Domains of composite functions

The domain of composite functions

In this lesson we’ll learn to find the domain of composite functions.

The domain of a function is the set of $x$-values that make the function true.

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Sometimes if you create a composite function you need to consider the domain of the new function. Often the domain of the new function is dependent on any domain restrictions on the original functions.

Remember a composite function of $f(x)$ and $g(x)$ is written as $f \circ g$ or $f(g(x))$, and is found by plugging $g(x)$ into $f(x)$.

The domain of a composite must exclude all values that make the “inside” function undefined, and all values that make the composite function undefined. In other words, given the composite $f(g(x))$, the domain will exclude all values where $g(x)$ is undefined, and all values where $f(g(x))$ is undefined.

How to find the domain of a composite function step-by-step

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Example of finding the domain of a composite

Example

What is the domain of $f \circ g$?

$f(x) = x^2 -3$

$g(x) = \sqrt{x+9}$

First, find the domain of $g(x)$. The expression $\sqrt{x+9}$ is undefined where $x+9$ is negative. For example, if $x=-10$, then $x+9$ is $-1$. Likewise, if $x$ is any number less than $-9$, $x+9$ will be negative. However, $-9$ itself is okay because $\sqrt{x+9}$ will then be $0$, which is not undefined.

The domain of $g(x)$ then is all real numbers $x$ such that $x\geq-9$.

The composite function is

$f \circ g = f(g(x)) = \left(\sqrt{x+9}\right)^2-3$

$f \circ g = f(g(x)) = x+9-3$

$f \circ g = f(g(x)) =x+6$

For this simple binomial, no real numbers are excluded, so its domain is all real numbers. But because the domain of $g(x)$ excludes $x<-9$, those values also have to be excluded from the composite $f(g(x))$.

That means the domain of $f(g(x))$ is $x\geq-9$.

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Let’s try another example.

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no real numbers are excluded, so its domain is all real numbers.

Example

What is the domain of $f \circ g$?

$f(x) = \frac{2}{2x+4}$

$g(x) = \frac{3}{x-5}$

First, find the domain of $g(x)$. The expression $3/(x-5)$ is undefined if the denominator is $0$. That means $x =5$ isn’t in the domain of $g(x)$. Therefore, the domain of $g(x)$ is all real numbers $x$ such that $x \neq 5$.

The composite function is

$f \circ g = f(g(x)) = \frac{2}{2\left(\frac{3}{x-5}\right)+4}$

$f \circ g = f(g(x)) = \frac{2}{\frac{6}{x-5}+4\left(\frac{x-5}{x-5}\right)}$

$f \circ g = f(g(x)) = \frac{2}{\frac{6+4x-20}{x-5}}$

$f \circ g = f(g(x)) = \frac{2}{\frac{4x-14}{x-5}}$

$f \circ g = f(g(x)) = 2\left(\frac{x-5}{4x-14}\right)$

$f \circ g = f(g(x)) = \frac{2x-10}{4x-14}$

$f \circ g = f(g(x)) = \frac{2(x-5)}{2(2x-7)}$

$f \circ g = f(g(x)) = \frac{x-5}{2x-7}$

For this rational function, any numbers that make the denominator $0$ are excluded from the domain.

$2x-7=0$

$2x=7$

$x=\frac{7}{2}$

Putting both exclusions together, the domain of the composite is all real numbers except $7/2$ and $5$. So,

$x \neq \frac{7}{2},\ 5$

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