A. The Composition of Functions
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Section 3.6 The Composition of Functions and the Difference Quotient
EXAMPLE 1
ᮣ
353
Evaluating a Function
For g1x2 ϭ x2 Ϫ 3, find
a. g1Ϫ52
b. g15t2
c. g1t Ϫ 42
Solution
ᮣ
g1x2 ϭ x2 Ϫ 3
a.
input Ϫ5
g1Ϫ52 ϭ 1Ϫ52 2 Ϫ 3
ϭ 25 Ϫ 3
ϭ 22
simplify
result
g1x2 ϭ x Ϫ 3
b.
original function
g15t2 ϭ 15t2 2 Ϫ 3
ϭ 25t2 Ϫ 3
square input, then subtract 3
result
g1x2 ϭ x2 Ϫ 3
c.
input t Ϫ 4
It’s important to note that t and
t Ϫ 4 are two different, distinct
values—the number represented
by t, and a number four less than t.
Examples would be 7 and 3, 12
and 8, as well as Ϫ10 and Ϫ14.
There should be nothing awkward
or unusual about evaluating g(t)
versus evaluating g1t Ϫ 42 as in
Example 1(c).
square input, then subtract 3
2
input 5t
WORTHY OF NOTE
original function
original function
g1t Ϫ 42 ϭ 1t Ϫ 42 2 Ϫ 3
ϭ t2 Ϫ 8t ϩ 16 Ϫ 3
ϭ t2 Ϫ 8t ϩ 13
square input, then subtract 3
expand binomial
result
Now try Exercises 7 and 8
ᮣ
When the input value is itself a function (rather than a single number or variable),
this process is called the composition of functions. The evaluation method is exactly
the same, we are simply using a function input. Using a general function g(x) and a
function diagram as before, we illustrate the process in Figure 3.70.
Figure 3.70
Input x
g specifies
operations on x
g(x)
Input g(x)
Output
g(x)
f specifies
operations on
g(x)
f(x)
Output ( f Њ g)(x) = f [g(x)]
The notation used for the composition of f with g is an open dot “ ” ؠplaced between them, and is read, “f composed with g.” The notation 1 f ؠg21x2 indicates that
g(x) is an input for f: 1 f ؠg2 1x2 ϭ f 3g1x2 4 . If the order is reversed, as in 1g ؠf 21x2, f 1x2
becomes the input for g: 1g ؠf 21x2 ϭ g 3 f 1x2 4 . Figure 3.70 also helps us determine the
domain of a composite function, in that the first function g can operate only if x is a
valid input for g, and the second function f can operate only if g(x) is a valid input for f.
In other words, 1 f ؠg2 1x2 is defined for all x in the domain of g, such that g(x) is in the
domain of f.
CAUTION
ᮣ
Try not to confuse the new “open dot” notation for the composition of functions, with the
multiplication dot used to indicate the product of two functions: 1f # g21x2 ϭ 1fg21x2 while
1f ؠg21x2 ϭ f 3 g1x2 4 .
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The Composition of Functions
Given two functions f and g, the composition of f with g is defined by
1 f ؠg2 1x2 ϭ f 3g1x2 4
The domain of the composition is all x in the domain of g
for which g(x) is in the domain of f.
In Figure 3.71, these ideas are displayed using mapping notation, as we consider
the simple case where g1x2 ϭ x and f 1x2 ϭ 1x.
Figure 3.71
fЊg
f
g
Domain of f Њ g
x2
f [g(x2)]
g(x1)
x1
Range of f Њ g
g(x2)
g
Range of g
Domain of f
Domain of g
Range of f
The domain of g (all real numbers) is shown within the red border, with g taking
the negative inputs represented by x1 (light red), to a like-colored portion of the
range—the negative outputs g(x1). The nonnegative inputs represented by x2 (dark
red) are also mapped to a like-colored portion of the range—the nonnegative outputs
g(x2). While the range of g is also all real numbers, function f can only use the nonnegative inputs represented by g(x2). This restricts the domain of 1 f ؠg21x2 to only the
inputs from g, where g(x) is in the domain of f.
EXAMPLE 2
ᮣ
Finding a Composition of Functions
Given f 1x2 ϭ 1x Ϫ 4 and g1x2 ϭ 3x ϩ 2, find
a. 1 f ؠg21x2
b. 1g ؠf 21x2
Also determine the domain for each.
Solution
ᮣ
a. f 1x2 ϭ 1x Ϫ 4 says “decrease inputs by 4, and take the square root of the
result.”
