Objectives

1 .Compute operations on functions2 .Find the composition of two functions and the domain of the composition

Operation on functions

2 5 1h x x x

h x

2 and 5 1f x x g x x

Functions are often defined using sums, differences, products and quotients of various expressions.

For example, if

We may regard as a sum of values of functions f and g given by

We may call h the sum of f and g and denote it by f + g, i.e,

h = f + g

Thus,

2 5 1h x f g x x x Therefore,

In general, if f and g are any two functions, we use the terminology and notation given by the following chart

Quotient

f g ( x ) = f ( x) g ( x )Product f g

)f – g ) ( x = (f ( x ) – g ( x )Difference f – g

)f + g ) ( x = ( f (x ) + g (x )Sum f + g

Function ValueTerminology

f

g and 0

f xfx g x

g g x

Example 1.

3If 3 2 and ,

find 2 , 2 , 2 and / 2

f x x g x x

f g f g fg f g

Solution 32 3 2 2 4 and 2 2 8f g

2 4 8 12

2 4 8 4

2 4 8 32

4 12

8 2

f g

f g

fg

fg

Class Work 1

If f( x ) = - x2 and g ( x ) = 2x – 1. Find

a 3

3

3

d / 3

f g

b f g

c fg

f g

:

3 9, 3 5

3 4

3 9 5 14

3 9 5 45

93

5

Answer

f g

f g

f g

fg

f

g

Domain of f + g, f – g, f g, and f / g

FunctionDomainf + g

)Domain of f ) ∩ ( Domain of g( f – g

f g

f / g )Domain of f ) ∩ ( Domain of g ( such that g ( x ) ≠ 0

Example 2.

2

1Let and 2 . Find the domain of

9

and

xf x g x x

xf

a f g b f g c fg dg

\ 3, 3 and [ 2, )Domain of f Domain of g R\( ) ( ) ( ), and have the same answera b c

Solution:

0-3 2 3

,33,2 gofDomainfofDomain

,33,2

g

fofDomain(d)

Class Work 2

2Let 3 4 and 2 . Find the domain of

f x x x g x x

fa f g b f g c fg d

g

( ) ( )

( ) ( )

: [ 2, )

( ), [ 2, )

2,

Solution Domain of f and domain of g

a b and c D

fd Domain of

g

R= = - ¥

= - ¥

= - ¥

Composite Functions

Definition: The composite function f ◦ g of two functions f and g is defined

by ( f ◦ g )( x ) = f ( g(x) )

( ) ( ){ } ( ) \Domain of f g x Domain g g x Domain f= Î Îo

x g( x ) f(g(x)

gf

f ◦ g

Domain of g Domain of f

Solution:

Example 3: Let f (x ) = x2 -1 and g ( x ) = 3x + 5.

)a (Find ( f ◦ g )( x ) and the domain of f ◦ g.

)b ( Find ( g ◦ f )( x ) and the domain of g ◦ f.

)c (Is f ◦ g = g ◦ f

( ) ( )( ) ( )( )a f g x f g x= =o ( )3 5f x+ = ( )2 23 5 1 9 30 24x x x+ - = + +

Domain of g = R, Range of g = R, and Domain of f = R

Domain of f ◦ g = R

( ) ( )( )b g f x =o ( )( )g f x = ( )2 1g x - = ( )2 23 1 5 3 2x x- + = +

In a similar way as in part )a(, domain of g◦ f = R

( ) ,c NO f g g f¹o o

Example 4: Let f (x ) = x2 -1 and g ( x ) = 3x + 5.

)a (Find f ( g(2) ) in two different ways: first using the functions f and g separately and second using the composite function f ◦ g

)b (Find ( f ◦ f ) ( x )

Solution:

First Method g(2) = 3(2) + 5 =11, therefore

f (g(2) ) = f ( 11)= )11(2 – 1= 121 – 1 = 120

Second Method

)f ◦ g ) ( x = ( 9 x2 +30 x + 24. Therefore ,

f ( g (2 ) ) = ( f ◦ g ) ( 2 ) = 9 ) 2(2 + 30 ) 2 + ( 24 = 120

Same Answer

(a)

(b)

)f ◦ f ) ( x = ( f (f ( x ) ) = f( x2 – 1 ) = ( x2 – 1 )2 - 1= x4 -2x2

Example 5: ) Finding values of composite functions using tables(

Several values of two functions f and g are listed in the following tables.

x1234

f ( x )3421

x1234

g(x)4132

Find ( f◦g)(2) = ( g ◦f ) ( 2 ) = ( f ◦ f ) (2 ) = ( g ◦ g )( 2 ) =

Solution:

)f◦g)(2= ( f(g(2))= f ( 1) =

)g ◦f ) ( 2 = ( g( f(2) )= g( 4 )=

3

2

Try to find the rest by yourself

)f ◦ f) ( 2= ( 1

)g ◦ g )( 2= ( 4

Example 6: ) Finding a composite function form (

Express y = ( 2x + 5 )8 in a composite function form

Solution: Function ValueChoice for

u = g(x)

Choice for

y = f( u)

y = ( 2x + 1 )82 1u x= +

8y u=

Inner function = u

Note: y = ( f ◦ g ) ( x ) = f ( g (x) ) = f ( u ) = f ( 2x +1 ) = ( 2x + 1) 8

Class Work Express the following functions in a composite function form

Choice for y = f( x )Choice for u = g(x)Function Value

( )3 45 1y x x= - +

2 4y x= -

2

3 7y

x=

+

Word Problem using composite Functions

Example 7: ) Dimensions of a balloon ( A spherical balloon is being inflated at a rate of 4.5 π ft 3 / min. Express its radius r as a function of time t ) t in minutes (, assuming that r = 0 when t = 0.

Solution: 34( )

3V r Volume of a spherep=

At time t , V(t) = 4.5 π t ft3 / min. And r = r ( t ). Therefore,

( ) ( )34

3V t r tp= Substitute V(t) = 4.5 π t

( )344.5

3t r tp p= ( )

( )3 3 4.5

4r t t= ( ) 3

13.5

4r t t=

( ) 33

2r t t=

The End