# 20 Pre-Requisites for Calculus

AP Calculus Summer Project Solutions

1. Show all work on separate paper.
2. Write your name in pen at the top of each page.
3. Use graph paper for all graphs. Do all examples.
4. Prepare to be tested on all topics on the first day of class.
5. If you have any questions during the summer, email the instructor at w at least 48 hours for a response.

Topic: Slope, Lines, and Linear Equations

1. Find slope from two given points. Example: (4 , –7) and (–5 , 8)

Slope (m) of the line through (x1,y1) and (x2,y2) is m =

Slope of the line through (4 , –7) and (–5 , 8) is m = = =

2. Write the slope and an equation for a horizontal line through (4, 7). m = 0
y = 7

3. Write the slope and an equation for a vertical line through (-3 , 5).

no slope or undefined slope ; equation: x = –3

Point-Slope Form: y – y1 = m(x – x1)

4. Use point-slope form to write an equation for the line through (6 , –5) with

slope 2.

y – (–5) = 2(x – 6)

y + 5 = 2x – 12

y = 2x – 17

5. Use point-slope form to write an equation for the line through (4 , –7) and (–5 , 8).

First find the slope: m =

Then choose either point and use point-slope form: y – y1 = m(x – x1)

y – 8 = ( x – (–5) )

y – 8 = ( x + 5)

y – 8 =

y = or 5x + 3y = –1

6. Convert from point slope form to slope-intercept form: y – 7 =

Distribute and simplify: y – 7 =

y =

7. Define and illustrate a tangent line.

A tangent line is a line that touches a curve at exactly one point.

8. Define and illustrate a secant line.

A secant line is a line that crosses a curve at two points.

9. Define “average rate of change” algebraically and graphically.

The ARC on [ a , b ] is the slope of the secant line through (a , f (a)) and (b , f (b)).

Algebraically: ARC = or

Geometrically: Average rate of change is slope of a secant line.

10. Define “instantaneous rate of change” algebraically and graphically.

Algebraically: IRC is the derivative or the limit as h approaches zero of the DQ.

Geometrically: IRC is the instantaneous slope of a tangent line

Do the following AP released free response questions found online at ap central.

1. 2010 FRQ #2(a)ARC = hundred entries per hour
2. 2011 FRQ #2(a)ARC@ 3.5 = ° Celsius/minute

13. 2008 FRQ #2(a)ARC@(5.5) =

14. 2005 FRQ #3(a)ARC@(7) = ° Celsius/minute

Topic: Functions

Graph the 15 parent functions listed below. Use graph paper. State the domain and

range for each.

DomainRange

15. f (x) = x2all Ry ≥ 0

16. f (x) = x3all Rall R

17. f (x) =x ≥ 0y ≥ 0

18. f (x) = all Rall R

19. f (x) = |x|all Ry ≥ 0

20. f (x) = [x]all Rall integers

21. f (x) = sin(x)all R[ –1 , 1 ]

22. f (x) = cos(x)all R[ –1 , 1 ]

23.f (x) = tan(x)all R except odd multiples of all R

24. f (x) =all R except x = 0all R except y = 0

Topic: Semi-Circles

25. Write the general function for a semi-circle: f (x) = , where r is the radius.

26. Give examples of three semi-circle functions and graphs. (Answers may vary.)

y = ; y = ; y = ; y =

Be able to instantly recognize the equation and graph of a semi-circle!

Topic: Special Right Triangles

1. Describe and illustrate the ratios for the sides of a 30º–60º–90ºtriangle.

In a 30º-60º-90º triangle, the hypotenuse is double the length of the short side

and the longer leg is times the length of the short side.

The ratio of the sides is 1 : 2 :(short leg : hypotenuse : long leg)

1. Describe and illustrate the ratios for the sides of a 45º–45º–90º right triangle.

In a 45º-45º-90º triangle, the hypotenuse is times the length of either leg.

The ratio of the sides is 1 : 1 :(leg : leg : hypotenuse)

1. Label all three sides for 30º–60º–90º triangles with hypotenuse of length

1 , 2 , 5 , 8 , 10, and x.

Short Leg / Long Leg / Hypotenuse
/ / 1
1 / / 2
/ / 5
4 / 4 / 8
5 / 5 / 10
/ / x

30. Label all three sides for 45º–45º–90º triangles with hypotenuse of length

1 , 2 , 5 , 8 , 10, and x.

