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.
( 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)
=
 g  ( x ) g( x ) , g( x ) ≠ 0
 
Quotient
2.8 - 2
Example 1
USING OPERATIONS ON
FUNCTIONS
Let ƒ(x) = x2 + 1 and g(x) = 3x + 5. Find the
following.
a. ( 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 )
= 2+8
= 10
2.8 - 3
Example 1
USING OPERATIONS ON
FUNCTIONS
Let ƒ(x) = x2 + 1 and g(x) = 3x + 5. Find the
following.
b. ( f − g ) ( −3 )
Solution Since ƒ(–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 - 4
Example 1
USING OPERATIONS ON
FUNCTIONS
Let ƒ(x) = x2 + 1 and g(x) = 3x + 5. Find the
following.
c. ( fg ) ( 5 )
Solution Since ƒ(5) = 26 and g(5) = 20, use
the definition to get
( fg ) ( 5 ) = f (5) g (5)
= 26 20
= 520
2.8 - 5
Example 1
USING OPERATIONS ON
FUNCTIONS
Let ƒ(x) = x2 + 1 and g(x) = 3x + 5. Find the
following.
 f 
d.   ( 0 )
g
Solution Since ƒ(0) = 1 and g(0) = 5, use
the definition to get
 f 
f (0) 1
(0) =
 g =
g (0) 5
 
2.8 - 6
Example 2
USING OPERATIONS OF FUNCTIONS
AND DETERMINING DOMAINS
Let
f (x) =
8 x − 9 and g ( x ) =
2 x − 1. Find the following.
a. ( f + g ) ( x )
Solution
(f
+ g ) ( x ) = f ( x ) + g (x ) = 8 x − 9 + 2 x − 1
2.8 - 7
Example 2
USING OPERATIONS OF FUNCTIONS
AND DETERMINING DOMAINS
Let
f (x) =
8 x − 9 and g ( x ) =
2 x − 1. Find the following.
b. ( f − g ) ( x )
Solution
(f
− g ) ( x ) = f ( x ) − g( x ) = 8 x − 9 − 2x − 1
2.8 - 8
Example 2
USING OPERATIONS OF FUNCTIONS
AND DETERMINING DOMAINS
Let
f (x) =
8 x − 9 and g ( x ) =
2 x − 1. Find the following.
c.
( fg ) ( x )
Solution
f ( x ) g ( x ) =
( fg ) ( x ) =
(8x − 9)
2x − 1
2.8 - 9
Example 2
USING OPERATIONS OF FUNCTIONS
AND DETERMINING DOMAINS
Let
f (x) =
8 x − 9 and g ( x ) =
2 x − 1. Find the following.
 f 
d.   ( x )
g
Solution
 f 
f (x)
(x) =
 g =
g( x )
 
8x − 9
2x − 1
2.8 - 10
Example 2
USING OPERATIONS OF FUNCTIONS
AND DETERMINING DOMAINS
Let
f (x) =
8 x − 9 and g ( x ) =
2 x − 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
Example 2
USING OPERATIONS OF FUNCTIONS
AND DETERMINING DOMAINS
Let
f (x) =
8 x − 9 and g ( x ) =
2 x − 1. Find the following.
e. Give the domains of the functions.
(x)
2 x − 1 , the domain
Solution Since g=
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
Example 2
USING OPERATIONS OF FUNCTIONS
AND DETERMINING DOMAINS
Let
f (x) =
8 x − 9 and g ( x ) =
2 x − 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
Example 2
USING OPERATIONS OF FUNCTIONS
AND DETERMINING DOMAINS
Let
f (x) =
8 x − 9 and g ( x ) =
2 x − 1. Find the following.
e. Give the domains of the functions.
f
Solution The domains of g includes those
real numbers in the intersection for which
g ( x=
)
2 x − 1 ≠ 0;
1 
f
that is, the domain of is  , ∞  .
g
2 
2.8 - 14
Example 3
EVALUATING COMBINATIONS
OF FUNCTIONS
If possible, use the given representations of
functions ƒ and g to evaluate …
(f
+ g )( 4 ) ,
(f
− g )( −2 ) ,
( fg )(1) ,
and
 f 
 g  (0).
 
