DOCSLIB.ORG

Injective, Surjective and Bijective Functions

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Injective, Surjective and Bijective Functions DCS Module 1 Part 3 FUNCTION Functions :- Injective, Surjective and Bijective functions - Inverse of a function- Composition ​ :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: A function is a special case of relation. ​ ​ Definition: Let X and Y be any two sets. A relation f from X to Y is called a function if for ​ ​ ​ every x∈ X, there is a unique element y ∈ Y such that (x, y) ∈ f. ​ ​ For every x there should be unique y .i.e. x has only one image. A relation must satisfy two additional conditions to become a function. These conditions are: ​ ​ 1. Every x∈ X must be related to y ∈ Y. ○ Domain of f must be X 2. Uniqueness . Every x has only one y image ○ (x,y)∈ f ∧ (x,z)∈ f ⇒ y=z Function f from set X to Y can be expressed as: f: X→Y or If (x,y) ∈ f then x is called an argument and Y is called image of x under f. ​ ​ ​ ​ (x,y) ∈ f is same as y=f(x) ​ f: x→y or Domain of function f from X to Y is denoted as D ​ ​f . Df=X ​ ​ Range of function f is denoted as Rf is defined as:- ​ ​ ​ Rf=f(X) ​ ​ Range of f, Rf is defined as:- ​ ​ Rf={y| ∃x∊X ∧ y=f(x) } ∃means “there exists” ​ ​ Rf ⊆Y ​ ​ Codomain f:X → Y. Here set Y is called codomain of function f ​ ​ E.g. Let X={1,3,4} Y={2,5,6} Check whether following are functions from X to Y or not ​ (1)R1={(1,2),(3,5),(4, 2)} (2)R2={(1,2)(1,5),(3,2)(4,6)} (3)R3={(1,2),(3,2)} R1 is a function because every x has unique image , Domain of R={1,3,4}=X R2 is not a function because one of the x value 1 has two images since (1,2) and (1,5) R3 is not a function because domain of R3={1,3}≠X E.g. Find the domain,codomain and range of the function ​ S={1,2,3} T={a,b} f:S->T f={(1,a),(2,b),(3,a)} Domain Df={1,2,3} ​ ​ Range Rf={a,b} ​ ​ Codomain=T={a,b} E.g. 1 Let X = {1, 2, 3}, Y = {p, q, r}, f : X → Y ​ and f = {(1, p), (2, q), (3,q)} then f(1) = p f(2) = q f(3)= r. Df={1,2,3} ​ ​ ​ Rf={p,q} ​ ​ ​ Codomain=Y={p,q,r} ​ E.g. 2: 2 If the function f is defined by f(x)=x +​ 1 on the set {−2, −1, 0, 1, 2}, find the range of f. ​ Ans. 2 f(−2) = (−2) +​ 1=5 2​ f(−1) = (−1) +​ 1 = 2 ​ f(0) = 0 + 1 = 1 f(1) = 1 + 1 = 2 f(2) = 4 + 1 = 5 f={(-2,5),(-1,2),(0,1),(1,2),(2,5) } So the range of f = {1, 2, 5}. Domain of f=x= {-2,-1,0,1,2} 2 E.g. Let X=Y=R and f(x)=x +​ 2 Df=R and Rf⊆R. Values are different and lie on a parabola. So ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ this is a function. E.g. Let X=Y=R set of real numbers(set of positive negative integers,rational numbers zero etc) ​ and let 2 f={(x,x )​ | x∊R} 2 ​ g={(x ,x)​ | x∊R} ​ 2 2 Here f is a function from X to Y f(x)= x ,​ because for every x there is only one x .​ ​ 2 ​ 2 But g is not a function from X to Y because for any real number a, we have (a ,a)​ and (a ,-a)​ in 2 2 ​ ​ g. This means that a ​ has two images a and –a. If a ​ i s 4 then a can be 2 and -2. But in a function ​ ​ for every x ϵX, there should be a unique image y ϵY. So g is not a function E.g. Let P is a set of all positive integers and f:P→P such that f(n)=n+1 where nϵP f(1)=2, f(2)=3 … … The function f is called Peano’s successor function. ​ ​ E.g. Let ​ f={(x,⌊x⌋)| x∊R ∧ ⌊x⌋ = the greatest integer less than or equal to x} ​ ​ ​ ​ g={{(x,⌈x⌉)| x∊R ∧ ⌈x⌉ = the least integer greater than or equal to x} ​ ​ ​ ​ Function f(x)=⌊x⌋is called floor of x. (whole number just before that ) ​ ​ ​ Function g(x)= ⌈x⌉ is called ceil of x. (whole number just after that ) ​ ​ ​ ​ f(3.6)=3 g(3.6)=4 f(-3.6)=-4 g(-3.6)=-3 E.g. If R is a set of real numbers identify which of the following relations define a function? 