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Page 1: AP Calculus Summer Packet Answer Key - Weeblygoniocalc.weebly.com/uploads/2/8/7/2/28725613/ap...Find the domain and range of each function. Write your answer in interval notation

AP Calculus Summer

Packet Answer Key

AP Calculus AB students โ€“ Your packet stops at #98.

Page 2: AP Calculus Summer Packet Answer Key - Weeblygoniocalc.weebly.com/uploads/2/8/7/2/28725613/ap...Find the domain and range of each function. Write your answer in interval notation

1

Simplifying Complex Fractions

Simplify each of the following.

1. ๐‘ฅ3โˆ’9๐‘ฅ

๐‘ฅ2โˆ’7๐‘ฅ+12

๐‘ฅ(๐‘ฅ+3)

๐‘ฅโˆ’4 2.

๐‘ฅ2โˆ’2๐‘ฅโˆ’8

๐‘ฅ3+๐‘ฅ2โˆ’2๐‘ฅ

๐‘ฅโˆ’4

๐‘ฅ(๐‘ฅโˆ’1)

3. 1

๐‘ฅโˆ’

1

51

๐‘ฅ2โˆ’1

25

5๐‘ฅ

5+๐‘ฅ 4.

25

๐‘Žโˆ’๐‘Ž

5+๐‘Ž

5โˆ’๐‘Ž

๐‘Ž

Laws of Exponents

Write each of the following in the form ๐‘๐‘Ž๐‘๐‘๐‘ž where c, p, and q are constants (numbers).

5. (2๐‘Ž2)

2

๐‘ 4๐‘Ž4๐‘โˆ’1 6. โˆš9๐‘Ž๐‘33

91/3๐‘Ž1/3๐‘

7. ๐‘Ž๐‘โˆ’๐‘Ž

๐‘2โˆ’๐‘ ๐‘Ž๐‘โˆ’1 8.

๐‘Žโˆ’1

(๐‘โˆ’1)โˆš๐‘Ž ๐‘Žโˆ’3/2๐‘

9. (๐‘Ž2/3

๐‘1/2)

2

(๐‘3/2

๐‘Ž1/2) ๐‘Ž5/6๐‘1/2

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2

Laws of Logarithms

Simplify each of the following:

10. log2 5 + log2(๐‘ฅ2 โˆ’ 1) โˆ’ log2(๐‘ฅ โˆ’ 1) log2 5(๐‘ฅ + 1)

11. 32 log3 5 25 12. log101

10๐‘ฅ โˆ’๐‘ฅ

Solving Exponential and Logarithmic Equations

Solve for x. (DO NOT USE A CALCULATOR.)

13. 5(๐‘ฅ+1) = 25 14. 1

3= 32๐‘ฅ+2 15. log2 ๐‘ฅ2 = 3 16. log3 ๐‘ฅ2 = 2 log3 4 โˆ’ 4 log3 5

๐‘ฅ = 1 ๐‘ฅ = โˆ’3

2 ๐‘ฅ = 2โˆš2 ๐‘ฅ = ยฑ

4

25

Literal Equations

Solve for the indicated variables.

17. ๐‘‰ = 2(๐‘Ž๐‘ + ๐‘๐‘ + ๐‘๐‘Ž), ๐‘“๐‘œ๐‘Ÿ ๐‘Ž 18. ๐ด = ๐‘ƒ + ๐œ‹๐‘Ÿ๐‘ƒ, ๐‘“๐‘œ๐‘Ÿ ๐‘ƒ

๐‘Ž =๐‘‰โˆ’2๐‘๐‘

2๐‘+2๐‘ ๐‘ƒ =

๐ด

1+๐œ‹๐‘Ÿ

19. 2๐‘ฅ

4๐œ‹+

1โˆ’๐‘ฅ

2= 0, ๐‘“๐‘œ๐‘Ÿ ๐‘ฅ

๐‘ฅ =๐œ‹

๐œ‹โˆ’1

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3

Real Solutions

Find all real solutions.

