Until now, given a function
f(x),
you would plug a number or another variable in for x.
You could even
get fancy and plug in an entire expression for x.
For example, given f(x)
= 2x + 3, you
could find f(y^{2} – 1)
by plugging
y^{2} – 1
in for x
to get f(y^{2}
– 1) = 2(y^{2} – 1) + 3 = 2y^{2} – 2 + 3
= 2y^{2} + 1.

In function composition,
you're plugging entire functions in for the x.
In other words, you're always getting "fancy". But let's start
simple. Instead of dealing with functions as formulas, let's deal with
functions as sets of
(x, y)
points:

Let
f = {(–2, 3), (–1,
1), (0, 0), (1, –1), (2, –3)} and
let g = {(–3,
1), (–1, –2), (0, 2), (2, 2), (3, 1)}.

Find
(i)
f
(1), (ii)
g(–1), and
(iii)
(g o
f )(1).

(i)
This type of exercise is meant to emphasize that the (x,
y) points
are really (x,
f (x))
points. To find f
(1), I need to find
the (x,
y) point in
the set of (x,
f (x))
points that has a first coordinate of x = 1.
Then f
(1) is the y-value
of that point. In this case, the point with x
= 1 is (1,
–1), so:

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f
(1) = –1

(ii)
The point in the g(x)
set of point with x
= –1 is the point
(–1, –2),
so:

g(–1)
= –2

(iii)
What is "(g
o
f )(1)"?
This is read as "g-compose-f
of 1",
and means "plug 1
into f,
evaluate, and then plug the result into g".
The computation can feel a lot easier if I use the following, more intuitive,
formatting:

Working from the right
back toward the left, I am now plugging x
= –1 (from "f(1)
= –1") into
g(x),
so I look in the set of g(x)
points for a point with x
= –1. That point
is (–1,
–2). This tells me
that g(–1)
= –2, so now I have
my answer:

(g
o
f )(1) = g( f(1)) = g(–1) = –2

Note that they never told
us what were the formulas, if any, for f(x)
or g(x);
we were only given a list of points. But this list was sufficient for
answering the question, as long as we keep track of our x-
and y-values.

Let
f = {(–2, 3), (–1,
1), (0, 0), (1, –1), (2, –3)} and
let g = {(–3,
1), (–1, –2), (0, 2), (2, 2), (3, 1)}.

Find (i)
( f o
g)(0), (ii) (
fog)(–1),
and (iii)(gof )(–1).

(i)
To find (
f o
g)(0), ("f-compose-
g of zero"),
I'll rewrite the
expression as:

( f
o
g)(0) = f(g(0))

This tells me that I'm
going to plug zero into g(x),
simplify, and then plug the result into f(x).
Looking at the list of g(x)
points, I find (0,
2), so g(0)
= 2, and I need now
to find f(2).
Looking at the list of f(x)
points, I find (2,
–3), so f(2)
= –3. Then:

(
f o
g)(0) = f(g(0)) = f(2) = –3

(ii)
The second part works the same way:

(
f o
g)(–1) = f(g(–1)) = f(–2) = 3

(iii)
I can rewrite the composition as (g
o
f )(–1) = g( f(–1)) = g(1).

Uh-oh; there is no
g(x) point
with x
= 1, so it is nonsense
to try to find the value of g(1).
In math-speak, g(1)
is "not defined"; that is, it is nonsense.Then (g
o
f )(–1) is
also nonsense, so the answer is:

(g
o
f )(–1) is undefined.

Part (iii)
of the above example points out an important consideration regarding domains
and ranges. It
may be that your composed function (the result you get after composing
two other functions) will have a restricted domain, or at least a domain
that is more restricted than you might otherwise have expected. This will
be more important when we deal with composing functions symbolically later.

Another exercise of this
type gives you two graphs, rather than two sets of points, and has you
read the points (the function values) from these graphs.

Given f(x)
and g(x)
as shown below, find (
f o
g)(–1).

f(x):

g(x):

In this case, I will
read the points from the graph. I've been asked to find (
f o
g)(–1) = f(g(–1)).
This means that I first need to find g(–1).
So I look on the graph of g(x),
and find x
= –1. Tracing up
from x
= –1 to the graph
of g(x),
I arrive at y
= 3. Then the point
(–1, 3)
is on the graph of g(x),
and g(–1)
= 3.

Now I plug this value,
x =
3, into f(x).
To do this, I look at the graph of f(x)
and find x
= 3. Tracing
up from x
= 3 to the graph
of f(x),
I arrive at y
= 3. Then the point
(3, 3)
is on the graph of f(x),
and f(3)
= 3.

Then
( f o
g)(–1) = f(g(–1))
= f(3) = 3.

Given f(x)
and g(x)
as shown in the graphs below, find (g
o
f
)(x) for
integral values of x
on the interval –3
<x<
3.

f(x):

g(x):

This is asking me for
all the values of (g
o
f )(x) = g( f(x))
for x
= –3, –2,
–1, 0, 1, 2, and
3.
So I'll just follow the points on the graphs and compute all the values:

(g
o
f )(–3) = g( f(–3)) = g(1) = –1

I got this answer by
looking at x
= –3 on the f(x)
graph, finding the corresponding y-value
of 1
on the f(x)
graph, and using this answer as my new x-value
on the g(x)
graph. That is, I looked at x
= –3 on the f(x)
graph, found that this led to y
= 1, went to x
= 1 on the g(x)
graph, and found that this led to y
= –1. Similarly:

You aren't generally given
functions as sets of points or as graphs, however. Generally, you have
formulas for your functions. So let's see what composition looks like
in that case...