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DuVal High School Summer Review Packet

AP Calculus

Welcome to AP Calculus AB. This packet contains background skills you need to know for your AP Calculus. My suggestion is, you read the information and work through the assigned problems during the summer break. Do not wait until the week before school starts to begin this assignment. Work on it little by little each day, and before you know it, you are done doing the work. Write your answer on a separate paper and keep it in a folder together with this packet. Folders will be collected on September 6, 2017, first day of the school year 2017-2018. I just want to emphasize the importance of these background skills. Calculus is easy, what difficult are the Algebraic concepts that you will use in Calculus. Most of the time you will understand the Calculus concept taught, but you will struggle because you do not know the background skills needed in the problem. You need to be diligent and be mindful of the concepts/skills that you need to practice and master. This extra work during the summer will help you a lot next school year. I will count this packet as three homework grades and an assessment measuring your knowledge of the skills included in this packet will be given in the first week of school. If you have any problems you can e-mail me at jesusan.sixson@pgcps.org or post your question in Google Classroom (Google Classroom code: ysz4utc). I will answer all e-mails as soon as possible. Remember: 1. Do a little each day. 2. Have a separate answer sheet. 3. Show work. 4. Be very neat. 5. E-mail me with questions. 6. Do all problems in the packet. DON’T SIMPLIFY (unless instructed) DON’T RATIONALIZE 7. Have it all put together and ready to turn in on September 6, 2017. I am looking forward to seeing you next school year. Have a great summer.

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Summer Review Packet for Students Entering AP Calculus AB Complex Fractions - is a fraction where the numerator, denominator, or both contain a fraction

One way to simplify complex fraction is to simplify the numerator and denominator separately, and then simplify the resulting expressions. Another way of simplifying complex fraction is to find the LCD of all the denominators. Then multiply by a fraction equal to 1, which has a numerator and denominator composed of the common denominator of all the denominators in the complex fraction. Example: Using the first method:

1 − 𝑦𝑥1𝑦 +

1𝑥=

𝑥 − 𝑦𝑥

𝑥 + 𝑦𝑥𝑦

= 𝑥 − 𝑦𝑥 ∙

𝑥𝑦𝑥 + 𝑦 =

𝑥𝑦 − 𝑦)

𝑥 + 𝑦

Using the second Method *

+,-./

+01-

+0123

+,- ∙ 4.5 426

4.5 426= 7 426 .8 4.5

5 4.5 29 426= 742)7.84.75

54.)5294.79= :;4.):

2;4.<;

Simplify each of the following.

1. 12x− x

3+ x 2.

5− 3x +7

3+ 6x +7

3. 2− xx −3

5+ xx +3

3

4.

xx − 2

−3xx2 − 4

xx + 2

+4xx2 − 4

5.

-=,>-

@=>=-A

=>=-A

. @=,>

Functions To evaluate a function for a given value, simply plug the value into the function for x. Recall:

f ! g( )(x) = f (g(x)) OR f [g(x)] read “f of g of x”. Means to substitute the values in place

of each x in the equation. Example: Given f (x) =3x2 + 2 and g(x) = 2x −3 find f(x - 1), g(2x + 5) and f(g(x)).

𝑓 𝑥 − 1 = 3(𝑥 − 1)) + 2 = 3 𝑥) − 2𝑥 + 1 = 3𝑥) − 6𝑥 + 3 + 2 = 3𝑥) − 6𝑥 + 5 𝑔 2𝑥 + 5 = 2 2𝑥 + 5 − 3 = 4𝑥 + 10 − 3 = 4𝑥 + 7

f (g(x)) = f (2x −3)= 3(2x −3)2 + 2= 3(4x2 −12x + 9)+ 2=12x2 −36x + 27+ 2

f (g(x)) =12x2 −36x + 29

Let f (x) = 5x − 2 and g(x) = 3x2 −5. Find each. 6. f (3) = __________ 7. g(−2) =___________ 8. f (m+ 2) = __________

9. f g(−1)"# $%=________ 10. g f (n− 4)"# $%=________ 11. f (x + h)− f (x)h

= ______

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Let f (x) = cos x . Find each exactly.

