Business Calculus Summer Assignment 2015

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Business Calculus
Summer Assignment 2015
The following packet contains topics and definitions that you will be required to know in order to
succeed in Business Calculus this year. You are advised to be familiar with each of the concepts and to
complete the included problems by the September 3, 2015. All of these topics were discussed in
either Algebra II or Precalculus and will be used frequently throughout the year. All problems that
you are to complete are marked in bold. All problems are expected to be completed. You will be
tested on this material.
A Precalculus Review
Things to Remember:
Rational numbers are any real numbers that can be expressed as the ration of two integers. This
includes all terminating and all repeating decimals.
Examples:
2
5
= 0.4
7
1
7
8
=7
= 0.875
1
3
= 0.333 …
Irrational numbers are any real numbers that cannot be expressed as the ratio of two integers.
This will include all non-terminating, non-repeating decimals.
Examples:
√2 ≈ 1.4142135623 …
πœ‹ ≈ 3.1415926535 …
Properties of Inequalities:
Let a, b, c and d be real numbers.
1. Transitive property: a<b and b<c οƒ  a < c
2. Adding inequalities: a<b and c<d οƒ  ac < bc, c > 0
3. Multiplying by a positive constant: a < b οƒ  ac < bc, c > 0
4. Multiplying by a negative constant: a<b οƒ  ac > bc, c < 0
5. Adding a constant: a < b οƒ  a + c < b + c
6. Subtracting a constant: a < b οƒ  a – c < b – c
Definition of Absolute Value:
The absolute value of a real number a is:
π‘Ž 𝑖𝑓 π‘Ž ≥0
−π‘Ž 𝑖𝑓 π‘Ž<0
|π‘Ž| = {
1
𝑒 ≈ 2.7182818284
Properties of Absolute Value:
1. Multiplication: |π‘Žπ‘| = |π‘Ž||𝑏|
π‘Ž
|π‘Ž|
2. Division: |𝑏 | = |𝑏| , 𝑏 ≠ 0
3. Power: |π‘Žπ‘› | = |π‘Ž|𝑛
4. Square Root: √π‘Ž2 = |π‘Ž|
Distance Between Two Points:
The distance d between any two points x1 and x2 on a real number line is:
𝑑 = |π‘₯1− π‘₯2 | = √(π‘₯2 − π‘₯1 )2
The distance d between any two points (x1, y1) and (x2, y2) on a Cartesian plane is:
𝑑 = √(π‘₯2 − π‘₯1 )2 + (𝑦2 − 𝑦1 )2
Find the distance between each pair of points:
1) (6, -2) and (-5, 12)
2) (5.8, -1) and (7, -4)
Midpoints:
The midpoint of the interval with endpoints a and b is found by taking the average of the
endpoints.
𝑀=
π‘Ž+𝑏
2
The midpoint of a segment with endpoints at (x1, y1) and (x2, y2) is found by taking the average of
the two coordinate values.
π‘₯1 + π‘₯2 𝑦1 + 𝑦2
𝑀=(
,
)
2
2
Find the midpoint of each segment with the indicated endpoints:
3) (-4, 3) and (7, 12)
4) (10, -3) and (-14, -9)
2
Properties of Exponents:
1. Whole number exponents:
2. Zero exponents:
π‘₯ 𝑛 = π‘₯ βˆ™ π‘₯ βˆ™ π‘₯ βˆ™ … βˆ™ π‘₯ (n factors of x)
