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2.8Function Operations and CompositionArithmetic Operations on FunctionsThe Difference QuotientComposition of Functions and Domain
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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.( )( ) ( ) ( )x x x+ = +f g f g Sum
( )( ) ( ) ( )x x x− = −f g f g Difference
( )( ) ( ) ( )x x x= fg f g Product
( ) ( ) , ( ) 0( )xx xx
= ≠
f f gg g
Quotient
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Example 1 USING OPERATIONS ON FUNCTIONS
Let ƒ(x) = x2 + 1 and g(x) = 3x + 5. Find the following.
Solution Since ƒ(1) = 2 and g(1) = 8, use the definition to get
a. ( )( )1+f g
( )( )1 (1) (1)+ = +f g f g ( )( ) ( ) ( )x x x+ = +f g f g
82= +
10=
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Example 1 USING OPERATIONS ON FUNCTIONS
Let ƒ(x) = x2 + 1 and g(x) = 3x + 5. Find the following.
Solution Since ƒ(–3) = 10 and g(–3) = –4, use the definition to get
b. ( )( )3− −f g
( )( ) 3 )3(3 ) (− −= −− −f g f g
10 ( 4)= − −
14=
( )( ) ( ) ( )x x x− = −f g f g
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Example 1 USING OPERATIONS ON FUNCTIONS
Let ƒ(x) = x2 + 1 and g(x) = 3x + 5. Find the following.
Solution Since ƒ(5) = 26 and g(5) = 20, use the definition to get
c. ( )( )5fg
( )( )5 (5) (5)= fg f g
26 20=
520=
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Example 1 USING OPERATIONS ON FUNCTIONS
Let ƒ(x) = x2 + 1 and g(x) = 3x + 5. Find the following.
Solution Since ƒ(0) = 1 and g(0) = 5, use the definition to get
d. ( )0
fg
( ) (0) 10(0) 5
= =
f fg g
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Example 2 USING OPERATIONS OF FUNCTIONS AND DETERMINING DOMAINS
Let( ) 8 9 and ( ) 2 1. Find the following.x x x x= − = −f g
Solutiona. ( )( )x+f g
( )( ) ) 2 18 9)( (xx xx x−+ = + = + −ff g g
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Example 2 USING OPERATIONS OF FUNCTIONS AND DETERMINING DOMAINS
Let( ) 8 9 and ( ) 2 1. Find the following.x x x x= − = −f g
Solutionb. ( )( )x−f g
( )( ) ( ) ( ) 8 9 2 1x x x x x− = − = − − −f g f g
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Example 2 USING OPERATIONS OF FUNCTIONS AND DETERMINING DOMAINS
Let( ) 8 9 and ( ) 2 1. Find the following.x x x x= − = −f g
Solutionc. ( )( )xfg
( )( ) ( )( ) ( ) 8 9 2 1x x x x x= = − −fg f g
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Example 2 USING OPERATIONS OF FUNCTIONS AND DETERMINING DOMAINS
Let( ) 8 9 and ( ) 2 1. Find the following.x x x x= − = −f g
Solution
d. ( )x
fg
( ) ( ) 8 9( ) 2 1x xxx x
−= = −
f fg g
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Example 2 USING OPERATIONS OF FUNCTIONS AND DETERMINING DOMAINS
Let( ) 8 9 and ( ) 2 1. Find the following.x x x x= − = −f g
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 (–∞, ∞).
e. Give the domains of the functions.
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Example 2 USING OPERATIONS OF FUNCTIONS AND DETERMINING DOMAINS
Let( ) 8 9 and ( ) 2 1. Find the following.x x x x= − = −f g
Solution Since , 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
e. Give the domains of the functions.
( ) 2 1x x= −g
1,2 ∞
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Example 2 USING OPERATIONS OF FUNCTIONS AND DETERMINING DOMAINS
Let( ) 8 9 and ( ) 2 1. Find the following.x x x x= − = −f g
Solution The domains of ƒ + g, ƒ – g, ƒg are the intersection of the domains of ƒ and g,which is
e. Give the domains of the functions.