Leg / Hypotenuse
or / 1
/ 2
or / 5
4 / 8
10 / 10
or / x

Topic: Trigonometric Ratios

31. Know sine, cosine, and tangent of the following angles to automaticity.

θ =

If necessary, make 48 flashcards. Identify all 48 trigonometric ratios from randomly-ordered flashcards to 99% accuracy in four minutes or less.

Prove these seven identities and their corollaries and memorize the results.

32. Pythagorean Identity #1: sin2θ + cos2θ = 1

Proof: From a right triangle diagram with acute angle θ, adjacent leg labeled x,

opposite leg labeled y, and hypotenuse h:

x2 + y2 = h2 by the Pythagorean Theorem

Divide through by h2:

cos2θ + sin2θ = 1 or sin2θ+ cos2θ = 1

Remember: “This sign costs one dollar.”

33. Pythagorean Identity #2: tan2θ + 1 = sec2θ

sin2θ + cos2θ= 1

Divide through by sin2θ:

Remember: “One vampire got caught in a casket.” or

You cannot buy just “one cotton ball at Costco”.

34. Pythagorean Identity #3: 1 + cot2θ = csc2θ

sin2θ +cos2θ = 1

Divide through by cos2θ:

Remember: You should only “tan for one second”.

Note: You need to prove the addition and subtraction identities before the double angle identities. I use a geometric proof starting with a rectangle ABCD. Right triangle ∆AED is inscribed in the rectangle with right angle AEF and hypotenuse AF = 1.

35. Sine Addition and Sine Subtraction: sin(α ± β) = sin α cos β ± cos α sin β

36. Cosine Addition and Cosine Subtraction: cos(α ± β) = cos α cos β ± sin α sin β

Next we can prove these:

37. Sine Double Angle Identity: sin 2θ = 2sinθ cosθ

= sin Ө cos Ө + cos Ө sin Ө

= 2 sin Ө cos Ө

38. Cosine Double Angle Identities

(a) cos 2θ = cos2θ – sin2θ

= cosӨ cos Ө – sin Ө sin Ө

= cos2Ө – sin2Ө

(b) cos 2θ = 2cos2θ – 1

Use the first Pythagorean Identity: sin2Ө = 1 – cos2Ө

Substitute: cos 2Ө = cos2Ө – (1 – cos2Ө)

cos 2Ө = cos2Ө – 1 + cos2Ө

cos 2Ө = 2cos2Ө – 1

(c): cos 2θ = 1 – 2sin2θ

Use the second Pythagorean Identity: cos2Ө = 1 – sin2Ө

Substitute: cos 2Ө = (1 – sin2Ө) – sin2Ө

cos 2Ө = 1 – 2sin2Ө

Topic: Sinusoidal Waves

Know the amplitude, period, horizontal shift, and vertical shift for trigonometric functions in the following forms. f (x) = a sin b(x – c) + d and f (x) = a cos b(x – c) + d

Graph and label these sinusoidal waves. Use a separate graph for each.

39. f (x) = 3 sin 2 + 1 40.f (x) = –2 cos 3 – 3

Topic: Absolute Value Functions

41. Define the absolute value function, f (x) = | x | as a piecewise function.

Algebraically: | x | = x when x ≥ 0 and | x | = –x when x < 0.

Graphically: | x | is the distance from 0 to x on the number line.

Graph and label the following absolute value functions on four separate graphs.

42.f (x) = | x + 3 | – 5 V-shape with vertex at ( –3 , –5 )

43. f (x) = 2| x | – 4 Skinny V-shape with vertex at ( 0 , –4 )

44. f (x) = –| x – 2 | + 1 A-shape with vertex at ( 2 , 1 )

45. f (x) = | 3 – x |V-shape with vertex at ( 3 , 0 )

Topic: Greatest Integer Function

46. Define the greatest integer function.

[[ x ]] is the greatest integer that is less than or equal to x. It is a rounding DOWN function.

Graph and label the following on four separate graphs. These are all “stair step” functions with an open circle on the left and a closed circle on the right.

47. f (x) = 2[x]49. f (x) = [x + 3]

48. f (x) = –[x]50. f (x) = [0.5x]

Topic: Piecewise Functions Graph the following piecewise functions.

51. f (x) =52. f (x) =

53. f (x) =

Topic: Exponential Functions

Graph. Label the y-intercept, the horizontal asymptote, and one anchor point.