12. f π4

!

"#

$

%&= ___________ 13. f 3π

2!

"#

$

%&=______________

Let f (x) = 4x2 , g(x) = 3x +1, and h(x) = 2x2 − 4 . Find each. 14. h f (−5)"# $%= _______ 15. f g(t +1)!" #$= _______ 16. g h(m4 )!

"#$= _______

Find f (x + h) − f (x)

h for the given function f.

17. f (x) = 3x − 2 18. f (x) =1−8x Intercepts and Points of Intersection To find the x-intercepts, let y = 0 in your equation and solve. To find the y-intercepts, let x = 0 in your equation and solve. Example: y = x2 −12x +35

x − int. (Let y = 0)0 = x2 −12x +350 = (x − 5)(x − 7)x = 5or x = 7x − intercepts (5, 0) and (7, 0)

y− int. (Let x = 0)y = 02 −12(0)+35y = 35y− intercept (0,35)

Find the x and y intercepts for each. 19. y = 8x − 4 20. y = x2 +13x − 48 21. y = x 25− x2 22. 𝑦) = 2𝑥5 − 18𝑥;

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Use substitution or elimination method to solve the system of equations.

Substitution Method Solve one equation for one variable 𝑦) = −𝑥) + 16𝑥 − 19(First Equation solve for 𝑦)) 2𝑥) − (−𝑥) + 16𝑥 − 19) + 4𝑥 − 10 = 0 (Plug in the value of 𝑦) to the second equation) 2𝑥) + 𝑥) − 16𝑥 + 19 + 4𝑥 − 10 = 0 Simplify. 3𝑥) − 12𝑥 + 9 = 0 (The rest is the same as on the left.) 3(𝑥) − 4𝑥 + 3) = 0 3(𝑥 − 3)(𝑥 − 1) = 0 𝑥 = 3𝑎𝑛𝑑𝑥 = 1

Since the value of y is imaginary at x = 1, there is no intersection at x = 1. The points of intersection are R3, 2√5U𝑎𝑛𝑑R3, −2√5U

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Find the solutions of the following systems of equations.

23. 4𝑥 − 5𝑦 = −13𝑥 + 2𝑦 = 5 24. 0.2𝑥 + 0.3𝑦 = 0.6

0.5𝑥 + 0.1𝑦 = 0.2 25. 𝑥) + 𝑦 = 72𝑥 + 𝑦 = −1

Interval Notation 26. Complete the table with the appropriate notation or graph.

Solution Interval Notation Graph −4 ≤ 𝑥 < 5

[-6, 4]

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Solve each equation. State your answer in BOTH interval notation and graphically. 27. 2𝑥 − 4 ≤ 4𝑥 + 12 28. −19 ≤ 3𝑥 − 7 < 8 29. 4.)

5− 4.7

;≥ 2

Domain and Range Find the domain and range of each function. Write your answer in INTERVAL notation.

30. 𝑓 𝑥 = 2𝑥) − 4 31. 𝑓 𝑥 = 𝑥; + 4 32. 𝑓 𝑥 = 𝑠𝑖𝑛𝑥 + 2 33. 𝑓 𝑥 = 𝑥+2𝑥−3

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Inverses

To find the inverse of a function, simply switch the x and the y and solve for the new “y” value.

Example: 𝑓 𝑥 = 2𝑥 + 6 Rewrite f(x) in terms of y 𝑦 = 2𝑥 + 6 Switch x and y 𝑥 = 2𝑦 + 6Solve for new y (𝑥)) = ( 2𝑦 + 6))Square both sides 𝑥) = 2𝑦 + 6Simplify 𝑦 = 4

@

)− 3Solve for y

𝑓2: 𝑥 = 4@

)− 3Rewrite in inverse notation

Find the inverse for each function. 34. 𝑓 𝑥 = 3𝑥 − 1 35.

