2.8. Function Operations and Composition. Arithmetic Operations on Functions The Difference Quotient Composition of Functions and Domain

2.8 Function Operations and Composition Arithmetic Operations on Functions The Difference Quotient Composition of Functions and Domain 2.8-1

Operations of Functions Given two functions ƒ and g, then for all values of x for which both ƒ(x) and g(x) are defined, the functions ƒ + g, ƒ g, ƒg, and ƒ/g are defined as follows. ( + )( x ) = ( x ) + ( x ) f g f g Sum ( )( x) = ( x) ( x) f g f g Difference ( )( x) = ( x) ( x) fg f g Product f g ( x) f( x) =, g ( x ) 0 g( x) Quotient 2.8-2

Example 1 USING OPERATIONS ON FUNCTIONS Let ƒ(x) = x 2 + 1 and g(x) = 3x + 5. Find the following. f + g 1 Solution Since ƒ(1) = 2 and g(1) = 8, use the definition to get f + g 1 = f(1) + g (1) ( f + g)( x) = f( x) + g( x) a. ( )( ) ( )( ) = 2 + = 10 8 2.8-3

Example 1 USING OPERATIONS ON FUNCTIONS Let ƒ(x) = x 2 + 1 and g(x) = 3x + 5. Find the following. f g 3 Solution Since ƒ( 3) = 10 and g( 3) = 4, use the definition to get f g 3 = f( ) g( 3 ( f g)( x) = f( x) g( x) b. ( )( ) ( )( ) 3 ) = 10 ( 4) = 14 2.8-4

Example 1 USING OPERATIONS ON FUNCTIONS Let ƒ(x) = x 2 + 1 and g(x) = 3x + 5. Find the following. fg 5 Solution Since ƒ(5) = 26 and g(5) = 20, use the definition to get fg 5 = f (5) g(5) c. ( )( ) ( )( ) = 26 20 = 520 2.8-5

Example 1 USING OPERATIONS ON FUNCTIONS Let ƒ(x) = x 2 + 1 and g(x) = 3x + 5. Find the following. f 0 g Solution Since ƒ(0) = 1 and g(0) = 5, use the definition to get d. ( ) f g ( 0) f(0) 1 = = g(0) 5 2.8-6

Example 2 USING OPERATIONS OF FUNCTIONS AND DETERMINING DOMAINS Let f( x) = 8x 9 and g( x) = 2x 1. Find the following. a. ( f + g)( x) Solution ( f + g)( x) = f ( x) + g(x) = 8x 9+ 2x 1 2.8-7

Example 2 USING OPERATIONS OF FUNCTIONS AND DETERMINING DOMAINS Let f( x) = 8x 9 and g( x) = 2x 1. Find the following. b. ( f g)( x) Solution ( f g)( x) = f( x) g( x) = 8x 9 2x 1 2.8-8

f Example 2 USING OPERATIONS OF FUNCTIONS AND DETERMINING DOMAINS Let ( x) = 8x 9 and g( x) = 2x 1. Find the following. c. ( fg )( x) Solution ( fg )( ) f g ( ) x = ( x) ( x) = 8x 9 2x 1 2.8-9

Example 2 USING OPERATIONS OF FUNCTIONS AND DETERMINING DOMAINS Let f( x) = 8x 9 and g( x) = 2x 1. Find the following. f x g Solution d. ( ) f f( x) 8x 9 ( x) = = g g( x) 2x 1 2.8-10

f Example 2 USING OPERATIONS OF FUNCTIONS AND DETERMINING DOMAINS Let ( x) = 8x 9 and g( x) = 2x 1. Find the following. e. Give the domains of the functions. Solution To find the domains of the functions, we first find the domains of ƒ and g. The domain of ƒ is the set of all real numbers (, ). 2.8-11

f Example 2 USING OPERATIONS OF FUNCTIONS AND DETERMINING DOMAINS Let ( x) = 8x 9 and g( x) = 2x 1. Find the following. e. Give the domains of the functions. Solution Since g( x) = 2x 1, the domain of g includes just the real numbers that make 2x 1 nonnegative. Solve 2x 1 0 to get x ½. The domain of g is 1, 2 2.8-12

f Example 2 USING OPERATIONS OF FUNCTIONS AND DETERMINING DOMAINS Let ( x) = 8x 9 and g( x) = 2x 1. Find the following. e. Give the domains of the functions. Solution The domains of ƒ + g, ƒ g, ƒg are the intersection of the domains of ƒ and g, which is 1 1 (, ), =, 2 2 2.8-13

f Example 2 USING OPERATIONS OF FUNCTIONS AND DETERMINING DOMAINS Let ( x) = 8x 9 and g( x) = 2x 1. Find the following. e. Give the domains of the functions. Solution The domains of g includes those real numbers in the intersection for which g( x) = 2x 1 0; f that is, the domain of f g is 1,. 2 2.8-14

