Example

Solution

For 2( ) 2 and ( ) 3 1,f x x x g x x

a) ( f + g)(4) b) ( f – g)(x)

c) ( f /g)(x) d)

find the following.

( )( 1)f g

a) Since f (4) = 2(4) – (4)2 = 8 - 16 = -8 and g(4) = 3(4) + 1 = 13, we have

( f + g)(4) = f (4) + g(4) = –8 + 13 = 5

b) We have, ( )( ) ( ) ( )f g x f x g x

22 (3 1)x x x

2 1x x

Solution continued

d) Since f (–1) = 2(-1) – (-1)2 = –3 and g(–1) = 3(-1) + 1 = –2, we have,

( )( 1) ( 1) ( 1)

( 3)( 2)

6

f g f g

c) We have, ( )

( / )( )( )

f xf g x

g x

22

3 1

x x

x

1

3x We assume

Domains and Graphs of Combinations of Functions

Example

D : { | , 0}f x x x Notation

D : { | , 1}g x x x

Or use, ( ,0) (0, )x

Or use, ( , 1) ( 1, )x

D : { | , 1, 0}f g x x x x Or use, ( , 1) ( 1,0) (0, )x

/D : { | , 1, 0, 3}f g x x x x x

Or use, ( , 1) ( 1,0) (0,3) (3, )x

[1,5]x [1,3) (3,5]x

Example

3Given ( ) and ( ) 2,

1f x g x x

x

find the domains of

( )( ), ( )( ),( )( ) and ( / )( ). f g x f g x f g x f g x

Thus the domain of f + g, f – g, and is { | 1}.f g x x

Solution

The domain of f is { | 1}.x x

The domain of g is .x

Or use, ( , 1) ( 1, )x

Solution continued

To find the domain of f /g, note that

3( ) 1( / )( ) ( ) 2

f x xf g xg x x

can not be evaluated if x + 1 = 0 or x – 2 = 0.

Thus the domain of f /g is { | 1 or 2}.x x x

( , 1) ( 1,2) (2, ) x

Example

a.

b.