AP Calculus Summer
Packet Answer Key
AP Calculus AB students β Your packet stops at #98.
1
Simplifying Complex Fractions
Simplify each of the following.
1. π₯3β9π₯
π₯2β7π₯+12
π₯(π₯+3)
π₯β4 2.
π₯2β2π₯β8
π₯3+π₯2β2π₯
π₯β4
π₯(π₯β1)
3. 1
π₯β
1
51
π₯2β1
25
5π₯
5+π₯ 4.
25
πβπ
5+π
5βπ
π
Laws of Exponents
Write each of the following in the form πππππ where c, p, and q are constants (numbers).
5. (2π2)
2
π 4π4πβ1 6. β9ππ33
91/3π1/3π
7. ππβπ
π2βπ ππβ1 8.
πβ1
(πβ1)βπ πβ3/2π
9. (π2/3
π1/2)
2
(π3/2
π1/2) π5/6π1/2
2
Laws of Logarithms
Simplify each of the following:
10. log2 5 + log2(π₯2 β 1) β log2(π₯ β 1) log2 5(π₯ + 1)
11. 32 log3 5 25 12. log101
10π₯ βπ₯
Solving Exponential and Logarithmic Equations
Solve for x. (DO NOT USE A CALCULATOR.)
13. 5(π₯+1) = 25 14. 1
3= 32π₯+2 15. log2 π₯2 = 3 16. log3 π₯2 = 2 log3 4 β 4 log3 5
π₯ = 1 π₯ = β3
2 π₯ = 2β2 π₯ = Β±
4
25
Literal Equations
Solve for the indicated variables.
17. π = 2(ππ + ππ + ππ), πππ π 18. π΄ = π + πππ, πππ π
π =πβ2ππ
2π+2π π =
π΄
1+ππ
19. 2π₯
4π+
1βπ₯
2= 0, πππ π₯
π₯ =π
πβ1
3
Real Solutions
Find all real solutions.
20. π₯4 β 1 = 0 21. π₯6 β 16π₯4 = 0
π₯ = Β±1 π₯ = 0, Β±4
22. 4π₯3 β 8π₯2 β 25π₯ + 50 = 0
π₯ = Β±5
2, 2
Solving Equations
Solve the equations for x.
23. 4π₯2 + 12π₯ + 3 = 0 24. 2π₯ + 1 =5
π₯+2
π₯ =β3Β±β6
2 π₯ =
1
2, β3
25. π₯+1
π₯β
π₯
π₯+1= 0
π₯ = β1
2
Polynomial Division
26. (π₯5 β 4π₯4 + π₯3 β 7π₯ + 1) Γ· (π₯ + 2)
π₯4 β 6π₯3 + 13π₯2 β 26π₯ + 45 β89
π₯+2
27. (π₯6 + 2π₯4 + 6π₯ β 9) Γ· (π₯3 + 3)
π₯3 + 2π₯ β 3
28. The equation 12π₯3 β 23π₯2 β 3π₯ + 2 = 0 has a solution π₯ = 2. Find all other solutions.
π₯ =1
4, β
1
3
4
Interval Notation
29. Complete the table with the appropriate notation or graph.
Solution Interval Notation Graph
β2 < π₯ β€ 4
(β2, 4]
-2 4
π₯ β€ 8
(ββ, 8]
8
β1 β€ π₯ < 7
[β1, 7)
β1 7
Solving Inequalities
Solve the inequalities. Write the solution in interval notation.
30. π₯2 + 2π₯ β 3 β€ 0 31. 2π₯β1
3π₯β2β€ 1 32.
2
2π₯+3>
2
π₯β5
[β3, 1] (ββ,2
3) π[1, β) (ββ, β8)π (β
3
2, 5)
Solving Equations with Absolute Value
Solve for x. Give the solution for inequalities in interval notation.
33. |βπ₯ + 4| β€ 1 34. |5π₯ β 2| = 8 35. |2π₯ + 1| > 3
[3, 5] π₯ = β6
5, 2 (ββ, β2)π(1, β)
5
Functions
Let π(π₯) = 2π₯ + 1 and π(π₯) = 2π₯2 β 1. Find each of the following.
36. π(2) = 5 37. π(β3) = 17 38. π(π‘ + 1) = 2π‘ + 3
39. π(π(β2)) = 15 40. π(π(π + 2)) = 8π2 + 40π + 49
Let π(π₯) = π₯2, π(π₯) = 2π₯ + 5, and β(π₯) = π₯2 β 1. Find each of the following.
41. β(π(β2)) = 15
42. π(π(π₯ β 1)) = 4π₯2 + 12π₯ + 9
43. π(β(π₯3)) = 2π₯6 + 3
6
Intercepts and Points of Intersection
Find the π₯ and π¦ intercepts of each.
44. π¦ = 2π₯ β 5 45. π¦ = π₯2 + π₯ β 2
x-int.: (5
2, 0) x-int.: (β2, 0) πππ (1, 0)
y-int.: (0, β5) y-int.: (0, β2)
46. π¦ = β16 β π₯2
x-int.: (β4, 0) πππ (4, 0)
y-int.: (0, 4)
Domain and Range
Find the domain and range of each function. Write your answer in interval notation.
47. π(π₯) = π₯2 β 5
Domain: (ββ, β) Range: [β5, β)
48. π(π₯) = 3 sin π₯
Domain: (ββ, β) Range: [β3, 3]
49. π(π₯) =2
π₯β1
Domain: (ββ, 1)π(1, β) Range: (ββ, 0)π(0, β)
7
Systems
Find the point(s) of intersection of the graphs for the given equations.
