4 46a b b. 2 45a b c. 3 86a b d. 2 46a b

30. Simplify

32 5

3 6

2

3

x y

x y

a. 3

3

2

3

y

x b.

3

3

8

27

y

x c.

33

15

27

8

y

x d.

15

33

3

2

x

y

31. Simplify, assuming that the variables represent positive numbers:3 28 4a a

a. 4 2a a b. 32 2a c. 64 2a a d. Not possible

32. If 2 1

3 4

125

5

x

x

then ?x

a. 7

5 b. 6 c. 2

7 d. 5

33. If

2

5

5

a

a

axx

x

for all real values of x such that 0x , find the value(s) of a.

a. 5a b. 0a c. 5a d. 5, 1a a

4

34. Rewrite using logarithmic notation: xM y

a. logx y M b. logy M x c. logM x y d. logM y x

35. If 12 4

log x , then x ?

a. 1

2 b. 2 c. 2 d. 1

16

36. Express as a single logarithm: 10 10 10 102log 3 4log 6log 8logy z t

a. 106log (3 )yzt b. 10log (6 4 6 8 )y z t c. 4

10 6 8

9log

y

z t

d. 108log (3 )y z t

37. If 2 110 50 50x , find the value of x.

a. 3

2 b. 0 c. 1

2 d. 3

38. If ln(3 4) 0x , find the value of x.

a. 5

3 b. 4

3 c. 4

3

e d. 1

39. The function 0.14( ) 70 28.6 tT t e gives the temperature of an object in a room t hours after midnight.

Find t when T = 85. Give your answer rounded to the nearest tenth.

a. 1.1 b. 0.2 c. 70.0 d. 4.6

40. Simplify: 2237 2 3 4 1 2i i i i .

a. 3 3i b. 5 17i c. 5 13i d. 3 i

41. The product of two complex numbers is 2 3i . One of the complex numbers is 1 4i . Find the other

complex number.

a. 14 5i b. 10 1117 17

i c. 10 1113 13

i d. 3 i

42. If z is a complex number with z its complex conjugate, and 2 6z z i , find z .

a. 6 b. 12

3i c. 2i d. 2i

43. If the nth term of a sequence is given by 1

2

1( 1)n n

n

, find the ( 1)thn term.

a. 1

2

1( 1)n n

n

b.

2

1n

n

c. 2

2

2( 1)

2 1

n n

n n

d. 2

2

2( 1)

1

n n

n

44. A certain arithmetic sequence has 1 56a and 11 26a . Find 17a .

a. 21 b. 18 c. 3 d. 8

5

45. If 2, ,7x are the first three terms of a geometric sequence of positive real numbers, find the common ratio r.

a. 72

b. 72

c. 52

d. 72

46. Which of the following could be the seventh term in a geometric sequence with 2 3a and 4 9a ?

a. 3 b. 2187 c. 27 3 d. 729

47. A recursive sequence is given by 1 2 3a i , and 1n na i a for 2n , where 1i . Find 5a .

a.3 2i b. 2 3i c. 2 3i d. 32 243i

48. Evaluate 23

1

(6 3)n

n

.

a. 1587 b. 60 c. 1518 d. 135

49. A stack of cans has 60 cans in the bottom row, and each row above the bottom row has two less cans than

the row below it. Find the total number of cans in the first 10 rows, starting with the bottom row.

a. 600 b. 580 c. 560 d. 510

50. Evaluate 9!

3! 6!

a. 1 b. 84 c. 1

2 d.

45

126

51. If

8 2

3 7

5 0

A

and

3 4

0 3

2 6

B

, find the entry in the first row and the first column for 5A B .

a. 11 b. 5 c. 7 d. 47

52. If 2 3 1 6 8

5 6 0 2 15 17

k

, then k =?

a. 3 b. 4 c. 8 d. 3

2

53. If a b

ad bcc d

and 1

4 2

k=10, then k = ?

a. 5

4 b. 3 c. 10 d. 7

54. Give the z-value of the solution to the following system:

4 5 11

2 7

2 1

x y

x z

y z

a. 2 b. 1 c. 1 d. 4

6

55. Which of the following is an augmented matrix for the given system? 4 5 11

2 7

x y

y x

a. 4 5 11

1 2 7

b.

4 5 11

2 1 7

c.

4 2

5 1

11 7

d. 11 4 5

7 1 2

56. Which of the following gives the solution to 2 4

3

xx

?

a. b. [1,3] c. [ 4,1] d. ( , 4] [1, )

57. Which of the following graphs are functions?

I.