20. ๐‘ฅ4 โˆ’ 1 = 0 21. ๐‘ฅ6 โˆ’ 16๐‘ฅ4 = 0

๐‘ฅ = ยฑ1 ๐‘ฅ = 0, ยฑ4

22. 4๐‘ฅ3 โˆ’ 8๐‘ฅ2 โˆ’ 25๐‘ฅ + 50 = 0

๐‘ฅ = ยฑ5

2, 2

Solving Equations

Solve the equations for x.

23. 4๐‘ฅ2 + 12๐‘ฅ + 3 = 0 24. 2๐‘ฅ + 1 =5

๐‘ฅ+2

๐‘ฅ =โˆ’3ยฑโˆš6

2 ๐‘ฅ =

1

2, โˆ’3

25. ๐‘ฅ+1

๐‘ฅโˆ’

๐‘ฅ

๐‘ฅ+1= 0

๐‘ฅ = โˆ’1

2

Polynomial Division

26. (๐‘ฅ5 โˆ’ 4๐‘ฅ4 + ๐‘ฅ3 โˆ’ 7๐‘ฅ + 1) รท (๐‘ฅ + 2)

๐‘ฅ4 โˆ’ 6๐‘ฅ3 + 13๐‘ฅ2 โˆ’ 26๐‘ฅ + 45 โˆ’89

๐‘ฅ+2

27. (๐‘ฅ6 + 2๐‘ฅ4 + 6๐‘ฅ โˆ’ 9) รท (๐‘ฅ3 + 3)

๐‘ฅ3 + 2๐‘ฅ โˆ’ 3

28. The equation 12๐‘ฅ3 โˆ’ 23๐‘ฅ2 โˆ’ 3๐‘ฅ + 2 = 0 has a solution ๐‘ฅ = 2. Find all other solutions.

๐‘ฅ =1

4, โˆ’

1

3

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4

Interval Notation

29. Complete the table with the appropriate notation or graph.

Solution Interval Notation Graph

โˆ’2 < ๐‘ฅ โ‰ค 4

(โˆ’2, 4]

-2 4

๐‘ฅ โ‰ค 8

(โˆ’โˆž, 8]

8

โˆ’1 โ‰ค ๐‘ฅ < 7

[โˆ’1, 7)

โˆ’1 7

Solving Inequalities

Solve the inequalities. Write the solution in interval notation.

30. ๐‘ฅ2 + 2๐‘ฅ โˆ’ 3 โ‰ค 0 31. 2๐‘ฅโˆ’1

3๐‘ฅโˆ’2โ‰ค 1 32.

2

2๐‘ฅ+3>

2

๐‘ฅโˆ’5

[โˆ’3, 1] (โˆ’โˆž,2

3) ๐‘ˆ[1, โˆž) (โˆ’โˆž, โˆ’8)๐‘ˆ (โˆ’

3

2, 5)

Solving Equations with Absolute Value

Solve for x. Give the solution for inequalities in interval notation.

33. |โˆ’๐‘ฅ + 4| โ‰ค 1 34. |5๐‘ฅ โˆ’ 2| = 8 35. |2๐‘ฅ + 1| > 3

[3, 5] ๐‘ฅ = โˆ’6

5, 2 (โˆ’โˆž, โˆ’2)๐‘ˆ(1, โˆž)

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5

Functions

Let ๐‘“(๐‘ฅ) = 2๐‘ฅ + 1 and ๐‘”(๐‘ฅ) = 2๐‘ฅ2 โˆ’ 1. Find each of the following.

36. ๐‘“(2) = 5 37. ๐‘”(โˆ’3) = 17 38. ๐‘“(๐‘ก + 1) = 2๐‘ก + 3

39. ๐‘“(๐‘”(โˆ’2)) = 15 40. ๐‘”(๐‘“(๐‘š + 2)) = 8๐‘š2 + 40๐‘š + 49

Let ๐‘“(๐‘ฅ) = ๐‘ฅ2, ๐‘”(๐‘ฅ) = 2๐‘ฅ + 5, and โ„Ž(๐‘ฅ) = ๐‘ฅ2 โˆ’ 1. Find each of the following.

