Slide 1

10TH

EDITION

COLLEGE
ALGEBRA
LIAL
HORNSBY
SCHNEIDER

2.8 - 1


Slide 2

2.8

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

2.8 - 2


Slide 3

Operations of Functions
Given two functions  and g, then for all values
of x for which both (x) and g(x) are defined, the
functions  + g,  – g, g, and /g are defined as
follows.
 f  g   x   f ( x )  g ( x ) Sum

f

 g   x   f ( x )  g ( x ) Difference

 fg   x  

f ( x ) g ( x ) Product

 f 
f (x)
,
 x 
g(x)
 g 

g ( x )  0 Quotient
2.8 - 3


Slide 4

Note The condition g(x) ≠ 0 in the
definition of the quotient means that the
domain of (/g)(x) is restricted to all values
of x for which g(x) is not 0. The condition
does not mean that g(x) is a function that is
never 0.

2.8 - 4


Slide 5

Example 1

USING OPERATIONS ON
FUNCTIONS

Let (x) = x2 + 1 and g(x) = 3x + 5. Find the
following.
a.  f  g   1
Solution Since (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)

28
 10

2.8 - 5


Slide 6

Example 1

USING OPERATIONS ON
FUNCTIONS

Let (x) = x2 + 1 and g(x) = 3x + 5. Find the
following.
b.  f  g    3 
Solution Since (–3) = 10 and g(–3) = –4,
use the definition to get

f

 g   3   f (3 )  g (3 )  f

 g  x   f (x)  g(x)

 10  (4 )

 14

2.8 - 6


Slide 7

USING OPERATIONS ON
FUNCTIONS

Example 1

Let (x) = x2 + 1 and g(x) = 3x + 5. Find the
following.
c.  fg   5 
Solution Since (5) = 26 and g(5) = 20, use
the definition to get

 fg   5  

f (5 ) g (5 )

 26 20
 520

2.8 - 7


Slide 8

Example 1

USING OPERATIONS ON
FUNCTIONS

Let (x) = x2 + 1 and g(x) = 3x + 5. Find the
following.
 f 
d.    0 
 g 

Solution Since (0) = 1 and g(0) = 5, use
the definition to get
 f 
f (0 ) 1

  0  
g (0 ) 5
 g 

2.8 - 8


Slide 9

y

Domains
For functions  and g, the domains of
 + g,  – g, and g include all real
numbers in the intersections of the
domains of  and g, while the domain
of /g includes those real numbers in
the intersection of the domains of 
and g for which g(x) ≠ 0.
2.8 - 9


Slide 10

Example 2

USING OPERATIONS OF FUNCTIONS
AND DETERMINING DOMAINS

Let
f ( x )  8 x  9 and g( x ) 

2 x  1. F in d th e fo llo w in g .

a.  f  g   x 
Solution

f

 g  x   f (x)  g(x)  8x  9 

2x  1

2.8 - 10


Slide 11

Example 2

USING OPERATIONS OF FUNCTIONS
AND DETERMINING DOMAINS

Let
f ( x )  8 x  9 and g( x ) 

2 x  1. F in d th e fo llo w in g .

b.  f  g   x 
Solution

f

 g  x   f (x)  g(x)  8x  9 

2x  1

2.8 - 11


Slide 12

Example 2

USING OPERATIONS OF FUNCTIONS
AND DETERMINING DOMAINS

Let
f ( x )  8 x  9 and g( x ) 

2 x  1. F in d th e fo llo w in g .

c.  fg   x 
Solution

 fg   x  

f (x ) g(x )  8 x  9  2x  1

2.8 - 12


Slide 13

Example 2

USING OPERATIONS OF FUNCTIONS
AND DETERMINING DOMAINS

Let
f ( x )  8 x  9 and g( x ) 

d.

2 x  1. F in d th e fo llo w in g .

 f 
 x
 g 

Solution
 f 
f (x)
8x  9

 x 
g(x)
2x  1
 g 

2.8 - 13


Slide 14

Example 2

USING OPERATIONS OF FUNCTIONS
AND DETERMINING DOMAINS

Let
f ( x )  8 x  9 and g( x ) 

2 x  1. F in d th e fo llo w in g .

e. Give the domains of the functions.
Solution To find the domains of the functions,
we first find the domains of  and g.
The domain of  is the set of all real numbers
(–, ).
2.8 - 14


Slide 15

Example 2

USING OPERATIONS OF FUNCTIONS
AND DETERMINING DOMAINS

Let
f ( x )  8 x  9 and g( x ) 

2 x  1. F in d th e fo llo w in g .

e. Give the domains of the functions.
Solution Since g ( x )  2 x  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 - 15