II.

III.

IV.

a. I only b. I and II only c. I, III, and IV only d. all of I, II, III and IV

58. For the vectors 1,3 and 2, 5 a b , find 3a b .

a. 5,18 b. 13 c. 13 d. 11

59. For the vectors 1,3 and 2, 5 a b , find a b .

a. 5,18 b. 13 c. 13 d. 11

60. If the point (2, 4) is reflected across the x-axis, then translated by the vector 3, 1 , what will be the

coordinates of the resulting point?

a. (5, 5) b. (1,3) c. (1, 5) d. (5,3)

61. Given sets 1,3,7A and 2,8,20B , which of the following is a function from A to B?

a. ( ) 3 23f x x b. ( ) 6 2f x x c. 2( ) 2f x x x d. ( ) 3 1f x x

62. Data is collected on two variables, x and y, shown in the table below. Which of the following equations

describes the relationship between x and y?

x y

20 6

40 16

80 26

a. 14y x b. 2 1010log 4xy c. 2 10y x d. 1

10 10log 6xy

7

Answers to Advanced Algebra Topics COMPASS Review

1. c

2. b

3. d

4. c

5. a

6. a

7. d

8. a

9. b

10. b

11. d

12. d

13. a

14. b

15. c

16. b

17. b

18. b

19. c

20. d

21. d

22. b

23. c

24. d

25. b

26. c

27. b

28. b

29. d

30. c

31. c

32. c

33. d

34. d

35. c

36. c

37. a

38. a

39. d

40. b

41. b

42. c

43. c

44. d

45. d

46. c

47. b

48. a

49. d

50. b

51. c

52. a

53. d

54. b

55. b

56. c

57. b

58. a

59. c

60. d

61. d

62. b

Solutions to Advanced Algebra Topics COMPASS Review

1. 3 62

(2) 2(2) 2(2) 16 4 3 15f , and ( 3) ( 3 1)( 3 3) ( 2)( 6) 12g ,

so (2) ( 3) 15 12 27f g

2. 2 2 2 3 2( ) ( ) ( ) ( 3 ) (2 1)(2 5) ( 3 ) (4 2 10 5)f x g x h x x x x x x x x x x

2 3 2 3 23 4 2 10 5 4 3 13 5x x x x x x x x

3. 3 2( ) 12 8 9 6

( ) 3 2

g x x x x

f x x

, then use long division to simplify:

Quotient is 24 3x

4. ))(( xfg = ( ( ))g f x = (3 2)g x = 2(3 2) 4(3 2) 1x x = 2(9 12 4) 12 8 1x x x =

29 24 13x x

5. 2 2 2( 3) 4( 3) 2( 3) 1 4( 6 9) 2( 3) 1 4 24 36 2 6 1f x x x x x x x x x

24 22 31x x

6. 2( ) ( ) ( )(3( ) 2) ( (3 2)) ( )(3 3 2) (3 2 )f x h f x x h x h x x x h x h x x

2 2 2 23 3 2 3 3 2 3 2 6 3 2x xh x xh h h x x xh h h

2

3 2

3 2

4 3

3 212 8 9 6

(12 8 )

9 6

(9 6)

0

x

x x x x

x x

x

x

8

7. 2(2) 2 4(2) 1 4 8 1 3g , so [ (2)]f g = ( 3)f = 3(–3) – 2 = – 9 – 2 = – 11

8. 3 4 1y x Reverse the roles of the x and y to form the inverse relation.

3 4 1x y This is the inverse relation. Now solve for y to express as a function.

333 4 1x y 3 4 1x y

3 1

4

xy

31 1( )

4

xf x

9. ( )g x is the inverse function for ( )f x , so follow the same procedure as in question 8.

6 3y x Now reverse the roles of the x and y to form the inverse relation.

6 3x y This is the inverse relation. Now solve for y to express as a function.

3

6

xy

can also be written as 1 1

6 2x y , so 1 1 1

6 2( ) ( )g x f x x

10. The x and y values swap, so an ordered pair in 1( )f x is (1,4).

11. The input and output values swap, so 1(50)V will be the value of the input that will have an output of 50

for the function ( )V t .

50 7250 125t 50 7250

57.6125

t

months

12. The domain is all real values of x, except those that would make the denominator equal zero.

2 4 0x ( 2)( 2) 0x x 2, 2x x , so the domain is ( , 2) ( 2,2) (2, ) .

