Domain and Range

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Textbook: OpenStax Precalculus 2e • Chapter: 1 • Section: 2

Quick Reference

Textbook: OpenStax Precalculus 2e Chapter: Ch 1: Functions Section: 1.2 Domain and Range Subsection: 1.2 Domain and Range

Page numbers pending faculty verification.

Try This First

Here is a rule that takes a number, subtracts 4 from it, and then takes the square root of the result:

$$f(x) = \sqrt{x - 4}$$

Before any vocabulary, try feeding it three numbers and watch what happens.

Input $x$	Compute $x - 4$	Then $\sqrt{x-4}$	Allowed?
$13$	$9$	$3$	yes
$4$	$0$	$0$	yes
$1$	$-3$	$\sqrt{-3}$	?

The third row asks for the square root of a negative number, which is not a real number. So $x = 1$ is an input the machine cannot accept.

Predict before reading on: which inputs CAN this machine accept?

Every input from $x = 4$ upward works, because $x - 4$ is then zero or positive. Every input below $4$ fails. So the allowed inputs are exactly the numbers $x \geq 4$. The set of allowed inputs has a name: the domain. The set of values that come back out has a name: the range.

The Function as a Process

A function is a process that takes an input and gives back an output. Picture a vending machine. You press a button (the input), and a specific item drops (the output). Two questions follow naturally:

Which buttons are wired to actually dispense something? That collection of valid inputs is the domain.
Across all the working buttons, which items can ever come out? That collection of possible outputs is the range.

The domain is on the horizontal axis (the inputs you are allowed to use). The range is on the vertical axis (the outputs that result). Reading a graph, you sweep left to right to find the domain and bottom to top to find the range.

A function does not have to accept every real number. Some inputs break the rule, the way $x = 1$ broke $\sqrt{x-4}$ above. Identifying which inputs break a rule is most of the work in finding a domain.

Prerequisite Hub

graph LR
    subgraph BuildsOn["Builds On"]
        A["Functions and<br/>Function Notation"]
        B["Solving<br/>Inequalities"]
    end

    subgraph ThisSkill["This Skill"]
        C["Domain and<br/>Range"]
    end

    subgraph Unlocks
        D["Rates of<br/>Change"]
        E["Function<br/>Composition"]
        F["Inverse<br/>Functions"]
    end

    A --> C
    B --> C
    C --> D
    C --> E
    C --> F

    style C fill:#d1fae5,stroke:#a565f0,stroke-width:3px

Builds on (master these first):

Skill	skill_id	Why it is needed
Functions and Function Notation	`m141-functions-and-function-notation`	Domain and range describe the inputs and outputs of a function, so the function process must be clear first.
Solving Inequalities	`solving-inequalities`	Finding a domain often reduces to solving an inequality, such as requiring a radicand to be nonnegative.

Unlocks (these depend on this skill):

Skill	skill_id	What it needs from here
Rates of Change	`m141-rates-of-change`	A rate is measured over an interval inside the domain.
Function Composition	`function-composition`	The composition $f(g(x))$ is defined only where the output of $g$ lands inside the domain of $f$.
Inverse Functions	`m141-inverse-functions`	The domain of a function becomes the range of its inverse, and the reverse.

No cross-course prerequisites are listed for this node.

Quick Reference

Property	Value
Textbook	OpenStax Precalculus 2e
Section	1.2 Domain and Range
Course	MATH141
Difficulty	Introductory
Time	~30 minutes

Domain and Range at a Glance

Term	Plain meaning	Where to find it on a graph
Domain	The set of input values the function accepts	Sweep left to right; project the curve onto the $x$-axis
Range	The set of output values the function produces	Sweep bottom to top; project the curve onto the $y$-axis

Official Definitions

Domain and range. The domain of a function is the set of input values; the range is the set of possible output values. Restrictions exclude any input that would require an even root of a negative number or division by zero. (OpenStax Precalculus 2e, Section 1.2 Domain and Range.)

Interval notation. A way of describing sets that include all real numbers between a lower limit that may or may not be included and an upper limit that may or may not be included. (OpenStax Precalculus 2e, Section 1.2 Domain and Range.)

Set-builder notation. A method of specifying a set of elements that satisfy a certain condition. It takes the form $\{x \mid \text{statement about } x\}$, which is read as “the set of all $x$ such that the statement about $x$ is true.” (OpenStax Precalculus 2e, Section 1.2 Domain and Range.)

