121
Section 5.1 – Composite Functions
Objective #1:
Form a Composite Function.
In many cases, we can create a new function by taking the composition of
two functions. For example, suppose f(x) = x and g(x) = x3 + 1. If we
evaluate f(x) for when x = g(x), this will give us a new function.
f(g(x)) = f(x3 + 1) = (x 3 + 1) = x 3 + 1
€
This new function is called the composition
of f and g or we can say it is f
compose g. The innermost function, g, acts on x first; cubing x and then
adding 1. The output of the innermost function becomes the input or
€ function, f. Then outermost function acts
domain element €
of the outermost
on that input; take the square root. So, suppose we want to find f(g(2)), the
innermost function would cube 2 and then add 1: 23 + 1 = 8 + 1
= 9. Then, the outermost function would take the square root of that
answer: 9 = 3. Thus, f(g(2)) = 3.
Definition
The composition of f and g or (f compose g), denoted f O g, is defined by
€
(f O g)(x) = f(g(x)) provided that g(x) is in the domain of f.
The composition of g and f or (g compose f), denoted g O f, is defined by
(g O f)(x) = g(f(x)) provided that f(x) is in the domain of g.
Caution: (f O g)(x) and (g O f)(x) are not always equal.
Given f(x) = 2x – 5, g(x) = 3x2 + 1, and h(x) =
Ex. 1a (f O g)(– 1)
Ex. 1d (g O h)(7)
Solution:
a)
b)
Ex. 1b (g O f)(2)
Ex. 1e (f O f)(2)
€
1
,
x−5
find:
Ex. 1c (h O g)(3)
Ex. 1f (h O h)(4)
(f O g)(– 1) = f(g(– 1))
= f(4)
= 2(4) – 5 = 3
But, g(– 1) = 3(– 1)2 + 1 = 4.
Now, plug in 4 for x in f.
(g O f)(2) = g(f(2))
= g(– 1)
= 3(– 1)2 + 1 = 4
But, f(2) = 2(2) – 5 = – 1.
Now, plug in – 1 for x in g.
122
c)
(h O g)(3) = h(g(3))
= h(28)
1
=
=
(28)−5
d)
But, g(3) = 3(3)2 + 1 = 28.
Now, plug in 28 for x in h.
1
23
(g O h)(7) = g(h(7))
€
But, h(7) =
=€
g(0.5)
1
(7)−5
=
1
2
or 0.5.
Now, plug in 0.5 for x in g.
= 3(0.5)2 + 1 = 1.75
e)
(f O f)(2) = f(f(2))
= f(– 1)
= 2(– 1) – 5 = – 7
€ f(2) =€2(2) – 5 = – 1.
But,
Now, plug in – 1 for x in f.
f)
(h O h)(4) = h(h(4))
But, h(4) =
= h(– 1)
=
1
(−1)−5
1
(4) −5
= – 1.
Now, plug in – 1 for x in h.
=–
1
6
€
Finding the Domain of a Composite Function.
Objective #2:
Given f(x) = 3x – 2, g(x)€= x2, and h(x) = x + 7 , find:
€
Ex. 2a (f O g)(x)
Ex. 2b (g O f)(x)
Ex. 2d (g O h)(x)
Ex. 2e (h O f)(x)
€
Solution:
Ex. 2c (h O g)(x)
Ex. 2f (f O h)(x)
a)
(f O g)(x) = f(g(x))
= f(x2)
= 3(x2) – 2
= 3x2 – 2
Replace g(x) by x2.
Replace x by x2 in the function f.
Simplify.
The domain is (– ∞, ∞).
b)
(g O f)(x) = g(f(x))
= g(3x – 2)
= (3x – 2)2
= 9x2 – 12x + 4
Replace f(x) by 3x – 2.
Replace x by 3x – 2 in g
Simplify.
The domain is (– ∞, ∞).
c)
(h O g)(x) = h(g(x))
= h(x2)
Replace g(x) by x2.
