ap calculus 2020 summer review packet - prince george's
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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 3, 2019 first day of school year 2019-2020. 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 on the skills included in this packet will be given in the first week of school. If you have problems or concerns you can e-mail me at [email protected] or post your question in Google Classroom [ Code: (lyygwm)] and I will answer all 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. DONβT SIMPLIFY (unless instructed) DONβT RATIONALIZE 7. Have it all put together and ready to turn in on September 3, 2019. 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 β π¦π₯1π¦ +
1π₯=
π₯ β π¦π₯
π₯ + π¦π₯π¦
= π₯ β π¦π₯ β
π₯π¦π₯ + π¦ =
π₯π¦ β π¦)
π₯ + π¦
Using the second Method *
+,-./
+01-
+0123
+,- β 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. 12xβ x
3+ x 2.
5β 3x +7
3+ 6x +7
3. 2β xx β3
5+ xx +3
3
4.
xx β 2
β3xx2 β 4
xx + 2
+4xx2 β 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 βf of g of xβ. Means to substitute the values in place
of each x in the equation. Example: Given f (x) =3x2 + 2 and g(x) = 2x β3 find f(x - 1), g(2x + 5) and f(g(x)).
π π₯ β 1 = 3(π₯ β 1)) + 2 = 3 π₯) β 2π₯ + 1 = 3π₯) β 6π₯ + 3 + 2 = 3π₯) β 6π₯ + 5 π 2π₯ + 5 = 2 2π₯ + 5 β 3 = 4π₯ + 10 β 3 = 4π₯ + 7
f (g(x)) = f (2x β3)= 3(2x β3)2 + 2= 3(4x2 β12x + 9)+ 2=12x2 β36x + 27+ 2
f (g(x)) =12x2 β36x + 29
Let f (x) = 5x β 2 and g(x) = 3x2 β5. Find each. 6. f (3) = __________ 7. g(β2) =___________ 8. f (m+ 2) = __________
9. f g(β1)"# $%=________ 10. g f (nβ 4)"# $%=________ 11. f (x + h)β f (x)h
= ______
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Let f (x) = cos x . Find each exactly.
12. f Ο4
!
"#
$
%&= ___________ 13. f 3Ο
2!
"#
$
%&=______________
Let f (x) = 4x2 , g(x) = 3x +1, and h(x) = 2x2 β 4 . Find each. 14. h f (β5)"# $%= _______ 15. f g(t +1)!" #$= _______ 16. g h(m4 )!
"#$= _______
Find f (x + h) β f (x)
h for the given function f.
17. f (x) = 3x β 2 18. f (x) =1β8x 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 β12x +35
x β int. (Let y = 0)0 = x2 β12x +350 = (x β 5)(x β 7)x = 5or x = 7x β intercepts (5, 0) and (7, 0)
yβ int. (Let x = 0)y = 02 β12(0)+35y = 35yβ intercept (0,35)
Find the x and y intercepts for each. 19. y = 8x β 4 20. y = x2 +13x β 48 21. y = x 25β x2 22. π¦) = 2π₯5 β 18π₯;
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Use substitution or elimination method to solve the system of equations.
Substitution Method Solve one equation for one variable π¦) = βπ₯) + 16π₯ β 19(First Equation solve for π¦)) 2π₯) β (βπ₯) + 16π₯ β 19) + 4π₯ β 10 = 0 (Plug in the value of π¦) to the second equation) 2π₯) + π₯) β 16π₯ + 19 + 4π₯ β 10 = 0 Simplify. 3π₯) β 12π₯ + 9 = 0 (The rest is the same as on the left.) 3(π₯) β 4π₯ + 3) = 0 3(π₯ β 3)(π₯ β 1) = 0 π₯ = 3ππππ₯ = 1
Since the value of y is imaginary at x = 1, there is no intersection at x = 1. The points of intersection are R3, 2β5UπππR3, β2β5U
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Find the solutions of the following systems of equations.
23. 4π₯ β 5π¦ = β13π₯ + 2π¦ = 5 24. 0.2π₯ + 0.3π¦ = 0.6
0.5π₯ + 0.1π¦ = 0.2 25. π₯) + π¦ = 72π₯ + π¦ = β1
Interval Notation 26. Complete the table with the appropriate notation or graph.
Solution Interval Notation Graph β4 β€ π₯ < 5
[-6, 4]
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Solve each equation. State your answer in BOTH interval notation and graphically. 27. 2π₯ β 4 β€ 4π₯ + 12 28. β19 β€ 3π₯ β 7 < 8 29. 4.)
5β 4.7
;β₯ 2
Domain and Range Find the domain and range of each function. Write your answer in INTERVAL notation.
30. π π₯ = 2π₯) β 4 31. π π₯ = π₯; + 4 32. π π₯ = π πππ₯ + 2 33. π π₯ = π₯+2π₯β3
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Inverses
To find the inverse of a function, simply switch the x and the y and solve for the new βyβ value.
Example: π π₯ = 2π₯ + 6 Rewrite f(x) in terms of y π¦ = 2π₯ + 6 Switch x and y π₯ = 2π¦ + 6Solve for new y (π₯)) = ( 2π¦ + 6))Square both sides π₯) = 2π¦ + 6Simplify π¦ = 4
@
)β 3Solve for y
π2: π₯ = 4@
)β 3Rewrite in inverse notation
Find the inverse for each function. 34. π π₯ = 3π₯ β 1 35.
π π₯ = 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: π π₯ = 2π₯ β 4ππππ π₯ = :
)π₯ + 2show f(x) and g(x) are inverses of each other.