41. โ„Ž(๐‘“(โˆ’2)) = 15

42. ๐‘“(๐‘”(๐‘ฅ โˆ’ 1)) = 4๐‘ฅ2 + 12๐‘ฅ + 9

43. ๐‘”(โ„Ž(๐‘ฅ3)) = 2๐‘ฅ6 + 3

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6

Intercepts and Points of Intersection

Find the ๐‘ฅ and ๐‘ฆ intercepts of each.

44. ๐‘ฆ = 2๐‘ฅ โˆ’ 5 45. ๐‘ฆ = ๐‘ฅ2 + ๐‘ฅ โˆ’ 2

x-int.: (5

2, 0) x-int.: (โˆ’2, 0) ๐‘Ž๐‘›๐‘‘ (1, 0)

y-int.: (0, โˆ’5) y-int.: (0, โˆ’2)

46. ๐‘ฆ = โˆš16 โˆ’ ๐‘ฅ2

x-int.: (โˆ’4, 0) ๐‘Ž๐‘›๐‘‘ (4, 0)

y-int.: (0, 4)

Domain and Range

Find the domain and range of each function. Write your answer in interval notation.

47. ๐‘“(๐‘ฅ) = ๐‘ฅ2 โˆ’ 5

Domain: (โˆ’โˆž, โˆž) Range: [โˆ’5, โˆž)

48. ๐‘“(๐‘ฅ) = 3 sin ๐‘ฅ

Domain: (โˆ’โˆž, โˆž) Range: [โˆ’3, 3]

49. ๐‘“(๐‘ฅ) =2

๐‘ฅโˆ’1

Domain: (โˆ’โˆž, 1)๐‘ˆ(1, โˆž) Range: (โˆ’โˆž, 0)๐‘ˆ(0, โˆž)

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7

Systems

Find the point(s) of intersection of the graphs for the given equations.

50. {๐‘ฅ + ๐‘ฆ = 8

4๐‘ฅ โˆ’ ๐‘ฆ = 7 51. {

๐‘ฅ2 + ๐‘ฆ = 6๐‘ฅ + ๐‘ฆ = 4

(3, 5) (โˆ’1, 5) ๐‘Ž๐‘›๐‘‘ (2, 2)

Inverses

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8

Find the inverse for each function.

52. ๐‘“(๐‘ฅ) = 2๐‘ฅ + 3 53. ๐‘“(๐‘ฅ) =๐‘ฅ2

3

๐‘ฆ =๐‘ฅโˆ’3

2 ๐‘ฆ = โˆš3๐‘ฅ

Prove ๐‘“ and ๐‘” are inverses of each other.

54. ๐‘“(๐‘ฅ) =๐‘ฅ3

2 ๐‘”(๐‘ฅ) = โˆš2๐‘ฅ

3

Simplify: ( โˆš2๐‘ฅ

3)

2

2 and โˆš2 (

๐‘ฅ3

2)

3

55. ๐‘“(๐‘ฅ) = 9 โˆ’ ๐‘ฅ2 ๐‘”(๐‘ฅ) = โˆš9 โˆ’ ๐‘ฅ

Simplify: 9 โˆ’ (โˆš9 โˆ’ ๐‘ฅ)2 and โˆš9 โˆ’ (9 โˆ’ ๐‘ฅ2)

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9

Vertical Asymptotes

Determine the vertical asymptotes for each function. Set the denominator equal to zero to find

the x-value for which the function is defined. This will be the vertical asymptote.

56. ๐‘“(๐‘ฅ) =1

๐‘ฅ2 57. ๐‘“(๐‘ฅ) =

๐‘ฅ2

๐‘ฅ2โˆ’4. 58. ๐‘“(๐‘ฅ) =

2+๐‘ฅ

๐‘ฅ2(1โˆ’๐‘ฅ)

๐‘ฅ = 0 ๐‘ฅ = 2, โˆ’2 ๐‘ฅ = 0, 1

Horizontal Asymptotes

Determine the horizontal asymptotes using the three cases below.

Case I: Degree of the numerator is less than the degree of the denominator. The asymptote is

๐‘ฆ = 0.

Case II: Degree of the numerator is the same as the degree of the denominator. The asymptote

is the ratio of the lead coefficients.

