Relations and Functions

Transcription

1 Relations and Functions. RELATION Mathematical Concepts Any pair of elements (, y) is called an ordered pair where is the first component (abscissa) and y is the second component (ordinate). Relations between these two variables can be revealed when written as a set of ordered pair. Thus, any set of ordered pair is called a relation. We call the the set of all the first component, as the domain and the set of all the second component y, as the range. We can also define the domain as the set of all permissible values of and the range is the set of all permissible values of y. The y is dependent while is independent. This means that y depends upon the value of. A mapping or arrow diagram shows how each member of the domain is paired with a member in range. There are basically three kinds of relations as shown below. y y y one to one one to many many to one Describing Relations. Listing Method or Roster Form A = { (-, 4), (-, ), (0, 0), (, ), (, 4) }. Tabular Method y Graph Graph of a Quadratic Function y = a + b + c Graph of a Linear Function y = m + b Prepared by: Mr. Charles D. Chavez

2 4. Arrow Diagram Rule or Set Builder Notation { (, y) y = } read as set of all ordered pairs (, y) such that y equals squared { (, y) y = 3 } read as set of all ordered pairs (, y) such that y equals 3 minus 6. Formula or Equation y = y = 6 y = y = EXAMPLES A. Identify the domain and the range of the following. { (, 3), (, 4), (3, 6), (5, 8) } Domain = {,, 3, 5} Range = {3, 4, 6, 8}. { (-, 4), (-, 5), (-, 0), (-, ) } Domain: { - } Range = {4, 5, 0, } 3. { (love, joy), (love, peace), (love, happiness) } Domain: {love} Range = { joy, peace, happiness } B. Given the equation below, describe the relation by means of the following. a. ordered pairs b. table c. set builder notation Prepared by: Mr. Charles D. Chavez

3 . y = y = Answer:. y = a. (, y) = { (-, -), (-, ), (0, 5), (, 8) } b y c. { (, y) y = }. y = a. (, y) = { (-, 3), (-, 0), (0, -), (, 0) } b y c. { (, y) y = } C. Classify if the relation is one to one, one to many or many to one.. A = { (, -), (, -), (3, -3), (4, -4), (5, -5) } one to one. B = { (3, ), (3, 0), (3, -5), (, ) } one to many 3. C = { (0, ), (, 0), (0, -), (0, 0.5) } many to many D. Write a formula or equation for each relation.. C = { (-3, 9), (-, 4), (-, ), (0, 0), (, ), (, 4), (3, 9) } y =. A = { (, ), (, 3), (3, 4), (4, 5) } y = + 3. D = { (-, -), (, ), (, /), (3, /3), (4, /4) } y = EXERCISES A. Identify the domain and the range of the following { (, -3), (0, 4), (0, 6), (5, ) } { (-, 4), (-, 5), (-, 0), (-, ) } 3. { (short, long), (thick, thin), (black, white) } Prepared by: Mr. Charles D. Chavez 3

4 4. { (a, b), (c, d), (e, f) } 5. { (0, -), (, 0), (-, 0), (0, 7) } Odd Even Positive Negative B. Given the equation below, describe the relation by means of the following. a. y = b. y = 3 + c. y = + 4. ordered pair. table 3. set builder notation C. Classify if the relation is one to one, one to many or many to one.. K = { (0, 5), (3, ), (3, -4), (, -6) }. P = { (-5, 7), (, 7), (0, 7), (9, 7), (-7, 7) } 3. G = { (, -7), (, ), (, 7), (, -) } D. Find an equation that satisfies the given relation.. FUNCTION. L = { (-4, -), (-3, -8), (-, -5), (-, -), (0, ), (, 4), }. A = { (9, -9), (, -), (0, 0), (-, ), (-4, 4), } 3. P = { (-3, 0), (-, 5), (-, ), (0, ), (, ), (, 5),.} 4. Z = { (-, 4), (-, ), (, ), (, 4), } 5. H = { (-5, 5), (-4, 4), (0, 0), (, ), (-4, 4), (0, 0), } Definition A function is a relation in which element in the domain corresponds to one and only one element in the range. A function is a relation in which no two ordered pairs have the same first coordinate. EXAMPLES Consider the following relations.. A = { (, ), (, 3), (3, 4), (4, 5) } Observe that the first coordinates of set A do not repeat, therefore set A is a function because it satisfies the definition of a function. Notice that set A is a one to one relation is a function. Prepared by: Mr. Charles D. Chavez 4

