AP Calculus BC Summer Assignment Name:_______________ Due Date: First day of school. The purpose of this assignment is to have you practice the mathematical skills necessary to be successful in Calculus BC. All of the skills covered in this packet are skills from Algebra 2/Trigonometry and Pre-Calculus. Each question was carefully selected and every one of the questions in this packet is extremely important, it is necessary that you master them. I will assume you know all the material covered in this packet. Due to time management, we will not review the topics contained in this assignment. If you need to, you may use reference materials to assist you and refresh your memory (old notes, textbooks, online resources, etc.). While graphing calculators will be used in class, you can use calculators for some parts on this packet. AP Calculus BC is a fast paced course that is taught at the college level. There is a lot of material in the curriculum that must be covered before the AP exam in May. Therefore, we cannot spend a lot of class time re-teaching prerequisite skills. This is why you have this packet. Spend some time with it and make sure you are clear on everything covered in the packet so that you will be successful in Calculus. This assignment will be collected on our first day of class. Be sure to show all appropriate work to support your answers. You may use loose paper to show any required work. In addition, there may be a quiz on this material during the first week of school. You will be tested on this material on September 11th.

Name: ______________________________________ Show all work – no credit will be awarded for answers missing appropriate work. No calculators! Section I: Algebra Review (ln is the natural log function) 1. Solve xy + 2x + 1 = y for y. 2. Factor:

€

x 2(x −1) − 4(x −1) 3. Solve ln (y – 1) – ln 2 = x + ln x for y 4. Factor: 3x3/2 − 9x1/2 + 6x−1/2 Simplify each expression.

5.

€

(x 2)3 xx 7

6.

€

x ⋅ x3 ⋅ x16

7.

€

5(x + h)2 − 5x 2

h 8.

€

1x

+4x 2

3 − 1x

Simplify, using factoring of binomial expressions. Leave answers in factored form. Example:

[ ]

)34)(1()1)(6(

)6)(34()1()1)(6(

)18214()1()1)(6(

)916954()1()1)(6(

)916()94)(1()1()1)(6(

)1)(916()94()1(

2

22

22

223

++=

+−

−++=

+−

−−+=

+−

−−−−+=

+−

+−−++=

+−

++=−+

xxxx

xxxxx

xxxxx

xxxxxx

xxxxxx

xxxx

9.

€

(x −1)3(2x − 3) − (2x +12)(x −1)2

10.

€

(x −1)2(3x −1) − 2(x −1)(x −1)4

Simplify by rationalizing the numerator. Example:

€

x + 4 − 2x

=x + 4 − 2x

•x + 4 + 2x + 4 + 2

=x + 4 − 4

x x + 4 + 2( )=

xx x + 4 + 2( )

=1

x + 4 + 2

11.

€

x + 9 − 3x

12.

€

x + h − xh

Solve each equation or inequality for x over the set of real numbers.

13.

€

2x 4 + 3x 3 − 2x 2 = 0 14.

€

2x − 7x +1

=2xx + 4

15.

€

x 2 − 9 = x −1 16.

€

2x − 3 =14

17.

€

x 2 − 2x − 8 < 0 [Your answer should be interval(s)] 18.

€

3x + 5(x −1)(x 4 + 7)

= 0

Solve each of the systems algebraically and graphically .

€

19. x + y = 8 ____________________ 20. y = x 2 − 3x _______________________2x − y = 7 y = 2x − 6

Section II: Trigonometry Review Use your knowledge of the unit circle to evaluate each of the following. You MUST know your unit circle. Leave your answers in radical form. DON’T use your Calculator.

21. sin(30) _____ 22. cos 2π3_____ 23. tan45 _____

24. sin −π6

"

#$

%

&' _____ 25. tanπ ______ 26. csc 5π

6_____

27. cos 90( ) _____ 28. cos 3π4_____ 29. tan π

6_____

30. cos−1 12"

#$%

&' _____ 31. sin−1 2

2

"

#$

%

&' _____ 32. tan−1(1) _____

Solve each trigonometric equation for π20 ≤≤ x .

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33. sin x =32____________ 34. tan2 x =1 ___________

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35. cosx2

=22____________ 36. 2sin2 x + sin x −1 = 0 ____________

Solve each exponential or logarithmic equation.

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37. 5x =125 __________ 38. 8x+1 =16x __________ 39. 8134 = x _________

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40. 8−23 = x __________ 41. log2 32 = x __________ 42. logx

19

= −2 __________

43. log4 x = 3 __________ 44. log3(x + 7) = log3(2x −1) ____________

Expand each of the following using the laws of logs.

€

45. log3 5x2 ________________________ 46. ln

5xy 2_______________________

Section III: Graphing Review I. Symmetry – Even and Odd Functions

Quick Review

Even Function Symmetric about the y axis

( ) ( )f x f x− = for all x

Example: 2y x=

Odd Function

Symmetric about the origin (equivalent to a rotation of 180

degrees)

( ) ( )f x f x− = − for all x

Example: 3y x=

To determine algebraically if a function is even, odd, or neither, find ( )f x− and determine if it is equal

to ( )f x , ( )f x− , or neither.

