Chabot Mathematics
§9.2a
Composite Fcns
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Chabot College Mathematics
1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-58_sec_9-2a_Composite_Fcns.ppt
Review § 9.1
MTH 55
 Any QUESTIONS About
• §9.1 → The NATURAL Base, e
 Any QUESTIONS About HomeWork
• §9.1 → HW-43
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-58_sec_9-2a_Composite_Fcns.ppt
Composite Functions
 In the real world, functions
frequently occur in which some
quantity depends on a variable that,
in turn, depends on another
variable.
 Functions such as these are called
COMPOSITE FUNCTIONS
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-58_sec_9-2a_Composite_Fcns.ppt
Composing a Function
 Composition with sets A & B by fcns g & f
g
f
A
B
1
f ( x)  x  3
2
C
1
3
7
4
10
22
−1
2
8
g ( x)  3 x  1
h
Chabot College Mathematics
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h(x) = ?
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-58_sec_9-2a_Composite_Fcns.ppt
g ( x)  3x  1
A
Composing a Function
f ( x) 
B
4
-1
3
10
2
7
22
8
 From The Diagram notice that since
f takes the output from g we can
combine f and g to get a function h:
f (g (x)) = f (3x + 1)
1
 (3x  1)  3
2
3
5
 x
2
2
3
5
 This Yields an eqn for h: h( x)  x  .
2
2
5
C
1
h
Chabot College Mathematics
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x 3
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-58_sec_9-2a_Composite_Fcns.ppt
h(x) = ?
Composing a Function
g ( x)  3x  1
f ( x) 
1
x 3
2
A
B
1
4
-1
3
10
2
7
22
8
h
C
3
5
h( x )  x  .
2
2
h(x) = ?
 The function h is the composition of
f and g and is denoted f○g (read “the
composition of f and g,” “f composed
with g,” or “f circle g”).
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-58_sec_9-2a_Composite_Fcns.ppt
COMPOSITION OF FUNCTIONS
 If f and g are two functions, the
composition of function f with function g
is written as f○g and is defined by the
equation
 f og x   f g x ,
 The function where the domain of f○g
consists of those values x in the domain
of g for which g(x) is in the domain of f
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-58_sec_9-2a_Composite_Fcns.ppt
COMPOSITION OF FUNCTIONS
 Graphically the f○g Domain Chain
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-58_sec_9-2a_Composite_Fcns.ppt
COMPOSITION OF FUNCTIONS
 Conceptually the f○g Operation Chain
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-58_sec_9-2a_Composite_Fcns.ppt
Example  Evaluate Composites
 Given: f x   x 3 and g x   x  1.
 Find Each of the Following
a.  f og 1
b. g o f 1
c.  f o f 1
d. g og 1
 Solution a.a.  f og 1  f g 1
 f 2 
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 23
8
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-58_sec_9-2a_Composite_Fcns.ppt
Example  Evaluate Composites
 Solution b.
b. g o f 1  g  f 1
f x   x 3 and g x   x  1.
 g 1  1  1  2
 Solution c.  f o f 1  f  f 1
 f 1  1  1
3
 Solution d.
d. g og 1  g g 1
 g 0   0  1  1
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-58_sec_9-2a_Composite_Fcns.ppt
Example  Fcn Composition
 Given f(x) = 4x and g(x) = x2 + 2, find
f
g  ( x) and  g f  ( x).
 SOLUTION
f
g  ( x)  f ( g ( x)) = f (x2 + 2)
= 4(x2 + 2)
= 4x2 + 8
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12
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-58_sec_9-2a_Composite_Fcns.ppt
Example  Fcn Composition
 Given f(x) = 4x and g(x) = x2 + 2, find
f
g  ( x) and  g f  ( x).
 SOLUTION
g
f  ( x)  g ( f ( x)) = g(4x)
= (4x)2 + 2
= 16x2 + 2
 This example shows
that in general  f g  ( x)   g f  ( x).
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-58_sec_9-2a_Composite_Fcns.ppt
Example  Fcn Composition
 Given: f x   2x  1 and g x   x  3.
2
 Find Each Composite Function
a.  f og x 
b. g o f x 
c.  f o f x 
a.  f og x   f g x 