The two notations describe the same set in two languages. The set of all real numbers that are at least $4$ can be written either way:

Set-builder	Interval
$\{x \mid x \geq 4\}$	$[4, \infty)$

A square bracket $[$ or $]$ includes the endpoint. A parenthesis $($ or $)$ excludes it. Infinity always takes a parenthesis, because infinity is a direction, not a number that the set can reach.

Finding a Domain: The Two Restrictions

For the functions in this section, the domain is every real number except the inputs that trigger one of two forbidden operations.

Restriction	The rule	What to require	Where it comes from
Even root of a negative	A square root (or any even root) of a negative number is not a real number	Set the radicand $\geq 0$ and solve	$\sqrt{x-4}$ needs $x - 4 \geq 0$
Division by zero	Dividing by zero is undefined	Set the denominator $\neq 0$ and exclude those inputs	$\frac{1}{x-3}$ needs $x - 3 \neq 0$

A function that is a polynomial (only additions, subtractions, multiplications, and whole-number powers) has neither restriction, so its domain is all real numbers, written $(-\infty, \infty)$.

One way to see it, another way to see it

There is more than one valid path to a domain.

One way (algebra first): write down the restriction as an inequality or an exclusion, solve it, and translate the solution into interval notation.
Another way (graph first): sketch or picture the curve, then read off the shadow it casts on the horizontal axis.

Both give the same answer. Which one feels clearer to you, and why? For a square root the algebra is fast; for a piecewise curve the picture is often clearer.

Worked Examples

Example 1: Domain of a square-root function

Find the domain of $g(x) = \sqrt{2x + 6}$.

Predict first. The radicand grows as $x$ grows, so large inputs are safe and very negative inputs should fail. Expect a domain that starts at some cutoff and runs upward.

Compute. The even root forces the radicand to be nonnegative:

$$2x + 6 \geq 0$$ $$2x \geq -6$$ $$x \geq -3$$

Domain: $\{x \mid x \geq -3\}$, which is $[-3, \infty)$.

Check against the prediction. The cutoff is $x = -3$, and inputs run upward from there, matching the prediction. Testing $x = -3$ gives $\sqrt{0} = 0$ (allowed, so the bracket is square), and $x = -4$ gives $\sqrt{-2}$ (not real, correctly excluded).

Example 2: Domain of a rational function

Find the domain of $h(x) = \dfrac{x + 5}{x^2 - 9}$.

Predict first. The only danger here is a zero in the denominator, so expect the domain to be all real numbers with a few isolated points punched out.

Compute. Require the denominator to be nonzero:

$$x^2 - 9 \neq 0$$ $$(x - 3)(x + 3) \neq 0$$ $$x \neq 3 \quad \text{and} \quad x \neq -3$$

Domain: $\{x \mid x \neq 3, \ x \neq -3\}$, which in interval notation is

$$(-\infty, -3) \cup (-3, 3) \cup (3, \infty).$$

Check against the prediction. Two points are removed and everything else stays, matching the prediction. The numerator $x + 5$ never causes a problem; only the denominator restricts the domain.

Example 3: Range from a graph

The function $p(x) = x^2 - 1$ is a parabola opening upward with its lowest point at $(0, -1)$.

Predict first. A parabola opening upward has a bottom but no top, so expect a range that starts at the lowest output and climbs without bound.

Read the domain. No square root and no denominator means every real number is a valid input. Domain: $(-\infty, \infty)$.

Read the range. The smallest output is $-1$, reached at $x = 0$. From there the outputs increase forever. Range: $\{y \mid y \geq -1\}$, which is $[-1, \infty)$.

Check against the prediction. A floor of $-1$ with no ceiling matches the prediction. The endpoint is included (square bracket) because the value $-1$ is actually achieved at $x = 0$.

Two Representations of One Domain

The domain $x \geq -3$ from Example 1 shows up in four connected forms. Practice translating between them.

Representation	This domain
Words	All real numbers greater than or equal to $-3$
Set-builder	$\{x \mid x \geq -3\}$
Interval	$[-3, \infty)$
Number line	A filled dot at $-3$ with shading to the right

Translate in your head: if a problem hands you $(-\infty, 2)$, can you say it in words and draw it? It is all real numbers strictly less than $2$, an open dot at $2$ shaded to the left.

Common Misconceptions

Common misconception

the domain is the set of outputs and the range is the set of inputs. The names get swapped because both are sets of numbers. Anchor them to the process: input goes in first, so domain comes first alphabetically and holds the inputs; the range is the result that comes out. For $f(x) = \sqrt{x-4}$, the domain is the allowed inputs $x \geq 4$, and the range is the resulting outputs $y \geq 0$. Reversing them would claim the machine accepts $x = 0$, but $\sqrt{0 - 4}$ is not real, so $0$ cannot be an input.