π π π₯ = 2 :)π₯ + 2 β 4 π π π₯ = :
)2π₯ β 4 + 2
π π π₯ = π₯ + 4 β 4π π π₯ = π₯ β 2 + 2 π π π₯ = π₯π π π₯ = π₯ ππππππ π π₯ = π π π₯ = π₯, π‘βππππππππ‘βππ¦ππππππ£πππ ππ πππππβππ‘βππ Prove f(x) and g(x) are inverses of each other
36. π π₯ = 4π₯)ππππ π₯ = 2 π₯ 37. π π₯ = 2π₯) β 1ππππ π₯ = 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 β y1 = m(x β 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. β¦β¦..
π¦ = );π₯ β 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!
Ο radians to get rid of radians and Use
Ο 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. 125Β° b. β30Β° c. 840Β° Angles in Standard Position 48. Sketch the angle in standard position.
a.240Β° 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. 765Β°
c.
2e6
d. 1110Β°
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
Ο2= 0
ππππππΒ° = π
50. a. sin 450Β° b. cos 720Β°
c. sin(β270Β°) d. cos 2π
e. tan 90Β° f. tan π
Graphing Trig Functions y = sin x and y = cos x have a period of 2Ο and an amplitude of 1. Use the parent graphs above to help you sketch a graph of the functions below. For,π π₯ = π΄π ππ π΅π₯ + πΆ + π· A = amplitude, 2ΟB
= 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.π π₯ = π πππ₯ + 2 52. π π₯ = β5π ππ2π₯ 53. π π₯ = πππ (π₯ + e
)) 54. π π₯ = βπππ 3π₯ β 4
Inverse Trigonometric Functions: Recall: Inverse Trig Functions can be written in one of ways:
arcsin x( ) sinβ1 x( )
To evaluate inverse trigonometric functions remember that the following statements are equivalent.
π = πππ 2: π₯ β π₯ = cos(π) π = π ππ2: π₯ β π₯ = sin(π) π = π‘ππ2: π₯ β π₯ = tan(π)
Example: Express the value of βπβ in radians. π = arcsin(2:
)) Draw a reference triangle.
3
2 -1
This means the reference angle is 30Β° or Ο6
. So, π = β Ο6
so that it falls in the interval
from 2e)< π < e
)
Answer: π = βe6
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For each of the following, express the value for βπβ in radians. Use the given unit circle. 55. π = πππ 2:(β ;
)) 56. π = π ππ2:(:
)) 57. π = arctan(1)
58. π = π ππ2:(β )
)) 59. π = π‘ππ2:(β :
;) 60. π = πππ 2:(β )
))
Example: Find the value without a calculator.
tan(ππππππ 5;7
) 34 Draw the reference triangle in the correct quadrant first. 3 π Find the missing side using Pythagorean Theorem. 5
Find the ratio of the sine of the reference triangle. π‘πππ = ;
5
For each of the following give the value without a calculator.
61. cos(arc tan 5) 62. Csc (arc cos5<)
63. sec(πππsin 47) 64. sin(πππ tan 10)
65. cos(πππ 2:( ;)) 66. arcsin(π ππ e
7)
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Minor Axis
Major Axisab
c FOCUS (h + c, k)FOCUS (h - c, k)CENTER (h, k)
(x β h)2
a2+(y β 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 = (xβ h)2 + (yβ 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. π₯
) +π¦) = 25 68. π₯) +π¦) = 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
4β)
[email protected]:)4@2)4
x 1.9 1.99 1.999 2.001 2.01 2.1
f(x) 72. lim
οΏ½βοΏ½
:2οΏ½οΏ½οΏ½οΏ½οΏ½
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. π π₯ = 4
@2:42:
74. lim4β)
42)4@27
75. Below is the graph of f(x). For each of the given point determine the value of f(a) and
lim4βοΏ½
π(π₯). 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
4β;
4@2)42:4.;
77. lim4β;
π₯ + 24> 78. lim4β)
)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. π(π₯) = {π₯2β2π₯;π₯>1π₯+3;π₯β€1 80. lim
4β),|42)|42)
lim
4β:0π(π₯)
Vertical Asymptotes To find the vertical asymptote(s) of a rational function, simply set the denominator equal to 0 and solve for x. 81. π π₯ = )
4> 82.π π₯ 4.)
4@2)5 83. π π₯ = 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. π π₯ = 54>
4@274.) 85. π π₯ = ;4@24.:)
)4@264.< 86. π π₯ = 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.lim4βοΏ½
)4@.;4.74>.54
88. lim4β2οΏ½
4-.454>.<
89. lim4β.οΏ½
)4>2;4.:)274>.)42<
Limits to Infinity A rational function does not have a limit if it goes to + Β₯, 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 β or β β , please state which one the limit approaches.
90. lim4βοΏ½
:4@
91. lim4β)
;.4)24
92. lim4βοΏ½
;οΏ½οΏ½οΏ½4
For problems 96 and 97, express in terms of the other variables in the picture: 93. 94.
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95. Find the ratio of the area inside the square but outside the circle to the area of the square in the picture below.
96.
97.
98. Two cars start moving from the same point. One travels south at 100 km/hr, the other travels west at 50 km/hr. How far apart are they two hours later? 99. A kite is 100 m above the ground. If there are 200 m of string out, what is the angle of between the string and the horizontal. (Assume that the that string is perfectly straiignt) 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 β sin2 x
= 1β 2sin2 x
= 2cos2 x β1
Formula Sheet
Reciprocal Identities: csc x = 1sin x
sec x = 1cos x
cot x = 1tan 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 x1β tan2 x
Logarithms: y = loga x is equivalent to x = ay Product property: logb mn = logb m + logb n
Quotient property: logbmn= logb m β 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: limhββ
f (x + h) β f (x)h
Slope-intercept form: y = mx + b Point-slope form: y β y1 = m(x β x1) Standard form: Ax + By + C = 0