5 . B = { (3, -), (, -), (0, 5), (-4, 5) } Set B is a function because there is no ordered pairs have the same first coordinate. Notice also hat set B is an eample of many to one relation, hence many to one is a function. 3. C = { (, 8), (, 0), (0, 3), (, -8), (0, 7) } Set C is a one to many relation and is not a function. Why? Conclusion: All functions are relations but not all relations are functions. One to one and many to one are functions. One to many relation is NOT a function. EXERCISES A. Tell whether each of the following is a function X = { (, -3), (, 5), (, 0), (-5, ) } 4. Y = { (0, -3), (-, 0), (, 3), (, -8) } 5. Z = { (-6, ), (4, 0), (/, 3) } B. For what value/s of k will the relation NOT be a function?. P = { (k +, 3), (k + 3, -) }. Q = { (3k +, -4), (7 5k, ) } 3. N = { (-, ), (k 8k, 3) } 4. G = { (k + k, 3), (k + k + 6, -5) } 5. L = { (0, ), (3, -), (k, 0) } 6. T = { (3k 6, ), (-k, -7) } Prepared by: Mr. Charles D. Chavez 5

6 .3 FUNCTION NOTATION In Symbols How to say the symbols in words. f: 5 Under function f, maps to 5 or under f, is assigned to 5 or the image of under f is Same as in # 3. f() = 5 f of equals 5 4. f = {(, y) y = 5 } The function f is the set of all ordered pairs (, y) such that y equals 5 5. y = 5 y equals 5 a f y b 7 Pre image Image Study the diagram above. If f(a) = b, we say that b is the image of a under f and a is the pre-image of b. For eample, if f() = + 3, then f() = 7. Why? 7 is the image of under f and is the preimage of 7. When we write y = f(), read as y is a function of, we are relating two variables. In this notation, is called the independent variable, whose values are chosen and y is called the dependent variable whose values are computed. Instead of using f, we may use any letter of the alphabet to denote a function. Thus, we may write, g() = 9 4, read as g of equals 9 4. Eamples:. P() = h() =. Q() = EVALUATING A FUNCTION To find a value of a function at a given value of, we substitute the given value of into the given epression. EXAMPLES. Given f() = 3, find a. f(3) b. f(-5) Solution: a. f(3) = 3(3) b. f(-5) = 3(-5) Prepared by: Mr. Charles D. Chavez 6

7 = 9 = - 5 = 7 = - 7. Given g() =, find a. g(3) b. g(-) Solution: a. g(3) = (3) (3) b. g(-) = (-) (-) = 9 6 = + = 3 = 3 EXERCISES. If f() = 5 +, find a. f(-3) d. f(a + ) b. f(0) e. f(a + ) f(a) c. f(a). If h() = 6, find a. h() d. h(/) b. h(-) e. h(0) c. h(k ) 3. If p() =, find... - a. p() d. p(-) b. p(-4) e. p( y ) c. p(0) 4. A function p assigns to each positive integer n the nth prime number. Thus, p() =, find a. p(3) b. p(5) c. p(7) 5. Given f() = 5, find the value of such that a. f() = 3 d. f() = 0 b. f() = - 3 e. f() = c. f() = 5/ 6. Given g() = - +, find a. g() d. g(3) b. g(-) e. g(-5) c. g(0) 7. Given q() = 4-5, find a. q(-5) d. q() Prepared by: Mr. Charles D. Chavez 7

8 b. q(/) e. q(0) c. q(n + ).5 VERTICAL LINE TEST To determine the whether the graph of a relation is a function use the vertical line test. Vertical Line Test If each vertical line intersects the graph of a relation in no more than one point, then the relation is a function. EXAMPLE y.. y Vertical Line Vertical Line Each graph above defines a function because the vertical line intersects the graph at eactly one point. y y Each of the graphs is NOT a function since the vertical line intersects the graph in more than one point. EXERCISES Tell whether the graph defines a function... Prepared by: Mr. Charles D. Chavez 8