Replace x by x2 in the function f.
€
€
=
x2 + 7
Since x2 + 7 > 0,
=
x2 + 7
the domain is (– ∞, ∞).
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d)
(g O h)(x) = g(h(x))
= g(
=(
€
e)
x + 7 )2
=x+7
(h O€f)(x) = h(f(x))
= h(3x – 2)
=
=
f)
x+7 )
(3x – 2) + 7
3x + 5
Replace h(x) by
x+7 .
x + 7 in the
Replace x by
function g.
Simplify.€Be careful, for x + 7 to
be defined,
x + 7 ≥ 0 or x ≥ – 7.
€
The domain is [– 7, ∞).
Replace f(x) by€3x – 2.
Replace x by 3x – 2 in the
function h.
Simplify. But, 3x + 5 ≥ 0 or
x≥–
5
,
3
the domain is [–
5
3
, ∞).
(f O h)(x) = f(h(x))
Replace h(x) by x + 7
€
= f( x + 7 )
Replace x by x + 7 in f
€
€
€
= 3( x + 7 ) – 2
Simplify. Be careful, for x + 7 to
be defined,
€ x + 7 ≥ 0 or x ≥ – 7.
= 3 x+7 – 2
The domain
is [– 7, ∞).
€
€
€
Use the functions below to find:
€ g(x) = 16−2x
€
r(x) =
3
2
x −4
Ex. 3a
(r O g)(x)
Ex. 3b
(r O r)(x)
Solution:
To find the
€ domain of g, set 16 – 2x ≥ 0 and solve:
– 2x ≥ – 16 or x ≤ 8. Thus, the domain of g is { x| x ≤ 8}
To find the domain of r, set x2 – 4 = 0, solve and exclude those
values:
(x – 2)(x + 2) = 0 or x = 2 or – 2
Thus, the domain of r is { x| x ≠ – 2 or x ≠ 2}
a)
(r O g)(x) = r(g(x))
Note that { x| x ≤ 8} because of g. But, x ≠ ± 2 in r which implies
that g(x) = 16−2x ≠ ± 2. Solving yields:
16−2x ≠ ± 2 (square both sides)
16 – 2x ≠ 4
€ – 2x ≠ – 12 or x ≠ 6
Thus, the domain of r O g is { x | x ≤ 8, x ≠ 6}
€ Now, let's find the composition:
124
Replace g(x) by
= r( 16−2x )
Replace x by
=
=
€
=
€
b)
16−2x .
16−2x in r.
r(g(x))
3
( 16−2x )2 −4
3
(16−2x)−4
Simplify.
€
Simplify.
€
3
12−2x
The domain is { x | x ≤ 8, x ≠ 6}.
O r)(x) = r(r(x))
(r €
Note that { x| x ≠ ± 2} because of innermost r function. But,
x≠
€ ± 2 in the outermost function r which implies
3
that r(x) = 2 ≠ ± 2. Solving yields:
x −4
3
2
x −4
3
≠–2
2
x −4
2
3 ≠ 2x2 – 8
11 = 2x2
3 ≠ – 2x + 8
– 5 ≠ – 2x2
x2 ≠
5
2
€
€
x2 ≠
5
2
x≠±
10
2
=±
11
2
x≠±
So, the domain is { x| x ≠€±
10
2
11
2
=
=
€
=
€
€
3
x2 −4
3
)
€
€
Replace x by
3
2
x −4
3
x2 −4
.
in r.
Simplify.
 3 2

 −4
 x2 − 4 
3
9
4
x − 8x2 +16
22
2
22
}.
2
=±
, x ≠ ± 2, ±
Now, let's find the composition:
€
€
€
r(r(x))
Replace r(x) by
= r(
(Multiply by x2 – 4)
≠2
€
(Multiply top & bottom by x4 – 8x2 + 16)
−4
3x 4 −24x2 +48
9−4x 4 +32x2 −64
125
=
3x 4 −24x2 +48
4x 4 +32x2 −55
The domain is { x| x ≠ ±
10
2
, x ≠ ± 2, ±
22
2
}.
