Practice for Chapter 12Disclaimer. The actual exam is different. This is a study aid.SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

Given f(x) and g(x), find the indicated composition.1) f(x) = x2 + 7; g(x) = 5x + 3Find (f ∘ g)(x).

1)

2) f(x) = 4x2 + 6x + 8; g(x) = 6x - 4Find (g ∘ f)(x).

2)

3) f(x) = 9x2; g(x) = x - 4

Find (g ∘ f)(x).

3)

4) f(x) = 2 x + 7; g(x) = 7x + 10Find (f ∘ g)(x).

4)

5) f(x) = 1x; g(x) = 6x

Find (g ∘ f)(x).

5)

Given f(x) and g(x), find the indicated composition and evaluate.

6) f(x) = 3x + 6; g(x) = 2x

Find (g ∘ f)(3).

6)

7) f(x) = x; g(x) = 4x - 14Find (f ∘ g)(9).

7)

8) f(x) = x; g(x) = 4x - 14Find (f ∘ g)(14).

8)

9) f(x) = 3x + 8; g(x) = 2x2Find (g ∘ f)(2).

9)

Determine whether the function is one-to-one.10) f(x) = -8 - 7x 10)

11) f(x) = -4 - x2 11)

Determine whether the function is one-to-one. Give a valid reason.12) f(x) = 3x2 + 8 12)

1

Determine whether the function is one-to-one.13)

x

f (x)

f

x

f (x)

f

13)

14)

x

f (x)

f

x

f (x)

f

14)

15)

x

f (x)

f

x

f (x)

f

15)

Determine whether the function is one-to-one. If so, find a formula for the inverse.16) g(x) = 2x 16)

17) h(x) = 6x - 9 17)

18) g(x) = x2 - 3 18)

2

19) h(x) = 4x

19)

20) f(x) = -10 20)

21) h(x) = x3 - 7 21)

22) f(x) = x - 9 22)

Graph the function using a solid line and its inverse using a dashed line on the same set of axes.23) f(x) = 2x + 2

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

23)

Graph the function using a solid line and its inverse using a dashed line on the same set of axes. Label two points oneach line.

24) f(x) = 53x - 2

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

24)

3

Graph the function using a solid line and its inverse using a dashed line on the same set of axes.25) f(x) = x3 + 1

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

25)

26) f(x) = x + 4

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

26)

27) f(x) = x2 - 5, x ≥ 0

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

27)

For the function f, use composition of functions to show that f-1 is as given.

28) Let f(x) = 3 x + 27. Show that f-1(x) = x3 - 27. 28)

29) Let f(x) = 3 x + 3. Show that f-1(x) = x3 - 3. 29)

4

30) Let f(x) = (2 - x)/x. Show that f-1(x) = 2x + 1

. 30)

Find a formula for the inverse of the function described below.31) 32° Fahrenheit = 0° Celsius. A function that converts temperatures in Fahrenheit to those

in Celsius is f(x) = 59(x - 32) Fahrenheit.

31)

Graph. Label at least two points on each graph.32) y = 3x

x-4 -2 2 4 6

y

4

2

-2

-4

x-4 -2 2 4 6

y

4

2

-2

-4

32)

Graph. Label two points on the graph.33) y = 5x - 3

x-4 -2 2 4 6

y

4

2

-2

-4

x-4 -2 2 4 6

y

4

2

-2

-4

33)

5

34) y = 4(x + 1)

x-4 -2 2 4 6

y

4

2

-2

-4

x-4 -2 2 4 6

y

4

2

-2

-4

34)

35) y = 13

x

x-4 -2 2 4 6

y

4

2

-2

-4

x-4 -2 2 4 6

y

4

2

-2

-4

35)

36) y = 5-x

x-4 -2 2 4 6

y

4

2

-2

-4

x-4 -2 2 4 6

y

4

2

-2

-4

36)

6

Graph the pair of equations on the same set of axes. Graph the equation on the left using a solid line and the functionon the right with a dashed line. Label two points on each line.

37) y = 5x, x = 5y

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

37)

38) y = 12x, x = 1

2 y

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

38)

Solve the problem.39) The amount of particulate matter left in solution during a filtering process decreases by

the equation P = 600 (0.5). 6 n, where n is the number of filtering steps. Find the amountsleft for n = 0 and n = 5. (Round to the nearest whole number.)