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10 .6. Even and Odd Functions EXAMPLES Definition A function f is an even function if f() = f(-) for every in the domain of f. A function f is odd function if f(-) = - f() for every in the domain of f. Determine whether the function is odd, even or neither.. f() = 4. f() = 3 f(-) = (-) 4 (-) f(-) = (-) 3 = 4 = - 3 f() = f(-) Therefore, f() is an even function. 3. f() = + 3 f(-) = -f() Therefore, f() is an odd function. Is it even? Is it odd? f(-) = (-) + 3(-) f(-) = (-) + 3(-) = 3 = 3 f() = - ( ) Therefore, it is NOT even. f() Therefore, it is NOT odd. Thus, f() is neither even nor odd. EXERCISES Determine whether the function is odd, even or neither.. f() = f() =. f() = f() = 3. f() = 8. f() = f() = f() = f() = f() = 3 + Prepared by: Mr. Charles D. Chavez 0

11 .7 Some Types of Functions. LINEAR FUNCTION Any function of the form y = m + b or a + by + c = 0 of f() = m + b is a linear function. The graph of a linear function is a straight line. Eamples of linear functions:. y = y = 0 3. f() = 5 7 Domain: Set of Real Numbers Range: Set of Real Numbers. CONSTANT FUNCTION Constant functions are of the form y = c or f() = c. The graph of a constant function is a horizontal line (parallel to the ais) y = c. Eamples of constant functions:. y =. y = y = 7 Domain: Set of Real Numbers Range: { c } 3. IDENTITY FUNCTION The identity function is a special linear function. It is defined as y =. Domain: Set of Real Numbers Range: Set of Real Numbers 4. QUADRATIC FUNCTION Quadratic functions are of the form y = a + b + c, a 0 or f() = a + b + c, a 0. The graph of a quadratic function is a parabola. Eamples of quadratic functions:. y = y = y = 4. y = Domain: Set of Real Numbers Prepared by: Mr. Charles D. Chavez

12 5. POLYNOMIAL FUNCTION The foregoing four types of function are special cases of a general types of functions called polynomial function.the general equation is of the form f() = a n n + a n n + + a 0 where n > 0 and a n, a n and a 0 are real numbers. If n = 0, we have the constant function; if n =, the function is a linear function; if n =, the function is quadratic and so on. Eamples:. y = +. y = y = y = ABSOLUTE VALUE FUNCTION The absolute value function has of the form y = f() where f() is any polynomial with n > in the definition of polynomials. Eamples:. y =. y = + 3. y = 5 4. y = y = a + b + c 7. GREATEST INTEGER FUNCTION The greatest integer function is one of the step functions so called because the graphs look like a series of steps. It is defined by f() = [ ] = n <, that is n is the greatest integer equal to or less than. Eamples:. If f() = [ ], find Solution: a. f(3) c. f(-4. 5) b. f(3.) d. f(-6) a. If = 3, [ 3 ] = 3 b. If = 3., [.] 3 = 3 c. If = - 4.5, [ - 4.5] = - 5 d. If = - 6, [ - 6 ]= - 6. If f() = [ + 4], find a. f(-.3) b. f(3.7) c. f(-5.) Prepared by: Mr. Charles D. Chavez

13 EXERCISES I. Identify each function as being a linear, constant, quadratic, absolute value or a greatest integer function.. f() = f() = 3. f() = 5 7. f() = f() = 3 8. f() = / 4. f() = 5 9. f() = f() = 9 0. f() = II. Given g() = 5 -, find. g(.) 4. g(-7.6). g(0) 5. g(-3.3) 3. g(.4).8 Fundamental Operations on Functions Definitions. The sum/difference of two functions f and g is that function defined by (f ± g) = f() ± f(g). The product of two functions f and g is that function defined by (fg)() = f()g() 3. The quotient of two functions f and g is that function defined by EXAMPLES f() (f/g)() =,g() 0 g(). If f() = 8 3 and g() = 5, find a. (f + g)() b. (f g)() c. (fg)() d. (f/g)() Solution: a. (f + g)() = f() + g() = [8 3()] + [5 () ] = + = 3 b. (f g)() = f() g() = = = Prepared by: Mr. Charles D. Chavez 3