2 (1) f={(x,x )|x​ ∈R} ​ 2 Every real number has a unique square eg: -1 =1​ ​ So f is a function 2 (2) g={(x ,x)|x​ ∈R} ​ Real number can have more than one square root e.g. square root of 4 is -2 and 2 So (4,-2) ,(4,2) is in R here 4 has two images(y values) -2 and 2 So g is not a function RESTRICTION AND EXTENSION If f : X → Y and A ⊆ X then f ∩(A ×Y) is a function g from A→ Y . g is called restriction of f ​ ​ ​ to A denoted as f/A. If g is a restriction of f then f is called extension of g. ​ ​ ​ ​ E.g. Let f : R → R. The function f(x)=|x| (Here |x| denote absolute value of x) ​ and R+ ⊆ R.and g:R+ → R ​ ​ R is a set of real numbers. R+ is a set of positive real numbers. Here g is a restriction of f. f is the extension of g. TYPES OF FUNCTION 1. Onto(surjective): ​ 2. One-to-one(Injection) 3. Oneto-one onto(bijective) 1.Onto(surjective) ​ A mapping f : X → Y is called onto or surjective or surjection if range Rf=Y ​ ​ ​ ​ ​ Otherwise i.e . Rf≠Y then function is not into. ​ ​ ​ ​ ​ E.g. X={1,2,3} Y={a,b} f : X → Y f={(1,a),(2,b),(3,a)} Here range Rf={a,b}=Y ​ ​ ​ So f is onto. ​ ​ E.g. X={1,2,3} Y={a,b,c} f : X → Y ​ g={(1,a),(2,b),(3,a)} Here range Rf={a,b}≠Y So g is not onto but into. ​ ​ ​ ​ ​ 2. One-to-one(Injection): A mapping f : X → Y is called one-to-one if distinct elements of X ​ ​ are mapped into distinct elements of Y , ​ i.e., f is one-to-one if x1 ≠ x2 ⇒ f(x1) ≠ f(x2) Or f(x1) = f(x2) ⇒ x1 = x2 for x1, x2 ∈ X. ⇒means “implies” or “”it means” ​ ​ ​ f(x1) = f(x2) ⇒ x1 = x2 is read as “if f(x1)=f(x2) then x1 =x2” E.g. Let X={a,b,c} Y={1,2,3,4} f : X → Y f={(a,2),(b,3),(c,4)} Here f is one-to-one(injection) f is not onto but f is into. 3. One-to-one onto(Bijective) ​ or one-to-one correspondence between two sets A mapping f : X → Y is called one-to-one onto(bijective) if it is both onto and one-to-one. ​ ​ ​ ​ ​ i.e. distinct elements of X are mapped into distinct elements of Y and Range of f, Rf=Y ​ ​ ​ ​ ​ E.g. Let X={1,2,3} Y={a,b,c} f :X→Y. f={(1,b),(2,a),(3,c}) Domain of f={1,2,3}=X Each element in x is mapped to different element in Y. Range of f=(a,b,c}=Y So f is bijective(one-to-one onto) E.g. Let X={a,b,c,k} and Y={1,2,3,4}.The function f :X→Y. Identify which of the following are ​ functions.Check whether they are onto,one-to-one or bijective.. Find the domain, range and co-domain of the function. 1)f={(a,1),(b,2),(c,3),(k,4) } Ans. f is a function from X→Y because here every element in X has unique image in Y and every element in X has image. Domain={a,b,c,k} Range={1,2,3,4}=Y Co-domain={1,2,3,4} Here distinct elements of X are mapped into distinct elements of Y. So this is one to one. ​ ​ ​ ​ Here Range set=Y So function is onto. So this function f is bijective(both one-to-one and onto). 2) g={(a,1),(c,2)} Here g is not a function since every element in X is not related to Y.Here b and k inX has no image in Y. 3) h={(a,1),(b,2),(b,3),(c,4),(k,2)} Here h is not a function since element b in X has more than one image in Y .(b,2 )and (b,3). 4) j={(a,1),(b,1),(c,1),(k,1)} j is a function. Here every element in X has unique image in Y. But a,b,c and k has same image inY i.e, 1 j is into.Not one-to-one Not onto. ​ ​ Domain={a,b,c,k} Range={1} Co-domain={1,2,3,4} Note :Range of a function may not be same as its co-domain. ​ Composition of function Let f : X → Y and g : Y → Z be two functions. ​ ​ ​ The composition relation gof is a relation from X→ Z ​ ​ gof is possible if and only if range of f is subset of domain of g.