54. f (x) = 2xincreasing for all Ry = 0all Ry > 0(0 , 1)

55. f (x) = exincreasing for all Ry = 0all Ry > 0(0 , 1)

56. f (x) = 10xincreasing for all R y = 0all Ry > 0(0 , 1)

57. f (x) = ex+2– 4 same as ex shifted y = –4all Ry > –4(0, –3)

left 2 and down 4

58. f (x) = –5xdecreasing for all Ry = 0all Ry < 0(0 , –1)

59. f (x) = 4x + 3increasing for all Ry = 3all Ry > 3(0 , 4)

60. f (x) = 10x+2 + 3same as 10x shifted

left 2 and up 3y = 3all Ry > 3(0 , 4)

31. f (x) = log(x – 3) same as log(x) shifted right 3x > 3all R

32. f (x) = log(x + 5) – 1 same as ln(x) shifted left 5 and down 1x > –5all

Topic: Logarithmic Functions

61. Graph f (x) = ln(x) and g(x) = ex on one graph.

Label the x- and y-intercepts and one anchor point on each graph.

( 1 , 0 ) and (e , 1) are on f (x) = ln(x)

( 0 , 1 ) and (1 , e) are on g(x) = ex

The y-intercept for g(x) = exis y = 1.

The x-intercept for f (x) = ln(x) is ( 1 , 0 ).

62. Graph f (x) = log(x) and g(x) = 10x on one graph.

Label the x- and y-intercepts and one anchor point on each graph.

The y-intercept for g(x) = 10xis y = 1.

( 0 , 1 ) and (1 , 10) are on g(x) = 10x

The x-intercept for f (x) = log(x) is ( 1 , 0 ).

( 1 , 0 ) and (10 , 1) are on f (x) = log(x)

Topic: Transformations of Logarithmic Functions

Graph. Label the x-intercept, the vertical asymptote (VA) and one anchor point. Use four separate graphs.

1. f (x) = ln(x – 1) + 2VA: x = 1Anchor Point: ( 2 , –3)
1. f (x) = ln(x – 3) + 5VA: x = 3Anchor Point: ( 4 , 5 )
2. f (x) = ln(x + 2) + 4VA: x = –2Anchor Point: (–1 , 4)
1. f (x) = ln(x + 4) – 2VA: x = –4Anchor Point: ( –3 , –2)

Topic: Solving Logarithmic Equations

67. ln(x – 2) + 4 = ln x

4 = ln x – ln(x – 2)

4 = ln

e4 =

e4 (x – 2) = x

xe4 – 2e4 = x

xe4 – x = 2e4

x( e4 – 1 ) = 2e4

x =

68. 3(x+1) = 7(x–2)

ln 3(x+1) = ln 7(x–2)

( x + 1) ln 3 = ( x – 2 ) ln 7

x ln 3 + ln 3 = x ln 7 – 2 ln 7

ln 3 + 2 ln 7 = x ln 7 – x ln 3

ln 3 + 2 ln 7 = x ( ln 7 – ln 3 )

= x

= x

= x

69. log5(x–2) = log5x + log57

log5(x–2) – log5x = log57

log5 = log57

= 7

x – 2 = 7x

–2 = 6x

= x

70. ex(ex–4 ) = 1

ex–4 =

ex–4= e-x

x – 4 = –x

–4 = –2x

2 = x

Topic: Inverse Functions

1. What is inverse function notation for f (x)? f -1(x)
2. How do you find the inverse of f (x) algebraically? Switch x and y, then solve for x.

73. How do you find the inverse of f (x) graphically?

Reflect the graph across the diagonal line y = x.

74. Give 3 examples of points on a function and points on the inverse.

( 5 , –2 ) (–2 , 5 )

( 3 , 4 ) ( 4 , 3 )

(–7 , –10 ) (–10 , –7 )

75. If f (x) and g(x) are inverses, what is f (g(x))? What is g(f (x))?

f (g(x)) = x

g(f (x)) = x

76. Give 3 examples of slopes of functions and their inverses at corresponding points.

Slope of Function at ( x , y ) / Slope of Inverse at ( y , x )
5 / 1/5
–3 / –⅓
½ / 2

77. Inverse functions are reflections across what line? y = x

78. The slopes of inverse functions are reciprocals of each other at corresponding points reflected across the line y = x.