𝑓 𝑥 = 4

*

)

Also, recall that to PROVE one function is an inverse of another function, you need to show that:

f (g(x)) = g( f (x)) = x Example: If: 𝑓 𝑥 = 2𝑥 − 4𝑎𝑛𝑑𝑔 𝑥 = :

)𝑥 + 2show f(x) and g(x) are inverses of each other.

𝑓 𝑔 𝑥 = 2 :)𝑥 + 2 − 4 𝑔 𝑓 𝑥 = :

)2𝑥 − 4 + 2

𝑓 𝑔 𝑥 = 𝑥 + 4 − 4𝑔 𝑓 𝑥 = 𝑥 − 2 + 2 𝑓 𝑔 𝑥 = 𝑥𝑔 𝑓 𝑥 = 𝑥 𝑆𝑖𝑛𝑐𝑒𝑓 𝑔 𝑥 = 𝑔 𝑓 𝑥 = 𝑥, 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒𝑡ℎ𝑒𝑦𝑎𝑟𝑒𝑖𝑛𝑣𝑒𝑟𝑠𝑒𝑠𝑜𝑓𝑒𝑎𝑐ℎ𝑜𝑡ℎ𝑒𝑟 Prove f(x) and g(x) are inverses of each other

36. 𝑓 𝑥 = 4𝑥)𝑎𝑛𝑑𝑔 𝑥 = 2 𝑥 37. 𝑓 𝑥 = 2𝑥) − 1𝑎𝑛𝑑𝑔 𝑥 = 4.:)

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Equation of a line Slope intercept form: y = mx + b Vertical line: x = c (slope is undefined) Point-slope form: y − y1 = m(x − x1) Horizontal line: y = c (slope is 0) 38. Use slope-intercept form to find the equation of the line having a slope of 4 and a y-intercept of -6. 39. Determine the equation of a line passing through the point (3, -7) with a slope of 0. 40. Determine the equation of a line passing through the point (-5, 5) with an Undefined slope. 41. Use point-slope form to find the equation of the line passing through the point (-2, 7) with a

slope of ;5.

42. Find the equation of a line passing through the point (-3, 2) and parallel to the line 𝑦 =);𝑥 − 5.

43. Find the equation of a line perpendicular to the x- axis passing through the point (3,7). 44. Find the equation of a line passing through the points (4,7) and (-3, -6). 45. Find the equation of a line with an x-intercept (-3, 0) and a y-intercept (0, -7).

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Radian and Degree Measure

Use

180!

π radians to get rid of radians and Use

π radians

180! to get rid of degrees and

convert to degrees. convert to radians. 46. Convert to degrees: a.3𝜋4 b. 9e

5 c. 1.25 radians

47. Convert to radians: a. 125° b. −30° c. 840° Angles in Standard Position 48. Sketch the angle in standard position.

a.240° b. :8e6

c. 27e;

d. 1.5 radians Reference Triangles 49. Sketch the angle in standard position. Draw the reference triangle and label the sides, if possible.

a. )e;

b. 765°

c.

2e6

d. 1110°

10

11

2

- 2

- 5 5

f x( ) = sinx( )2

- 2

- 5 5

f x( ) = cosx( )

2

-2

(-1,0)

(0,-1)

(0,1)

(1,0)

2

-2

(-1,0)

(0,-1)

(0,1)

(1,0)

Unit Circle You can determine the sine or cosine of a quadrantal angle by using the unit circle. The x-coordinate of the circle is the cosine and the y-coordinate is the sine of the angle.

Example: sin90! = 1 cos

π2= 0

𝒕𝒂𝒏𝟏𝟖𝟎° = 𝟎

50. a. sin 450° b. cos 720°

c. sin(−270°) d. cos 2𝜋

e. tan 90° f. tan 𝜋

Graphing Trig Functions y = sin x and y = cos x have a period of 2π and an amplitude of 1. Use the parent graphs above to help you sketch a graph of the functions below. For,𝑓 𝑥 = 𝐴𝑠𝑖𝑛 𝐵𝑥 + 𝐶 + 𝐷 A = amplitude, 2πB

= period, CB

= phase shift (positive xy shift left, negative

xy shift right) and D = vertical

shift.