1 f ؠg21x2 ϭ f 3 g1x2 4
g (x ) is an input for f
ϭ 1g1x2 Ϫ 4
decrease input by 4, and take the square root of the result
ϭ 113x ϩ 22 Ϫ 4 substitute 3x ϩ 2 for g (x )
result
ϭ 13x Ϫ 2
While g is defined for all real numbers, f (x) represents a real number only
when x Ն 4. For f 3g1x2 4 , this means we need g1x2 Ն 4, giving 3x ϩ 2 Ն 4,
x Ն 23. In interval notation, the domain of 1 f ؠg2 1x2 is x ʦ 3 23, q2 .
WORTHY OF NOTE
Example 2 shows that 1f ؠg21x2 is
generally not equal to 1g ؠf 21x2 . On
those occasions when they are
equal, the functions have a unique
relationship that we’ll study in
Section 5.1.
b. The function g says “inputs are multiplied by 3, then increased by 2.”
f (x ) is an input for g
1g ؠf 21x2 ϭ g 3 f 1x2 4
multiply input by 3, then increase by 2
ϭ 3f 1x2 ϩ 2
ϭ 3 1x Ϫ 4 ϩ 2
substitute 1x Ϫ 4 for f (x )
For g[ f (x)], g can accept any real number input, but f can supply only those
where x Ն 4. The domain of 1g ؠf 21x2 is x ʦ 34, q 2.
Now try Exercises 9 through 18
ᮣ
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Most graphing calculators have the ability to evaluate a composition of functions at
a given point. From Example 2, enter f(x) as Y1 ϭ 1x Ϫ 4 and g(x) as Y2 ϭ 3x ϩ 2.
To evaluate the composition 1 f ؠg2 192 (since we know x ϭ 9 is in the domain), we can
(1) return to the home screen, enter Y1 1Y2 1922 and press
(Figure 3.72), or
(2) enter Y3 ϭ Y1 1Y2 1X2 2 on the Y= screen and use the TABLE feature (Figure 3.73).
See Exercises 19 through 24.
ENTER
Figure 3.72
EXAMPLE 3
ᮣ
Figure 3.73
Finding a Composition of Functions
For f 1x2 ϭ
3x
2
and g1x2 ϭ , analyze the domain of
x
xϪ1
a. 1 f ؠg21x2
b. 1g ؠf 2 1x2
c. Find the actual compositions and comment.
Solution
ᮣ
a. 1 f ؠg21x2 : For g to be defined, x
0 is our first restriction. Once g(x) is used as
3g1x2
the input, we have f 3g1x2 4 ϭ
, and additionally note that g(x) cannot
g1x2 Ϫ 1
2
1, so x 2. The domain of f ؠg is all real numbers
equal 1. This means
x
except x ϭ 0 and x ϭ 2.
b. 1g ؠf 21x2 : For f to be defined, x 1 is our first restriction. Once f(x) is used as
2
the input, we have g 3f 1x2 4 ϭ
, and additionally note that f(x) cannot be 0.
f 1x2
3x
0, so x 0. The domain of 1g ؠf 2 1x2 is all real numbers
This means
xϪ1
except x ϭ 0 and x ϭ 1.
c. For 1 f ؠg21x2 :
3g1x2
f 3g1x2 4 ϭ
composition of f with g
g1x2 Ϫ 1
3 2
a ba b
1 x
2
ϭ
substitute for g (x)
x
2
a bϪ1
x
6
x
x
6
simplify denominator; invert and multiply
ϭ
ϭ #
x
2Ϫx
2Ϫx
x
6
result
ϭ
2Ϫx
Notice the function rule for 1 f ؠg21x2 has an implied domain of x 2, but does
not show that g (the inner function) is undefined when x ϭ 0 (see Part a). The
domain of 1 f ؠg2 1x2 is actually all real numbers except x ϭ 0 and x ϭ 2.
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For 1g ؠf 21x2 we have:
g 3f 1x2 4 ϭ
2
f 1x2
2
ϭ
3x
xϪ1
2 xϪ1
ϭ #
1
3x
21x Ϫ 12
ϭ
3x
composition of g with f
substitute
3x
for f (x)
xϪ1
invert and multiply
result
Similarly, the function rule for 1g ؠf 21x2 has an implied domain of x 0, but
does not show that f (the inner function) is undefined when x ϭ 1 (see Part b).
The domain of 1g ؠf 21x2 is actually all real numbers except x ϭ 0 and x ϭ 1.
