Composite Functions

Table of Contents

In this article, we you will learn about composite functions, it means that a function can become input of another function. Before you learn about the composite functions, you must be familiar with the arithmetic of functions which means given two function , you will be able to perform basic arithmetic operations on the function itself. Identifying Domain

Sometimes the function expression is given, but domain is not specified. In such cases, you must identify the domain of given function. Suppose

$\begin{align*} f(x) = x^2 -1\end{align}$

is a function whose domain is $D_f$ which is “all real numbers”, because the function is true for all real numbers. Another function

$\begin{align*} g(x) = \sqrt{x - }3\end{align}$

has a domain $D_g$ which must have all numbers that are $x - 3 \geq 0$ otherwise the function is invalid. Therefore, $D_g$ contains all numbers in the interval $[3, -\infty]$ .

Why this is important ? because when we perform arithmetic operations on two or more functions the domain of new function is set of all real numbers that belong to both $D_f$ and $D_g$ , that is, $(D_f \cap D_g)$ .

Arithmetic Operations on Functions

Normal arithmetic operations are possible on functions too. If $f$ and $g$ are two functions, then there are four operations possible on these functions.

Sum $(f + g)$
Difference $(f - g)$
Product $(fg)$
Quotient $(\frac{f}{g})$

Now we discuss about each of these operations.

Sum of functions

The sum of functions is

$\begin{align*}(f + g)(x) = f(x) + g(x)\end{align}$

The expressions of both functions are added together to form a new function. For example, if $f(x) = 3x +1$ and $g(x) = 5x - 4$ , then

$\begin{align*}\\&(f + g)(x) = f(x) + g(x)\\ \\ & = 3x + 1 + 5x - 4\\ \\&=8x - 3\end{align}$

Difference of functions

The difference of functions is

$\begin{align*}(f + g)(x) = f(x) - g(x)\end{align}$

The expressions of function $g$ is subtracted from expression of function $f$ to form a new expression for $f-g$ . For example, if $f(x) = 3x +1$ and $g(x) = 5x - 4$ , then

$\begin{align*}\\&(f - g)(x) = f(x) - g(x)\\ \\ & = (3x + 1) - (5x - 4)\\ \\ & = 3x + 1 - 5x - 4 \\ \\&=-2x -3 \end{align}$

Product of functions

The product of the functions is

$\begin{align*}(f \cdot g)(x) = f(x) \cdot g(x)\end{align}$

The expressions of function $f$ and $g$ is multiplied to get the new product expression of $f \cdot g$ . Each term is of $f$ is multiplied with each term of $g$ . For example, if $f(x) = x - 1$ and $g(x) = x +2$ are two expressions, then

$\begin{align*}\\&(f \cdot g)(x) = f(x) \cdot g(x)\\ \\ & = (x - 1) \cdot (x + 2)\\ \\ & = x^2 + 2x -x -2\\ \\&=x^2 + x -2\end{align}$

Quotient of functions

The quotient of functions is obtained by dividing two functions, which is

$\begin{align*}\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)}\end{align}$

The functions are divided to get the quotient, however, there is one condition, that is, $g(x) \ne 0$ , otherwise the quotient of function will be “divide by 0” which is “undefined“.

Therefore, if $f(x) = x^2 + x - 2$ and $g(x) = x + 2$ , then

$\begin{align*}\\&\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)}\\ \\ & = \frac{x^2 + x -2}{x + 2}\\ \\ & Factor \hspace{2mm} the \hspace{2mm} numerator \\ \\ &=\frac{ (x-1)(x+2)}{x+2}\\ \\&=x - 1\end{align}$

The domain of $(f + g)$ , $(f - g)$ , $(fg)$ is set of all reals numbers that are common to domain of $f(x)$ and $g(x)$ which is $(D_f \cap D_g)$ except where $g(x) \ne 0$ for $(\frac{f}{g}$ .

The composite functions or the idea of composition of functions is simple. Suppose there are two functions, $f$ and $g$ , then if function $g$ becomes input for function $f$ , it is called “composition of function $f$ with function $g$ or simply a composite function.

Let us try to understand this with an example, suppose Nancy work as a maid, and get paid $x$ amount every week and after paying taxes she receives only $80%$ of her earnings. Each week she spends $\textdollar 50$ for grocery from her earning . The total earning of Nancy after paying taxes can be defined by function $g(x) = 0.80 \cdot x$ . Her savings after expense of $\textdollar50$ can be defined as function $f(x) = x - 50$ .

$\begin{align*}Each \hspace{2mm} week \hspace{2mm} Nancy's \hspace{2mm} saving = (f \circ g)(x) = f(g(x))\\ \\ f(0.80x) = 0.80x - 50\end{align}$

If her gross earning is $\textdollar 200$ , then her saving would be

$\begin{align*}&Weekly \hspace{2mm} Savings = 0.80 \cdot 200 - 50\\ \\&= 80 \cdot 2 - 50 \\ \\&=160 - 50\\ \\ &= 110 \end{align}$

Therefore, total saving after paying taxes and expenses of $\textdollar 50$ is $\textdollar 110$ .

What is the domain of a composite function?

If $(f \circ g)(x)$ is a composite function such that $(f \circ g)(x) = f(g(x))$ . Then the domain of composite function must be

$x \in D_g$ meaning “ $x$ must be in the domain of $g(x)$ .
$g(x) \in D_f$ meaning “ $g(x)$ must be in the domain of $f(x)$ .

Therefore, while finding the domain of a composite function, we must first exclude all values of $x$ than can make the function $g(x)$ “invalid”. Also, if $x$ happens to be valid, that is, $x \in D_g$ , then $g(x)$ must be valid value for the function $f(x)$ . If $g(x)$ happens to invalid value, then both $g(x)$ and $x$ must be excluded from the domain of composite function $(D_g \cap D_f)$ .

Decomposing Functions

Like composing two functions, it is possible to decompose a function because we know that “composition of two functions” creates a new function. Consider the following example,

$\begin{align*}h(x) = \sqrt{x - 1}\end{align}$

Here we can clearly see that there are two functions involved in the expression. Therefore, $h(x) = \sqrt{x -1} = \sqrt{c}$ where $c = x -1$ which implies that $c = g(x) = x -1$ .

We can write the function $h(x)$ and $g(x)$ as composition of function $h \circ g$ where $h(x) = \sqrt{x}$ and $g(x) = x -1$ .