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Business Calculus Summer Assignment 2016 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 September 1, 2016. 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 β¦ 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: β2 β 1.4142135623 β¦ π β 3.1415926535 β¦ Properties of Inequalities: Let a, b, c and d be real numbers. 1. Transitive property: a<b and b<c ο a < c 2. Adding inequalities: a<b and c<d ο ac < bc, c > 0 3. Multiplying by a positive constant: a < b ο ac < bc, c > 0 4. Multiplying by a negative constant: a<b ο ac > bc, c < 0 5. Adding a constant: a < b ο a + c < b + c 6. Subtracting a constant: a < b ο a β c < b β c Definition of Absolute Value: The absolute value of a real number a is: π ππ π β₯0 βπ ππ π<0 |π| = { 1 π β 2.7182818284 Properties of Absolute Value: 1. Multiplication: |ππ| = |π||π| π |π| 2. Division: |π | = |π| , π β 0 3. Power: |ππ | = |π|π 4. Square Root: βπ2 = |π| Distance Between Two Points: The distance d between any two points x1 and x2 on a real number line is: π = |π₯1β π₯2 | = β(π₯2 β π₯1 )2 The distance d between any two points (x1, y1) and (x2, y2) on a Cartesian plane is: π = β(π₯2 β π₯1 )2 + (π¦2 β π¦1 )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. π= π+π 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 π¦1 + π¦2 π=( , ) 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₯ π = π₯ β π₯ β π₯ β β¦ β π₯ (n factors of x) π₯ 0 = 1, π₯ β 0 3. Negative Exponents: 1 π₯ βπ = π₯ π π βπ₯ = π β π₯ = ππ 4. Radicals (principal nth root): 5. Rational exponents: π₯ 1β π 6. Rational exponents: π₯ πβ π π = βπ₯ π = βπ₯ π Operations with Exponents: 1. Multiplying like bases: π₯ π π₯ π = π₯ π+π 2. Dividing like bases: π₯π π₯π 3. Removing parentheses: (π₯π¦)π = π₯ π π¦ π = π₯ πβπ π₯ π π₯π (π¦) = π¦π (π₯ π )π = π₯ ππ Simplify each of the following expressions: 5) πππ πβπ β ππβπ ππ 6) πππ πβπ πππ πβπ πππ Special Products and Factorization Techniques Quadratic Formula: ππ₯ 2 + ππ₯ + π = 0 β π₯ = π 7) (πβπ) βπ±βπ2 β4ππ 2π Solve the following equations: 8) πππ β ππ β π = π 9) βπππ + ππ + π = π 3 Special Products: π₯ 2 β π2 = (π₯ β π)(π₯ + π) π₯ 3 β π3 = (π₯ β π)(π₯ 2 + ππ₯ + π2 ) π₯ 3 + π3 = (π₯ + π)(π₯ 2 β ππ₯ + π2 ) Factor each of the following: 10) πππ β πππ 11) πππ β ππ 12) ππππ + πππππ 13) ππ β πππ + ππ 14) πππ β πππ β π 15) πππ + πππ β ππ β ππ Binomial Theorem: (π₯ + π)2 = π₯ 2 + 2ππ₯ + π2 (π₯ β π)2 = π₯ 2 β 2ππ₯ + π2 (π₯ + π)3 = π₯ 3 + 3ππ₯ 2 + 3π2 π₯ + π3 (π₯ β π)3 = π₯ 3 β 3ππ₯ 2 + 3π2 π₯ β π3 Expand each of the following: 16) (π + π)π 17) (π β π)π 18) (π + π)π 4 19) (π β π)π Lines Slope: π¦2 βπ¦1 π₯2 βπ₯1 Slope Intercept Form: y = mx + b Standard Form: ax + by = c Point-Slope Form: y - y1 = m(x β 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: π¦ = π(π₯) ± π Horizontal Translations: π¦ = π(π₯ ± π) Y-axis flip: π¦ = π(βπ₯) X-axis flip: π¦ = βπ(π₯) Describe each of the following transformations: 21) π(π) = βππ + π 22) π(π) = |βπ| β π 5 23) π(π) = βπ β π 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 = βπ + π ππβπ 25) π = ππ+ππ 26) π = { π β π, π < 1 βπ β π, π β₯ π Even and Odd Functions: πΌπ π(βπ₯) = βπ(π₯), π‘βππ π‘βπ ππ’πππ‘πππ ππ πππ πΌπ π(βπ₯) = π(π₯), π‘βππ π‘βπ ππ’πππ‘πππ ππ ππ£ππ Determine if the function is even or odd: 27) π = πππ 28) π = πππ + ππ 29) π = πππ 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) π(π) = βπππ + ππ β π 31) π(π) = πππ β πππ + ππ β π 6 32) π(π) = πππ β π Functions Evaluate each of the following for the function π(π) = ππ β ππ + π: 34) π(ππ ) = 33) π(π) = 35) π(π + π) = Evaluate the following compositions of functions: Let π(π) = ππ β π and π(π) = ππ + π 36) π(π) β π(π) 37) π(π(π)) 38) π(π(π)) 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) π(π) = ππ + π 40) π(π) = πππ β π 7 41) π(π) = βπ + π Logarithms Natural Logarithmic Function: πππ₯ = π ππ πππ ππππ¦ ππ π π = π₯ Inverse Properties of Logarithms: πππ π₯ = π₯ π πππ₯ = π₯ Solve each of the following equations: 42) ππ + ππ.ππ = ππ 43) π + ππππ = π Graph the following equations without a calculator: πππ + π, π < 1 46) π(π) = { πβπ π ,π β₯ π 45) π(π) = βπ + ππ+π 44) π(π) = ππ(π β π) + π y y x y x 8 x Properties of Logarithms Product Property: ππππ π + ππππ π = ππππ (ππ) Quotient Property: ππππ π β ππππ π = ππππ ( π ) Power Property: ππππ ππ = π β ππππ π π Condense each of the following to a single logarithm: π 48) π [πππ(π + π) + πππ β ππ(ππ β π)] 47) ππππ + ππππ β πππ Rewrite the expression as a sum, difference or multiple of logarithms: ππ ππ(π+π) 49) ππβπ+π 50) ππ (ππ+π)π Compound Interest Let P be the amount deposited, t the number of years, A the balance and r the annual interest rate (in decimal form). π ππ‘ 1. Compounded n time per year: π΄ = π (1 + π) 2. Compounded continuously: π΄ = ππ ππ‘ Determine the balance A for P dollars invested at rate r for t years, compounded n times per year. 51) P=$1000, r=3%, t=10 years t P 1 10 20 52) P=$2500, r=5%, t=40 years 30 continuous t P 9 1 10 20 30 continuous