Load more

Recommended publications
  • The Matrix Calculus You Need for Deep Learning
    The Matrix Calculus You Need For Deep Learning Terence Parr and Jeremy Howard July 3, 2018 (We teach in University of San Francisco's MS in Data Science program and have other nefarious projects underway. You might know Terence as the creator of the ANTLR parser generator. For more material, see Jeremy's fast.ai courses and University of San Francisco's Data Institute in- person version of the deep learning course.) HTML version (The PDF and HTML were generated from markup using bookish) Abstract This paper is an attempt to explain all the matrix calculus you need in order to understand the training of deep neural networks. We assume no math knowledge beyond what you learned in calculus 1, and provide links to help you refresh the necessary math where needed. Note that you do not need to understand this material before you start learning to train and use deep learning in practice; rather, this material is for those who are already familiar with the basics of neural networks, and wish to deepen their understanding of the underlying math. Don't worry if you get stuck at some point along the way|just go back and reread the previous section, and try writing down and working through some examples. And if you're still stuck, we're happy to answer your questions in the Theory category at forums.fast.ai. Note: There is a reference section at the end of the paper summarizing all the key matrix calculus rules and terminology discussed here. arXiv:1802.01528v3 [cs.LG] 2 Jul 2018 1 Contents 1 Introduction 3 2 Review: Scalar derivative rules4 3 Introduction to vector calculus and partial derivatives5 4 Matrix calculus 6 4.1 Generalization of the Jacobian .
    [Show full text]
  • IVC Factsheet Functions Comp Inverse
    Imperial Valley College Math Lab Functions: Composition and Inverse Functions FUNCTION COMPOSITION In order to perform a composition of functions, it is essential to be familiar with function notation. If you see something of the form “푓(푥) = [expression in terms of x]”, this means that whatever you see in the parentheses following f should be substituted for x in the expression. This can include numbers, variables, other expressions, and even other functions. EXAMPLE: 푓(푥) = 4푥2 − 13푥 푓(2) = 4 ∙ 22 − 13(2) 푓(−9) = 4(−9)2 − 13(−9) 푓(푎) = 4푎2 − 13푎 푓(푐3) = 4(푐3)2 − 13푐3 푓(ℎ + 5) = 4(ℎ + 5)2 − 13(ℎ + 5) Etc. A composition of functions occurs when one function is “plugged into” another function. The notation (푓 ○푔)(푥) is pronounced “푓 of 푔 of 푥”, and it literally means 푓(푔(푥)). In other words, you “plug” the 푔(푥) function into the 푓(푥) function. Similarly, (푔 ○푓)(푥) is pronounced “푔 of 푓 of 푥”, and it literally means 푔(푓(푥)). In other words, you “plug” the 푓(푥) function into the 푔(푥) function. WARNING: Be careful not to confuse (푓 ○푔)(푥) with (푓 ∙ 푔)(푥), which means 푓(푥) ∙ 푔(푥) . EXAMPLES: 푓(푥) = 4푥2 − 13푥 푔(푥) = 2푥 + 1 a. (푓 ○푔)(푥) = 푓(푔(푥)) = 4[푔(푥)]2 − 13 ∙ 푔(푥) = 4(2푥 + 1)2 − 13(2푥 + 1) = [푠푚푝푙푓푦] … = 16푥2 − 10푥 − 9 b. (푔 ○푓)(푥) = 푔(푓(푥)) = 2 ∙ 푓(푥) + 1 = 2(4푥2 − 13푥) + 1 = 8푥2 − 26푥 + 1 A function can even be “composed” with itself: c.
