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Chapter 2 – Functions and Function Operations Mini-Lesson Section 2.1 – Combining Functions Function notation can be expanded to include notation for the different ways we can combine functions as described below. Basic Mathematical Operations on Functions The basic mathematical operations are: addition, subtraction, multiplication, and division. When working with function notation, these operations will look like this: Addition " # + 2(#) Subtraction Multiplication " # − 2(#) " # ∙2 # Division C D E D ; g(x) ≠ 0 Many of the problems we will work in this lesson are problems you may already know how to do. You will just need to get used to some new notation. We will start with the operations of addition and subtraction. Problem 1 WORKED EXAMPLE – Adding and Subtracting Functions Given f (x) = 2x2 + 3x – 5 and g(x) = –x2 + 5x + 1. a) Find f (x) + g(x) f ( x) + g ( x) = (2 x 2 + 3 x − 5) + (− x 2 + 5 x + 1) = 2 x 2 + 3x − 5 − x 2 + 5 x + 1 f ( x) + g ( x) = x 2 + 8 x − 4 b) Find f (x) – g(x) f ( x) − g ( x) = (2 x 2 + 3x − 5) − (− x 2 + 5 x + 1) = 2 x 2 + 3x − 5 + x 2 − 5 x − 1 f ( x) − g ( x) = 3 x 2 − 2 x − 6 c) Find f (1) – g(1) f (1) − g (1) = [2(1) 2 + 3(1) − 5] − [−(1) 2 + 5(1) + 1] = [2 + 3 − 5] − [−1 + 5 + 1] = [0] − [5] f (1) − g (1) = −5 Page 58 Chapter 2 – Functions and Function Operations Problem 2 Mini-Lesson MEDIA/CLASS EXAMPLE – Adding and Subtracting Functions Given f (x) = 3x2 + 2x – 1 and g(x)= x2 + 2x + 5: a) Find f (x) + g(x) b) Find f (x) – 2g(x) Problem 3 YOU TRY – Adding and Subtracting Functions Given f (x) = x2 + 4 and g(x)= x2 + 1, determine each of the following. Show complete work. a) Find f (2) + g(2) b) Find 5f (x) – 2g(x) c) Find 3f (2) – g(2) Page 59 Chapter 2 – Functions and Function Operations Mini-Lesson MULTIPLICATION PROPERTY OF EXPONENTS Let m and n be rational numbers. To multiply powers of the same base, keep the base and add the exponents: FG · FI = FGJI Problem 4 WORKED EXAMPLE – Function Multiplication a) Given " # = −8# 4 and 2 # = 5# 6 , find " # · 2(#) f ( x) ⋅ g ( x) = (−8 x 4 )(5 x 3 ) = (−8)(5)( x 4 )( x 3 ) Reorder using Commutative Property Simplify using the Multiplication Property of Exponents = (−40)( x 7 ) f ( x) ⋅ g ( x) = −40 x 7 Final Result b) Given " # = −3# and 2 # = 4# 1 − # + 8, find " # · 2(#) f ( x) ⋅ g ( x) = (−3x)(4 x 2 − x + 8) = (−3x)(4 x 2 ) + (−3x)(− x) + (−3x)(8) f ( x) ⋅ g ( x) = −12 x 3 + 3x 2 − 24 x Apply the Distributive Property Remember the rules of exponents −3# 4# 1 = −3 4 # 3 # 1 = −12# 6 Final Result c) Given f (x) = 3x + 2 and g(x) = 2x – 5, find " # · 2 # f ( x) ⋅ g ( x) = (3x + 2)(2 x − 5) = (3x)(2 x) + (3x)(−5) + (2)(2 x) + (2)(−5) = FIRST + OUTER + INNER + LAST = (6 x 2 ) + (−15 x) + (4 x) + (−10) f ( x) ⋅ g ( x) = 6 x 2 − 11x − 10 Use the Distributive Property Remember the rules of exponents 3# 2# ! = ! 3 2 # # = 6# 1 Combine Like Terms Final Result Page 60 Chapter 2 – Functions and Function Operations Problem 5 MEDIA/CLASS EXAMPLE – Function Multiplication Given " # = 3# + 2 and 2 # = 2# 1 + 3# + 1, find " # · 2(#) Problem 6 YOU TRY – Function Multiplication For each of the following, find " # · 2(#) a) f (x) = 3x – 2 and g(x) = 3x + 2 b) f (x) = 2x2 and g(x) = x3 – 4x + 5 c) f (x) = 4x3 and g(x) = –6x Page 61 Mini-Lesson Chapter 2 – Functions and Function Operations Mini-Lesson DIVISION PROPERTY OF EXPONENTS Let m, n be rational numbers. To divide powers of the same base, keep the base and subtract the exponents. KL KM Problem 7 = NOPQ where N ≠ 0 WORKED EXAMPLE – Function Division For each of the following, find C D E D . Use only positive exponents in your final answer. a) " # = 15# 3S and 2 # = 3# T f ( x) 15 x15 = g ( x) 3x 9 = 5 x15 − 9 = 5x 6 NEGATIVE EXPONENTS If a ≠ 0 and n is a rational number, then 1 a −n = n a b) " # = −4# S and 2 # = 2# U f ( x) − 4 x 5 = g ( x) 2 x8 = −2 x 5 − 8 = −2 x − 3 2 =− x3 Page 62 Chapter 2 – Functions and Function Operations Mini-Lesson DIVISION OF FUNCTIONS: When dividing two functions V D W D : If q(x) is a monomial, we can perform the division of each term of X(#) by Y(#). If q(x) is not a monomial, we will have to perform the long division algorithm..This is explained in problem 11 below. Problem 8 Determine WORKED EXAMPLE – DIVIDING MONOMIALS C D E D given f(x) = 4x2 + 5x – 3 and g(x) = 2x2. Use only positive exponents in your final answer. Since the denominator is a monomial, divide each term in the numerator by the denominator. 