2.13 Composition of Functions

Let $f (x) = \sqrt{x + 1}$ and $g (x) = x^{2}$ . We can define a new function $h$ as
$h (x) = f (g (x)) = f (x^{2}) = \sqrt{x^{2} + 1} .$ That is, to obtain $h (x)$ , we substitute $g (x)$ for $x$ in the expression $f (x)$ .

To generalize this process suppose $f$ and $g$ are any two given functions. We start with a number $x$ in the domain of $g$ and apply $g$ to it to get $g (x)$ , then we apply $f$ to $g (x)$ and thereby obtain the number $f (g (x))$ . Obviously this process presupposes that it makes sense to calculate $f$ at the point $g (x)$ . In other words, the new function is defined only if $g (x)$ is in the domain of $f$ (otherwise we cannot use $g (x)$ as the input for $f$ ).

Consecutive application of functions is known as composition of functions. The new function that takes $x$ and assigns to it the value $f (g (x))$ is often denoted by $f \circ g$ . The symbol $f \circ g$ is read “ $f$ circle $g$ .” Figure 1(a) shows the composition $f \circ g$ as a machine diagram and Figure 1(b) illustrates it as an arrow diagram.

(a)

(b)

Fig 1(a) Machine digram for

f \circ g

. A composite (or composition) function

f \circ g

applies the

g

machine to the input

x

and then uses the output

g (x)

as the input for the

f

machine.

Fig. 1(b) Arrow diagram for

f \circ g

. If

x

is in the domain of

g

and

g (x)

is in the domain of

f

, then we may compose

f

and

g

to form

f \circ g

Let

f

and

g

be two functions. For those points

x

in the domain of

g

for which

g (x)

lies in the domain of

f

, the composite function

f \circ g

is defined by

(f \circ g) (x) = f (g (x)) .

In set notation, the domain of $f \circ g$ is
$D o m (f \circ g) = {x | x \in D o m (g) and g (x) \in D o m (f)} .$
Note that the order of composition of two functions matters. In calculating $(f \circ g) (x)$ , we first evaluate $g$ at $x$ and then use $g (x)$ as the input to calculate $f$ of the result. But in calculating $(g \circ f) (x)$ , we first evaluate $f$ at $x$ and then calculate $g$ at the the point $f (x)$ . Therefore, $(f \circ g) (x)$ is often quite different from $(g \circ f) (x)$ .

Example

Let

f (x) = \sqrt{x}

and

g (x) = 1 + x^{2}

. Find the following functions and their domains.
(a)

f \circ g

(b)

g \circ f

Solution

We have

D o m (f) = {x | x \geq 0} = [0, \infty), D o m (g) = R .

(a)

\begin{aligned} (f \circ g) (x) & = f (g (x)) \\ = f (1 + x^{2}) \\ = \sqrt{1 + x^{2}} . \end{aligned}

Because

g (x) = 1 + x^{2} \geq 1

for all

x

g (x)

is always in the domain of

f

. Therefore

D o m (f \circ g) = R .

(b)

\begin{aligned} (g \circ f) (x) & = g (f (x)) \\ = g (\sqrt{x}) \\ = 1 + (\sqrt{x})^{2} \\ = 1 + x & \footnotesize (if x \geq 0) \end{aligned}

Because the domain of

g

is the entire set of real numbers,

f (x)

always lies in the domain of

g

. However,

f

is defined only for

x \geq 0

. Therefore, the domain of

g \circ f

is the same as the domain of

f

, namely

D o m (g \circ f) = D o m (f) = {x | x \geq 0} = [0, \infty) .

Example

f (x) = \frac{x + 1}{x - 1}

, find the function

f \circ f

and its domain.

Solution

Let’s first find

f \circ f

\begin{aligned} (f \circ f) (x) & = f (f (x)) \\ = f (\frac{x + 1}{x - 1}) \\ = \frac{\frac{x + 1}{x - 1} + 1}{\frac{x + 1}{x - 1} - 1} \\ = \frac{\frac{x + 1 + x - 1}{x - 1}}{\frac{x + 1 - (x - 1)}{x - 1}} \\ = \frac{2 x}{2} = x \end{aligned}

Just by looking at the above formula one may be tempted to say that the domain of

f \circ f

R

. However, the domain of

f \circ f

consists of all

x

that
(1)

x

lies in the domain of

f

,
(2)

f (x)

lies in the domain of

f

The domain of $f$ is all real numbers except $x = 1$ (recall that division by zero is not defined). So to satisfy the first condition we have to exclude $x = 1$ . To satisfy the second condition, we also have to exclude $x$ for which
$\frac{x + 1}{x - 1} = 1.$ It turns out that this equation has no solution because it is equivalent to $x + 1 = x - 1$ or $- 1 = 1$ . This means that $f (x)$ is always in the domain of $f$ , and the second conditions is automatically satisfied without any further restriction on $x$ . Thus in this specific example
$D o m (f \circ f) = D o m (f) = {x | x \neq 1} = R - {1} .$

Example

Find functions

f

and

g

such that

h = f \circ g

given that

h (x) = \cos (x^{2}) .

Solution

To calculate

h (x)

we first square

x

and then take the cosine of

x^{2}

. We can therefore set

Misplaced &

and

Misplaced &

We observe that

(f \circ g) (x) = f (g (x)) = f (x^{2}) = \cos (x^{2}) .

We can combine three or more functions by composing them two at a time. For example, the function

F

given by

F (x) = \frac{1}{2 + x^{2}},

is a composition of three functions

F = f \circ g \circ h

, where

h (x) = x^{2}, g (x) = x + 2, f (x) = \frac{1}{x} .

Notice that we can compose

f

and

g

first then compose

f \circ g

with

h

or we can first compose

g

and

h

and then compose

f

and

g \circ h

. In general, composition is associative. That is

f \circ g \circ h = (f \circ g) \circ h = f \circ (g \circ h)

Example

Find three functions

f, g,

and

h

such that

F = f \circ g \circ h

given that

F (x) = \sqrt{| x | + 3} .

Solution

To calculate

F (x)

, we fist take the absolute value of

x

, add 3, and then take the square root. We can therefore set

h (x) = | x |,

g (x) = x + 3,

and

f (x) = \sqrt{x} .

With this choice of

f, g,

and

h

, we observe that

(f \circ g \circ h) (x) = f (g (h (x))) = f (g (| x |)) = f (| x | + 3) = \sqrt{| x | + 3} .