How can I find the domain of a composition of functions?
Quick answer
For a composition (f∘g)(x), check two things: x must be allowed in the inner function g, and g(x) must be an allowed input to the outer function f. Find each restriction, then combine them and exclude any input that fails either check. Keep restrictions from the original functions even if they disappear when you simplify the composition. [S1]
The question, the details, the next step.
What is the basic rule for finding the domain of a composition? I understand that f(g(x)) means put g inside f, but I’m not sure which restrictions to check.
Check the composition in two stages. First, x has to be a legal input for the inner function g. Second, after you calculate g(x), that result has to be a legal input for the outer function f. So the domain of f∘g consists of inputs allowed by g whose outputs are allowed by f. [S1]
A useful way to keep the checks straight is to write them separately:
- Inner check: Is x in the domain of g?
- Outer check: Is g(x) in the domain of f?
Then keep only x-values that pass both. For example, if f has a denominator, its input cannot make that denominator zero. If g has a denominator, x cannot make that denominator zero either. The key is that the outer function’s restriction applies to the value g(x), not directly to x.
Could you show both checks matter in one example, and show how to solve the outer restriction?
Suppose f(u)=1/(u−1) and g(x)=(2x+1)/(x−2). We want the domain of (f∘g)(x)=f(g(x)). The letter u makes it clear that f’s input is the output of g.
First, check the inner function. In g(x), the denominator x−2 cannot equal zero, so x≠2.
Next, check what f allows. Its denominator is its input minus 1, so f accepts any input except 1. Therefore g(x) must not equal 1. Solve the excluded case: (2x+1)/(x−2)=1 2x+1=x−2 x=−3
So x=−3 must also be excluded. These are restrictions from two different stages: x≠2 because g is undefined there, and x≠−3 because g(−3)=1, which f does not accept. This follows the two-stage domain method. [S1]
When I simplify f(g(x)), I get (x−2)/(x+3). That expression only appears to exclude −3. Does that mean 2 is allowed after all?
No. The original composition still excludes x=2 because g(2) is undefined. Simplifying can cancel a factor and make an original restriction less visible, but it does not make an input valid in a function stage where it was undefined.
Here, substituting g(x) into f gives: 1/(((2x+1)/(x−2))−1)
Combining terms in the denominator produces (x+3)/(x−2), so the expression simplifies to (x−2)/(x+3). The simplified expression is undefined at −3, but the original composition also has the restriction x≠2 from g. The domain is therefore all real numbers except −3 and 2, written ℝ\{−3, 2}. The simplified formula alone does not preserve every restriction inherited from the original functions.
What’s a reliable layout I can use on homework so I don’t forget one of the restrictions?
Use a short two-check layout, and keep the exclusions visible until the end:
- Write the composition and identify the inner and outer functions.
- Find the inner restriction: x must be in the domain of g.
- Find the outer restriction: g(x) must be in the domain of f; solve for x-values that fail this test.
- Combine the exclusions, then check them against the original composition.
- State the domain clearly.
For this example, a copyable solution is:
- Inner: (2x+1)/(x−2) is defined only if x≠2.
- Outer: f excludes input 1, so set (2x+1)/(x−2)=1; this gives x=−3 to exclude.
- Domain: ℝ\{−3, 2}.
The checks identify why each value is excluded, not just the final answer. This method determines the domain from the stated functions; it does not depend on using a particular simplified form.
A practical checklist
Identify which function is inner and which is outer.
List the x-values that make the inner function undefined.
Find which outputs the outer function does not accept.
Solve for x-values where the inner output equals a forbidden outer input.
Combine exclusions and verify them against the original composition.
Write the final domain clearly, such as ℝ\{a, b} when appropriate.
A few more useful answers
Do I check the outer function’s domain using x or g(x)?
Use g(x), because that is the input passed to the outer function. Solve for x-values where g(x) equals an input the outer function does not allow.
Can simplifying a composition change its domain?
The simplified expression may look defined at an input where the original composition was not. Keep all restrictions found from the original inner and outer functions, even if simplification hides one.
What if neither function has any restrictions?
Then every real x passes both checks, so the composition’s domain is all real numbers. The two-stage check still provides a way to justify that result.
Sources
References used for this guide.
- [S1] OpenStax Precalculus 2e: Composition of Functions openstax.org · Checked