    [Show full text]
  • The Logic of Recursive Equations Author(S): A
    The Logic of Recursive Equations Author(s): A. J. C. Hurkens, Monica McArthur, Yiannis N. Moschovakis, Lawrence S. Moss, Glen T. Whitney Source: The Journal of Symbolic Logic, Vol. 63, No. 2 (Jun., 1998), pp. 451-478 Published by: Association for Symbolic Logic Stable URL: http://www.jstor.org/stable/2586843 . Accessed: 19/09/2011 22:53 Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at . http://www.jstor.org/page/info/about/policies/terms.jsp JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range of content in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new forms of scholarship. For more information about JSTOR, please contact [email protected]. Association for Symbolic Logic is collaborating with JSTOR to digitize, preserve and extend access to The Journal of Symbolic Logic. http://www.jstor.org THE JOURNAL OF SYMBOLIC LOGIC Volume 63, Number 2, June 1998 THE LOGIC OF RECURSIVE EQUATIONS A. J. C. HURKENS, MONICA McARTHUR, YIANNIS N. MOSCHOVAKIS, LAWRENCE S. MOSS, AND GLEN T. WHITNEY Abstract. We study logical systems for reasoning about equations involving recursive definitions. In particular, we are interested in "propositional" fragments of the functional language of recursion FLR [18, 17], i.e., without the value passing or abstraction allowed in FLR. The 'pure," propositional fragment FLRo turns out to coincide with the iteration theories of [1]. Our main focus here concerns the sharp contrast between the simple class of valid identities and the very complex consequence relation over several natural classes of models.
    [Show full text]
  • Monoids: Theme and Variations (Functional Pearl)
    University of Pennsylvania ScholarlyCommons Departmental Papers (CIS) Department of Computer & Information Science 7-2012 Monoids: Theme and Variations (Functional Pearl) Brent A. Yorgey University of Pennsylvania, [email protected] Follow this and additional works at: https://repository.upenn.edu/cis_papers Part of the Programming Languages and Compilers Commons, Software Engineering Commons, and the Theory and Algorithms Commons Recommended Citation Brent A. Yorgey, "Monoids: Theme and Variations (Functional Pearl)", . July 2012. This paper is posted at ScholarlyCommons. https://repository.upenn.edu/cis_papers/762 For more information, please contact [email protected]. Monoids: Theme and Variations (Functional Pearl) Abstract The monoid is a humble algebraic structure, at first glance ve en downright boring. However, there’s much more to monoids than meets the eye. Using examples taken from the diagrams vector graphics framework as a case study, I demonstrate the power and beauty of monoids for library design. The paper begins with an extremely simple model of diagrams and proceeds through a series of incremental variations, all related somehow to the central theme of monoids. Along the way, I illustrate the power of compositional semantics; why you should also pay attention to the monoid’s even humbler cousin, the semigroup; monoid homomorphisms; and monoid actions. Keywords monoid, homomorphism, monoid action, EDSL Disciplines Programming Languages and Compilers | Software Engineering | Theory and Algorithms This conference paper is available at ScholarlyCommons: https://repository.upenn.edu/cis_papers/762 Monoids: Theme and Variations (Functional Pearl) Brent A. Yorgey University of Pennsylvania [email protected] Abstract The monoid is a humble algebraic structure, at first glance even downright boring.