( ) 1 1, , ,2 2 −∞ ∞ ∩ ∞ = ∞
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Example 2 USING OPERATIONS OF FUNCTIONS AND DETERMINING DOMAINS
Let( ) 8 9 and ( ) 2 1. Find the following.x x x x= − = −f g
Solution The domains of includes those real numbers in the intersection for which
that is, the domain of is
e. Give the domains of the functions.fg
( ) 2 1 0;x x= − ≠g
fg
1, .2
∞
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Example 3 EVALUATING COMBINATIONS OF FUNCTIONS
If possible, use the given representations of functions ƒ and g to evaluate …
( )( ) ( )( ) ( )( ) ( )4 , 2 , 1 , and 0 . + − −
ff g f g fgg
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Example 3EVALUATING COMBINATIONS OF FUNCTIONS
( )( ) ( )( ) ( )( ) ( )4 , 2 , 1 , and 0 . + − −
ff g f g fgg
–4 –2 2
5
0
9
4
( )y x= f
( )y x= g
a.(4) 9=f (4) 2=g
( ) ( )2 14
9 14= +
= + =
gf
For (ƒ – g)(–2),although ƒ(–2) = –3, g(–2) is undefined because –2 is not in the domain of g.
x
y
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( )( ) ( )( ) ( )( ) ( )4 , 2 , 1 , and 0 . + − −
ff g f g fgg
Example 3
–4 –2 2
5
0
9
4
( )y x= f
( )y x= g
a.(4) 9=f (4) 2=g
( ) ( )2 14
9 14= +
= + =
gf
The domains of ƒand g include 1, so
( )( ) ( ) ( )1 11 31 3= = = gf fg
EVALUATING COMBINATIONS OF FUNCTIONS
x
y
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Example 3
–4 –2 2
5
0
9
4
( )y x= f
( )y x= g
a.(4) 9=f (4) 2=g
( ) ( )2 14
9 14= +
= + =
gf
The graph of gincludes the origin, so ( ) .0 0=g
Thus, is undefined.( )0
fg
EVALUATING COMBINATIONS OF FUNCTIONS
( )( ) ( )( ) ( )( ) ( )4 , 2 , 1 , and 0 . + − −
ff g f g fgg
x
y
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Example 3
If possible, use the given representations of functions ƒ and g to evaluate
( )( ) ( )( ) ( )( ) ( )4 , 2 , 1 , and 0 . + − −
ff g f g fgg
x ƒ(x) g(x)–2 –3 undefined0 1 01 3 11 1 undefined4 9 2
b. (4) 9=f (4) 2=g( ) ( )4 4
9 2 11= +
= + =
f g
In the table, g(–2) is undefined. Thus, (ƒ–g)(–2) is undefined.
EVALUATING COMBINATIONS OF FUNCTIONS
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Example 3
If possible, use the given representations of functions ƒ and g to evaluate
( )( ) ( )( ) ( )( ) ( )4 , 2 , 1 , and 0 . + − −
ff g f g fgg
x ƒ(x) h(x)–2 –3 undefined0 1 01 3 11 1 undefined4 9 2
b. (4) 9=f (4) 2=g( ) ( )4 4
9 2 11= +
= + =
f g
( )( ) ( ) ( ) ( )1 1 1 3 1 3= = =fg f
EVALUATING COMBINATIONS OF FUNCTIONS
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Example 3
If possible, use the given representations of functions ƒ and g to evaluate
( )( ) ( )( ) ( )( ) ( )4 , 2 , 1 , and 0 . + − −
ff g f g fgg
x ƒ(x) h(x)–2 –3 undefined0 1 01 3 11 1 undefined4 9 2
b. (4) 9=f (4) 2=g( ) ( )4 4
9 2 11= +
= + =
f g
and( ) ( )( )
( )
00
0
is undefined since 0 0
=
=
ffg g
g
EVALUATING COMBINATIONS OF FUNCTIONS
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Example 3
If possible, use the given representations of functions ƒ and g to evaluate
( )( ) ( )( ) ( )( ) ( )4 , 2 , 1 , and 0 . + − −
ff g f g fgg
c. ( ) 2 1, ( )x x x x= + =f g
( )( ) ( ) ( ) ( )4 4 4 2 4 1 4 9 2 11+ = + = + + = + =f g f g
( )( ) ( ) ( ) ( )2 2 2 2 2 1
is unde i e
.
2
f n d
− − = − + − = − + − −f g f g
( )( ) ( ) ( ) ( ) ( )1 1 1 2 1 1 1 3 1 3= = + = = fg f g
EVALUATING COMBINATIONS OF FUNCTIONS
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Example 3
c. ( ) 2 1, ( )x x x x= + =f g( )( ) ( ) ( ) ( )4 4 4 2 4 1 4 9 2 11+ = + = + + = + =f g f g
( )( ) ( ) ( ) ( )2 2 2 2 2 1
is unde i e
.