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Graph two complete periods of the function. 51.𝑓 𝑥 = 𝑠𝑖𝑛𝑥 + 2 52. 𝑓 𝑥 = −5𝑠𝑖𝑛2𝑥 53. 𝑓 𝑥 = 𝑐𝑜𝑠(𝑥 + e

)) 54. 𝑓 𝑥 = −𝑐𝑜𝑠3𝑥 − 4

Inverse Trigonometric Functions: Recall: Inverse Trig Functions can be written in one of ways:

arcsin x( ) sin−1 x( )

To evaluate inverse trigonometric functions remember that the following statements are equivalent.

𝜃 = 𝑐𝑜𝑠2: 𝑥 ↔ 𝑥 = cos(𝜃) 𝜃 = 𝑠𝑖𝑛2: 𝑥 ↔ 𝑥 = sin(𝜃) 𝜃 = 𝑡𝑎𝑛2: 𝑥 ↔ 𝑥 = tan(𝜃)

Example: Express the value of “𝜃” in radians. 𝜃 = arcsin(2:

)) Draw a reference triangle.

3

2 -1

This means the reference angle is 30° or π6

. So, 𝜃 = – π6

so that it falls in the interval

from 2e)< 𝜃 < e

)

Answer: 𝜃 = −e6

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For each of the following, express the value for “𝜃” in radians. Use the given unit circle. 55. 𝜃 = 𝑐𝑜𝑠2:(− ;

)) 56. 𝜃 = 𝑠𝑖𝑛2:(:

)) 57. 𝜃 = arctan(1)

58. 𝜃 = 𝑠𝑖𝑛2:(− )

)) 59. 𝜃 = 𝑡𝑎𝑛2:(− :

;) 60. 𝜃 = 𝑐𝑜𝑠2:(− )

))

Example: Find the value without a calculator.

tan(𝑎𝑟𝑐𝑐𝑜𝑠 5;7

) 34 Draw the reference triangle in the correct quadrant first. 3 𝜃 Find the missing side using Pythagorean Theorem. 5

Find the ratio of the sine of the reference triangle. 𝑡𝑎𝑛𝜃 = ;

5

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

61. cos(arc tan 5) 62. Csc (arc cos5<)

63. sec(𝑎𝑟𝑐sin 47) 64. sin(𝑎𝑟𝑐 tan 10)

65. cos(𝑐𝑜𝑠2:( ;)) 66. arcsin(𝑠𝑖𝑛 e

7)

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Minor Axis

Major Axisab

c FOCUS (h + c, k)FOCUS (h - c, k)CENTER (h, k)

(x − h)2

a2+(y − k)2

b2= 1

4

2

-2

- 4

- 5 5

4

2

-2

- 4

- 5 5

4

2

-2

- 4

- 5 5

4

2

-2

- 4

- 5 5

Circles and Ellipses r2 = (x− h)2 + (y− k)2 For a circle centered at the origin, the equation is x2 + y2 = r2 , where r is the radius of the circle.

For an ellipse centered at the origin, the equation is x2

a2+y2

b2= 1 , where a is the distance

from the center to the ellipse along the x-axis and b is the distance from the center to the ellipse along the y-axis. If the larger number is under the y2 term, the ellipse is elongated along the y-axis. For our purposes in Calculus, you will not need to locate the foci. Graph the circles and ellipses below: 67. 𝑥

) +𝑦) = 25 68. 𝑥) +𝑦) = 8 69. 4@

7+~

@

8= 1 70. 4

@

)5+ ~@

:6= 1

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Limits Finding limits numerically. Complete the table and use the result to estimate the limit. 71. lim

4→)

4@.742:)4@2)4

x 1.9 1.99 1.999 2.001 2.01 2.1

f(x) 72. lim

�→�

:2�����

x -0.001 -0.01 -0.1 0.1 .01 .001 f(x)

Finding limits graphically. Find each limit graphically. Use your calculator to assist in graphing. 73. 𝑓 𝑥 = 4

@2:42:

74. lim4→)

42)4@27

75. Below is the graph of f(x). For each of the given point determine the value of f(a) and

lim4→�

𝑓(𝑥). If any of the quantities do not exist explain why?