Example 3 EVALUATING COMBINATIONS OF FUNCTIONS If possible, use the given representations of functions ƒ and g to evaluate f + 4, 2, 1, and 0. g ( f g )( ) ( f g )( ) ( fg )( ) ( ) 2.8-15

a. Example 3 EVALUATING COMBINATIONS OF FUNCTIONS f + 4, 2, 1, and 0. y g ( f g )( ) ( f g )( ) ( fg )( ) ( ) 4 2 0 2 9 5 y 4 y = g( x) = f ( x) f (4) = 9 g(4) = 2 x ( 4) g ( 4) = f + = 9 + 2 = 11 For (ƒ g)( 2),although ƒ( 2) = 3, g( 2) is undefined because 2 is not in the domain of g. 2.8-16

a. Example 3 f + 4, 2, 1, and 0. y g ( f g )( ) ( f g )( ) ( fg )( ) ( ) 4 2 0 2 9 5 y = f y EVALUATING COMBINATIONS OF FUNCTIONS ( x) = g( x) 4 f (4) = 9 g(4) = 2 x ( 4) g ( 4) = f + = 9 + 2 = 11 The domains of ƒ and g include 1, so ( fg )( ) f ( ) g ( ) 1 = 1 1 = 3 1 = 3 2.8-17

a. Example 3 4 2 0 2 9 5 y = f y EVALUATING COMBINATIONS OF FUNCTIONS f + 4, 2, 1, and 0. y g ( f g )( ) ( f g )( ) ( fg )( ) ( ) ( x) = g( x) 4 f (4) = 9 g(4) = 2 ( 4) g ( 4) = f + = 9 + 2 = 11 The graph of g includes the origin, so g ( 0) = 0. f Thus, ( 0) is undefined. g x 2.8-18

Example 3 If possible, use the given representations of functions ƒ and g to evaluate f + 4, 2, 1, and 0. g ( f g )( ) ( f g )( ) ( fg )( ) ( ) b. f (4) = 9 g(4) = 2 x ƒ(x) g(x) 2 3 undefined 0 1 0 1 3 1 1 1 undefined 4 9 2 EVALUATING COMBINATIONS OF FUNCTIONS ( 4) g( 4) = f + = 9 + 2 = 11 In the table, g( 2) is undefined. Thus, (ƒ g)( 2) is undefined. 2.8-19

Example 3 If possible, use the given representations of functions ƒ and g to evaluate f + 4, 2, 1, and 0. g ( f g )( ) ( f g )( ) ( fg )( ) ( ) b. f (4) = 9 g(4) = 2 x ƒ(x) h(x) 2 3 undefined 0 1 0 1 3 1 1 1 undefined 4 9 2 EVALUATING COMBINATIONS OF FUNCTIONS ( 4) g( 4) = f + = 9 + 2 = 11 ( fg )( ) f ( ) ( ) ( ) 1 = 1 1 = 3 1 = 3 2.8-20

Example 3 If possible, use the given representations of functions ƒ and g to evaluate f + 4, 2, 1, and 0. g ( f g )( ) ( f g )( ) ( fg )( ) ( ) b. f (4) = 9 g(4) = 2 x ƒ(x) h(x) 2 3 undefined 0 1 0 1 3 1 1 1 undefined 4 9 2 EVALUATING COMBINATIONS OF FUNCTIONS ( 4) g( 4) = f + = 9 + 2 = 11 ( 0) ( 0) f f ( 0) = and g g is undefined since g 0 = 0 ( ) 2.8-21

Example 3 If possible, use the given representations of functions ƒ and g to evaluate f + 4, 2, 1, and 0. g ( f g )( ) ( f g )( ) ( fg )( ) ( ) c. f( x) = 2x + 1, g( x) = x EVALUATING COMBINATIONS OF FUNCTIONS ( f g)( ) f ( ) g( ) ( ) + 4 = 4 + 4 = 2 4 + 1 + 4 = 9 + 2 = 11 ( f g)( 2) = f ( 2) + g( 2) = 2( 2) + 1 is undefn i e d. ( fg )( ) f ( ) g ( ) ( ) ( ) 1= 1 1= 21+ 1 1= 31= 3 2 2.8-22

Example 3 EVALUATING COMBINATIONS OF FUNCTIONS c. f( x) = 2x + 1, g( x) = x ( f g)( ) f ( ) g( ) ( ) + 4 = 4 + 4 = 2 4 + 1 + 4 = 9 + 2 = 11 ( f g)( 2) = f ( 2) + g( 2) = 2( 2) + 1 2 is undefn i e d. fg 1= f 1 g 1= 21 + 1 1= 31= 3 ( )( ) ( ) ( ) ( ) ( ) f g is undefined. 2.8-23