    [Show full text]
  • Contents 3 Homomorphisms, Ideals, and Quotients
    Ring Theory (part 3): Homomorphisms, Ideals, and Quotients (by Evan Dummit, 2018, v. 1.01) Contents 3 Homomorphisms, Ideals, and Quotients 1 3.1 Ring Isomorphisms and Homomorphisms . 1 3.1.1 Ring Isomorphisms . 1 3.1.2 Ring Homomorphisms . 4 3.2 Ideals and Quotient Rings . 7 3.2.1 Ideals . 8 3.2.2 Quotient Rings . 9 3.2.3 Homomorphisms and Quotient Rings . 11 3.3 Properties of Ideals . 13 3.3.1 The Isomorphism Theorems . 13 3.3.2 Generation of Ideals . 14 3.3.3 Maximal and Prime Ideals . 17 3.3.4 The Chinese Remainder Theorem . 20 3.4 Rings of Fractions . 21 3 Homomorphisms, Ideals, and Quotients In this chapter, we will examine some more intricate properties of general rings. We begin with a discussion of isomorphisms, which provide a way of identifying two rings whose structures are identical, and then examine the broader class of ring homomorphisms, which are the structure-preserving functions from one ring to another. Next, we study ideals and quotient rings, which provide the most general version of modular arithmetic in a ring, and which are fundamentally connected with ring homomorphisms. We close with a detailed study of the structure of ideals and quotients in commutative rings with 1. 3.1 Ring Isomorphisms and Homomorphisms • We begin our study with a discussion of structure-preserving maps between rings. 3.1.1 Ring Isomorphisms • We have encountered several examples of rings with very similar structures. • For example, consider the two rings R = Z=6Z and S = (Z=2Z) × (Z=3Z).
    [Show full text]
  • Math 311 - Introduction to Proof and Abstract Mathematics Group Assignment # 15 Name: Due: at the End of Class on Tuesday, March 26Th
    Math 311 - Introduction to Proof and Abstract Mathematics Group Assignment # 15 Name: Due: At the end of class on Tuesday, March 26th More on Functions: Definition 5.1.7: Let A, B, C, and D be sets. Let f : A B and g : C D be functions. Then f = g if: → → A = C and B = D • For all x A, f(x)= g(x). • ∈ Intuitively speaking, this definition tells us that a function is determined by its underlying correspondence, not its specific formula or rule. Another way to think of this is that a function is determined by the set of points that occur on its graph. 1. Give a specific example of two functions that are defined by different rules (or formulas) but that are equal as functions. 2. Consider the functions f(x)= x and g(x)= √x2. Find: (a) A domain for which these functions are equal. (b) A domain for which these functions are not equal. Definition 5.1.9 Let X be a set. The identity function on X is the function IX : X X defined by, for all x X, → ∈ IX (x)= x. 3. Let f(x)= x cos(2πx). Prove that f(x) is the identity function when X = Z but not when X = R. Definition 5.1.10 Let n Z with n 0, and let a0,a1, ,an R such that an = 0. The function p : R R is a ∈ ≥ ··· ∈ 6 n n−1 → polynomial of degree n with real coefficients a0,a1, ,an if for all x R, p(x)= anx +an−1x + +a1x+a0.
    [Show full text]
  • List of Mathematical Symbols by Subject from Wikipedia, the Free Encyclopedia
    List of mathematical symbols by subject From Wikipedia, the free encyclopedia This list of mathematical symbols by subject shows a selection of the most common symbols that are used in modern mathematical notation within formulas, grouped by mathematical topic. As it is virtually impossible to list all the symbols ever used in mathematics, only those symbols which occur often in mathematics or mathematics education are included. Many of the characters are standardized, for example in DIN 1302 General mathematical symbols or DIN EN ISO 80000-2 Quantities and units – Part 2: Mathematical signs for science and technology. The following list is largely limited to non-alphanumeric characters. It is divided by areas of mathematics and grouped within sub-regions. Some symbols have a different meaning depending on the context and appear accordingly several times in the list. Further information on the symbols and their meaning can be found in the respective linked articles. Contents 1 Guide 2 Set theory 2.1 Definition symbols 2.2 Set construction 2.3 Set operations 2.4 Set relations 2.5 Number sets 2.6 Cardinality 3 Arithmetic 3.1 Arithmetic operators 3.2 Equality signs 3.3 Comparison 3.4 Divisibility 3.5 Intervals 3.6 Elementary functions 3.7 Complex numbers 3.8 Mathematical constants 4 Calculus 4.1 Sequences and series 4.2 Functions 4.3 Limits 4.4 Asymptotic behaviour 4.5 Differential calculus 4.6 Integral calculus 4.7 Vector calculus 4.8 Topology 4.9 Functional analysis 5 Linear algebra and geometry 5.1 Elementary geometry 5.2 Vectors and matrices 5.3 Vector calculus 5.4 Matrix calculus 5.5 Vector spaces 6 Algebra 6.1 Relations 6.2 Group theory 6.3 Field theory 6.4 Ring theory 7 Combinatorics 8 Stochastics 8.1 Probability theory 8.2 Statistics 9 Logic 9.1 Operators 9.2 Quantifiers 9.3 Deduction symbols 10 See also 11 References 12 External links Guide The following information is provided for each mathematical symbol: Symbol: The symbol as it is represented by LaTeX.
