Natural Domain and Range of a Function

The natural domain of a function is the largest set of real numbers for which the function is defined. Finding the domain requires identifying restrictions such as division by zero or square roots of negative numbers. The range is the set of all values the function actually outputs.

Quick Reference

Restriction	Rule	Example
Division by zero	Exclude values where denominator $= 0$	$f(x)=\frac{1}{x}$: exclude $x=0$
Even root of negative	Exclude values where radicand lt; 0$	$f(x)=\sqrt{x}$: require $x \geq 0$
Odd root	No restriction	$f(x)=\sqrt[3]{x}$: domain is $\mathbb{R}$
Polynomial	No restriction	$f(x)=x^2-4x$: domain is $\mathbb{R}$
Exponential $a^x$, $a>0$	No restriction	$f(x)=2^x$: domain is $\mathbb{R}$

What Is the Natural Domain?

We say a function $y = f(x)$ is defined for $x = a$ when $f(a)$ is a real number. When the domain is not specified, it is assumed to be the natural domain: the largest set of real numbers for which $f(x)$ is real.

The set of all real numbers for which $f(x)$ is real is called the natural domain (or simply the domain) of the function, denoted $\operatorname{Dom}(f)$.

Key rules that determine when a function is defined:

$f(x) = x^m$ for positive integer $m$: defined for every real $x$.
$f(x) = \dfrac{1}{x}$: defined for every $x \neq 0$ (division by zero is undefined).
$f(x) = \sqrt{x}$: defined for $x \geq 0$ (square roots of negative numbers are not real).
$f(x) = \sqrt[n]{x}$ for even integer $n$: defined for $x \geq 0$.
$f(x) = \sqrt[m]{x}$ for odd integer $m$: defined for every real $x$.
$f(x) = a^x$ for constant $a > 0$: defined for every real $x$.

If you want to restrict the domain beyond the natural domain, you must say so explicitly. For example, writing "$f(x) = x^2,\ x > 0

quot; restricts $f$ to positive inputs. Without that restriction, the domain is all of $\mathbb{R}$.

Finding Natural Domains

Determine the natural domains of the following functions:

$f(x) = x^2 - 4x$
$g(x) = \sqrt{x-1}$
$h(x) = \sqrt{2-x}$
$F(x) = \sqrt{x^2 - 6x}$
$G(x) = \sqrt{4 - x^2}$
$H(x) = \sqrt{x-1} + \sqrt{2-x}$
$u(x) = \dfrac{1}{x^2 - 3x}$

Solution

(a) $f(x) = x^2 - 4x$ is a polynomial, so it is real for all $x$: $\operatorname{Dom}(f) = (-\infty, \infty) = \mathbb{R}.$ (b) $\sqrt{x-1}$ is real when $x - 1 \geq 0$, i.e., $x \geq 1$: $\operatorname{Dom}(g) = [1, \infty).$ (c) $\sqrt{2-x}$ is real when $2 - x \geq 0$, i.e., $x \leq 2$: $\operatorname{Dom}(h) = (-\infty, 2].$ (d) We need $x^2 - 6x \geq 0$. Factoring: $x(x-6) \geq 0$. From a sign diagram, this product is nonnegative when $x \leq 0$ or $x \geq 6$:

A sign chart for the domain of F(x) = √(x² - 6x) showing the interval between 0 and 6 is negative. — Sign diagram for $x(x-6)$.

$\operatorname{Dom}(F) = (-\infty, 0] \cup [6, \infty).$ (e) We need $4 - x^2 = (2-x)(2+x) \geq 0$. From a sign diagram, this holds when $-2 \leq x \leq 2$:

A sign chart for the domain of G(x) = √(4 - x²) showing the interval [-2, 2] is nonnegative. — Sign diagram for $(2-x)(2+x)$.

$\operatorname{Dom}(G) = [-2, 2].$ (f) $\sqrt{x-1}$ requires $x \geq 1$ and $\sqrt{2-x}$ requires $x \leq 2$. Both must hold simultaneously: $\operatorname{Dom}(H) = [1, 2].$ (g) $u(x) = \frac{1}{x(x-3)}$ is undefined where $x = 0$ or $x = 3$: $\operatorname{Dom}(u) = (-\infty, 0) \cup (0, 3) \cup (3, \infty) = \mathbb{R} - \{0, 3\}.$

The Range of a Function

The range of a function $f : A \to B$ is the set of all values actually taken by the function:

$\operatorname{Rng}(f) = \{f(x) \mid x \in A\}.$

The range is always a subset of the codomain: $\operatorname{Rng}(f) \subseteq B$.

Let $f : \mathbb{R} \to \mathbb{R}$ and $f(x) = x^2$. Find the range of $f$.

Solution

The range consists of all possible outputs. Any $y \geq 0$ is a possible output because we can set $x = \sqrt{y}$ to get $f(x) = y$. No $y < 0$ is in the range because $x^2 \geq 0$ for all real $x$. Therefore: $\operatorname{Rng}(f) = [0, \infty).$

Finding the range is generally harder than finding the domain. In elementary calculus, powerful methods (derivatives, monotonicity) make it easier to determine ranges precisely. For now, sketching the graph is the most reliable approach.

Frequently Asked Questions

What happens if I plug in a number outside the domain?

The function is undefined there. For example, if $f:[0,\infty)\to \mathbb{R}$ with $f(x) = \sqrt{x}$, then f is undefined at $x = -1$ because $\sqrt{-1}$ is not a real number. Plugging in such values produces an imaginary result, which is outside the scope of real-valued functions.

How do I find the domain of a fraction?

Find all values of $x$ for which the denominator equals zero, then exclude those values from $\mathbb{R}$. For $f(x) = \frac{1}{x^2 - 3x} = \frac{1}{x(x-3)}$, the denominator is zero at $x = 0$ and $x = 3$, so $\operatorname{Dom}(f) = \mathbb{R} - \{0, 3\}$.

Is the range always an interval?

No. The range can be a single point (for a constant function), a union of intervals, or a discrete set. For example, the signum function $\operatorname{sgn}(x)=\left\{\begin{aligned} 1 &&\text{if } x>0\\ 0 &&\text{if } x=0\\ -1 && \text{if }x<0 \end{aligned} \right.$ has range $\{-1, 0, 1\}$, which is a finite set.

What is the difference between range and codomain?

The codomain is the set you declare outputs should come from (often $\mathbb{R}$). The range is the set of outputs the function actually produces. For $f(x) = x^2$ with codomain $\mathbb{R}$, the range is only $[0, \infty)$, a proper subset of $\mathbb{R}$.

Can the domain be empty?

Technically yes (a function with an empty domain is called the empty function), but in practice the functions you encounter in algebra and calculus always have nonempty domains. If your calculation gives an empty domain, double-check your work.