39)

40) The amount of particulate matter left in solution during a filtering process decreases bythe equation P = 300 (0.5). 4 n, where n is the number of filtering steps. Find the amountsleft for n = 0 and n = 5. (Round to the nearest whole number.)

40)

41) The half-life of a certain radioactive substance is 13 years. Suppose that at time t = 0,there are 30 g of the substance. Then after t years, the number of grams of the substanceremaining will be

N(t) = 30 12t/26

How many grams of the substance will remain after 52 years?Round to the nearest hundredth of a gram.

41)

7

Simplify.

42) log8 164

42)

43) log 10 0.01 43)

44) log6 65 44)

45) log7 72 45)

46) log8 512 46)

47) 10log 7 47)

Graph. Label two points on the graph.48) y = log1/8 x

x-4 -2 2 4 6

y

4

2

-2

-4

x-4 -2 2 4 6

y

4

2

-2

-4

48)

Graph. Label at least two points on the graph.49) y = log1/9 x

x-4 -2 2 4 6

y

4

2

-2

-4

x-4 -2 2 4 6

y

4

2

-2

-4

49)

8

Graph. Label two points on the graph.50) y = log5 (x - 2)

x6

y

x6

y

50)

Graph both functions using the same set of axes. Label two points on each graph.51) f(x) = 3x, f-1(x) = log3 x

x6

y

x6

y51)

Rewrite as an equivalent exponential equation. Do not solve.52) t = log 3 81 52)

53) log 9 81 = 2 53)

54) log 3 19 = -2 54)

55) log w Q = 12 55)

Rewrite as an equivalent logarithmic equation. Do not solve.56) 32 = 9 56)

57) 2-2 = 14

57)

58) 100.8451 = 7 58)

9

59) yz = 8 59)

Solve the problem.60) log 4 16 = x 60)

61) log 2 18

= x 61)

62) log x 32 = 5 62)

63) log 5 x = -2 63)

64) log 3 x = -4 64)

65) log 64 x = 23

65)

Express as a sum of logarithms.66) log 2 (16 ∙ 8) 66)

67) logr (4T) 67)

68) log 6 (xy) 68)

69) logx (5yz) 69)

Express as a single logarithm.70) log 6 13 + log 6 12 70)

71) log c m + log c n 71)

Express as a product.72) log c Z-3 72)

73) log c Z-7 73)

Express as a difference of logarithms.

74) log gM61

74)

75) log a qr

75)

10

Express as a single logarithm.76) log b y - log b z 76)

Express as a sum, difference, and product of logarithms, without using exponents.

77) log7 15 mn

77)

78) log4 9 xy

78)

79) logb m5p9

n6b879)

80) logb m2p4

n6b380)

81) logb x4y7

z581)

82) logb 3 x9

y2z682)

83) logba6b9

a2b383)

Express as a single logarithm, and, if possible, simplify.

84) 14

loga x + 6 loga y - 2 loga x 84)

85) 13

loga x + 7 loga y - 4 loga x 85)

86) logb 4x + 3(logb x - logb y) 86)

87) logw (x2 - 64) - logw (x - 8) 87)

Solve the problem. Decimal answers are expected.88) Given log b A = 3.940 and log b B = 0.253.

Find log b AB.

88)

11

Solve the problem.89) Given log b 5 = 2.5179 and log b 7 = 1.4873.

Find log b 57.

89)

90) Given log b 5 = 1.2301 and log b 7 = 1.4873.

Find log b b7 .

90)

Simplify.91) logaa4059 91)

92) log e e5 92)

93) log e e9 93)

Graph and state the domain and the range of the function. Label two points on the graph.94) f(x) = -e-x

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

94)

95) f(x) = ex + 1

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

95)

Solve. Where appropriate, include approximations to the nearest thousandth. If no solution exists, state this.96) 4(x + 3) = 16 96)

12

97) 4(2x + 1) = 64 97)

98) 8x = 32(2x + 5) 98)

99) 2(x2 + 4x) = 18

99)

100) 42x ∙ 4x2 = 64 100)

101) 94x = 81(2x + 9) 101)

102) ln x = 1 102)

103) log2 (9x - 8) = 3 103)

104) log x - log(x + 7) = 1 104)

105) log4 (x - 4) + log4 (x - 4) = 1 105)

106) log 3 x = log 4 + log (x - 3 ) 106)

107) ln (x - 6) + ln (x + 5) = ln 42 107)

108) ln (x - 9) + ln (x + 5) = ln 51 108)

Solve the problem.109) A college loan of $24,000 is made at 3% interest, compounded annually. After t years, the

amount due, A, is given by the functionA(t) = 24,000(1.03)t .