Example: Determine if ( ) 2

41xf x

x=

+ is even or odd.

( ) ( )( )

( )2 2 2

4 4 41 11

x x xf x f xx xx

− −− = = = − = −

+ +− + Therefore, ( )f x is an odd function.

Determine if the following functions are even, odd, or neither.

47. ( )2

4 3xf xx

=+

48. ( )1xf xx

=+

49. ( ) 2 41 3 3f x x x= + + 50. ( ) 3 51 3 3f x x x= + + II. Essential Graphs Sketch each graph. You should know the graphs of these functions. 51. ( )f x x=

-4 -2 2 4

4

2

-2

-4

52. ( ) 3f x x=

-4 -2 2 4

4

2

-2

-4

53. ( ) sinf x x=

-2π -π π 2π

54. ( ) xf x e=

-4 -2 2 4

4

2

-2

-4

55. ( ) ( )lnf x x=

-4 -2 2 4

4

2

-2

-4

56. ( ) cosf x x=

-2π -π π 2π

57. For each graph above, state the domain, range, x-intercept(s), y-intercept(s), and any asymptote(s). 58.

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Graphing Skill #1: You should be able to graph a function in a viewing window that shows the important features. You should be familiar with the built-in zoom options for setting the window such as zoom-decimal and zoom-standard. You should also be able to set the window conditions to values you choose.

59. Graph = −2 3y x using the built in zoom-decimal and zoom-standard options. Draw each. 60. Find the appropriate viewing window to see the intercepts and the vertex defined by = − +2 11 10y x x . Use the window editor to enter the x and y values.

Window: Xmin = _______ Xmax = _______ Xscl = _______ Ymin = _______ Ymax = _______ Yscl = _______

Graphing Skill #2: You should be able to graph a function in a viewing window that shows the x-intercepts (also called roots and zeros). You should be able to accurately estimate the x-intercepts to 3 decimal places. Use the built-in root or zero command. [You should use your graphing calculator] 61. Find the x-intercepts of = − −2 1y x x . Window [-4.7, 4.7] x [-3.1, 3.1] (Write intercepts as points)

x-intercepts: ______________ 62. Find the x-intercepts of = − −3 2 1y x x .

x-intercepts: ______________

Graphing Skill #3: You should be able to graph two functions in a viewing window that shows the intersection points. Sometimes it is impossible to see all points of intersection in the same viewing window. You should be able to accurately estimate the coordinates of the intersection points to 3 decimal places. Use the built-in intersection command. 63. Find the coordinates of the intersection points for the functions:

= + = − − +2( ) 3 ( ) 7f x x g x x x .

Intersection points: ______________ 64. Find the coordinates of the intersection points of: = =2( ) 4 ( ) 2xf x x g x . Intersection points: ______________

Graphing Skill #4: You should be able to graph a function and estimate the local maximum or minimum values to 3 decimal places. Use the built-in max/min command. 65. Find the maximum and minimum values of the function = − −3 4 1y x x . (Value means the y-value)

Minimum value: ______________

Maximum value: ______________

66. Find the maximum and minimum values of the function = − +3 24 4y x x x .

67. Find the x-intercepts, relative maximum, and relative minimum of 3 22 1y x x= + − 68. Find the coordinates of the intersection points for the functions 2( ) 2 9f x x x= + − and

3( ) 34

g x x= − +

Section IV: Linear equations Write the equation for the line in both forms given a slope and a point. 69. m = 2/3 and P(3,5) 70. m = -4/5 and P(1,2) Point-Slope: Point-Slope: Slope-Intercept: Slope-Intercept: Write the equation for the line in both forms given two points. 71. P(2,2) and Q(4,2) 72. P(3,-2) and Q(3,7) Point-Slope: Point-Slope: Slope-Intercept: Slope-Intercept: Section V: Polynomial Functions

The Intermediate-Value Theorem

If f is continuous on [a, b] and v lies between f(a) and f(b), then there exists c between a and b such that f(c)

= v.

The intermediate value theorem states the following: If the function y = f(x) is continuous on the interval [a, b], and u is a number between f(a) and f(b), then there is a c ∈ [a, b] such that f(c) = u.

Example 1: Suppose that we want to know if f(x) = x4 – 7x3 – 4x + 8 is ever 0. Solution: Since this function is a polynomial, we know that it’s continuous everywhere. At x = -1, we get f(-1) = 20. At x = 1, we get f(1) = -2. So at the two endpoints of the interval [-1, 1] the function has values 20 and -2. Therefore, it must take on all values between -2 and 20 as x varies between -1 and 1. In particular, it must take on the value 0 for some x in [-1, 1]. The Intermediate Value Theorem doe not tell us exactly where it equals 0, but it does tell us it is 0 somewhere on the interval [-1, 1]. 72. Show that p(x) = 2x3 – 5x2 – 10x + 5 has a root somewhere between -1 and 2. 73. Use the Intermediate Value Theorem to prove that the equation x3 = x + 8 has at least one solution.