€
Use the graphs
of the functions below to find the following:
5
€
4
€
3
2
g(x)
1
0
-5
-4
-3
-2
-1 0
-1
-2
1
2
3
4
5
f(x)
-3
-4
-5
Ex. 4a
f(– 3)
Ex. 4b
g(2)
O
Ex. 4c
(g f)(– 3)
Ex. 4d
(f O g)(2)
Ex. 4e
(g O f)(4)
Ex. 4f
(f O f)(– 4)
Solution:
a)
Looking at the graph of f (solid line), the point that has xcomponent of – 3 is (– 3, – 1). Thus, f(– 3) = – 1.
b)
Looking at the graph of g (dashed line), the point that has xcomponent of 2 is (2, – 2). Thus, g(2) = – 2.
c)
(g O f)(– 3) = g(f(– 3)) = g(– 1) from part a.
The point on g that has x-component of – 1 is (– 1, 1), so
(g O f)(– 3) = g(f(– 3)) = g(– 1) = 1.
d)
(f O g)(2) = f(g(2)) = f(– 2) from part b.
The point on f that has x-component of – 2 is (– 2, – 1), so
(f O g)(2) = f(g(2)) = f(– 2) = – 1.
e)
(g O f)(4) = g(f(4)). The point on f that has x-component of 4 is
(4, – 2). So, f(4) = – 2. Thus, g(f(4)) = g(– 2). But, the point on
g that has x-component of – 2 is (– 2, 1.5) which means
g(– 2) = 1.5. Hence, (g O f)(4) = g(f(4)) = 1.5.
126
f)
(f O f)(– 4) = f(f(– 4)). The point on f that has x-component of
– 4 is (– 4, 3). So, f(– 4) = 3. Thus, f(f(– 4)) = f(3). But, the
point on f that has x-component of 3 is (3, – 1) which means
f(3) = – 1. Hence, (f O f)(– 4) = f(f(– 4)) = – 1.
Objective #3: Writing Functions as a Composition of Two Simpler Functions.
Find f(x) and g(x) such that the given function h(x) = (f o g)(x):
5a)
h(x) =
5c)
h(x) =
x −7
3
x+6
5b)
h(x) = 5(2x – 7)2
5d)
h(x) = |x3 – 5x + 11|
Solution:
€ a)
The outside function f is the square root function and the
inside function g is x – 7. Thus, f(x) = x and g(x) = x – 7.
€
b)
The outside function f is the quadratic function and the inside
function g is 2x – 7. Thus, f(x) = 5x2 and g(x) = 2x – 7.
€
c)
The outside function f is the inverse
function and the inside
function g is x + 6. Thus, f(x) =
d)
3
x
and g(x) = x + 6.
The outside function f is the absolute value function and
the inside function g is x3 – 5x + 11. Thus, f(x) = |x| and
g(x) = x3 – 5x + 11. €
Given f(x) =
2
3
x – 5 and g(x) = 1.5x + 7.5:
Ex. 6
Show (f o g)(x) = (g o f)(x) = x for all x
Solution:
€ Since f and g are linear polynomial functions, their domains are all
real numbers and hence the domain of the respective compositions is
all real numbers.
(f o g)(x) = f(g(x))
and
(g o f)(x) = f(g(x))
= f(1.5x + 7.5)
=
2
3
(1.5x + 7.5) – 5
= g(
2
3
= 1.5(
x – 5)
2
3
x – 5) + 7.5
=x+5–5
= x – 7.5 + 7.5
€
=x
=x
Hence, (f o g)(x) = (g o f)(x) = x for all x.
€
€
We will see in the next section that f and g are inverse functions.