    [Show full text]
  • Written Homework # 5 Solution
    Math 516 Fall 2006 Radford Written Homework # 5 Solution 12/12/06 Throughout R is a ring with unity. Comment: It will become apparent that the module properties 0¢m = 0, ¡(r¢m) = (¡r)¢m, and (r ¡ r0)¢m = r¢m ¡ r0¢m are vital details in some problems. 1. (20 total) Let M be an (additive) abelian group and End(M) be the set of group homomorphisms f : M ¡! M. (a) (12) Show End(M) is a ring with unity, where (f+g)(m) = f(m)+g(m) and (fg)(m) = f(g(m)) for all f; g 2 End(M) and m 2 M. Solution: This is rather tedious, but not so unusual as a basic algebra exercise. The trick is to identify all of the things, large and small, which need to be veri¯ed. We know that the composition of group homomorphisms is a group homomorphism. Thus End (M) is closed under function composition. Moreover End (M) is a monoid since composition is an associative op- eration and the identity map IM of M is a group homomorphism. Let f; g; h 2 End (M). The sum f + g 2 End (M) since M is abelian as (f + g)(m + n) = f(m + n) + g(m + n) = f(m) + f(n) + g(m) + g(n) = f(m) + g(m) + f(n) + g(n) = (f + g)(m) + (f + g)(n) for all m; n 2 M. Thus End (M) is closed under function addition. 1 Addition is commutative since f + g = g + f as (f + g)(m) = f(m) + g(m) = g(m) + f(m) = (g + f)(m) for all m 2 M.
    [Show full text]
  • Remarks About the Square Equation. Functional Square Root of a Logarithm
    M A T H E M A T I C A L E C O N O M I C S No. 12(19) 2016 REMARKS ABOUT THE SQUARE EQUATION. FUNCTIONAL SQUARE ROOT OF A LOGARITHM Arkadiusz Maciuk, Antoni Smoluk Abstract: The study shows that the functional equation f (f (x)) = ln(1 + x) has a unique result in a semigroup of power series with the intercept equal to 0 and the function composition as an operation. This function is continuous, as per the work of Paulsen [2016]. This solution introduces into statistics the law of the one-and-a-half logarithm. Sometimes the natural growth processes do not yield to the law of the logarithm, and a double logarithm weakens the growth too much. The law of the one-and-a-half logarithm proposed in this study might be the solution. Keywords: functional square root, fractional iteration, power series, logarithm, semigroup. JEL Classification: C13, C65. DOI: 10.15611/me.2016.12.04. 1. Introduction If the natural logarithm of random variable has normal distribution, then one considers it a logarithmic law – with lognormal distribution. If the func- tion f being function composition to itself is a natural logarithm, that is f (f (x)) = ln(1 + x), then the function ( ) = (ln( )) applied to a random variable will give a new distribution law, which – similarly as the law of logarithm – is called the law of the one-and -a-half logarithm. A square root can be defined in any semigroup G. A semigroup is a non- empty set with one combined operation and a neutral element.
