Domain of a Composition

Learning Outcomes

Find the domain of a composite function.
Decompose a composite function.

As we discussed previously, the domain of a composite function such as $f\circ g$ is dependent on the domain of $g$ and the domain of $f$ . It is important to know when we can apply a composite function and when we cannot, that is, to know the domain of a function such as $f\circ g$ . Let us assume we know the domains of the functions $f$ and $g$ separately. If we write the composite function for an input $x$ as $f\left(g\left(x\right)\right)$ , we can see right away that $x$ must be a member of the domain of $g$ in order for the expression to be meaningful, because otherwise we cannot complete the inner function evaluation. However, we also see that $g\left(x\right)$ must be a member of the domain of $f$ , otherwise the second function evaluation in $f\left(g\left(x\right)\right)$ cannot be completed, and the expression is still undefined. Thus the domain of $f\circ g$ consists of only those inputs in the domain of $g$ that produce outputs from $g$ belonging to the domain of $f$ . Note that the domain of $f$ composed with $g$ is the set of all $x$ such that $x$ is in the domain of $g$ and $g\left(x\right)$ is in the domain of $f$ .

A General Note: Domain of a Composite Function

The domain of a composite function $f\left(g\left(x\right)\right)$ is the set of those inputs $x$ in the domain of $g$ for which $g\left(x\right)$ is in the domain of $f$ .

How To: Given a function composition $f\left(g\left(x\right)\right)$ , determine its domain.

Find the domain of $g$ .
Find the domain of $f$ .
Find those inputs, $x$ , in the domain of $g$ for which $g(x)$ is in the domain of $f$ . That is, exclude those inputs, $x$ , from the domain of $g$ for which $g(x)$ is not in the domain of $f$ . The resulting set is the domain of $f\circ g$ .

Example: Finding the Domain of a Composite Function

Find the domain of

$\left(f\circ g\right)\left(x\right)\text{ where}f\left(x\right)=\dfrac{5}{x - 1}\text{ and }g\left(x\right)=\dfrac{4}{3x - 2}$

Show Solution

The domain of $g\left(x\right)$ consists of all real numbers except $x=\frac{2}{3}$ , since that input value would cause us to divide by 0. Likewise, the domain of $f$ consists of all real numbers except 1. So we need to exclude from the domain of $g\left(x\right)$ that value of $x$ for which $g\left(x\right)=1$ .

$\begin{align}&\dfrac{4}{3x - 2}=1\hspace{5mm}&&\text{Set}\hspace{2mm}g(x)\hspace{2mm}\text{equal to 1} \\[2mm]& 4=3x - 2 &&\text{Multiply by}\hspace{2mm} 3x-2\\[2mm]& 6=3x&&\text{Add 2 to both sides}\\[2mm]& x=2&&\text{Divide by 3} \end{align}$

So the domain of $f\circ g$ is the set of all real numbers except $\frac{2}{3}$ and $2$ . This means that

$x\ne \frac{2}{3}\hspace{2mm}\text{or}\hspace{2mm}x\ne 2$

We can write this in interval notation as

$\left(-\infty ,\frac{2}{3}\right)\cup \left(\frac{2}{3},2\right)\cup \left(2,\infty \right)$

Example: Finding the Domain of a Composite Function Involving Radicals

Find the domain of

$\left(f\circ g\right)\left(x\right)\text{ where}f\left(x\right)=\sqrt{x+2}\text{ and }g\left(x\right)=\sqrt{3-x}$

Show Solution

Because we cannot take the square root of a negative number, the domain of $g$ is $\left(-\infty ,3\right]$ . Now we check the domain of the composite function

$\left(f\circ g\right)\left(x\right)=\sqrt{\sqrt{3-x}+2}$

For $\left(f\circ g\right)\left(x\right)$ , we need $\sqrt{3-x}+2\ge{0}$ , since the radicand of a square root must be positive. Since square roots are positive, $\sqrt{3-x}\ge{0}$ , or $3-x\ge{0}$ , which gives a domain of $\left(f\circ g\right)\left(x\right) = (-\infty,3]$ .

Analysis of the Solution

This example shows that knowledge of the range of functions (specifically the inner function) can also be helpful in finding the domain of a composite function. It also shows that the domain of $f\circ g$ can contain values that are not in the domain of $f$ , though they must be in the domain of $g$ .

You cannot rely on an algorithm to find the domain of a composite function. Rather, you will need to first ask yourself “what is the domain of the inner function”, and determine whether this set will comply with the domain restrictions of the outer function. In this case, the set $(-\infty,3]$ ensures a non-negative output for the inner function, which will in turn ensure a positive input for the composite function.

Try It

Find the domain of

$\left(f\circ g\right)\left(x\right)\text{ where}f\left(x\right)=\dfrac{1}{x - 2}\text{ and }g\left(x\right)=\sqrt{x+4}$

Show Solution

$\left[-4,0\right)\cup \left(0,\infty \right)$

Try It

We can use graphs to visualize the domain that results from a composition of two functions.
Graph the two functions below with an online graphing calculator.

$f(x)=\sqrt{3-x}$
$g(t) = \sqrt{x+4}$

Next, create a new function, $h(x) = g(f(x))$ . Based on the graph, what is the domain of this function? Explain why $g(f(x))$ and $f(x)$ have the same domain.

Now define another composition, $p(x) = f(g(x)$ . What is the domain of this function? Explain why you can evaluate $g(10)$ , but not $p(10)$ .

Decompose a Composite Function

In some cases, it is necessary to decompose a complicated function. In other words, we can write it as a composition of two simpler functions. There is almost always more than one way to decompose a composite function, so we may choose the decomposition that appears to be most obvious.

Example: Decomposing a Function

Write $f\left(x\right)=\sqrt{5-{x}^{2}}$ as the composition of two functions.

Show Solution

We are looking for two functions, $g$ and $h$ , so $f\left(x\right)=g\left(h\left(x\right)\right)$ . To do this, we look for a function inside a function in the formula for $f\left(x\right)$ . As one possibility, we might notice that the expression $5-{x}^{2}$ is the inside of the square root. We could then decompose the function as

$h\left(x\right)=5-{x}^{2}\hspace{2mm}\text{and}\hspace{2mm}g\left(x\right)=\sqrt{x}$

We can check our answer by recomposing the functions.

$g\left(h\left(x\right)\right)=g\left(5-{x}^{2}\right)=\sqrt{5-{x}^{2}}$

Analysis of the Solution

For every composition there are infinitely many possible function pairs that will work. In this case, another function pair where $g\left(h\left(x\right)\right)=\sqrt{5-{x}^{2}}$ is $h(x)=x^2$ and $g(x)=\sqrt{5-x}$

Try It

Write $f\left(x\right)=\dfrac{4}{3-\sqrt{4+{x}^{2}}}$ as the composition of two functions.

Show Solution

Possible answer:

$g\left(x\right)=\sqrt{4+{x}^{2}}$

$h\left(x\right)=\dfrac{4}{3-x}$

$f=h\circ g$

Unit 1