 2 x

 3 1
 f x2  3
 Solution a.
2
 2x  6  1
2
 2x  5
2
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-58_sec_9-2a_Composite_Fcns.ppt
Example  Fcn Composition
 Given: f x   2x  1 and g x   x  3.
2
 Solution
 f og
x  b.b. g o f x  c.  f o f x 
b. g o f x   g  f x 
 g 2x  1
2
2
 2x  1  3  4x  4x  2
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-58_sec_9-2a_Composite_Fcns.ppt
Example  Fcn Composition
 Given: f x   2x  1 and g x   x  3.
2
g o f Solution
x  c.c.  f o f x 
c.  f o f x   f  f x 
 f 2x  1
 2 2x  1  1  4 x  3
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-58_sec_9-2a_Composite_Fcns.ppt
Example  Composite Domain
1
Let f x   x  1 and g x   .
 Given:
x
a. Find  f og 1.
b. Find g o f 1.
c. Find  f og x  and its domain.
d. Find g o f x  and its domain.
 Solution
a.  f og 1  f g 1
a.
 f 1  1  1  0
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-58_sec_9-2a_Composite_Fcns.ppt
Example  Composite Domain
1
Let f x   x  1 and g x   .
 Given:
x
 Solution
b. g o f 1  g  f 1
b.
 g 0  not defined
 1 1
 Soln
c.  f og x   f g x   f     1
 x x
c.
• Domain: (−∞, 0)U(0, ∞) or {x|x ≠ 0}
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-58_sec_9-2a_Composite_Fcns.ppt
Example  Composite Domain
1
Let f x   x  1 and g x   .
 Given:
x
1
 Soln
d. g o f x   g  f x   g x  1 
x 1
d.
• Domain: (−∞, −1)U(−1, ∞) or {x|x ≠ −1}
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-58_sec_9-2a_Composite_Fcns.ppt
DEcomposing a Function
Let H x  
 Given:
1
.
2x  1
 Show that each of the following
provides a DEcomposition of H(x)
2
a. Express H x  as f g x ,
1
where f x  
and g x   2x 2  1.
x
b. Express H x  as f g x ,
1
where f x   and g x   2x 2  1.
x
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-58_sec_9-2a_Composite_Fcns.ppt
Decomposing a Function
Let
 Solution: a. Express H x  as f g x ,
1
H x  
.
1
2
2x  1 where f x  

and g x   2x 2  1.
x

a. f g x   f 2x  1

2
1
2x  1
 H x 
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2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-58_sec_9-2a_Composite_Fcns.ppt
Decomposing a Function
Let
 Solution: b. Express H x  as f g x ,
1
H x  
.
1
2
2x  1
where f x  
b. f g x   f

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x
and g x   2x 2  1.
 2x  1
2
1
2x  1
 H x 
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-58_sec_9-2a_Composite_Fcns.ppt
Example  Automobile Sales

A car dealer offers an 8% discount off
the manufacturer’s suggested retail
price (MSRP) of x dollars for any new
car on his lot. At the same time, the
manufacturer offers a $4000 rebate on
the purchase.
a. Write a function f(x) that represents the
price after the rebate.
b. Write a function g(x) that represents the
price after the dealer’s discount.
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-58_sec_9-2a_Composite_Fcns.ppt
Example  Automobile Sales
c. Write the Functions (f○g)(x) & (g○f)(x).
What do these Functions Represent?
d. Calculate (g○f)(x) − (f○g)(x). Interpret this
odd-looking expression
 Solution a.
The MSRP is x dollars, rebate is $4k, so
f(x) = x – 4000
represents the price of the car after
the rebate.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-58_sec_9-2a_Composite_Fcns.ppt
Example  Automobile Sales
 Solution b.
The dealer’s discount is 8% of x, or 0.08x,
so:
g(x) = x – 0.08x = 0.92x
represents the price of the car after the
dealer’s discount.
c. c.(i)  f og x   f g x   f 0.92x 
 Soln
 0.92x  4000
• This represents the price when the
DEALER’S discount is is applied first.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-58_sec_9-2a_Composite_Fcns.ppt
Example  Automobile Sales
 Solution c. (cont.)
(ii)
g o f x   g  f x  g x  4000 
 0.92 x  4000 
 0.92x  3680
• This represents the price when the
MANUFACTURER’S rebate is applied first.
Chabot College Mathematics
26
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-58_sec_9-2a_Composite_Fcns.ppt
Example  Automobile Sales
 Solution d.
g o f x    f og x   g  f x  f g x 
 0.92x  3680   0.92x  4000 
d.
 320 dollars
• This equation shows that it will cost $320
MORE for any car, regardless of its price, if
you apply the rebate first and then the
discount second.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-58_sec_9-2a_Composite_Fcns.ppt
WhiteBoard Work
 Problems From §9.2 Exercise Set
• 10, 12, 56, 58, 70
 Composition of Functions Corresponds
to a Production Line
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-58_sec_9-2a_Composite_Fcns.ppt
All Done for Today
Function
Machines
&
CoDomain
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29
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-58_sec_9-2a_Composite_Fcns.ppt
Chabot Mathematics
Appendix
r  s  r  s r  s 
2
2
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
–
Chabot College Mathematics
30
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-58_sec_9-2a_Composite_Fcns.ppt