Section VI: Average Rate of Change Review

Calculus - 2.1 Rates of Change and Limits

Calculus - 2.1 Rates of Change and Limits – 17

Example 1 Find the average speed of a car that has traveled 350 miles in 7 hours.

Initial: 0 miles and 0 hours

Final: 350 miles and 7 hours

Speed

350 0 milesSpeed 50 mi/hr

7 0 hours

AVG

AVG

y

t

!=

!

"= =

"

Example 2

Given the equation 24.9y t= , which represents the free fall equation in relationship to meters per

second squared. Find the average from 0t = to 3t = and instantaneous speed at 3t = seconds. Average Speed Instantaneous Speed

Initial: 0 meters and 0 seconds

Final: 44.1 meters and 3 seconds

Speed

44.1 0 metersSpeed

3 0 seconds

Speed 14.7 meters/second

AVG

AVG

AVG

y

t

!=

!

"=

"

=

( ) ( )

( ) ( )

( ) ( )

Instant

2 2

Instant

2

Instant

2

Instant

2

Instant

metersSpeed

second

4.9 3 4.9 3 metersSpeed

second

4.9 9 6 4.9 9 metersSpeed

second

44.1 29.4 4.9 44.1 metersSpeed

second

29.4 4.9Speed 29.4 4.9 meter

f t h f t

h

h

h

h h

h

h h

h

h hh

h

+ "=

+ "=

+ + "=

+ + "=

+= = +

( )Instant

s/second

as (the change in time) approaches zero

Speed 29.4 4.9 0 29.4 meters/second

h

= + =

Definition: Average Speed Average speed is found by dividing the distance covered by the elapsed time.

total distance traveled final position initial position

time elapsed final time initial time

y

t

! "= =

! "

Definition: Instantaneous Speed Instantaneous speed is found by dividing the distance covered by the elapsed time. For this case we want the elapsed time to be “zero.” However, if the time is zero, the change in position will also be zero, which gives 0/0. So to alleviate this mathematical error, we want the time to be very small, as close to zero as we can make it.

total distance traveled final position initial position

time elapsed final time initial time

y

t

! "= =

! "

( ) ( )f t h f ty

t h

+ "!=

!, where h is the change in time from initial to final. We want h to

approach zero.

74. Find the average speed of a car that has traveled 350 miles in 7 hours. 75. Given the equation y = -4.9t2, which represents the free fall equation in relationship to meter per second squared. Find the average from t = 0 to t = 3 and instantaneous speed at t = 3 seconds.

Calculus - 2.1 Rates of Change and Limits

Calculus - 2.1 Rates of Change and Limits – 17

Example 1 Find the average speed of a car that has traveled 350 miles in 7 hours.

Initial: 0 miles and 0 hours

Final: 350 miles and 7 hours

Speed

350 0 milesSpeed 50 mi/hr

7 0 hours

AVG

AVG

y

t

!=

!

"= =

"

Example 2

Given the equation 24.9y t= , which represents the free fall equation in relationship to meters per

second squared. Find the average from 0t = to 3t = and instantaneous speed at 3t = seconds. Average Speed Instantaneous Speed

Initial: 0 meters and 0 seconds

Final: 44.1 meters and 3 seconds

Speed

44.1 0 metersSpeed

3 0 seconds

Speed 14.7 meters/second

AVG

AVG

AVG

y

t

!=

!

"=

"

=

( ) ( )

( ) ( )

( ) ( )

Instant

2 2

Instant

2

Instant

2

Instant

2

Instant

metersSpeed

second

4.9 3 4.9 3 metersSpeed

second

4.9 9 6 4.9 9 metersSpeed

second

44.1 29.4 4.9 44.1 metersSpeed

second

29.4 4.9Speed 29.4 4.9 meter

f t h f t

h

h

h

h h

h

h h

h

h hh

h

+ "=

+ "=

+ + "=

+ + "=

+= = +

( )Instant

s/second

as (the change in time) approaches zero

Speed 29.4 4.9 0 29.4 meters/second

h

= + =

Definition: Average Speed Average speed is found by dividing the distance covered by the elapsed time.

total distance traveled final position initial position

time elapsed final time initial time

y

t

! "= =

! "

Definition: Instantaneous Speed Instantaneous speed is found by dividing the distance covered by the elapsed time. For this case we want the elapsed time to be “zero.” However, if the time is zero, the change in position will also be zero, which gives 0/0. So to alleviate this mathematical error, we want the time to be very small, as close to zero as we can make it.

total distance traveled final position initial position

time elapsed final time initial time

y

t

! "= =

! "

( ) ( )f t h f ty

t h

+ "!=

!, where h is the change in time from initial to final. We want h to

approach zero.

76. Find f (x + h)− f (x)

h if:

a. f(x) = 2x – 3 b. f(x) = x2 c. f(x) = 3x2 – 4x d. f(x) = 2x3 + 5x

Section VII: Limits

Section VIII: Optimization 77. A farmer has 2400 ft of fencing and wants to fence off a rectangular field that borders a straight river. He needs no fence along the river. What are the dimensions of the field that has the largest area?