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

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

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

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 $c = g(x) = x -1$ .