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Objectives
1 .Compute operations on functions2 .Find the composition of two functions and the domain of the composition
Operation on functions
2 5 1h x x x
h x
2 and 5 1f x x g x x
Functions are often defined using sums, differences, products and quotients of various expressions.
For example, if
We may regard as a sum of values of functions f and g given by
We may call h the sum of f and g and denote it by f + g, i.e,
h = f + g
Thus,
2 5 1h x f g x x x Therefore,
In general, if f and g are any two functions, we use the terminology and notation given by the following chart
Quotient
f g ( x ) = f ( x) g ( x )Product f g
)f – g ) ( x = (f ( x ) – g ( x )Difference f – g
)f + g ) ( x = ( f (x ) + g (x )Sum f + g
Function ValueTerminology
f
g and 0
f xfx g x
g g x
Example 1.
3If 3 2 and ,
find 2 , 2 , 2 and / 2
f x x g x x
f g f g fg f g
Solution 32 3 2 2 4 and 2 2 8f g
2 4 8 12
2 4 8 4
2 4 8 32
4 12
8 2
f g
f g
fg
fg
Class Work 1
If f( x ) = - x2 and g ( x ) = 2x – 1. Find
a 3
3
3
d / 3
f g
b f g
c fg
f g
:
3 9, 3 5
3 4
3 9 5 14
3 9 5 45
93
5
Answer
f g
f g
f g
fg
f
g
Domain of f + g, f – g, f g, and f / g
FunctionDomainf + g
)Domain of f ) ∩ ( Domain of g( f – g
f g
f / g )Domain of f ) ∩ ( Domain of g ( such that g ( x ) ≠ 0
Example 2.
2
1Let and 2 . Find the domain of
9
and
xf x g x x
xf
a f g b f g c fg dg
\ 3, 3 and [ 2, )Domain of f Domain of g R\( ) ( ) ( ), and have the same answera b c
Solution:
0-3 2 3
,33,2 gofDomainfofDomain
,33,2
g
fofDomain(d)
Class Work 2
2Let 3 4 and 2 . Find the domain of
f x x x g x x
fa f g b f g c fg d
g
( ) ( )
( ) ( )
: [ 2, )
( ), [ 2, )
2,
Solution Domain of f and domain of g
a b and c D
fd Domain of
g
R= = - ¥
= - ¥
= - ¥
Composite Functions
Definition: The composite function f ◦ g of two functions f and g is defined
by ( f ◦ g )( x ) = f ( g(x) )
( ) ( ){ } ( ) \Domain of f g x Domain g g x Domain f= Î Îo
x g( x ) f(g(x)
gf
f ◦ g
Domain of g Domain of f
Solution:
Example 3: Let f (x ) = x2 -1 and g ( x ) = 3x + 5.
)a (Find ( f ◦ g )( x ) and the domain of f ◦ g.
)b ( Find ( g ◦ f )( x ) and the domain of g ◦ f.
)c (Is f ◦ g = g ◦ f
( ) ( )( ) ( )( )a f g x f g x= =o ( )3 5f x+ = ( )2 23 5 1 9 30 24x x x+ - = + +
Domain of g = R, Range of g = R, and Domain of f = R
Domain of f ◦ g = R
( ) ( )( )b g f x =o ( )( )g f x = ( )2 1g x - = ( )2 23 1 5 3 2x x- + = +
In a similar way as in part )a(, domain of g◦ f = R
( ) ,c NO f g g f¹o o
Example 4: Let f (x ) = x2 -1 and g ( x ) = 3x + 5.
)a (Find f ( g(2) ) in two different ways: first using the functions f and g separately and second using the composite function f ◦ g
)b (Find ( f ◦ f ) ( x )
Solution:
First Method g(2) = 3(2) + 5 =11, therefore
f (g(2) ) = f ( 11)= )11(2 – 1= 121 – 1 = 120
Second Method
)f ◦ g ) ( x = ( 9 x2 +30 x + 24. Therefore ,
f ( g (2 ) ) = ( f ◦ g ) ( 2 ) = 9 ) 2(2 + 30 ) 2 + ( 24 = 120
Same Answer
(a)
(b)
)f ◦ f ) ( x = ( f (f ( x ) ) = f( x2 – 1 ) = ( x2 – 1 )2 - 1= x4 -2x2
Example 5: ) Finding values of composite functions using tables(
Several values of two functions f and g are listed in the following tables.
x1234
f ( x )3421
x1234
g(x)4132
Find ( f◦g)(2) = ( g ◦f ) ( 2 ) = ( f ◦ f ) (2 ) = ( g ◦ g )( 2 ) =
Solution:
)f◦g)(2= ( f(g(2))= f ( 1) =
)g ◦f ) ( 2 = ( g( f(2) )= g( 4 )=
3
2
Try to find the rest by yourself
)f ◦ f) ( 2= ( 1
)g ◦ g )( 2= ( 4
Example 6: ) Finding a composite function form (
Express y = ( 2x + 5 )8 in a composite function form
Solution: Function ValueChoice for
u = g(x)
Choice for
y = f( u)
y = ( 2x + 1 )82 1u x= +
8y u=
Inner function = u
Note: y = ( f ◦ g ) ( x ) = f ( g (x) ) = f ( u ) = f ( 2x +1 ) = ( 2x + 1) 8
Class Work Express the following functions in a composite function form
Choice for y = f( x )Choice for u = g(x)Function Value
( )3 45 1y x x= - +
2 4y x= -
2
3 7y
x=
+
Word Problem using composite Functions
Example 7: ) Dimensions of a balloon ( A spherical balloon is being inflated at a rate of 4.5 π ft 3 / min. Express its radius r as a function of time t ) t in minutes (, assuming that r = 0 when t = 0.
Solution: 34( )
3V r Volume of a spherep=
At time t , V(t) = 4.5 π t ft3 / min. And r = r ( t ). Therefore,
( ) ( )34
3V t r tp= Substitute V(t) = 4.5 π t
( )344.5
3t r tp p= ( )
( )3 3 4.5
4r t t= ( ) 3
13.5
4r t t=
( ) 33
2r t t=
The End
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