Slide 16

Example 2

USING OPERATIONS OF FUNCTIONS
AND DETERMINING DOMAINS

Let
f ( x )  8 x  9 and g( x ) 

2 x  1. F in d th e fo llo w in g .

e. Give the domains of the functions.
Solution The domains of  + g,  – g, g are
the intersection of the domains of  and g,
which is
1
 1

 ,     ,     ,  
2
 2

2.8 - 16


Slide 17

Example 2

USING OPERATIONS OF FUNCTIONS
AND DETERMINING DOMAINS

Let
f ( x )  8 x  9 and g( x ) 

2 x  1. F in d th e fo llo w in g .

e. Give the domains of the functions.
f

Solution The domains of g includes those
real numbers in the intersection for which
g(x) 

that is, the domain of

2 x  1  0;
f
g

is

1

 , .
2

2.8 - 17


Slide 18

Example 3

EVALUATING COMBINATIONS
OF FUNCTIONS

If possible, use the given representations of
functions  and g to evaluate …

f

 g   4 ,

f

 g   2 ,

 fg   1 ,

and

 f 
  0 .
 g 

2.8 - 18


Slide 19

EVALUATING COMBINATIONS
Example 3
OF FUNCTIONS
 f 
 f  g   4  ,  f  g    2  ,  fg   1 , a n d    0  .
y

 g 

y  f (x)

9

a.

f (4)  9

 f 4  g 4

5

y  g(x)

 9  2  11
x

–4

–2

0

g (4)  2

2

4

For ( – g)(–
2),although (–2) = –
3, g(–2) is undefined
because –2 is not in
the domain of g.
2.8 - 19


Slide 20

EVALUATING COMBINATIONS
Example 3
OF FUNCTIONS
 f 
 f  g   4  ,  f  g    2  ,  fg   1 , a n d    0  .
y

9

a.

 g 

y  f (x)

f (4)  9

 f 4  g 4

5

y  g(x)

 9  2  11
x

–4

–2

0

g (4)  2

2

4

The domains of 
and g include 1, so

 f g   1 

f  1 g  1  3 1  3

2.8 - 20


Slide 21

EVALUATING COMBINATIONS
Example 3
OF FUNCTIONS
 f 
 f  g   4  ,  f  g    2  ,  fg   1 , a n d    0  .
y

9

a.

 g 

y  f (x)

f (4)  9

 f 4  g 4

5

y  g(x)
x
–4

–2

0

g (4)  2

2

4

 9  2  11
The graph of g
includes the origin, so

g 0 
 f
Thus,
 g

 0.

  0  is


undefined.
2.8 - 21


Slide 22

Example 3

EVALUATING COMBINATIONS
OF FUNCTIONS

If possible, use the given representations of
functions  and g to evaluate

f
b.

 g   4 ,

x
–2
0
1
1
4

(x)
–3
1
3
1
9

f

 g   2 ,

g(x)
undefined
0
1
undefined
2

 fg   1 ,

and

f (4)  9

 f 
  0 .
 g 

g (4)  2

 f 4  g 4
 9  2  11
In the table, g(–2)
is undefined.
Thus, (–g)(–2) is
undefined.

2.8 - 22


Slide 23

Example 3

EVALUATING COMBINATIONS
OF FUNCTIONS

If possible, use the given representations of
functions  and g to evaluate

f
b.

 g   4 ,

x
–2
0
1
1
4

(x)
–3
1
3
1
9

f

 g   2 ,

 fg   1 ,

and

 f 
  0 .
 g 

g (4)  2
f (4)  9
h(x)
 f 4  g 4
undefined
0
 9  2  11
1
 fg   1  f  1  1  3  1  3
undefined
2
2.8 - 23


Slide 24

Example 3

EVALUATING COMBINATIONS
OF FUNCTIONS

If possible, use the given representations of
functions  and g to evaluate

f
b.

 g   4 ,

x
–2
0
1
1
4

(x)
–3
1
3
1
9

f

 g   2 ,

h(x)
undefined
0
1
undefined
2

 fg   1 ,

and

f (4)  9

 f 
  0 .
 g 

g (4)  2

 f 4  g 4
 9  2  11
f 0 
 f 
  0  
g 0 
 g 

and

is u n d e fin e d sin ce g  0   0
2.8 - 24


Slide 25

Example 3

EVALUATING COMBINATIONS
OF FUNCTIONS

If possible, use the given representations of
functions  and g to evaluate
 f 
 f  g   4  ,  f  g    2  ,  fg   1 , a n d    0  .
c. f ( x )  2 x  1, g ( x ) 

f

x

 g   4   f  4   g  4    2 4  1 

f

 g 

4  9  2  11

 g    2   f   2   g   2    2   2   1 

2

is u n d e f i n e d .