Now try Exercises 25 through 30
ᮣ
As Example 3 illustrates, the domain of h1x2 ϭ 1 f ؠg2 1x2 cannot simply be taken from
the new function rule for h. It must be determined from the functions composed to
obtain h. The graph of 1 f ؠg21x2 is shown in Figure 3.74. We can easily see x ϭ 2 is not
in the domain as there a vertical asymptote at x ϭ 2. The fact that x ϭ 0 is also
excluded is obscured by the y-axis, but the table shown in Figure 3.75 confirms that
x ϭ 0 is likewise not in the domain.
Figure 3.74
Figure 3.75
10
Ϫ10
10
Ϫ10
To further explore concepts related to the domain of a composition, see Exercises 69 and 70.
Decomposing a Composite Function
WORTHY OF NOTE
The decomposition of a function is
not unique and can often be done
in many different ways.
Based on Figure 3.76, would you say that the circle is inside the
square or the square is inside the circle? The decomposition of a
composite function is related to a similar question, as we ask
ourselves what function (of the composition) is on the “inside”—
the input quantity — and what function is on the “outside.” For
instance, consider h1x2 ϭ 1x Ϫ 4, where we see that x Ϫ 4 is
“inside” the radical. Letting g1x2 ϭ x Ϫ 4 and f 1x2 ϭ 1x, we have
h1x2 ϭ 1 f ؠg21x2 or f [g(x)].
Figure 3.76
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EXAMPLE 4
ᮣ
Decomposing a Composite Function
Solution
ᮣ
3
Noting that 1
x ϩ 1 is inside the squaring function, we assign g(x) as this inner
3
function: g1x2 ϭ 1 x ϩ 1. The outer function is the squaring function decreased
by 3, so f 1x2 ϭ x2 Ϫ 3.
A. You’ve just seen how we
can compose two functions
and determine the domain, and
decompose a function
3
Given h1x2 ϭ 1 1 x ϩ 12 2 Ϫ 3, identify two functions f and g so that
1 f ؠg21x2 ϭ h1x2, then check by composing the functions to obtain h(x).
Check: 1 f ؠg21x2 ϭ f 3g1x2 4
ϭ 3g1x2 4 2 Ϫ 3
3
ϭ 31
x ϩ 142 Ϫ 3
ϭ h1x2 ✓
g(x ) is an input for f
f squares inputs, then decreases the result by 3
substitute 13 x ϩ 1 for g (x)
Now try Exercises 31 through 34
ᮣ
B. A Numerical and Graphical View
of the Composition of Functions
Just as with the sum, difference, product, and quotient of functions, the composition of
functions can also be interpreted and understood graphically. For 1 f ؠg21x2 , once the
value of g(x) is known (read from the graph), the value of 1 f ؠg21x2 ϭ f 3g1x2 4 , can also
be determined.
EXAMPLE 5
ᮣ
Interpreting the Composition of Functions Numerically
For f(x) and g(x) as shown, use the graph given to
determine the value of each expression.
a. f(4), g(2), and 1 f ؠg2 (2)
b. g(6), f(8), and 1g ؠf 2 (8)
c. 1 f ؠg2 (8) and 1g ؠf 2 (0)
d. 1 f ؠf 2 (9) and 1g ؠg2 (0)
Solution
ᮣ
y
6
f (x)
Ϫ2
10 x
g(x)
Ϫ4
a. For f(4), we go to x ϭ 4 and note that 14, Ϫ12 is a
point on the red graph: f 142 ϭ Ϫ1. For g(2), go to x ϭ 2 and note that (2, 4)
is a point on the blue graph: g122 ϭ 4. Since g122 ϭ 4 and f 142 ϭ Ϫ1,
1 f ؠg2122 ϭ f 3g122 4 ϭ f 142 ϭ Ϫ1: 1 f ؠg2122 ϭ Ϫ1.
b. For g162, x ϭ 6 and we find that 16, Ϫ42 is a point on the blue graph: g162 ϭ Ϫ4.