    [Show full text]
  • Unit 5 Function Operations (Book Sections 7.6 and 7.7)
    Unit 5 Function Operations (Book sections 7.6 and 7.7) NAME ______________________ PERIOD ________ Teacher ____________________ 1 Learning Targets Function 1. I can perform operations with functions. Operations 2. I can evaluate composite functions Function 3. I can write function rules for composite functions Composition Inverse 4. I can graph and identify domain and range of a Functions function and its inverse. 5. I can write function rules for inverses of functions and verify using composite functions . 2 Function Operations Date: _____________ After this lesson and practice, I will be able to… • perform operations with functions. (LT1) • evaluate composite functions. (LT2) Having studied how to perform operations with one function, you will next learn how to perform operations with several functions. Function Operation Notation Addition: (f + g) = f(x) + g(x) Multiplication: (f • g) = f(x) • g(x) Subtraction: (f - g) = f(x) - g(x) € € " f % f (x) Division $ '( x) = ,g(x) ≠ 0 # g & g(x) The domain€ of the results of each of the above function operation are the _____-values that are in the domains of both _____ and _____ (except for _____________, where you must exclude any _____-values that cause ____________. (Remember you cannot divide by zero) Function Operations (LT 1) Example 1: Given � � = 3� + 8 and � � = 2� − 12, find h(x) and k(x) and their domains: a) ℎ � = � + � � and b) � � = 2� � − � � Example 2: Given � � = �! − 1 and � � = � + 5, find h(x) and k(x) and their domains: !(!) a) ℎ � = � ∙ � � b) � � = !(!) 3 Your Turn 1: Given � � = 3� − 1, � � = 2�! − 3, and ℎ � = 7�, find each of the following functions and their domains.
    [Show full text]
  • Function Arithmetic Composition of Functions
    Nick Egbert MA 158 Lesson 6 Function arithmetic Recall that we define +; −; ·; = on functions by performing these operations on the out- puts. So we have • (f + g)(x) = f(x) + g(x) • (f − g)(x) = f(x) − g(x) • (fg)(x) = f(x)g(x) f f(x) • g (x) = g(x) ; provided that g(x) 6= 0: As an example, let f(x) = x2 + 1 and g(x) = 2x − 4: Then • (f + g)(x) = (x2 + 1) + (2x − 4) = x2 + 2x − 3 • (f − g)(x) = (x2 + 1) − (2x − 4) = x2 − 2x + 5 • (fg)(x) = (x2 + 1)(2x − 4) = 2x3 − 4x2 + 2x − 4; note that it is not necessary to multiply this out. You could leave it in factored form. f x2+1 • g (x) = 2x−4 We could also ask for something like (f + g)(3) or (fg)(3); etc. So, we could say (f + g)(3) = f(3) + g(3) and find f(3) and g(3) separately. This is beneficial when we don't actually care what (f + g)(x) is in general and only care about a specific value. Composition of functions Composition of functions is another way of saying that we perform a function on a function. There is an \inside" function and an \outside" function. The notation we use is f ◦ g; and we read this as \f composed of g" or \f composed with g." What this notation means in practice is (f ◦ g)(x) = f(g(x)): Here, first we apply the function g to our input x, then we apply the function f to g(x); so the output of g is the input for f.
    [Show full text]
  • MATH 436 Notes: Functions and Inverses
    MATH 436 Notes: Functions and Inverses. Jonathan Pakianathan September 12, 2003 1 Functions Definition 1.1. Formally, a function f : A → B is a subset f of A × B with the property that for every a ∈ A, there is a unique element b ∈ B such that (a, b) ∈ f. The set A is called the domain of f and the set B is the codomain of f. While the above definition provides a definition of a function purely in terms of set theory, it is usually not a useful picture to work with. However it does emphasize the important point that the domain and codomain of a function are an intrinsic part of any function f. Less formally, we usually think of a function f : A → B as a rule of assignment which assigns a unique output f(a) ∈ B for every input a ∈ A. The graph of f, denoted Graph(f) = {(a, f(a))|a ∈ A} ⊂ A × B then recovers the more formal set theoretic definition of the function. Definition 1.2. Let f : A → B be a function. Given S ⊂ A we define f(S)= {f(s)|s ∈ S}. Note that f(S) ⊆ B. f(S) is called the image of the set S under f. f(A) is called the image of f, and is denoted Im(f). Given T ⊂ B we define f −1(T )= {a ∈ A|f(a) ∈ T }. Note that f −1(T ) ⊆ A. f −1(T ) is called the preimage of the set T under f. Fix a function f : A → B, then it is easy to see that for all S ⊂ A, S ⊂ f −1(f(S)) and for all T ⊆ B, we have f(f −1(T )) ⊆ T .
    [Show full text]