13. 3 9x must be greater than or equal to zero in order to get real-valued outputs.

3 9 0x 3x , so the domain is [ 3, ) , or using set notation, | 3x x .

14. The graph of ( 2) 1f x will be the graph of ( )f x , but shifted right 2 units and down 1 unit. Therefore,

the x-coordinate of the minimum will be shifted right 2 units, so it will occur at 5x .

15. One way to answer this is to examine the graph to determine the range of ( )f x . The graph has a horizontal

asymptote at 1y , and when 2x or 2x , the graph is above the horizontal asymptote, so 1y . When

x is between 2 and 2, the highest output occurs at 0x , with 14

(0)y f . Therefore the range, or set

of real values of y, is 14

| 1 or y y y .

16. The outputs of the radical expression 3 9x will be 0 or larger, and then the 1 term will decrease these

values by 1. Therefore the range is 1y , or in interval notation, 1, .

17. Use ( 2) 6f to find k : 36 ( 2) 2( 2)k 6 8 4k 14

k

Then 314

(4) (4) 2(4) 16 8 24f

18. Substitute 4 1y in place of x in the equation 3 5z x , then simplify: 3(4 1) 5z y 12 2z y

9

19. Substitute 9000

31p

in place of d in the equation R pd , then simplify:

2 29000 9000 9000 3 ( 1) 9000 3 3 3 89973 3

1 1 1 1 1 1

p p p p p p p p pR p p

p p p p p p

20. Method 1: Complete the square to obtain the form 2( ) ( )f x a x h k , where the vertex is ( , )h k :

2( ) 3 6 5f x x x 23( 2 ) 5x x 23( 2 1) 5 3(1)x x 23( 1) 8x

The vertex is ( 1, 8)

Method 2: Use x = 2

b

a

to find the x-coordinate, then substitute into ( )f x to find the y-coordinate:

61

2(3)x

, and 2( 1) 3( 1) 6( 1) 5y f = – 8. The vertex is ( 1, 8) .

21. Since the equation is quadratic, write it in standard form and use the quadratic formula:

20 2 8 3x x , and 2 4

2

b b acx

a

2( 8) ( 8) 4(2)(3) 8 40 8 2 10

2(2) 4 4x

simplifies to 4 10

2x

or 1

22 10x

22. Many equations could have this solution set, but the choices given are all polynomials in factored form, set

equal to zero. This allows us to use the fact that if x a is a zero of a polynomial function, then ( )x a is a

factor of the polynomial. This allows us to write the equation ( 2)( 4) 0x x x .

23. This function is factorable by grouping: 2 2( ) ( 2) 9( 2) ( 9)( 2) ( 3)( 3)( 2)f x x x x x x x x x .

Set ( )f x = 0 to get zeros of 3, – 3, and – 2.

24. The zeros will create factors of ( )( 2)( 5)x x x . Multiply these out to get 3 2( ) 3 10f x x x x .

25. Set 2 20 9 9x x and solve to get 2 9 0x or 2 9 0x . The solutions to 2 9 0x are nonreal,

and since the domain is specified as real numbers, we discard the nonreal solutions. This means we have

only 2 9 0 ( 3)( 3) 0 3x x x x .

26. Factoring as a difference of squares we get 2 2( ) 9 9f x x x . The zeroes of 2 9 are 3x x , as in

question 25, and the zeroes of 2 2 29 satisfy 9 0 9 9 3x x x x i . Using the fact

that if x a is a zero of a polynomial function, then ( )x a is a factor of the polynomial, we have the

factored polynomial ( ) ( 3)( 3)( 3 )( 3 )f x x x x i x i .

10

27. Use the form of a parabola: 2( ) ( )f x a x h k .

Insert the coordinates of the vertex for h and k: 2( ) ( 2) ( 3)f x a x

Now use the other ordered pair to find a: 25 ( 2 2) ( 3)a 1

2a

Now create the function using the a value: 1 2

2( ) ( 2) ( 3)f x x or

1 2

2( ) 2 1f x x x

28. Multiply the exponents together to get 4

3a .