After what amount of time will the amount reach $36,000?

109)

110) A loan of $18,000 is made at 2% interest, compounded annually. After t years, theamount due, A, is given by the function

A(t) = 18,000(1.02)t .Find the doubling time.

110)

111) The hydrogen ion concentration of a substance is about 8.0 x 10 -10 moles per liter.Find the pH. Round to the nearest tenth. Use the formula pH = -log [H+].

111)

Solve the problem. Write answer in scientific notation.112) Find the hydrogen ion concentration of a solution whose pH is 5.9. Use the formula

pH = -log [H+].112)

13

Solve the problem.

113) Use the formula L = 10 ∙ log II0, where the loudness of a sound in decibels is determined

by I, the number of watts per square meter produced by the soundwave, andI0 = 10-12 W/m2. A certain noise measures 115 dB. If the intensity is multiplied by 1000,how many decibels will the new noise measure?

113)

114) Use the formula L = 10 ∙ log II0, where the loudness of a sound in decibels is determined

by I, the number of watts per square meter produced by the soundwave, andI0 = 10-12 W/m2. What is the intensity of a noise measured at 34 dB?

114)

115) Suppose that $4000 is invested at an interest rate of 5.4% per year, compoundedcontinuously. What is the balance after 2 years?

115)

116) Suppose that $7000 is invested at an interest rate of 5.6% per year, compoundedcontinuously. What is the doubling time?

116)

117) How long will it take for $7000 to grow to $16,000 at an interest rate of 3.9% if theinterest is compounded continuously?

117)

118) When interest is compounded continuously, the balance in an account after t years isgiven by P(t) = P0ekt, where P0 is the initial investment and k is the interest rate.Suppose that P0 is invested in a savings account where interest is compoundedcontinuously at 8% per year. Express P(t) in terms of P0 and 0.08.

118)

Solve.119) How long will it take for the population of a certain country to double if its annual

growth rate is 6.3 %? (Round to the nearest year.)119)

120) There are currently 78 million cars in a certain country, decreasing by 1.5 % annually.How many years will it take for this country to have 65 million cars? (Round to thenearest year.)

120)

Solve the problem.121) The population growth of an animal species is described by F(t) = 400 + 50 log3 (2t + 1),

where t is measured in months. Find the population of this species in an area 4 month(s)after the species is introduced.

121)

122) Use the formula N = Iekt, where N is the number of items in terms of the initialpopulation I, at time t, and k is the growth constant equal to the percent of growth perunit of time. An artifact is discovered at a certain site. If it has 80% of the carbon-14 itoriginally contained, what is the approximate age of the artifact? (carbon-14 decays atthe rate of 0.0125% annually.)

122)

14

123) Use the formula N = Iekt, where N is the number of items in terms of the initialpopulation I, at time t, and k is the growth constant equal to the percent of growth perunit of time. A certain radioactive isotope decays at a rate of 0.3 % annually. Determinethe half-life of this isotope, to the nearest year.

123)

15

Answer KeyTestname: 125CH12P

1) (f ∘ g)(x) = 25x2 + 30x + 162) (g ∘ f)(x) = 24x2 + 36x + 44

3) (g ∘ f)(x) = 9x2

- 4

4) (f ∘ g)(x) = 2 7x + 17

5) (g ∘ f)(x) = 6x

6) 215

7) 228) 429) 2810) Yes11) No12) No13) No14) Yes15) Yes

16) g-1(x) = x2

17) h-1(x) = x + 96

18) Not a one-to-one function

19) h-1(x) = 4x

20) Not a one-to-one function

21) h-1(x) = 3x + 7

22) f-1(x) = x2 + 9, x ≥ 023)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