14 c. (fg)() = f()g() = ()() = d. (f/g)() = f()/g() = / = For each of the given combinations of functions, we can find a general definiton of the function. a. (f + g)() = f() + g() = (8 3) + (5 ) = = 3 3 b. (f g)() = f() g() = (8 3) - (5 ) = = 3 3 c. (fg)() = (8 3)(5 ) = d. (f/g)() = 5 -. f() = 7 and g() = + Solution: a. (f + g)(-) b. (f g)(3) c. (fg)() d. (f/g)(4) a. (f + g)(-) = f(-) + g(-) = 7(-) + (-) + = = b. (f g)(3) = f(3) g(3) = 7(3) + (3) + = + 0 = c. (fg)() = f()g() = [7() ] [ () + ] = 3 5 f(4) d. (f/g)(4) = g(4) 7(4) - = (4) + = = = Prepared by: Mr. Charles D. Chavez 4

15 EXERCISES. Given: f() = g() = 3 h() = 3 4 Find: a. (f + g)(-4) b. (g h)() c. (fg)(0) d. (h/f)(-). Given: p() = + 5 q() = + r() = 3 Find: a. (p + q + r)(3) b. (pqr)(-) c. (pq)() r(5) d. (pq)(4)/r(-) e. (qr)(6) (pr)() 3. Given: k() = + 3 n() = 4 Find: m() = a. 3k() 5m(-) b. [n(-3)] (fm)(-) c. (nm)(4) d. m() + m() e. (knm)(-).9 Composition of Functions EXAMPLES The composition of two functions f and g is (f o g)() = f(g()) read as f of g. The functions can be combined to give a composite function (sometimes called a function of a function). Find the values of the following composite functions.. Given: h() = + 3 g() = 4 Find: (h o g)() Solution: (h o g)() = h(g()) Prepared by: Mr. Charles D. Chavez 5

16 EXERCISES Replace in h() with 4. (h o g)() = ( 4) + 3 = = Given: f() = 5 g() = + Find: (g o f)() Solution: (g o f)() = g(f()) = (5 ) + = Given: f() = 6 Find: (f o f)() Solution: (f o f)() = f(f()) = 6(6 + ) = = Given: f() = 3 4 g() = + 3 Find: (f o g)() Solution: (f o g)() = f(g()) Compute first g(). g() = () + 3 = 7 f(g()) = 3 4[g()] = 3 4(7) = 3 8 = - 5. Given: f() = 4 + g() = Find: a. (f o g)() b. (g o f)() c. (f o f)() d. (g o g)(). Given: f() = g() = + a. (f o g)(-3) b. (g o g)(4) c. (g o f)(0) *d. ((f o f) o f)() 3. Given: f() = + + g() = a. (f o g)() Prepared by: Mr. Charles D. Chavez 6

17 b. (f o g)() c. (g o g)() 4. If f(g()) = ( ) and f() =, find g(). 5. Let f() = + and g() =. Find a. (f o g)() b. (g o g)(-) c. (g o f)(4) 6. Let f() = + 3 and h() =. Find a. 3f(g()) + g(g(-)) b. [f(f())] g(f(-4)) g(f(3)) c. f(g(0).0 The Domain of a Function Recall that the domain is the permissible values of in a function. A. Set of Ordered Pairs Find the domain of A = { (, 3), (-, 7), (0, 4), (-7, ) } Domain of A = {, -, 0, - 7} B. Some Types of Functions The domain of linear function, quadratic function, absolute value function, greatest integer function and polynomial function is a set of real numbers.. Eamples of Linear function ( f() = m + b ) y = y 6 = 0 y = f() = 3 9 = -y + y =. Eamples of Quadratic Function ( f() = a + b + c, a 0 ) y = q() = 5 7 y = 7 p() = Eamples of absolute value function y = y = 3 - y = + 6 f() = Eamples of greatest integer function y = f() = Eamples of polynomial function P() = H() = Prepared by: Mr. Charles D. Chavez 7

18 C. Rational Function Find the domain of the following.. y = Recall that if the denominator is zero then the given epression is undefined. Therefore, the domain of y = is a set of real numbers ecept zero. In symbols, D = { R, 0 }, read as set of all such that is an element of real numbers ecept zero.. y = The denominator must not be equal to zero. Hence, / Therefore, the domain is D = { R, - 3/ } 3. y = Recall that is defined if the radicand () is greater than or equal to zero. But the denominator must not be equal to zero. Therefore, 4. f() = D = { > 0 } Again, the denominator must not be equal to zero. 0 ( + )( ) 0, - Thus. The domain is D = { R,, - }. 5. p() = ( 3)( + ) 0 3, - Thus. The domain is D = { R, 3, - }. Prepared by: Mr. Charles D. Chavez 8

