Domain

Table of Contents

When working with rational functions, the idea of domain becomes especially important because rational functions involve division, and division by zero is undefined.

A rational function (as introduced in the parent chapter) is any function that can be written in the form
$$
f(x) = \frac{P(x)}{Q(x)},
$$
where $P(x)$ and $Q(x)$ are polynomials and $Q(x)\neq 0$.

In this chapter, we focus on how to determine which $x$-values are allowed in such a function—that is, how to find its domain.

What makes the domain of rational functions special?

For any function, the domain is the set of all input values $x$ for which the function is defined.

For a rational function $f(x) = \dfrac{P(x)}{Q(x)}$:

The numerator $P(x)$ can be any polynomial; polynomials are defined for all real numbers.
The denominator $Q(x)$ cannot be zero, because division by zero is undefined.

So the main restriction for rational functions is:

Exclude all $x$ such that $Q(x) = 0$.

Thus, the domain is:

all real numbers
except the solutions to the equation $Q(x) = 0$.

Basic procedure: finding the domain

Given a rational function
$$
f(x) = \frac{P(x)}{Q(x)},
$$
to find its domain:

Set the denominator equal to zero:
$$
Q(x) = 0.
$$
Solve this equation for $x$. These are the forbidden values.
State the domain as:

all real numbers except those forbidden values.

You can express the domain in words, set notation, or interval notation, depending on context.

Example 1: Linear denominator

Let
$$
f(x) = \frac{2x + 1}{x - 3}.
$$

Denominator: $Q(x) = x - 3$.
Solve $x - 3 = 0$:
$$
x = 3.
$$
Domain: all real numbers except $x = 3$.

In interval notation:
$$
(-\infty, 3) \cup (3, \infty).
$$

Example 2: Quadratic denominator

Let
$$
g(x) = \frac{5}{x^2 - 4}.
$$

Denominator: $Q(x) = x^2 - 4$.
Solve $x^2 - 4 = 0$:
$$
x^2 = 4 \quad \Rightarrow \quad x = 2 \text{ or } x = -2.
$$
Domain: all real numbers except $x = -2$ and $x = 2$.

In interval notation:
$$
(-\infty, -2) \cup (-2, 2) \cup (2, \infty).
$$

Example 3: Denominator with no real zeros

Let
$$
h(x) = \frac{3x - 1}{x^2 + 1}.
$$

Denominator: $Q(x) = x^2 + 1$.
Solve $x^2 + 1 = 0$:
$$
x^2 = -1.
$$
There is no real solution to this equation.
This means the denominator is never zero for any real $x$.

So the domain is all real numbers:
$$
(-\infty, \infty).
$$

Domains and factored denominators

Often, denominators are easier to analyze if you factor them.

Example 4: Factored quadratic

Let
$$
f(x) = \frac{x + 1}{x^2 - x - 6}.
$$

Factor the denominator:
$$
x^2 - x - 6 = (x - 3)(x + 2).
$$
Set each factor to zero:

$x - 3 = 0 \Rightarrow x = 3$,
$x + 2 = 0 \Rightarrow x = -2$.

Domain: all real numbers except $x = -2$ and $x = 3$:
$$
(-\infty, -2) \cup (-2, 3) \cup (3, \infty).
$$

Example 5: Higher powers in the denominator

Let
$$
g(x) = \frac{1}{x^3(x - 4)}.
$$

The denominator is zero when any factor is zero:

$x^3 = 0 \Rightarrow x = 0$,
$x - 4 = 0 \Rightarrow x = 4$.

So the domain is all real numbers except $x = 0$ and $x = 4$:
$$
(-\infty, 0) \cup (0, 4) \cup (4, \infty).
$$

Note that it does not matter that the factor $x$ is cubed; if $x = 0$, the entire denominator is zero.

Canceled factors vs. domain restrictions

Sometimes a rational function can be simplified by canceling common factors in the numerator and denominator. However, domain restrictions come from the original form, not the simplified one.

Example 6: Hole vs. simplification

Consider
$$
f(x) = \frac{x^2 - 1}{x - 1}.
$$

Factor the numerator:
$$
x^2 - 1 = (x - 1)(x + 1).
$$
So
$$
f(x) = \frac{(x - 1)(x + 1)}{x - 1}.
$$
Algebraically, you can cancel $(x - 1)$ and write
$$
f(x) = x + 1,
$$
but the original denominator $x - 1$ is zero when $x = 1$.

So:

The simplified formula $x + 1$ is valid only for $x \neq 1$.
The domain of the original rational function is all real numbers except $x = 1$.

In other words:

Domain: $(-\infty, 1) \cup (1, \infty)$.
At $x = 1$, the function is not defined, even though the simplified expression $x + 1$ gives a number there.

This situation corresponds to a “hole” in the graph at $x = 1$, but the key point for domain is: never re-include a value that made the original denominator zero, even if it cancels later.

Example 7: Multiple canceled factors

Let
$$
g(x) = \frac{(x - 2)(x + 3)}{(x - 2)(x - 5)}.
$$

The denominator is zero at $x = 2$ and $x = 5$.
You can cancel $(x - 2)$ to get
$$
g(x) = \frac{x + 3}{x - 5},
$$
but $x = 2$ is still not allowed because it made the original denominator zero.
Domain: all real numbers except $x = 2$ and $x = 5$.

Summary of domain rules for rational functions