29. Multiply the coefficients, and add the exponents on the like variables:

1 3 3 5

2 2 2 26a b

= 2 46a b

30. Simplify the outside power first, then the negative exponents, and then combine the like variables. 3 6 15

3 9 18

2

3

x y

x y

=

3 15 18

3 6 9

3

2

y y

x x =

33

15

27

8

y

x

31. Write the radicals as fractional exponents, then use properties of exponents to simplify.

1 1 13 73 2 2123 6 63 63 3 6 62 22 3 2 2 13 7 2 1 1 18 4 2 2 2 2 2 2 2 2 4 2a a a a a a a a a a a a

32. Start by writing both sides of the equation as powers of 5, then solve for x.

2 1

2 (3 4)5 5x

x

4 2 3 45 5x x 4 2 3 4x x 7 2x 2

7x

33. Start by simplifying both sides as powers of x, then solve for a. 2 5 5a a ax x 2 5 5a a a 2 6 5 0a a ( 5)( 1) 0a a 5, 1a a

34. Use the definition logxaa y y x to get logM y x

35. 12 4

log x 1

42x 22 2x 2x

36. Use the properties log( ) log( ) log( )ab a b , log log( ) log( )a

ba b , and log logax a x

10 10 10 102log 3 4log 6log 8logy z t = 2 4 6 810 10 10 10log 3 log log logy z t =

4 6 810 10 10 10log 9 log log logy z t = 4 6 8

10 10log 9 logy z t = 4

10 6 8

9log

y

z t

37. 2 110 50 50x 2 110 100x 2 1 210 10x 2 1 2x 2 3x 32

x

38. ln(3 4) 0x 03 4x e 3 4 1x 3 5x 53

x

39. Solve 0.1485 70 28.6 te for t:

0.1485 70 28.6 te 0.1415 28.6 te 15 0.14

28.6

te 15

28.6ln 0.14t

1 15

0.14 28.6ln 4.6t

hours

11

40. Remember 1i so 2 1i . Then 2237 2 3 4 1 2i i i i = 23 27 2 3 4 1 4 4i i i i i

= 23 2 27 6 8 1 4 4i i i i i = 11

2 1 2 27 6 8 1 4 4i i i i i i =

117 1 6 8( 1) 1 4 4( 1)i i i = 7 6 8 1 4 4i i i = 5 17i

41. Write ( 1 4 )( ) 2 3i a bi i , then find a bi .

2 3 2 3 1 4

1 4 1 4 1 4

i i ia bi

i i i

=

2

2

2 11 12

1 16

i i

i

=

10 11

17

i = 10 11

17 17i

42. Let z a bi , so z a bi , and we have ( ) 2( ) 6a bi a bi i .

Simplify the left side to get 3 6a bi i 0 and 2a b , so 2z i .

43. Substitute n + 1 in place of n in the expression 1

2

1( 1)n n

n

to get

2

( 1) 1 2

2 2

( 1) 1 2( 1) ( 1)

( 1) 2 1n

n nn na

n n n

44. You can use the formula for the nth term in an arithmetic sequence: 1 ( 1)( )na a n d

26 56 (11 1)( ) 26 56 10 3d d d

Since we now have 1a and d, use the formula to find 17a : 17 56 (16)( 3) 8a

45. Use the formula for the nth term in a geometric sequence: 1

1n

na a r , with 1 2a and 3 7a :

3 1 27 7

2 27 2 r r r

But since we are told that all the terms in the sequence are positive, this rules out the possibility of a

negative value for r, so 72

r .

46. Use the formula for the nth term in a geometric sequence: 1

1n

na a r to get a system of two equations:

1 1

1 1

2 1

4 1 3

3 3 , and

9 9

a r a r

a r a r

We can substitute 3 in place of 1a r in the second equation to get 29 3 3r r . Since all of the

choices given are positive values for the seventh term of the sequence, r must be positive, so we use 3 .

Now we can use 3r to find 1a , then use the formula for the nth term to find 7a :

1 1

3

33 3a a , and

7

7 1 53

33 3 3 27 3a

12

47. 1 2 3a i

2 1

2(2 3 ) 2 3 3 2a i a i i i i i

3 2

2( 3 2 ) 3 2 2 3a i a i i i i i

4 3

2( 2 3 ) 2 3 3 2a i a i i i i i

5 4

2(3 2 ) 3 2 2 3a i a i i i i i

48. 23

1

(6 3) (6 1 3) (6 2 3) (6 3 3) ... (6 23 3)n

n

= 3 + 9 + 15 + …+ 135.