.

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Evaluating Limits Analytically Solve by direct substitution whenever possible. If needed, rearrange the expression so that you can do direct substitution. 76. lim

4→;

4@2)42:4.;

77. lim4→;

𝑥 + 24> 78. lim4→)

)244@27

One-Sided Limits Find the limit if it exists. First, try to solve for the overall limit. If an overall limit exists, then the one-sided limit will be the same as the overall limit. If not, use the graph and/or a table of values to evaluate one-sided limits.

79. 𝑓(𝑥) = {𝑥2−2𝑥;𝑥>1𝑥+3;𝑥≤1 80. lim

4→),|42)|42)

lim

4→:0𝑓(𝑥)

Vertical Asymptotes To find the vertical asymptote(s) of a rational function, simply set the denominator equal to 0 and solve for x. 81. 𝑓 𝑥 = )

4> 82.𝑓 𝑥 4.)

4@2)5 83. 𝑓 𝑥 = 42:

4@(4.;)

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 y = 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.

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84. 𝑓 𝑥 = 54>

4@274.) 85. 𝑓 𝑥 = ;4@24.:)

)4@264.< 86. 𝑓 𝑥 = 94@254.:

74@2;

Determine each limit as x goes to infinity. RECALL: This is the same process you used to find Horizontal Asymptotes for a rational function. ** In a nutshell

1. Find the highest power of x. 2. How many of that type of x do you have in the numerator? 3. How many of that type of x do you have in the denominator? 4. That ratio is your limit!

87.lim4→�

)4@.;4.74>.54

88. lim4→2�

4-.454>.<

89. lim4→.�

)4>2;4.:)274>.)42<

Limits to Infinity A rational function does not have a limit if it goes to + ¥, however, you can state the direction the limit is headed if both the left and right hand side go in the same direction. Determine each limit if it exists. If the limit approaches ∞ or − ∞ , please state which one the limit approaches.

90. lim4→�

:4@

91. lim4→)

;.4)24

92. lim4→�

;���4

For problems 96 and 97, express in terms of the other variables in the picture: 93. 94.

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95.

96.

97.

98.

99.

100. A soap company changes the design of its soap from a cone to a sphere. The cone had a height of 3cm and a radius of 2cm . The sphere has a diameter of 3cm. How much more soap, in cubic centimeters, does the new design contain than the old design?

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cos2x = cos2 x − sin2 x

= 1− 2sin2 x

= 2cos2 x −1

Formula Sheet

Reciprocal Identities: csc x = 1

sin x sec x = 1

cos x cot x = 1

tan x

Quotient Identities: tan x = sin xcos x

cot x = cos xsin x

Pythagorean Identities: sin2 x + cos2 x = 1 tan2 x +1 = sec2 x 1+ cot2 x = csc2 x Double Angle Identities: sin2x = 2sin x cos x

tan2x = 2 tan x1− tan2 x

Logarithms: y = loga x is equivalent to x = ay Product property: logb mn = logb m + logb n

Quotient property: logbmn= logb m − logb n

Power property: logb m

p = p logb m Property of equality: If logb m = logb n , then m = n

Change of base formula: loga n =logb nlogb a

Derivative of a Function: Slope of a tangent line to a curve or the derivative: limh→∞

f (x + h) − f (x)h

Slope-intercept form: y = mx + b Point-slope form: y − y1 = m(x − x1) Standard form: Ax + By + C = 0