2.8 - 15
Example 3
(f
+ g )( 4 ) ,
y
(f
EVALUATING COMBINATIONS
OF FUNCTIONS
− g )( −2 ) ,
( fg )(1) ,
y = f (x)
9
a.
f (4) = 9
y = g( x )
= 9 + 2 = 11
x
–2
0
g (4) = 2
= f ( 4) + g ( 4)
5
–4
and
 f 
 g  (0).
 
2
4
For (ƒ – g)(–
2),although ƒ(–2) = –
3, g(–2) is undefined
because –2 is not in
the domain of g.
2.8 - 16
Example 3
(f
+ g )( 4 ) ,
9
a.
y
(f
EVALUATING COMBINATIONS
OF FUNCTIONS
− g )( −2 ) ,
( fg )(1) ,
y = f (x)
f (4) = 9
y = g( x )
= 9 + 2 = 11
x
–2
0
g (4) = 2
= f ( 4) + g ( 4)
5
–4
and
 f 
 g  (0).
 
2
4
The domains of ƒ
and g include 1, so
1)
( fg )(=
f (1) g (=
1) 3=
1 3
2.8 - 17
Example 3
(f
+ g )( 4 ) ,
9
a.
y
(f
EVALUATING COMBINATIONS
OF FUNCTIONS
− g )( −2 ) ,
and
y = f (x)
f (4) = 9
y = g( x )
–2
0
g (4) = 2
= f ( 4) + g ( 4)
5
–4
( fg )(1) ,
 f 
 g  (0).
 
2
4
= 9 + 2 = 11
The graph of g
x includes the origin, so
g ( 0 ) = 0.
 f 
Thus,  ( 0 ) is undefined.
g
2.8 - 18
Example 3
EVALUATING COMBINATIONS
OF FUNCTIONS
If possible, use the given representations of
functions ƒ and g to evaluate
(f
+ g )( 4 ) ,
b.
ƒ(x)
–3
1
3
1
9
x
–2
0
1
1
4
(f
− g )( −2 ) ,
g(x)
undefined
0
1
undefined
2
( fg )(1) ,
and
f (4) = 9
 f 
 g  (0).
 
g (4) = 2
= f ( 4) + g ( 4)
= 9 + 2 = 11
In the table, g(–2)
is undefined.
Thus, (ƒ–g)(–2) is
undefined.
2.8 - 19
Example 3
EVALUATING COMBINATIONS
OF FUNCTIONS
If possible, use the given representations of
functions ƒ and g to evaluate
(f
+ g )( 4 ) ,
b.
ƒ(x)
–3
1
3
1
9
x
–2
0
1
1
4
(f
− g )( −2 ) ,
h(x)
undefined
0
1
undefined
2
( fg )(1) ,
and
f (4) = 9
 f 
 g  (0).
 
g (4) = 2
= f ( 4) + g ( 4)
= 9 + 2 = 11
1)
( fg )(=
f (1) =
(1) 3=
(1) 3
2.8 - 20
Example 3
EVALUATING COMBINATIONS
OF FUNCTIONS
If possible, use the given representations of
functions ƒ and g to evaluate
(f
+ g )( 4 ) ,
b.
ƒ(x)
–3
1
3
1
9
x
–2
0
1
1
4
(f
− g )( −2 ) ,
h(x)
undefined
0
1
undefined
2
( fg )(1) ,
and
f (4) = 9
 f 
 g  (0).
 
g (4) = 2
= f ( 4) + g ( 4)
= 9 + 2 = 11
f (0)
 f 
and
 g  (0) = g 0
( )
 
is undefined since g ( 0 ) = 0
2.8 - 21
Example 3
EVALUATING COMBINATIONS
OF FUNCTIONS
If possible, use the given representations of
functions ƒ and g to evaluate
 f 
(f
+ g )( 4 ) ,
(f
− g )( −2 ) ,
( fg )(1) ,
2 x 1, g ( x ) =
x
c. f ( x ) =+
(f
and
 g  (0).
 