Example 4 FINDING THE DIFFERENCE QUOTIENT Let ƒ(x) = 2x 2 3x. Find the difference quotient and simplify the expression. Solution Step 1 Find the first term in the numerator, ƒ(x + h). Replace the x in ƒ(x) with x + h. f 2 ( x + h) = 2( x + h) 3( x + h) 2.8-24

Example 4 FINDING THE DIFFERENCE QUOTIENT Let ƒ(x) = 2x 3x. Find the difference quotient and simplify the expression. Solution Step 2 Find the entire numerator f( x + h) f( x). f 2 2 ( x + h) ( x) = 2( x + h) 3( x + h) (2x 3 x) f Substitute = + + + 2 2 2 2( x 2 xh h ) 3( x h) (2x 3 x) Remember this term when squaring x + h Square x + h 2.8-25

Example 4 FINDING THE DIFFERENCE QUOTIENT Let ƒ(x) = 2x 3x. Find the difference quotient and simplify the expression. Solution Step 2 Find the entire numerator f( x + h) f( x). = + + + 2 2 2 2( x 2 xh h ) 3( x h) (2x 3 x) 2 2 2 = 2x + 4xh + 2h 3x 3h 2x + 3x Distributive property 2 = 4xh + 2h 3h Combine terms. 2.8-26

Example 4 FINDING THE DIFFERENCE QUOTIENT Let ƒ(x) = 2x 3x. Find the difference quotient and simplify the expression. Solution Step 3 Find the quotient by dividing by h. 2 f( x + h) f( x) 4xh + 2h 3h = Substitute. h h h(4x + 2h 3) h = Factor out h. = 4x + 2h 3 Divide. 2.8-27

Caution Notice that ƒ(x + h) is not the same as ƒ(x) + ƒ(h). For ƒ(x) = 2x 2 3x in 2 Example 4. f ( x + h) = 2( x + h) 3( x + h) 2 2 = 2x + 4xh + 2h 3x 3h but f 2 2 ( x) + f( h) = (2x 3 x) + (2h 3 h) 2 2 = 2x 3x + 2h 3h These expressions differ by 4xh. 2.8-28

Composition of Functions and Domain If ƒ and g are functions, then the composite function, or composition, of g and ƒ is defined by ( g f )( x ) = g ( f x ) g f ( ). The domain of is the set of all numbers x in the domain of ƒ such that ƒ(x) is in the domain of g. 2.8-29

Example 5 EVALUATING COMPOSITE FUNCTIONS 4 Let ƒ(x) = 2x 1 and g(x) = x 1 a. Find ( f g)( 2 ). 4 Solution First find g(2). Since g ( x) =, x 1 4 4 g(2) = = = 4 2 1 1 f g 2 = f g 2 = f 4 : Now find ( )( ) ( ) ( ) ( ) f g 2 = f 4 = 2 4 1= 7 ( ( )) ( ) ( ) 2.8-30

b. Example 5 Let ƒ(x) = 2x 1 and g(x) Find g f ( 3). Solution Don t confuse composition with multiplication ( ) EVALUATING COMPOSITE FUNCTIONS 4 = x 1 f g 3 = g f 3 = g 7 : ( )( ) ( ) ( ) ( ) = 4 4 7 1 = 8 = 1 2. 2.8-31

Example 8 SHOWING THAT ( g f )( x) ( f g)( x) Let ƒ(x) = 4x + 1 and g(x) = 2x 2 + 5x. Show that g f x g f x in general. ( )( ) ( )( ) Solution ( g f )( x) First, find. g f x = g f x = g 4x + 1 f ( x) = 4x + 1 ( )( ) ( ) ( ) ( ) Square 4x + 1; distributive property. = 2 4 + 1 + 5( 4x + 1) g ( ) ( x ) 2 ( 2 ) = 2 16x + 8x + 1 + 20x + 5 2 x = 2x + 5x 2.8-32

Example 8 SHOWING THAT ( g f )( x) ( f g)( x) Let ƒ(x) = 4x + 1 and g(x) = 2x 2 + 5x. Show that g f x g f x in general. ( )( ) ( )( ) Solution ( g f )( x) First, find. Distributive property. ( 2 ) = 2 16x + 8x + 1 + 20x + 5 = + + + + 2 32x 16x 2 20x 5 2 = 32x + 36x + 7 Combine terms. 2.8-33

Example 8 SHOWING THAT ( g f )( x) ( f g)( x) Let ƒ(x) = 4x + 1 and g(x) = 2x 2 + 5x. Show that g f x g f x in general. ( )( ) ( )( ) Solution ( f g)( x) ( f g)( x) = f g( x) Now find. ( ) ( 2 2x 5x) = + f g ( ) ( 2x 2 5x) 2 x = 2x + 5x = 4 + + 1 f ( x) = 4x + 1 = + + 2 8x 20x 1 Distributive property So... g f x f g x. ( )( ) ( )( ) 2.8-34