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

Transcription

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

2 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

3 Example 1 USING OPERATIONS ON FUNCTIONS Let ƒ(x) = x 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 + =

4 Example 1 USING OPERATIONS ON FUNCTIONS Let ƒ(x) = x 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) =

5 Example 1 USING OPERATIONS ON FUNCTIONS Let ƒ(x) = x 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. ( )( ) ( )( ) = =

6 Example 1 USING OPERATIONS ON FUNCTIONS Let ƒ(x) = x 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)

7 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

8 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

9 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

10 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

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 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 (, )

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 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,

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 ƒ + g, ƒ g, ƒg are the intersection of the domains of ƒ and g, which is 1 1 (, ), =,

14 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,

15 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 )( ) ( )

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

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

18 a. Example 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 + = = 11 The graph of g includes the origin, so g ( 0) = 0. f Thus, ( 0) is undefined. g x

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) g(x) 2 3 undefined undefined EVALUATING COMBINATIONS OF FUNCTIONS ( 4) g( 4) = f + = = 11 In the table, g( 2) is undefined. Thus, (ƒ g)( 2) is undefined

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 undefined EVALUATING COMBINATIONS OF FUNCTIONS ( 4) g( 4) = f + = = 11 ( fg )( ) f ( ) ( ) ( ) 1 = 1 1 = 3 1 =

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 )( ) ( ) b. f (4) = 9 g(4) = 2 x ƒ(x) h(x) 2 3 undefined undefined EVALUATING COMBINATIONS OF FUNCTIONS ( 4) g( 4) = f + = = 11 ( 0) ( 0) f f ( 0) = and g g is undefined since g 0 = 0 ( )

22 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 = = = = 11 ( f g)( 2) = f ( 2) + g( 2) = 2( 2) + 1 is undefn i e d. ( fg )( ) f ( ) g ( ) ( ) ( ) 1= 1 1= = 31=

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

24 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)

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). f 2 2 ( x + h) ( x) = 2( x + h) 3( x + h) (2x 3 x) f Substitute = ( x 2 xh h ) 3( x h) (2x 3 x) Remember this term when squaring x + h Square x + h

26 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). = ( x 2 xh h ) 3( x h) (2x 3 x) = 2x + 4xh + 2h 3x 3h 2x + 3x Distributive property 2 = 4xh + 2h 3h Combine terms

27 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

28 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

29 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

30 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 g(2) = = = f g 2 = f g 2 = f 4 : Now find ( )( ) ( ) ( ) ( ) f g 2 = f 4 = 2 4 1= 7 ( ( )) ( ) ( )

31 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 : ( )( ) ( ) ( ) ( ) = = 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. g f x = g f x = g 4x + 1 f ( x) = 4x + 1 ( )( ) ( ) ( ) ( ) Square 4x + 1; distributive property. = ( 4x + 1) g ( ) ( x ) 2 ( 2 ) = 2 16x + 8x x x = 2x + 5x

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 ( g f )( x) First, find. Distributive property. ( 2 ) = 2 16x + 8x x + 5 = x 16x 2 20x 5 2 = 32x + 36x + 7 Combine terms

34 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 = f ( x) = 4x + 1 = x 20x 1 Distributive property So... g f x f g x. ( )( ) ( )( )