+ g ) ( 4 ) = f ( 4 ) + g ( 4 ) = ( 2 4 + 1) + 4 = 9 + 2 = 11
(f
− g ) ( −2 ) = f ( −2 ) + g ( −2 ) = 2 ( −2 ) + 1 − −2
is undefined.
( fg ) (1) = f (1) g (1) = ( 2 1 + 1)
1 = 3 (1) = 3
2.8 - 22
Example 3
EVALUATING COMBINATIONS
OF FUNCTIONS
c. f ( x ) =+
2 x 1, g ( x ) =
x
( f + g ) ( 4 ) = f ( 4 ) + g ( 4 ) = ( 2 4 + 1) + 4 = 9 + 2 = 11
( f − g ) ( −2 ) = f ( −2 ) + g ( −2 ) = 2 ( −2 ) + 1 − −2
is undefined.
( fg ) (1) = f (1) g (1) = ( 2 1 + 1) 1 = 3 (1) = 3
 f 
 g  is undefined.
 
2.8 - 23
Example 4
FINDING THE DIFFERENCE
QUOTIENT
Let ƒ(x) = 2x2 – 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 ( x + h ) = 2( x + h )2 − 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 ).
Substitute
f ( x + h ) − f ( x )= 2( x + h )2 − 3( x + h ) − (2 x 2 − 3 x )
= 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 - 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( x + 2 xh + h ) − 3( x + h ) − (2 x − 3 x )
2
2
2
= 2 x 2 + 4 xh + 2h 2 − 3 x − 3h − 2 x 2 + 3 x
Distributive property
= 4 xh + 2h 2 − 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 ) 4 xh + 2h − 3h
Substitute.
=
h
h
h(4 x + 2h − 3)
=
h
= 4 x + 2h − 3
Factor out h.
Divide.
2.8 - 27
Caution Notice that ƒ(x + h) is not the
same as ƒ(x) + ƒ(h). For ƒ(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 2 − 3 x ) + (2h 2 − 3h )
= 2 x − 3 x + 2h − 3h
These expressions differ by 4xh.
2
2
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 )) .
The domain of g  f 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
= = 4
g (2)=
2 −1 1
Now find ( f=
 g )( 2 ) f=
( g ( 2)) f ( 4 ) :
1 7
f ( g ( 2=
) ) f ( 4=) 2 ( 4 ) −=
2.8 - 30
Example 5
EVALUATING COMPOSITE
FUNCTIONS
4
Let ƒ(x) = 2x – 1 and g(x) =
x −1
b. Find ( g  f ) ( −3).
Solution
Don’t confuse
composition
with
multiplication
( 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) = 2x2 + 5x.
Show that ( g  f )( x ) ≠ ( g  f )( x ) in general.
Solution First, find ( g  f )( x ) .
 f )( x ) g=
( g=
( f ( x ) ) g ( 4 x + 1)
= 2 ( 4 x + 1) + 5( 4 x + 1)
2
Square 4x + 1;
distributive=
property.
f ( x=
) 4x + 1
g (=
x ) 2x 2 + 5 x
2 (16 x + 8 x + 1) + 20 x + 5
2
2.8 - 32
Example 8
SHOWING THAT ( g  f
)( x ) ≠ ( f  g )( x )
Let ƒ(x) = 4x + 1 and g(x) = 2x2 + 5x.
Show that ( g  f )( x ) ≠ ( g  f )( x ) in general.
Solution First, find ( g  f )( x ) .
= 2 (16 x 2 + 8 x + 1) + 20 x + 5
Distributive
=
property.
32 x 2 + 16 x + 2 + 20 x + 5
= 32 x + 36 x + 7
2
Combine terms.
2.8 - 33
Example 8
SHOWING THAT ( g  f
)( x ) ≠ ( f  g )( x )
Let ƒ(x) = 4x + 1 and g(x) = 2x2 + 5x.
Show that ( g  f )( x ) ≠ ( g  f )( x ) in general.
Solution Now find ( f  g )( x ) .
( f  g )( x ) = f ( g ( x ) )
= f ( 2x + 5 x )
2
g (=
x ) 2x 2 + 5 x
= 4 ( 2x 2 + 5 x ) + 1
= 8 x 2 + 20 x + 1
f ( x=
) 4x + 1
Distributive
property
So... ( g  f )( x ) ≠ ( f  g )( x ) .
2.8 - 34