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10TH
EDITION
COLLEGE
ALGEBRA
LIAL
HORNSBY
SCHNEIDER
2.8 - 1
2.8
Function Operations and
Composition
Arithmetic Operations on Functions
The Difference Quotient
Composition of Functions and Domain
2.8 - 2
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)
 g   x   g( x ) , g( x )  0
 
Quotient
2.8 - 3
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
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
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
Example 1
USING OPERATIONS ON
FUNCTIONS
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
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
 g   0   g (0)  5
 
2.8 - 8
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
Example 2
USING OPERATIONS OF FUNCTIONS
AND DETERMINING DOMAINS
Let
f ( x )  8 x  9 and g( x )  2x  1. Find the following.
a.  f  g  x 
Solution
f
 g  x   f ( x )  g (x )  8 x  9  2 x  1
2.8 - 10
Example 2
USING OPERATIONS OF FUNCTIONS
AND DETERMINING DOMAINS
Let
f ( x )  8 x  9 and g( x )  2x  1. Find the following.
b.  f  g  x 
Solution
f
 g  x   f ( x )  g ( x )  8 x  9  2 x  1
2.8 - 11
Example 2
USING OPERATIONS OF FUNCTIONS
AND DETERMINING DOMAINS
Let
f ( x )  8 x  9 and g( x )  2x  1. Find the following.
c.
 fg  x 
Solution
 fg  x  
f ( x ) g( x )  8 x  9  2x  1
2.8 - 12
Example 2
USING OPERATIONS OF FUNCTIONS
AND DETERMINING DOMAINS
Let
f ( x )  8 x  9 and g( x )  2x  1. Find the following.
 f 
d.    x 
g
Solution
 f 
f ( x ) 8x  9
 g   x   g( x ) 
2x  1
 
2.8 - 13
Example 2
USING OPERATIONS OF FUNCTIONS
AND DETERMINING DOMAINS
Let
f ( x )  8 x  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  and g.
The domain of  is the set of all real numbers
(–, ).
2.8 - 14
Example 2
USING OPERATIONS OF FUNCTIONS
AND DETERMINING DOMAINS
Let
f ( x )  8 x  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 
 2 ,  
2.8 - 15
Example 2
USING OPERATIONS OF FUNCTIONS
AND DETERMINING DOMAINS
Let
f ( x )  8 x  9 and g( x )  2x  1. Find the following.
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
Example 2
USING OPERATIONS OF FUNCTIONS
AND DETERMINING DOMAINS
Let
f ( x )  8 x  9 and g( x )  2x  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 )  2x  1  0;
1 
f
that is, the domain of is  ,   .
g
2 
2.8 - 17
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 
 g  0 .
 
2.8 - 18
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 ( – g)(–
2),although (–2) = –
3, g(–2) is undefined
because –2 is not in
the domain of g.
2.8 - 19
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 
and g include 1, so
 fg 1  f 1 g 1  3
1 3
2.8 - 20
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 - 21
Example 3
EVALUATING COMBINATIONS
OF FUNCTIONS
If possible, use the given representations of
functions  and g to evaluate
f
 g  4  ,
b.
(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, (–g)(–2) is
undefined.
2.8 - 22
Example 3
EVALUATING COMBINATIONS
OF FUNCTIONS
If possible, use the given representations of
functions  and g to evaluate
f
 g  4  ,
b.
(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
 fg 1  f 1 1  3 1  3
2.8 - 23
Example 3
EVALUATING COMBINATIONS
OF FUNCTIONS
If possible, use the given representations of
functions  and g to evaluate
f
 g  4  ,
b.
(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 
 g  0  g 0
 
 
and
is undefined since g  0   0
2.8 - 24
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,
c. f ( x )  2x  1, g( x )  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 - 25
Example 3
EVALUATING COMBINATIONS
OF FUNCTIONS
c. f ( x )  2x  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 - 26
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 )2  3( x  h )
2.8 - 27
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)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 - 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 ).
 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 - 29
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.
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 - 30
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 2  4 xh  2h 2  3 x  3h
but
f ( x )  f (h )  (2 x  3 x )  (2h  3h )
2
2
 2 x  3 x  2h  3h
These expressions differ by 4xh.
2
2
2.8 - 31
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
Example 5
EVALUATING COMPOSITE
FUNCTIONS
4
Let (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
g (2) 
 4
2 1 1
Now find  f g  2  f  g  2   f  4  :
f  g  2   f  4   2  4   1  7
2.8 - 33
Example 5
EVALUATING COMPOSITE
FUNCTIONS
4
Let (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 - 34
Example 7
DETERMINING COMPOSITE
FUNCTIONS AND THEIR DOMAINS
6
1
Given that f ( x ) 
and g ( x )  ,find the following.
x 3
x
a.
 f g  x  and its domain
Solution
( f g )( x ) 
6
1
3
x
6 x Multiply the numerator and
( f g )( x ) 
1  3 xdenominator by x.
2.8 - 35
Example 7
DETERMINING COMPOSITE
FUNCTIONS AND THEIR DOMAINS
6
1
Given that f ( x ) 
and g ( x )  ,find the following.
x 3
x
a.
 f g  x  and its domain
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
Example 7
DETERMINING COMPOSITE
FUNCTIONS AND THEIR DOMAINS
6
1
Given that f ( x ) 
and g ( x )  ,find the following.
x 3
x
a.
 f g  x  and its domain
Solution
1
 3 The solution must be excluded.
x
1  3x
1
x
3
Multiply by x.
Divide by 3.
2.8 - 37
Example 7
DETERMINING COMPOSITE
FUNCTIONS AND THEIR DOMAINS
6
1
Given that f ( x ) 
and g ( x )  ,find the following.
x 3
x
a.
 f g  x  and its domain
Solution
1
x
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
Example 7
DETERMINING COMPOSITE
FUNCTIONS AND THEIR DOMAINS
6
1
Given that f ( x ) 
and g ( x )  ,find the following.
x 3
x
b.  g f  x  and its domain
 6 
Solution  g f  x   g  f  x    g 

 x 3
1

Note that this is
6
meaningless if x = 3
x 3
x 3

6
2.8 - 39
Example 7
DETERMINING COMPOSITE
FUNCTIONS AND THEIR DOMAINS
6
1
Given that f ( x ) 
and g ( x )  ,find the following.
x 3
x
b.  g f  x  and its domain
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
Example 7
DETERMINING COMPOSITE
FUNCTIONS AND THEIR DOMAINS
6
1
Given that f ( x ) 
and g ( x )  ,find the following.
x 3
x
b.  g f  x  and its domain
Solution Therefore, the domain of g f
is the set of all real numbers except 3, written
 ,3    3,  
2.8 - 41
Example 8
SHOWING THAT  g f
 x    f g  x 
Let (x) = 4x + 1 and g(x) = 2x2 + 5x.
Show 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
 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 - 42
Example 8
SHOWING THAT  g f
 x    f g  x 
Let (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 - 43
Example 8
SHOWING THAT  g f
 x    f g  x 
Let (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 - 44
Example 9
FINDING FUNCTIONS THAT FORM
A GIVEN COMPOSITE
Find functions  and g such that
 f g  x    x  5   4  x  5   3.
3
2
2
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 


There are other
 f  x 2  5
pairs of functions
3
2
2
 and g that also   x  5   4  x  5   3
work.
2.8 - 45