2
f n d
− − = − + − = − + − −f g f g
( )( ) ( ) ( ) ( ) ( )1 1 1 2 1 1 1 3 1 3= = + = = fg f g
is undefined.
fg
EVALUATING COMBINATIONS OF FUNCTIONS
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Example 4 FINDING THE DIFFERENCE QUOTIENT
Let ƒ(x) = 2x2 – 3x. Find the difference quotient and simplify the expression.SolutionStep 1 Find the first term in the numerator, ƒ(x + h). Replace the x in ƒ(x) with x + h.
2( ) 2( ) 3( )x h x h x h+ + − +=f
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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 ( ) ( ).x h x+ −f f
2 2( ) ( ) 2( ) 3( ) (2 3 )x h x x h x h x x + − = + − + − − f fSubstitute
2 2 22( 2 ) 3( ) (2 3 )x xh h x h x x= + + − + − −Remember this
term when squaring x + h
Square x + h
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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 ( ) ( ).x h x+ −f f2 2 22( 2 ) 3( ) (2 3 )x xh h x h x x= + + − + − −
2 2 22 4 2 3 3 2 3x xh h x h x x= + + − − − +
Distributive property24 2 3xh h h+= − Combine terms.
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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.
Substitute.2( ) 2( ) 4 3x h x
hh
hxh h+ − = + −f f
(4 2 3)h x hh+ −= Factor out h.
4 2 3x h= + − Divide.
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Caution Notice that ƒ(x + h) is not the same as ƒ(x) + ƒ(h). For ƒ(x) = 2x2 – 3x in Example 4. 2
2 2
( ) 2( ) 3( )2 24 3 3x h x h x hx h xh hx+ = + − +
= + + − −
f
but 2 2
2 2
( ) ( ) (2 3 ) (2 3 )2 3 2 3x h x x h hx x h h+ = − + −
= − + −
f f
These expressions differ by 4xh.
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Composition of Functions and DomainIf ƒ and g are functions, then the composite function, or composition, of g and ƒ is defined by ( )( ) ( )( ) .x x=g f g f
The domain of is the set of all numbers x in the domain of ƒ such that ƒ(x)is in the domain of g.
g f
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Example 5 EVALUATING COMPOSITE FUNCTIONS
Let ƒ(x) = 2x – 1 and g(x) 41x
=−
( )( )Find 2 .f ga.
Solution First find g(2). ( ) 4Since ,1
xx
=−
g4 4
2(2) 4
1 1= = =
−g
Now find ( )( ) ( )( ) ( )2 42 := = gf g f f
( )( ) ( ) ( )2 72 4 14= = − =f fg
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Example 5 EVALUATING COMPOSITE FUNCTIONS
Let ƒ(x) = 2x – 1 and g(x) 41x
=−
( )Find ( 3).−g fb.
Solution 4 4
17 8= =
−− −
( )( ) ( )( ) ( )3 73 :− = =− − ff g g g
Don’t confuse composition
with multiplication
2.1= −
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Example 8 SHOWING THAT ( )( ) ( )( )x x≠ g f f g
Let ƒ(x) = 4x + 1 and g(x) = 2x2 + 5x.( )( ) ( )( )Show that in general.x x≠ g f g f
Solution ( )( )First, find .xg f
( )( ) ( )( ) ( )4 1x x x= = +g f g f g ( ) 4 1x x= +f
( )24 1 ( )42 15x x+ += + ( ) 22 5x x x= +g
( )22 16 8 1 20 5x x x= + + + +Square 4x + 1; distributive property.
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Example 8
Let ƒ(x) = 4x + 1 and g(x) = 2x2 + 5x.( )( ) ( )( )Show that in general.x x≠ g f g f
Solution ( )( )First, find .xg f
( )22 16 8 1 20 5x x x= + + + +Distributive property.
232 16 2 20 5x x x= + + + +
232 36 7x x= + + Combine terms.
SHOWING THAT ( )( ) ( )( )x x≠ g f f g
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Example 8
Let ƒ(x) = 4x + 1 and g(x) = 2x2 + 5x.( )( ) ( )( )Show that in general.x x≠ g f g f
Solution ( )( )Now find .xf g
Distributive property
( )( ) ( )( )x x=f g f g
( )22 5x x= +f ( ) 22 5x x x= +g
( )24 12 5x x= + + ( ) 4 1x x= +f
28 20 1x x= + +
( )( ) ( )( )So... .x x≠ g f f g
SHOWING THAT ( )( ) ( )( )x x≠ g f f g