 fg   1 

f  1 g  1    2 1  1  1  3  1   3
2.8 - 25


Slide 26

Example 3

EVALUATING COMBINATIONS
OF FUNCTIONS

c. f ( x )  2 x  1, g ( x ) 

f

x

 g   4   f  4   g  4    2 4  1 

f

4  9  2  11

 g    2   f   2   g   2    2   2   1 

2

is u n d e f i n e d .
 fg   1  f  1 g  1   2 1  1 1  3 1  3
 f 
  is u n d e fin e d.
 g 
2.8 - 26


Slide 27

Example 4

FINDING THE DIFFERENCE
QUOTIENT

Let (x) = 2x2 – 3x. Find the difference
quotient and simplify the expression.
Solution
Step 1 Find the first term in the numerator,
(x + h). Replace the x in (x) with x + h.
f ( x  h )  2( x  h )  3( x  h )
2

2.8 - 27


Slide 28

Example 4

FINDING THE DIFFERENCE
QUOTIENT

Let (x) = 2x – 3x. Find the difference quotient
and simplify the expression.
Solution
Step 2 Find the entire numerator f ( x  h )  f ( x ).
Substitute

f ( x  h )  f ( x )   2( x  h )  3( x  h )   (2 x  3 x )
2

2

 2( x  2 xh  h )  3( x  h )  (2 x  3 x )
2

2

2

Remember this
term when
squaring x + h

Square x + h
2.8 - 28


Slide 29

FINDING THE DIFFERENCE
QUOTIENT

Example 4

Let (x) = 2x – 3x. Find the difference quotient
and simplify the expression.
Solution
Step 2 Find the entire numerator f ( x  h )  f ( x ).
 2( x  2 xh  h )  3( x  h )  (2 x  3 x )
2

2

2

 2 x  4 xh  2 h  3 x  3 h  2 x  3 x
2

2

2

Distributive property

 4 xh  2 h  3 h
2

Combine terms.
2.8 - 29


Slide 30

Example 4

FINDING THE DIFFERENCE
QUOTIENT

Let (x) = 2x – 3x. Find the difference quotient
and simplify the expression.
Solution
Step 3 Find the quotient by dividing by h.
f (x  h)  f (x)
h

4 xh  2 h  3 h
2





Substitute.

h

h(4 x  2h  3 )

Factor out h.

h

 4 x  2h  3

Divide.
2.8 - 30


Slide 31

Caution Notice that (x + h) is not the
same as (x) + (h). For (x) = 2x2 – 3x in
Example 4. f ( x  h )  2( x  h ) 2  3( x  h )
 2 x  4 xh  2h  3 x  3h
2

but

2

f ( x )  f ( h )  (2 x  3 x )  (2 h  3 h )
2

2

 2 x  3 x  2h  3h
2

2

These expressions differ by 4xh.
2.8 - 31


Slide 32

Composition of Functions and
Domain
If  and g are functions, then the composite
function, or composition, of g and  is
defined by
 g f   x   g  f ( x ).

The domain of g f is the set of all
numbers x in the domain of  such that (x)
is in the domain of g.
2.8 - 32


Slide 33

EVALUATING COMPOSITE
FUNCTIONS
4
Let (x) = 2x – 1 and g(x) 
x 1

Example 5

a. Find  f g   2  .
Solution First find g(2). S in ce g  x  
g (2 ) 

4
2 1



4
1

4
x 1

,

 4

Now find  f g   2   f  g  2    f  4  :
f

 g 2 

f 4  24  1  7
2.8 - 33


Slide 34

EVALUATING COMPOSITE
FUNCTIONS
4
Let (x) = 2x – 1 and g(x) 
x 1

Example 5

b. Find  g

f  (  3 ).

Solution  f g    3   g  f   3    g   7  :
Don’t confuse
composition
with
multiplication



4
7  1

 

1
2



4
8

.