50. {π₯ + π¦ = 8
4π₯ β π¦ = 7 51. {
π₯2 + π¦ = 6π₯ + π¦ = 4
(3, 5) (β1, 5) πππ (2, 2)
Inverses
8
Find the inverse for each function.
52. π(π₯) = 2π₯ + 3 53. π(π₯) =π₯2
3
π¦ =π₯β3
2 π¦ = β3π₯
Prove π and π are inverses of each other.
54. π(π₯) =π₯3
2 π(π₯) = β2π₯
3
Simplify: ( β2π₯
3)
2
2 and β2 (
π₯3
2)
3
55. π(π₯) = 9 β π₯2 π(π₯) = β9 β π₯
Simplify: 9 β (β9 β π₯)2 and β9 β (9 β π₯2)
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. π(π₯) =1
π₯2 57. π(π₯) =
π₯2
π₯2β4. 58. π(π₯) =
2+π₯
π₯2(1βπ₯)
π₯ = 0 π₯ = 2, β2 π₯ = 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
π¦ = 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. π(π₯) =π₯2β2π₯+1
π₯3+π₯β7 π¦ = 0
60. π(π₯) =5π₯3β2π₯2+8
4π₯β3π₯3+5 π¦ = β
5
3
61. π(π₯) =4π₯5
π₯2β7 ππ βππππ§πππ‘ππ ππ π¦πππ‘ππ‘π
10
Equation of a Line
Slope Intercept Form: π¦ = ππ₯ + π Vertical Line: π₯ = π (slope is undefined)
Point-slope Form: π¦ β π¦1 = π(π₯ β π₯1) Horizontal Line: π¦ = π (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.
π¦ = 3π₯ + 5
63. Determine the equation of a line passing through the point (5, β3) with an undefined
slope.
π₯ = 5
64. Determine the equation of a line passing through the point (β4, 2) with a slope of 0.
π¦ = 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.
π¦ β 5 =2
3π₯
66. Find the equation of a line passing through the point (6, 8) and parallel to the line
π¦ =5
6π₯ β 1.
π¦ =5
6π₯ + 3
67. Find the equation of a line passing through points (β3, 6) and (1, 2).
π¦ β 2 = β1(π₯ β 1)
ππ π¦ β 6 = β1(π₯ + 3) ππ π¦ = β1π₯ + 3
68. Find the equation of a line with an x-intercept of (2, 0) and a y-intercept (0, 3).
π¦ = β3
2π₯ + 3
11
Parent Functions
For 69 β 78, identify the parent function associated with each graph.
69. 70.
π¦ = π₯ π¦ = π₯2
71. 72.
π¦ = βπ₯ π¦ = π₯3
73. 74.
π¦ = βπ₯3
π¦ = ln π₯
12
75. 76.
π¦ = ππ₯ π¦ = |π₯|
77. 78.
π¦ =1
π₯ π¦ =
1
π₯2
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 180Β° b) cos 270Β° c) sin π d) cos(βπ)
0 0 0 β1
e) sin5π
4 f) cos
9π
4 g) tan
7π
6
ββ2
2
β2
2 β
1
β3
14
Inverse Trigonometric Functions
For each of the following, find the value in radians.
81. π¦ = sinβ1 ββ3
2 β
π
3
82. π¦ = arccos(β1) π
83. π¦ = tanβ1(β1) βπ
4
84. π¦ = cosβ1 (sin (βπ
4))
3π
4
15
For each of the following give the value without a calculator.
85. tan (arccos2
3)
β5
2 86. sec (sinβ1 12
13)
13
5
87. sin (arcsin7
8)
7
8
Trigonometric Equations
Solve each of the equations for 0 β€ π₯ < 2π. 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 π₯ = β1
2 89. 2 cos π₯ = β3
π₯ =7π
6,
11π
6 π₯ =
π
6,
11π
6
16
90. sin2 π₯ =1
2 91. sin 2π₯ = β
β3
2
π₯ =π
4,
3π
4 π₯ =
2π
3,
5π
6
92. 2 cos2 π₯ β 1 β cos π₯ = 0 93. 4 cos2 π₯ β 3 = 0
π₯ = 0,2π
3,
4π
3 π₯ =
π
6,
11π
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 βπ
4
95. Find the formula for the perimeter of the window of the shape in the picture below.
4π + ππ
17
96. A water tank has the shape of a cone (where π =π
3π2β). The tank is 10 π high and has a
radius of 3 π at the top. When the water is 5 π deep (in the middle of the tank) what is the
surface area of the top of the water?
9π
4
97. Two cars start moving from the same point. One travels south at 100 ππ/βπ, the other
west at 50 ππ/βπ. How far apart are they two hours later?
100β5 ππ
98. A kite is 100 π above ground. If there is 200 π 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.)
π
6 ππ 30Β°
Limits
Using the graphs, find the following limits.
99. limπ₯β3+ π(π₯) = β 100. limπ₯ββ π(π₯) = 2 101. limπ₯β3 π(π₯) = π·ππΈ
18
102. limπ₯β2 π(π₯) = β3 103. π(2) = 0
Find π(π₯+β)βπ(π₯)
β for the given function π.
104. π(π₯) = 2π₯ + 3
2
105. π(π₯) = 3π₯2 β π₯ + 5
3β + 6π₯ β 1
106. π(π₯) = 5 β 2π₯
β2