Case III: Degree of the numerator is greater than the degree of the denominator. There is no

horizontal asymptote. The function increases without bound. (If the degree of the numerator is

exactly 1 more than the degree of the denominator, then there exists a slant asymptote, which

is determined by long division.)

Determine all horizontal asymptotes.

59. ๐‘“(๐‘ฅ) =๐‘ฅ2โˆ’2๐‘ฅ+1

๐‘ฅ3+๐‘ฅโˆ’7 ๐‘ฆ = 0

60. ๐‘“(๐‘ฅ) =5๐‘ฅ3โˆ’2๐‘ฅ2+8

4๐‘ฅโˆ’3๐‘ฅ3+5 ๐‘ฆ = โˆ’

5

3

61. ๐‘“(๐‘ฅ) =4๐‘ฅ5

๐‘ฅ2โˆ’7 ๐‘๐‘œ โ„Ž๐‘œ๐‘Ÿ๐‘–๐‘ง๐‘œ๐‘›๐‘ก๐‘Ž๐‘™ ๐‘Ž๐‘ ๐‘ฆ๐‘š๐‘๐‘ก๐‘œ๐‘ก๐‘’

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10

Equation of a Line

Slope Intercept Form: ๐‘ฆ = ๐‘š๐‘ฅ + ๐‘ Vertical Line: ๐‘ฅ = ๐‘ (slope is undefined)

Point-slope Form: ๐‘ฆ โˆ’ ๐‘ฆ1 = ๐‘š(๐‘ฅ โˆ’ ๐‘ฅ1) Horizontal Line: ๐‘ฆ = ๐‘ (slope is 0)

62. Use slope-intercept form to find the equation of the line having slope of 3 and a y-intercept

of 5.

๐‘ฆ = 3๐‘ฅ + 5

63. Determine the equation of a line passing through the point (5, โˆ’3) with an undefined

slope.

๐‘ฅ = 5

64. Determine the equation of a line passing through the point (โˆ’4, 2) with a slope of 0.

๐‘ฆ = 2

65. Use point-slope form to find the equation of a line passing through the point (0, 5) with a

slope of 2/3.

๐‘ฆ โˆ’ 5 =2

3๐‘ฅ

66. Find the equation of a line passing through the point (6, 8) and parallel to the line

๐‘ฆ =5

6๐‘ฅ โˆ’ 1.

๐‘ฆ =5

6๐‘ฅ + 3

67. Find the equation of a line passing through points (โˆ’3, 6) and (1, 2).

๐‘ฆ โˆ’ 2 = โˆ’1(๐‘ฅ โˆ’ 1)

๐‘œ๐‘Ÿ ๐‘ฆ โˆ’ 6 = โˆ’1(๐‘ฅ + 3) ๐‘œ๐‘Ÿ ๐‘ฆ = โˆ’1๐‘ฅ + 3

68. Find the equation of a line with an x-intercept of (2, 0) and a y-intercept (0, 3).

๐‘ฆ = โˆ’3

2๐‘ฅ + 3

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11

Parent Functions

For 69 โ€“ 78, identify the parent function associated with each graph.

69. 70.

๐‘ฆ = ๐‘ฅ ๐‘ฆ = ๐‘ฅ2

71. 72.

๐‘ฆ = โˆš๐‘ฅ ๐‘ฆ = ๐‘ฅ3

73. 74.

๐‘ฆ = โˆš๐‘ฅ3

๐‘ฆ = ln ๐‘ฅ

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12

75. 76.

๐‘ฆ = ๐‘’๐‘ฅ ๐‘ฆ = |๐‘ฅ|

77. 78.

๐‘ฆ =1

๐‘ฅ ๐‘ฆ =

1

๐‘ฅ2

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13

Unit Circle

79. Identify all parts of the unit circle, including degree, radian, and coordinates of each point.

Without using a calculator, evaluate the following.

80. a) sin 180ยฐ b) cos 270ยฐ c) sin ๐œ‹ d) cos(โˆ’๐œ‹)

0 0 0 โˆ’1

e) sin5๐œ‹

4 f) cos

9๐œ‹

4 g) tan

7๐œ‹

6

โˆ’โˆš2

2

โˆš2

2 โˆ’

1

โˆš3

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14

Inverse Trigonometric Functions

For each of the following, find the value in radians.