π‘₯ 0 = 1, π‘₯ ≠ 0
3. Negative Exponents:
1
π‘₯ −𝑛 = π‘₯ 𝑛
𝑛
√π‘₯ = π‘Ž → π‘₯ = π‘Žπ‘›
4. Radicals (principal nth root):
5. Rational exponents:
π‘₯
1⁄
𝑛
6. Rational exponents:
π‘₯
π‘š⁄
𝑛
𝑛
= √π‘₯
𝑛
= √π‘₯ π‘š
Operations with Exponents:
1. Multiplying like bases:
π‘₯ 𝑛 π‘₯ π‘š = π‘₯ π‘š+𝑛
2. Dividing like bases:
π‘₯π‘š
π‘₯𝑛
3. Removing parentheses:
(π‘₯𝑦)𝑛 = π‘₯ 𝑛 𝑦 𝑛
= π‘₯ π‘š−𝑛
π‘₯ 𝑛
π‘₯𝑛
(𝑦) = 𝑦𝑛
(π‘₯ 𝑛 )π‘š = π‘₯ π‘›π‘š
Simplify each of the following expressions:
5) πŸπ’‚πŸ 𝒃−πŸ’ βˆ™ πŸ’π’‚−πŸ– π’ƒπŸ”
6)
πŸ–π’‚πŸ 𝒃−𝟐
πŸ’π’‚πŸ’ 𝒄−πŸ“
πŸ‘π’™πŸ’
Special Products and Factorization Techniques
Quadratic Formula:
π‘Žπ‘₯ 2 + 𝑏π‘₯ + 𝑐 = 0 → π‘₯ =
πŸ‘
7) (π’š−𝟐)
−𝑏±√𝑏2 −4π‘Žπ‘
2π‘Ž
Solve the following equations:
8) πŸ‘π’™πŸ − πŸ’π’™ − 𝟐 = 𝟎
9) −πŸ“π’™πŸ + πŸ”π’™ + πŸ’ = 𝟎
3
Special Products:
π‘₯ 2 − π‘Ž2 = (π‘₯ − π‘Ž)(π‘₯ + π‘Ž)
π‘₯ 3 − π‘Ž3 = (π‘₯ − π‘Ž)(π‘₯ 2 + π‘Žπ‘₯ + π‘Ž2 )
π‘₯ 3 + π‘Ž3 = (π‘₯ + π‘Ž)(π‘₯ 2 − π‘Žπ‘₯ + π‘Ž2 )
Factor each of the following:
10) πŸ’π’™πŸ − πŸπŸ’πŸ’
11) πŸ–π’™πŸ‘ − πŸπŸ•
12) πŸ”πŸ’π’™πŸ‘ + πŸπŸπŸ“π’šπŸ‘
13) π’™πŸ − πŸπŸ‘π’™ + πŸ’πŸ
14) πŸ–π’™πŸ − πŸπŸŽπ’™ − πŸ‘
15) πŸ’π’™πŸ‘ + πŸ–π’™πŸ − πŸ“π’™ − 𝟏𝟎
Binomial Theorem:
(π‘₯ + π‘Ž)2 = π‘₯ 2 + 2π‘Žπ‘₯ + π‘Ž2
(π‘₯ − π‘Ž)2 = π‘₯ 2 − 2π‘Žπ‘₯ + π‘Ž2
(π‘₯ + π‘Ž)3 = π‘₯ 3 + 3π‘Žπ‘₯ 2 + 3π‘Ž2 π‘₯ + π‘Ž3
(π‘₯ − π‘Ž)3 = π‘₯ 3 − 3π‘Žπ‘₯ 2 + 3π‘Ž2 π‘₯ − π‘Ž3
Expand each of the following:
16) (𝒙 + πŸ’)𝟐
17) (𝒙 − πŸ”)𝟐
18) (𝒙 + πŸ‘)πŸ‘
4
19) (𝒙 − 𝟐)πŸ‘
Lines
Slope:
𝑦2 −𝑦1
π‘₯2 −π‘₯1
Slope Intercept Form:
y = mx + b
Standard Form:
ax + by = c
Point-Slope Form:
y - y1 = m(x – x1)
Use the two points to write the equation of the line in all three forms:
20) (2, -5) and (7, 12)
Transformations
Vertical Translations:
𝑦 = 𝑓(π‘₯) ± 𝑐
Horizontal Translations:
𝑦 = 𝑓(π‘₯ ± 𝑐)
Y-axis flip:
𝑦 = 𝑓(−π‘₯)
X-axis flip:
𝑦 = −𝑓(π‘₯)
Describe each of the following transformations:
21) 𝒇(𝒙) = −π’™πŸ + πŸ’
22) 𝒇(𝒙) = |−𝒙| − πŸ’
5
23) 𝒇(𝒙) = √𝒙 − πŸ‘
Functions
Domain:
a set of all possible values for the independent variable
Range:
a set of all possible values for the dependent variable
Find the domain and range of the following functions:
24) π’š = √𝒙 + 𝟏
πŸ‘π’™−πŸ’
25) π’š = πŸ’π’™+𝟏𝟎
26) π’š = {
𝟏 − 𝒙, 𝒙 < 1
√𝒙 − 𝟏, 𝒙 ≥ 𝟏
Even and Odd Functions:
𝐼𝑓 𝑓(−π‘₯) = −𝑓(π‘₯), π‘‘β„Žπ‘’π‘› π‘‘β„Žπ‘’ π‘“π‘’π‘›π‘π‘‘π‘–π‘œπ‘› 𝑖𝑠 π‘œπ‘‘π‘‘
𝐼𝑓 𝑓(−π‘₯) = 𝑓(π‘₯), π‘‘β„Žπ‘’π‘› π‘‘β„Žπ‘’ π‘“π‘’π‘›π‘π‘‘π‘–π‘œπ‘› 𝑖𝑠 𝑒𝑣𝑒𝑛
Determine if the function is even or odd:
27) π’š = πŸ‘π’™πŸ
28) π’š = πŸπ’™πŸ + πŸ’π’™
29) π’š = πŸ’π’™πŸ‘
End Behavior:
-If the degree of f is even and the lead term coefficient is positive, then the left and right
ends of the function both approach positive infinity.