16

Answer KeyTestname: 125CH12P

24)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

25)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

26)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

17

Answer KeyTestname: 125CH12P

27)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

28) 1. (f-1 ∘ f)(x) = f-1(f(x)) = f-1(3x + 27) =

3x + 27

3 - 27 = x;

2. (f ∘ f-1)(x) = f(f-1(x)) = f(x3 - 27) = 3(x3 - 27) + 27= x

29) 1. (f-1 ∘ f)(x) = f-1(f(x)) = f-1(3x + 3) =

3x + 3

3 - 3 = x;

2. (f ∘ f-1)(x) = f(f-1(x)) = f(x3 - 3) = 3(x3 - 3) + 3= x

30) 1. (f-1 ∘ f)(x) = f-1(f(x)) = f-1 2 - xx

= 22 - xx

+ 1 = 22 - x + xx

= 22x

= 2x2

= x;

2. (f ∘ f-1)(x) = f(f-1(x)) = f 2x + 1

=

2 - 2x + 1

2x + 1

=

2(x + 1) - 2x + 1

2x + 1

=

2x + 2 - 2x + 1

2x + 1

2x2

= x

31) f-1(x) = 95x + 32

32)

x-4 -2 2 4 6

y

4

2

-2

-4

x-4 -2 2 4 6

y

4

2

-2

-4

18

Answer KeyTestname: 125CH12P

33)

x-4 -2 2 4 6

y6

4

2

-2

-4

-6

x-4 -2 2 4 6

y6

4

2

-2

-4

-6

34)

x-4 -2 2 4 6

y6

4

2

-2

-4

-6

x-4 -2 2 4 6

y6

4

2

-2

-4

-6

35)

x-4 -2 2 4 6

y

4

2

-2

-4

x-4 -2 2 4 6

y

4

2

-2

-4

19

Answer KeyTestname: 125CH12P

36)

x-4 -2 2 4 6

y

4

2

-2

-4

x-4 -2 2 4 6

y

4

2

-2

-4

37)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

38)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

39) 600 , 7540) 300 , 7541) 7.5 g42) -243) -244) 545) 246) 347) 7

20

Answer KeyTestname: 125CH12P

48)

x-4 -2 2 4 6

y

4

2

-2

-4

x-4 -2 2 4 6

y

4

2

-2

-4

49)

x-4 -2 2 4 6

y

4

2

-2

-4

x-4 -2 2 4 6

y

4

2

-2

-4

50)

x6

y

x6

y

21

Answer KeyTestname: 125CH12P

51) f(x) = 3x

x6

y

x6

y

f-1(x) = log3x

52) 3 t = 8153) 9 2 = 81

54) 3 -2 = 19

55) w12 = Q56) 2 = log

3 9

57) -2 = log 2

14

58) 0.8451 = log 10 7

59) z = log y 8

60) 261) -362) 2

63) 125

64) 181

65) 1666) log 2 16 + log 2 867) logr 4 + logr T

68) log 6 x + log 6 y69) logx 5 + logx y + logx z70) log 6 15671) log c m n72) -3 log c Z73) -7 log c Z74) logg M - logg 61

75) log a q - log a r

22

Answer KeyTestname: 125CH12P

76) log byz

77) log7 15 + 12

log7 m - log7 n

78) log4 9 + 12

log4 x - log4 y

79) 5logb m + 9logb p - 6logb n - 8

80) 2logb m + 4logb p - 6logb n - 3

81) 2logb x + 72logb y - 5

2logb z

82) 3logb x - 23logb y - 2logb z

83) 2logba + 3

84) loga y6x7/4

85) loga y7x11/3

86) logb 4x4y3

87) logw (x + 8)88) 3.68789) 1.0306

90) 72

91) 405992) 593) 994) Domain: ℛ

Range: (-∞, 0)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

23

Answer KeyTestname: 125CH12P

95) Domain: ℛRange: (1, ∞)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

96) -197) 1

98) - 257

99) -3, -1100) -3, 1101) No solution102) e

103) 169

104) No solution105) 6106) 12107) 9108) 12109) 13.7 yr110) 35.0 yr111) 9.1112) 1.26 × 10-6113) 145 dB114) 2.51 x 10-9 W/m2115) $4456.19116) 12.4 years117) 21.2 yr118) P(t) = P0e0.08t119) 11 years120) 12 years121) 500122) 1785 years123) 231 years

24