81. ๐‘ฆ = sinโˆ’1 โˆ’โˆš3

2 โˆ’

๐œ‹

3

82. ๐‘ฆ = arccos(โˆ’1) ๐œ‹

83. ๐‘ฆ = tanโˆ’1(โˆ’1) โˆ’๐œ‹

4

84. ๐‘ฆ = cosโˆ’1 (sin (โˆ’๐œ‹

4))

3๐œ‹

4

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For each of the following give the value without a calculator.

85. tan (arccos2

3)

โˆš5

2 86. sec (sinโˆ’1 12

13)

13

5

87. sin (arcsin7

8)

7

8

Trigonometric Equations

Solve each of the equations for 0 โ‰ค ๐‘ฅ < 2๐œ‹. Isolate the variable and find all the solutions within

the given domain. Remember to double the domain when solving for a double angle. Use trig

identities, or rewrite the trig functions using substitution, if needed.

88. sin ๐‘ฅ = โˆ’1

2 89. 2 cos ๐‘ฅ = โˆš3

๐‘ฅ =7๐œ‹

6,

11๐œ‹

6 ๐‘ฅ =

๐œ‹

6,

11๐œ‹

6

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16

90. sin2 ๐‘ฅ =1

2 91. sin 2๐‘ฅ = โˆ’

โˆš3

2

๐‘ฅ =๐œ‹

4,

3๐œ‹

4 ๐‘ฅ =

2๐œ‹

3,

5๐œ‹

6

92. 2 cos2 ๐‘ฅ โˆ’ 1 โˆ’ cos ๐‘ฅ = 0 93. 4 cos2 ๐‘ฅ โˆ’ 3 = 0

๐‘ฅ = 0,2๐œ‹

3,

4๐œ‹

3 ๐‘ฅ =

๐œ‹

6,

11๐œ‹

6

94. Find the ratio of the area inside the square but outside the circle to the area of the square in

the picture below.

1 โˆ’๐œ‹

4

95. Find the formula for the perimeter of the window of the shape in the picture below.

4๐‘Ÿ + ๐œ‹๐‘Ÿ

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17

96. A water tank has the shape of a cone (where ๐‘‰ =๐œ‹

3๐‘Ÿ2โ„Ž). The tank is 10 ๐‘š high and has a

radius of 3 ๐‘š at the top. When the water is 5 ๐‘š deep (in the middle of the tank) what is the

surface area of the top of the water?

9๐œ‹

4

97. Two cars start moving from the same point. One travels south at 100 ๐‘˜๐‘š/โ„Ž๐‘Ÿ, the other

west at 50 ๐‘˜๐‘š/โ„Ž๐‘Ÿ. How far apart are they two hours later?

100โˆš5 ๐‘˜๐‘š

98. A kite is 100 ๐‘š above ground. If there is 200 ๐‘š of string connecting the kite to the

horizontal, what is the angle between the string and the horizontal? (Assume that the string is

perfectly straight.)

๐œ‹

6 ๐‘œ๐‘Ÿ 30ยฐ

Limits

Using the graphs, find the following limits.

99. lim๐‘ฅโ†’3+ ๐‘“(๐‘ฅ) = โˆž 100. lim๐‘ฅโ†’โˆž ๐‘“(๐‘ฅ) = 2 101. lim๐‘ฅโ†’3 ๐‘“(๐‘ฅ) = ๐ท๐‘๐ธ

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18

102. lim๐‘ฅโ†’2 ๐‘“(๐‘ฅ) = โˆ’3 103. ๐‘“(2) = 0

Find ๐‘“(๐‘ฅ+โ„Ž)โˆ’๐‘“(๐‘ฅ)

โ„Ž for the given function ๐‘“.

104. ๐‘“(๐‘ฅ) = 2๐‘ฅ + 3

2

105. ๐‘“(๐‘ฅ) = 3๐‘ฅ2 โˆ’ ๐‘ฅ + 5

3โ„Ž + 6๐‘ฅ โˆ’ 1

106. ๐‘“(๐‘ฅ) = 5 โˆ’ 2๐‘ฅ

โˆ’2