function operations and 2.8 composition · pdf file2.8 - 1 2.8. function operations and ......

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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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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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c. ( )( )5fg

( )( )5 (5) (5)= fg f g

26 20=

520=

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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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Solutionb. ( )( )x−f g

( )( ) ( ) ( ) 8 9 2 1x x x x x− = − = − − −f g f g

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Solutionc. ( )( )xfg

( )( ) ( )( ) ( ) 8 9 2 1x x x x x= = − −fg f g

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Solution

d. ( )x

fg

( ) ( ) 8 9( ) 2 1x xxx x

−= = −

f fg g

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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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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


( ) 2 1x x= −g

1,2 ∞

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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 −∞ ∞ ∩ ∞ = ∞

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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


( )( ) ( )( ) ( )( ) ( )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.


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Example 3


( )( ) ( )( ) ( )( ) ( )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


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Example 3


( )( ) ( )( ) ( )( ) ( )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


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Example 3


( )( ) ( )( ) ( )( ) ( )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


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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


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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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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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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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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


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


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

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