We can see that this is a sum of the first 23 terms in an arithmetic sequence with common difference 6d ,

so use the formula for the sum of the first n terms in an arithmetic sequence:

23 1 23

23

211.5(3 135) 1587S a a

49. We want to find the sum of the first 10 terms in an arithmetic sequence, with first term 1 60a and common

difference 2d . Use the formula for the nth term in an arithmetic sequence to find the number of cans in

the 10th

row: 10 60 (10 1)( 2) 42a cans. Then use the formula for the sum of the first n terms in an

arithmetic sequence: 10 1 10

10

25(60 42) 510S a a cans

50. 9! 9 8 7 6! 9 8 7

843! 6! 3 2 1 6! 3 2 1

51. Multiply each entry in B by 5, then add the corresponding entries in A and 5B. The entry in the first row

first column is 8 15 7 .

52. Multiply the two matrices on the left side of the equation to get 2 8 6 8

5 17 15 17

k

k

. The entries in the

two matrices must all be the same, so we have 2 6k and 5 15k , which imply 3k .

53. Apply the definition a b

ad bcc d

to the left side of the equation to get 1

10 2 4 104 2

kk , then

solve for k to get 7k .

54. One method is to rewrite the system as:

4 5 11

2 7

2 1

x y

x z

y z

Then multiply the 2nd

equation by 2 and add the result to the 1st equation to eliminate x , which gives

5 2 3y z . Using this result with the original 3rd

equation, we have a system of two equations with two

variables: 5 2 3

2 1

y z

y z

We want eliminate y now, so multiply the top equation by 2 and the bottom equation by 5. When we add

them together and solve for z, we get z = 1 .

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55. First, write the terms in the same order to get 4 5 11

2 7

x y

x y

Then, write a matrix where each row gives the coefficients of an equation. One type of notation for an

augmented matrix uses a dashed line to indicate the augmentation of the coefficient matrix with the matrix

of constant terms: 4 5 11

2 1 7

56. One way to proceed is multiply both sides of the inequality by 3 , but this requires us to switch the

direction of the inequality symbol: 2 4 3x x . Then one method to finish the solution is to write the

inequality with zero on one side and examine the sign changes of the other side: 2 3 4 0 ( 4)( 1) 0x x x x . The product on the left side will be equal to 0 when either factor is

0, which occurs when 4x or when 1x . The product on the left side will be less than 0 when one of

the factors is negative and the other is positive. Since for all values of x, the factor 4x is larger than the

factor 1x , the positive factor with be 4x and the negative factor will be 1x . This means 4 0x

and 1 0x , which implies 4x and 1x , which means 4 1x . So ( 4)( 1) 0x x is true when

4 1x , or when x is in the interval 4,1 .

57. The only graphs that pass the vertical line test are I and II.

58. Distribute the 3 through the components of b and then add the corresponding components of a and 3 b :

3 1,3 3 2, 5 1,3 6,15 5,18 a b .

59. 1,3 2, 5 (1)(2) (3)( 5) 2 15 13 a b

60. When the point (2, 4) is reflected across the x-axis, we get the point (2,4) . Then, if we translate the point

(2,4) by the vector 3, 1 , we move right 3 and down 1 to end up at (5,3) .

61. A function from A to B must map each element of A to one element of B (but it is not necessary to use all of

the elements of B). There are many possible functions from A to B, so the only way to answer this question

is to find which of the choices works. We can try evaluating each function using the elements of A to

determine whether or not we get values that are in B.

a. (1) 3(1) 23 20f , which is in B

(3) 3(3) 23 14f , which is NOT in B, so this choice is incorrect.

b. (1) 6(1) 2 8f , which is in B

(3) 6(3) 2 20f , which is in B

(7) 6(7) 2 44f , which is NOT in B, so this choice is incorrect.

c. 2(1) 1 1 2 2f , which is in B

2(3) 3 3 2 8f , which is in B

2(7) 7 7 2 44f , which is NOT in B, so this choice is incorrect. Choice d had better work.

d. (1) 3(1) 1 2f , which is in B

(3) 3(3) 1 8f , which is in B

(7) 3(7) 1 20f , which is in B, so this choice is correct.

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62. As in question 61, there are many possible equations that could relate the variables. We can try the

equations to see which one works for each pair of data values.

a. 14y x works when 20x , but not for 40x since 40 14 26 16 . So this choice is

incorrect.

b. 2 1010log 4xy works for each pair of data values:

When 20x 2 2

2010

10log 4 10log 2 4 10(1) 4 6y , and

when 40x 2 2

4010

10log 4 10log 4 4 10(2) 4 16y , and

when 80x 2 2

8010

10log 4 10log 8 4 10(3) 4 26y . So this choice is correct.

(Choices c and d do not work for any of the pairs of data values.)