For f 182, x ϭ 8 and note that that (8, 6) is a point on the red graph: f 182 ϭ 6. Since
f 182 ϭ 6 and g162 ϭ Ϫ4, 1g ؠf 2 182 ϭ g3 f 182 4 ϭ g162 ϭ Ϫ4: 1g ؠf 2182 ϭ Ϫ4.
c. As illustrated in parts (a) and (b), for 1f ؠg2 182 ϭ f 3g182 4 we first determine
g(8), then substitute this value into f. From the blue graph g182 ϭ Ϫ2, and
f 1Ϫ22 ϭ 3. For 1g ؠf 2102 ϭ g3 f 102 4 we first determine f(0), then substitute
this value into g. From the red graph f 102 ϭ Ϫ1, and g1Ϫ12 ϭ 1.
d. To evaluate 1 f ؠf 2 192 ϭ f 3 f 192 4 we follow the same sequence as before. From
the graph of f, f 192 ϭ 5 and f 152 ϭ 1, showing 1 f ؠf 2192 ϭ 1. The same ideas
are applied to 1g ؠg2102 ϭ g 3g102 4 . Since g102 ϭ 4 and g142 ϭ Ϫ2,
1g ؠg2102 ϭ Ϫ2.
Now try Exercises 35 and 36
ᮣ
As with other operations on functions, a composition can be understood graphically at specific points, as in Example 5, or for all allowable x-values. Exploring the
second option will help us understand certain transformations of functions, and why
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the domain of a composition is defined so carefully. For instance, consider the functions
f 1x2 ϭ Ϫx2 ϩ 3 and g1x2 ϭ 1x Ϫ 22 . The graph of f is a parabola opening downward
with vertex (0, 3). For h1x2 ϭ 1 f ؠg2 1x2 we have h1x2 ϭ Ϫ1x Ϫ 22 2 ϩ 3, which
we recognize as the graph of f shifted 2 units to the right. Using Y1 ϭ ϪX2 ϩ 3,
Y2 ϭ X Ϫ 2, and Y3 ϭ Y1 1Y2 1X2 2 , we can deactivate Y2 so the only the graphs of Y1
and Y3 (in bold) are shown (see Figure 3.77). Here the domain of the composition was
not a concern, as both f and g have domain x ʦ ޒ. But now consider f 1x2 ϭ 41x ϩ 1,
3
g1x2 ϭ
, and h1x2 ϭ 1 f ؠg2 1x2 . The domain of g must exclude x ϭ 2 (which is
2Ϫx
also excluded from the domain of h), while the domain of f is x Ն Ϫ1. But after
3
assigning Y1 ϭ 4 2X ϩ 1, Y2 ϭ
, and Y3 ϭ Y1 1Y2 1X22 , the graph of Y3 shows
2ϪX
a noticeable gap between x ϭ 2 and x ϭ 5 (Figure 3.78). The reason is that
1 f ؠg21x2 ϭ f 3g1x2 4 uses g(x) as the input for f, meaning for the domain, x Ն Ϫ1
3
Ն Ϫ1. Solving this inequality graphically (Figure 3.79)
becomes g1x2 Ն Ϫ1 S
2Ϫx
shows g1x2 Ն Ϫ1 only for the intervals x ʦ 1Ϫq, 22 ´ 3 5, q 2 , leaving the gap seen
in Y3 for the interval (2, 5]. The domain of h is x ʦ 1Ϫq, 22 ´ 3 5, q2 .
Figure 3.77
5
Ϫ5
5
Ϫ5
EXAMPLE 6
Figure 3.78
Figure 3.79
10
10
Ϫ5
10
Ϫ5
ᮣ
Ϫ5
10
Ϫ5
Interpreting a Composition Graphically and Understanding the Domain
Given f 1x2 ϭ 3 11 Ϫ x , g1x2 ϭ
4
,
xϩ3
a. State the domains of f and g.
b. Use a graphing calculator to study the graph of h1x2 ϭ 1 f ؠg21x2 .
c. Algebraically determine the domain of h, and reconcile it with the graph.
Solution
ᮣ
a. The domain of g must exclude x ϭ Ϫ3, and for the domain of f we must have
x Յ 1.
4
b. For the graph of h, enter Y1 ϭ 3 21 Ϫ x, Y2 ϭ
, and Y3 ϭ Y1 1Y2 1X22 .
xϩ3
Graphing Y3 on the ZOOM 6:ZStandard screen shows a gap between x ϭ Ϫ3
and x ϭ 1 (Figure 3.80).
c. Since the composition 1 f ؠg2 1x2 ϭ f 3g1x2 4 uses g(x) as the input for f,
4
x Յ 1 (the domain for f ) becomes g1x2 Յ 1 S
Յ 1. Figure 3.81
xϩ3
shows a graphical solution to this inequality, which indicates g1x2 Յ 1 for
x ʦ 1Ϫq, Ϫ32 ´ 3 1, q 2 . The domain of h is x ʦ 1Ϫq, Ϫ32 ´ 31, q2 .