19 D. Square Root Function Recall is defined if > 0.. y = + + > 0 > - Therefore, the domain is D = { > - }.. f() = > 0-4 > - 5 < 5/4 Recall: Therefore, the domain is D = { < 5/4 }. Theorem Let r and s be the roots of a + b + c = 0. If a + b + c < 0 can be written as ( r)( s) < 0 and r < s, then r < < s. or written as ( r)( s) > 0 and r < s, then < r or > s. Eamples: Solve for the value/s of. a. < 0 Solution: Consider the equation = 0. Solving for the values of, ( )( + ) = 0 =, - and - <. Therefore, the solution set is < <. b > 0 Solution: Consider the equation + 35 = 0. Solving for the values of, ( + 7)( 5) = 0 = - 7, 5 and 7 < 5. Therefore, < - 7 or > 5. Prepared by: Mr. Charles D. Chavez 9

20 Find the domain of the following functions.. f() = - The radicand must be > to zero. Notice that the radicand is in quadratic form. Using the theorem above,. y = > 0 Factoring the left side, ( + )( ) > 0 = -, Hence, < - or >. Thus, the domain of f() is D = { < - or > }. 8 > 0 ( 4)( + ) > 0 = -, 4 Hence, < - or > 4. Therefore, the domain is D = { < - or > 4 }. 3. g() = > < 0 Multiply both side by (3 + )(3 ) < 0 = - /3, /3 Hence, Therefore, the domain of g() is D = { - }. 3 3 EXERCISES Find the domain of the following.. A = { (-5, 3), (4, -), (0, -4), (-5, 7), (3, ) }. K = { (-, 7), (-, 0), (-, ½), (-, 0) } 3. L = { (a, p), (b, h), (c, i), (d, q) } 4. y = f() = y = h() = g() = y = - 4 Prepared by: Mr. Charles D. Chavez 0

21 0. y = 3. y = f() = f() = 6-4. g() = p() = y = + 7. f() = y = f() = E. Domain in operations on functions D f + g = D f D g (common to f and g) D f - g = D f D g D fg = D f D g D f/g = (D f D g ) \ { g() = 0 } EXAMPLES. Let f() = + 3 and g() = Find the domains of a. f + g b. f g c. fg d. f/g D f = { > - 3/ } D g = { R } a. D f + g = D f D g = { > - 3/ } { R } Using number line, { R } { > - 3/ }. -3/ 0 Therefore, D f + g = { > - 3/ } b. D f - g = { > - 3/ } c. D fg = { > - 3/ } d. D f/g = (D f D g ) \ { g() = 0 } But g() = = ( )( 3) for =, 3. Prepared by: Mr. Charles D. Chavez

22 EXERCISES Hence, D f/g = { > - 3/ \ {, 3} }. Let f() = and g() = Find the domains of a. f + g b. f g c. fg d. f/g. Let f() = 3 4 and g() = Find the domains of a. f + g b. f g c. fg d. f/g F. Domain of composite functions Definition Given two functions f and g, then the domain of f o g and g o f are: D f o g = { D g g() D f } D g o f = { D f f() D g } EXAMPLES. Let f() = and g() = Find the domains of f o g and g o f. Solution: a. Domain of f o g Domain of f = { R,, - } Domain of g = { R, 0 } The domain of D f o g = { D g g() D f }. f(g()) = ( + 3 ) - 4 This means that [(+3)/] 4 0. The denominator must not be equal to zero. It follows that Solving for, Prepared by: Mr. Charles D. Chavez or Note: We need to consider also the domain of g() which is g() 0. Therefore the domain of D f o g = { R, 0, -, 3 }