Topic: Polynomials and Rational Expressions

79. X-Intercepts occur when the numerator of a rational expression equals zero.

80. Vertical Asymptotes occur when the denominator of a rational expression equals zero.

81. Horizontal Asymptotes occur when:

the degree of the numerator ≤ the degree of the denominator.

82. Holes occur when there is a:

zero of both the numerator and denominator with multiplicity ≥ in the numerator.

Non-Vertical Asymptotes:In a rational expression, let n = the degree of the numerator and

d = the degree of the denominator. Describe the end behavior.

83. If n < d, then the horizontal asymptote is y = 0 (the x-axis).

84. If n = d, then the horizontal asymptote is y =

where LCON is the leading coefficient of the numerator and

LCOD is the leading coefficient of the denominator.

85. If n > d, and n = d + 1, then the non-vertical asymptote is a line.

86. If n > d, and n = d + 2, then the non-vertical asymptote is a parabola.

87. If n > d, and n = d + 3, then the non-vertical asymptote is a cubic.

Find the x-intercepts, vertical asymptotes, horizontal asymptotes, and holes.

Graph each function separately on graph paper. Describe the end behavior.

Optional: Find and .

88.

HA: y = 0

VA: none

Holes: none

x-intercepts: x = 0 and x = –7

= 0 and = 0

89. =

HA: none, but there is a parabolic asymptote at approx. y = –2

VA: none

Holes: x = 0

x-intercepts: none

= -2and = -2

90. =

HA: none, but there is a parabolic asymptote at approx. y = x2

VA: none

Holes: x = 0

x-intercepts: x =

= ∞and = ∞

91. =

HA: y = 1

VA: none

Holes: none

x-intercepts: x = 4 and x = –3

= 1 and = 1

Topic: Area Formulas for Basic Geometric Shapes

92. Area of a Square with side s.A = s2

93. Area of a Semi-Circle with diameter DA = or A =

94. Area of an Isosceles Right Triangle with leg x.A =

95. Area of an Isosceles Right Triangle with Hypotenuse h. A =

96. Area of a TrapezoidA =

97. Area of a Rectangle with width x and length equal to three times the width. A = 3x2

98. Area of an Equilateral Triangle with side s. A =

Do the following AP released free response questions at

99. 2011B #6: Find the area of the triangle on [–2π , 4π].A = 6π2

100. 2011 #4: Find the area between the graph and the x-axis on the intervals

(a) [–4 , –3]:A = units2

(b) [–3 , 0]:A = units2

(c) [0 , 1.5]:A = units2

101. 2010 #5: Find the area between the graph and the x-axis on the intervals

[–7, –2]:A = units2

[–2 , 2]: A = units2

[2 , 4.5]:A = units2

[4.5, 5]:A = units2

102. 2010B #4: Find the area of the three trapezoids on [0 , 18].

1st Trapezoid on [0 , 9]:A = = 140 units2

2nd Trapezoid on [9 , 15]:A = = 50 units2

3rd Trapezoid on [15 , 18]:A = = 25 units2

Topic: Difference Quotient and Derivative

1. Find the difference quotient.
2. Find the derivative, f '(x). Use the limit of the difference quotient as h approaches 0.

103. f (x) = –x – 4

DQ =

DQ =

DQ =

DQ = = –1f '(x) =

104. f (x) = x2

DQ =

DQ =

DQ =

DQ =

DQ = 2x + hf '(x) =

105. f (x) = x3 + x

DQ =

DQ =

DQ =

DQ =

DQ = f '(x) =

DQ = 3x2 + 3xh + h2 + 1f '(x) = 3x2 + 1

106. f (x) = 4x2 – 3x – 2

DQ =

DQ =

DQ =

DQ =

DQ = f '(x) =

DQ = 8x + 4h – 3 f '(x) = 8x – 3

107. f (x) =

DQ =

DQ =

DQ = .

DQ = f '(x) =

DQ = f '(x) =

DQ = f '(x) =

108. f (x) =

DQ =

DQ =

DQ =

DQ =

DQ = f '(x) =

DQ = f '(x) =

109. f (x) = sin(x)

DQ =

DQ =

DQ = You can STOP here.

We will find the derivative, f '(x) = cos(x) soon.

110. f (x) = cos(x)

DQ =

DQ =

DQ = You can STOP here.

We will find the derivative, f '(x) = –sin(x) soon.