4 x 2 + 5x − 3 4 x 2 5x 3 = 2 + 2 − 2 2 2x 2x 2x 2x 5 3 = 2 + − 2 2x 2x Problem 9 MEDIA/CLASS EXAMPLE – DIVIDING MONOMIALS For each of the following, determine 4 3 2 C D E D . Use only positive exponents in your final answer. a) f(x) = 12x – 8x – 2x + 4x and g(x) = 4x b) f(x) = 5x3 + 15x2 – 25x and g(x) = 5x3 Page 63 Chapter 2 – Functions and Function Operations Problem 10 YOU TRY – DIVIDING MONOMIALS For each of the following, determine 3 Mini-Lesson 2 C D E D . Use only positive exponents in your final answer. a) f(x) = 8x – 6x + 4x – 12 and g(x) = 4x b) f(x) = 9x3 + 6x2 + 3x + 12 and g(x) = 3x2 Problem 11 WORKED EXAMPLE – LONG DIVISION Given f(x) = x3 – 6x2 – 5 and g(x) = x – 2. Determine C D E D by long division. STEP 1: Rewrite the terms in the numerator and denominator in descending order, putting x3 − 6x 2 + 0x − 5 a coefficient of 0 for the missing term: x−2 STEP 2: Divide the first term of the dividend by the first term of the divisor. x2 x − 2 x3 − 6x 2 + 0x + 5 Page 64 Chapter 2 – Functions and Function Operations STEP 3: Multiply the divisor by the result just obtained. x2 x − 2 x3 − 6x 2 + 0x + 5 x3 − 2x 2 STEP 4: Subtract like terms. x2 x3 − 6x 2 + 0x + 5 x−2 ( − x3 − 2x 2 ) − 4x 2 + 0x STEP 5: Repeat the previous three steps. x 2 − 4x x−2 x3 − 6x 2 + 0x + 5 ( − x3 − 2x 2 ) − 4x 2 + 0x − 4 x 2 + 8x STEP 6: Repeat the previous step. x 2 − 4x − 8 x−2 x3 − 6x 2 + 0x + 5 ( − x3 − 2x 2 ) 2 − 4x + 0x ( − − 4 x 2 + 8x ) − 8x + 5 − (− 8 x + 16 ) –11 (this is the remainder) Therefore, x3 − 6x 2 + 5 11 = x 2 − 4x − 8 − x−2 x−2 This is the long division algorithm for division of polynomials. Page 65 Mini-Lesson Chapter 2 – Functions and Function Operations Problem 12 MEDIA/CLASS EXAMPLE – LONG DIVISION Divide the following using long division. 4x3 + 2x 2 − 2x − 3 a) 2x + 1 Problem 13 Mini-Lesson b) (2x3 + 3x2 + 27) ÷!(x + 3) YOU TRY – LONG DIVISION Divide the following using long division. x3 − x 2 − 2x + 6 a) x−2 b) Page 66 (9x3 – 7x + 2) ÷ (3x + 2) Chapter 2 – Functions and Function Operations Mini-Lesson Functions can be presented in multiple ways including: equations, data sets, graphs, and applications. If you understand function notation, then the process for working with functions is the same no matter how the information if presented. Problem 14 MEDIA/CLASS EXAMPLE – Working with Functions in Table Form x f (x) –3 8 –2 6 0 3 1 2 4 5 5 8 8 11 10 15 12 20 x g(x) 0 1 2 3 3 5 4 10 5 4 8 2 9 0 11 –2 15 –5 Functions f (x) and g(x) are defined in the tables above. Find the following: a) f (1) = b) g(9) = c) f (0) + g(0) = d) g(5) – f(8) = e) " 0 ·2 3 = Problem 15 YOU TRY – Working with Functions in Table Form x f (x) 0 4 1 3 2 –2 3 0 4 1 x g(x) 0 6 1 –3 2 4 3 –2 4 2 Given the above two tables. Complete the table below. Show your work in the table cell for each column. The first one is done for you. x f (x)+ g(x) 0 f (0) + g(0) 1 2 =4+6 = 10 Page 67 3 4 Chapter 2 – Functions and Function Operations Mini-Lesson Remember that graphs are just infinite sets of ordered pairs. If you do a little work ahead of time (as in the example below) then the graphing problems are a lot easier to work with. Problem 16 YOU TRY – Working with Functions in Graph Form Use the graph to determine each of the following. Assume integer answers. The graph of g is the graph in bold. Complete the following ordered pairs from the graphs above. Use the information to help you with the problems below. The first ordered pair for each function has been completed for you. f: (–7, 2), (–6, (3, ), (4, g: (–7, 3), (–6, (3, a) ), (4, ), (–5, ), (5, ), (–4, ), (6, ), (–5, ), (5, ), (–3, ), (7, ), (–4, ), (6, ), (–2, ), (–1, ), (0, ), (1, ), (2, ), ), (–2, ), (–1, ), (0, ), (1, ), (2, ), ) ), (–3, ), (7, ) g(4) = ________________________ b) f (2) =________________ c) g(0) = ________________________ d) f (–6) = ______________ e) If f (x) = 0, x = ________________ f) If g(x) = 0, x = ___________ g) If f (x) = 1, x =____________________ h) If g(x) = – 4, x =__________ i) f (–1) + g(–1) =____________________ j) g(–6) – f (–6) =__________ k) " 1 ∗ 2 −2 =____________________ l) E \ C P3 Page 68 _______________