23 b. Domain of g o f Domain of f = { R,, - } Domain of g = { R, 0 } D g o f = { D f f() D g } g(f()) = The denominator must not be equal to zero D g o f = { R, 0,, - } EXERCISES Find the domains of f o g and g o f.. f() = - g() = 5 -. f() = g() = f() = g() =. Inverse of a Function Definition If f and g are two functions which are so related that f(g()) = for every in the domain of g and g(f()) = for every in the domain of f, then f and g are called inverses of each other. Theorem If the function f is one to one, then f has an inverse. Theorem If f and g are inverses, then f(g()) = g(f()) = or f(f ()) = f (f()) =. ***We can also write the inverse of f() as f (). To find the inverse of f, use the following procedure.. Test to see that f is one to one.. Write the function in the form y = f(). 3. Interchange the roles of and y. 4. Solve for y in terms of. 5. Replace y by f (). Prepared by: Mr. Charles D. Chavez 3

24 EXAMPLES Find the inverse of the following functions.. f() = 3 y = 3 Write in the form y = f(). = 3y Interchange the role of and y. 3y = 3y = + + y = 3 Solve for y in terms of. f () = + 3 Replace y by f (). Check: f(f - ()) = 3( + 3 ) f (f()) = 3 = + = 3 = = ( 3 -) + 3 Since f(f - ()) = f (f()) =, then f () = + 3 is the inverse of f(). Concept: The domain of the inverse of f() is the range of f(). The range of the inverse of f() is the domain of f().. f() = 4 y = 4 Write in the form y = f(). = 4 y Interchange the role of and y. 4 y = -y = y = f 4 - () = Solve for y in terms of. Change y to f (). 3. g() = 3 - y = = 3-3 y - Write in the form y = g(). Interchange the role of and y. (y ) = 3 Cross - multiply Prepared by: Mr. Charles D. Chavez 4

25 y = 3 y = 3 + DPMA Transposition y = g () = Solve for y in terms of. Replace y by g (). 4. h() = + 5 y = + 5 Write in the form y = h(). = y + 5 Interchange the role of and y = y + 5 Square both sides of the equation. y = 5 Solve for y in terms of. h () = + 5 Replace y by h (). 5. g() = y = = y + y - 3 Write in the form y = g(). Interchange the role of and y. (y 3) = y + y 3 = y + y y = + 3 y( ) = y = - g 3 + () = - Cross multiply. DPMA Transposition Isolate y (Factor out y). Solve for y in terms of. Replace y by g (). 6. h() = 9 y = 9 Write in the form y = h(). = y 9 Interchange the role of and y. y 9 = y = + 9 Refleive property of equality Transposition y = + 9 Taking square roots of both sides of the equation. Prepared by: Mr. Charles D. Chavez 5

26 EXERCISES Solve for y In terms of. h () = + 9 Replace y by h (). Find the inverse of the following functions.. f() = h() = y = + 7. g() = g() = y = f() = 3 9. f() = y = - 0. y = - EXERCISES IN RELATIONS AND FUNCTIONS. Given the function f() = +, find the following + 3 a. f(-) c. f() b. f(0) d. f(-3) f( + h) - f(). If f() = find,h 0. h 3. Given: f() = , g() = 3 and h() = 3 4. Find the following. a. (f + g)() b. (f h)(-) c. f(0) 3(g + h)(3) 4. Given p() = + 5, q() = 3 and r() = 3 +. Find the following. a. (pq)(5) b. (qr)(-) c. (p/q)(0) 5. Given: f() = + and g() =. Find the following. a. f(g()) d. g(g()) b. f(f()) e. f(f(f(-))) c. g(f()) f. g(f(4)) 6. Find the domain of the following functions. a. A = { (3, -), (0, ), (-, 5), (-, ) } b. K = { (a, b), (c, d), (e, f) } c. y = 5 d. f() = 6 + e. y = f. g() = + 7 Prepared by: Mr. Charles D. Chavez 6

27 g. h() = h. y = - i. y = j. y = k. f() = 6 - l. g() = + 4 m. y = n. h() = o. f() = p. y = Find the inverse of the following functions a. f() = f. k() = - b. h() = 4 5 g. f() = c. p() = 3 + h. g() = d. f() = i. f() = 5 e. g() = - 3 k. h() = Answer the following questions. a. Show that f() = is an even function. b. If g() = 3 +, find g(a + ) g(a). c. If A = { (, 3), (3, -), (, 0), (, ) is to be a function, list the integers that cannot represent. d. If f(g()) = ( ), and f() =, find g(). e. What is the range of f() = +? - Prepared by: Mr. Charles D. Chavez 7