2.8 - 34


Slide 35

Example 7

DETERMINING COMPOSITE
FUNCTIONS AND THEIR DOMAINS

G iv e n th a t f ( x ) 

a.  f

6
x 3

and g( x ) 

1
x

, fin d th e fo llo w in g .

g   x  and its dom ain

Solution
(f

g )( x ) 

6
1
x

(f

g )( x ) 

3

6 x Multiply the numerator and
1  3 xdenominator by x.
2.8 - 35


Slide 36

Example 7
G iv e n th a t f ( x ) 

a.  f

DETERMINING COMPOSITE
FUNCTIONS AND THEIR DOMAINS
6
x 3

and g( x ) 

1
x

, fin d th e fo llo w in g .

g   x  and its dom ain

Solution The domain of g is all real numbers
except 0, which makes g(x) undefined. The
domain of  is all real numbers except 3. The
expression for g(x), therefore cannot equal 3;
we determine the value that makes g(x) = 3
and exclude it from the domain of f g .
2.8 - 36


Slide 37

DETERMINING COMPOSITE
FUNCTIONS AND THEIR DOMAINS

Example 7
G iv e n th a t f ( x ) 

a.  f

6
x 3

and g( x ) 

1
x

, fin d th e fo llo w in g .

g   x  and its dom ain

Solution

1
x

 3 The solution must be excluded.

1  3x

x 

1
3

Multiply by x.
Divide by 3.
2.8 - 37


Slide 38

DETERMINING COMPOSITE
FUNCTIONS AND THEIR DOMAINS

Example 7
G iv e n th a t f ( x ) 

a.  f

6
x 3

and g( x ) 

1
x

, fin d th e fo llo w in g .

g   x  and its dom ain

Solution

x 

1
3

Divide by 3.

Therefore the domain of f g is the set of all
real numbers except 0 and ⅓, written in interval
notation as   , 0    0, 1    1 ,   .




3


3




2.8 - 38


Slide 39

DETERMINING COMPOSITE
FUNCTIONS AND THEIR DOMAINS

Example 7
G iv e n th a t f ( x ) 

b.  g

f

6
x 3

and g( x ) 

1
x

, fin d th e fo llo w in g .

  x  and its dom ain

Solution  g

f

  x   g  f  x 


1
6
x 3



 6 
 g

x 3

Note that this is
meaningless if x = 3

x 3
6
2.8 - 39


Slide 40

Example 7
G iv e n th a t f ( x ) 

b.  g

f

DETERMINING COMPOSITE
FUNCTIONS AND THEIR DOMAINS
6
x 3

and g( x ) 

1
x

, fin d th e fo llo w in g .

  x  and its dom ain

Solution The domain of  is all real numbers
except 3, and the domain of g is all real
numbers except
0.
The
expression
for
(x),
6
which is x  3 , is never zero, since the
numerator is the nonzero number 6.

2.8 - 40


Slide 41

DETERMINING COMPOSITE
FUNCTIONS AND THEIR DOMAINS

Example 7
G iv e n th a t f ( x ) 

b.  g

f

6
x 3

and g( x ) 

1
x

, fin d th e fo llo w in g .

  x  and its dom ain

Solution Therefore, the domain of g f
is the set of all real numbers except 3, written

  , 3    3,  

2.8 - 41


Slide 42

Example 8

SHOWING THAT  g

f

x    f g x 

Let (x) = 4x + 1 and g(x) = 2x2 + 5x.
S how that  g f   x    g f   x  in general.
Solution First, find  g f   x  .

 g f   x   g  f  x    g  4 x  1

f x   4x  1

2
 2  4 x  1  5 ( 4 x  1) g  x   2 x  5 x

2

Square 4x + 1;
2
distributive
 2 16 x  8 x  1  20 x  5
property.





2.8 - 42


Slide 43

Example 8

SHOWING THAT  g

f

x    f g x 

Let (x) = 4x + 1 and g(x) = 2x2 + 5x.
S how that  g f   x    g f   x  in general.
Solution First, find  g f   x  .
 2  1 6 x  8 x  1  2 0 x  5
2

Distributive
property.

 32 x  16 x  2  20 x  5
2

 3 2 x  3 6 x  7 Combine terms.
2

2.8 - 43


Slide 44

Example 8

SHOWING THAT  g

f

x    f g x 

Let (x) = 4x + 1 and g(x) = 2x2 + 5x.
S how that  g f   x    g f   x  in general.
Solution N ow find  f g   x  .
 f g   x   f  g  x 
 f 2x  5x 
2

g  x   2x  5x
2

 4 2x  5x   1
2

f x   4x  1

2
 8 x  2 0 x  1 Distributive

S o...  g

property

f

  x    f g   x .

2.8 - 44


Slide 45

Example 9

FINDING FUNCTIONS THAT FORM
A GIVEN COMPOSITE

Find functions  and g such that

 f g   x    x  5   4  x  5   3.
Solution Note the repeated quantity x2 – 5. If
we choose g(x) = x2 – 5 and (x) = x3 – 4x + 3,
then
 f g   x   f  g  x 
2

3

2

2
There are other
 f x  5
pairs of functions
3
2
2
 and g that also   x  5   4  x  5   3
work.
2.8 - 45