-If the degree of f is even and the lead term coefficient is negative, then the left and
right ends of the function both approach negative infinity.
-If the degree of f is odd and the lead term coefficient is positive, then the left end
approaches negative infinity and the right end approaches positive infinity.
-If the degree of f is odd and the lead term coefficient is negative, then the left end
approaches positive infinity and the right end approaches negative infinity.
Describe the end behavior for each of the following functions:
30) 𝒇(𝒙) = −πŸ‘π’™πŸ + πŸ’π’™ − 𝟐
31) 𝒇(𝒙) = πŸπ’™πŸ‘ − πŸ‘π’™πŸ + πŸ”π’™ − 𝟏
6
32) 𝒇(𝒙) = πŸ“π’™πŸ“ − πŸ”
Functions
Evaluate each of the following for the function 𝒇(𝒙) = π’™πŸ − πŸ’π’™ + πŸ•:
34) 𝒇(π’šπŸ‘ ) =
33) 𝒇(πŸ’) =
35) 𝒇(𝒙 + π’š) =
Evaluate the following compositions of functions:
Let 𝒇(𝒙) = πŸπ’™ − πŸ‘ and π’ˆ(𝒙) = π’™πŸ + 𝟏
36) 𝒇(𝒙) βˆ™ π’ˆ(𝒙)
37) 𝒇(π’ˆ(𝒙))
38) π’ˆ(𝒇(𝒙))
Inverse Functions
In order to calculate an inverse of a function algebraically, you must switch all of the x and y
variables and solve the new equation for y. The inverse only exists if the resulting equation is a
function.
Find the inverse (if it exists) of each of the following functions:
39) 𝒇(𝒙) = πŸ‘π’™ + 𝟐
40) 𝒇(𝒙) = πŸπ’™πŸ − πŸ’
7
41) 𝒇(𝒙) = √𝒙 + 𝟏
Logarithms
Natural Logarithmic Function:
𝑙𝑛π‘₯ = 𝑏 𝑖𝑓 π‘Žπ‘›π‘‘ π‘œπ‘›π‘™π‘¦ 𝑖𝑓 𝑒 𝑏 = π‘₯
Inverse Properties of Logarithms:
𝑙𝑛𝑒 π‘₯ = π‘₯
𝑒 𝑙𝑛π‘₯ = π‘₯
Solve each of the following equations:
42) 𝟏𝟎 + π’†πŸŽ.πŸπ’™ = πŸπŸ’
43) πŸ‘ + πŸπ’π’π’™ = πŸ•
Graph the following equations without a calculator:
𝒍𝒏𝒙 + 𝟏, 𝒙 < 1
46) 𝒇(𝒙) = { 𝒙−𝟏
𝒆 ,𝒙 ≥ 𝟏
45) 𝒇(𝒙) = −𝟐 + 𝒆𝒙+𝟏
44) 𝒇(𝒙) = 𝒍𝒏(𝒙 − 𝟐) + πŸ‘
y
y
x
y
x
8
x
Trigonometry
Reciprocal Identities:
1
𝑠𝑖𝑛𝑒 = 𝑐𝑠𝑐𝑒
1
π‘π‘œπ‘ π‘’ = 𝑠𝑒𝑐𝑒
1
π‘‘π‘Žπ‘›π‘’ = π‘π‘œπ‘‘π‘’
1
𝑐𝑠𝑐𝑒 = 𝑠𝑖𝑛𝑒
1
𝑠𝑒𝑐𝑒 = π‘π‘œπ‘ π‘’
1
π‘π‘œπ‘‘π‘’ = π‘‘π‘Žπ‘›π‘’
Quotient Identities:
𝑠𝑖𝑛𝑒
π‘π‘œπ‘ π‘’
π‘‘π‘Žπ‘›π‘’ = π‘π‘œπ‘ π‘’
π‘π‘œπ‘‘π‘’ = 𝑠𝑖𝑛𝑒
Pythagoren Identities:
𝑠𝑖𝑛2 𝑒 + π‘π‘œπ‘  2 𝑒 = 1
1 + π‘‘π‘Žπ‘›2 𝑒 = 𝑠𝑒𝑐 2 𝑒
1 + π‘π‘œπ‘‘ 2 𝑒 = 𝑐𝑠𝑐 2 𝑒
47) Fill in the following table:
Degree
Radians
SINE
COS
TAN
0
30
45
60
90
120
135
150
180
210
225
240
270
300
315
330
360
9
CSC
SEC
COT
Graph one period for each of the trig. functions.
y
y
x
y
x
x
48) y=sinx
49) y=cosx
50) y=tanx
Domain:
Domain:
Domain:
Range:
Range:
Range:
y
y
x
y
x
x
48) y=cscx
49) y=secx
50) y=cotx
Domain:
Domain:
Domain:
Range:
Range:
Range:
10
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