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

\[\bbox[#F2F2F2,5px,border:2px solid black]{Dom(f\circ g)=\{x|\ x\in Dom(g)\text{ and }g(x)\in Dom(f)\}.}\]
- 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 1

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

\[

Dom(f)=\{x|\ x\geq0\}=[0,\infty),\qquad Dom(g)=\mathbb{R}.

\]
**(a)**

\begin{align*}

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

& =f(1+x^{2})\\

& =\sqrt{1+x^{2}}.

\end{align*}

Because $g(x)=1+x^{2}\geq1$ for all $x$, $g(x)$ is always in the domain of $f$. Therefore

\[

Dom(f\circ g)=\mathbb{R}.

\]
**(b)**

\begin{align*}

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

& =g(\sqrt{x})\\

& =1+(\sqrt{x})^{2}\\

& =1+x & (\text{ if } x\geq 0)

\end{align*}

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 $x\geq0$. Therefore, the domain of $g\circ f$ is the same as the domain of $f$, namely

\[

Dom(g\circ f)=Dom(f)=\{x|\ x\geq0\}=[0,\infty).

\]

Example 2

If $f(x)=\dfrac{x+1}{x-1}$, find the function $f\circ f$ and its domain.

Solution

Let’s first find $f\circ f$

\begin{align*}

(f\circ f)(x) & =f(f(x))\\

& =f\left(\dfrac{x+1}{x-1}\right)\\

& =\dfrac{\dfrac{x+1}{x-1}+1}{\dfrac{x+1}{x-1}-1}\\

& =\dfrac{\dfrac{x+1+x-1}{x-1}}{\dfrac{x+1-(x-1)}{x-1}}\\

& =\frac{2x}{2}=x

\end{align*}

Just by looking at the above formula, one may be tempted to say that the domain of $f\circ f$ is $\mathbb{R}$. 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 condition is automatically satisfied without any further restriction on $x$. Thus in this specific example

\[

Dom(f\circ f)=Dom(f)=\{x|\ x\neq1\}=\mathbb{R}-\{1\}.

\]

Example 3

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

\[

g(x)=x^{2} \qquad (\text{square the input})

\]
and

\[

f(x)=\cos x \qquad (\text{ take cosine of the input})

\]
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},\qquad g(x)=x+2,\qquad 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 5

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

\]