Common misconception

every function accepts every real number. A square root or a fraction can reject inputs. Test it on a specific case: for $\frac{1}{x-3}$ at $x = 3$, the rule asks for $\frac{1}{0}$, which is undefined, so $x = 3$ is not in the domain even though it is an ordinary number. The single excluded point is invisible unless you check the denominator.

Common misconception

a square bracket and a parenthesis are interchangeable. They record whether the endpoint itself belongs to the set. For $\sqrt{x - 4}$, the value $x = 4$ gives $\sqrt{0} = 0$, a real output, so $4$ belongs and the notation is $[4, \infty)$ with a square bracket. Writing $(4, \infty)$ would wrongly drop a valid input. Always test the endpoint before choosing the bracket.

Practice Problems

Level 1 Translate to Interval Notation

Write the set $\{x \mid x \geq 7\}$ in interval notation.

Thought Process

The set includes $7$ itself and everything larger. An included endpoint takes a square bracket. There is no upper bound, so the right end is $\infty$ with a parenthesis.

Show Answer

$$[7, \infty)$$

The endpoint $7$ is included, so the bracket is square; infinity always takes a parenthesis.

Level 2 Domain of a Square Root

Find the domain of $f(x) = \sqrt{x - 10}$.

Thought Process

The only restriction is the even root. Set the radicand to be nonnegative and solve. Then check whether the endpoint is included.

Show Answer

Require $x - 10 \geq 0$, so $x \geq 10$.

Domain: $\{x \mid x \geq 10\}$, which is $[10, \infty)$.

The endpoint $x = 10$ gives $\sqrt{0} = 0$, a real output, so $10$ is included.

Level 3 Domain of a Rational Function

Find the domain of $g(x) = \dfrac{4}{x^2 - 5x + 6}$.

Thought Process

The danger is a zero denominator. Factor $x^2 - 5x + 6$, find the values that make it zero, and exclude them.

Show Answer

Factor: $x^2 - 5x + 6 = (x - 2)(x - 3)$.

Require $(x - 2)(x - 3) \neq 0$, so $x \neq 2$ and $x \neq 3$.

Domain: $\{x \mid x \neq 2, \ x \neq 3\}$, which is

$$(-\infty, 2) \cup (2, 3) \cup (3, \infty).$$

Level 4 Find the Flaw

A student finds the domain of $f(x) = \dfrac{\sqrt{x + 2}}{x - 1}$ and writes: “The square root needs $x + 2 \geq 0$, so $x \geq -2$. That is the whole domain: $[-2, \infty)$.”

What did the student miss, and what is the correct domain?

Thought Process

This function has both a square root in the numerator and a denominator. The student handled the square root but ignored the second restriction. Both restrictions must hold at the same time.

Show Answer

The flaw: the student checked the even-root restriction but forgot the division-by-zero restriction.

Restriction 1 (square root): $x + 2 \geq 0$, so $x \geq -2$.

Restriction 2 (denominator): $x - 1 \neq 0$, so $x \neq 1$.

Both must hold. Start with $x \geq -2$ and then remove $x = 1$:

$$[-2, 1) \cup (1, \infty).$$

The single point $x = 1$ is punched out of the interval the student wrote.

Level 5 Domain and Range Together

For $f(x) = \sqrt{x - 4}$, find both the domain and the range. Justify the range by describing how the output behaves as the input moves across the domain.

Thought Process

The domain comes from the even-root restriction. For the range, track the output as the input increases from its smallest allowed value: start at the cutoff, then let the input grow.

Show Answer

Domain. Require $x - 4 \geq 0$, so $x \geq 4$. Domain: $[4, \infty)$.

Range. At the smallest input $x = 4$, the output is $\sqrt{0} = 0$. As the input moves upward from $4$, the radicand grows, so the output grows and never returns to a value it already passed. The smallest output is $0$, reached at $x = 4$, and outputs climb without bound.

Range: $\{y \mid y \geq 0\}$, which is $[0, \infty)$.

Why the range starts at $0$ and not below: a square root of a real number is never negative, so no input can produce an output less than $0$. The output $0$ is achieved (at $x = 4$), so the endpoint is included.

Mastery Checklist

Novice (Level 1-2):

I can state in plain words what the domain and the range of a function are.
I can translate a domain among words, set-builder notation, and interval notation.
I can decide whether an endpoint takes a square bracket or a parenthesis.

Competent (Level 3-4):

I can find the domain of a square-root function by solving the radicand inequality.
I can find the domain of a rational function by excluding the zeros of the denominator.
I can apply both restrictions at once when a function has a root and a denominator.

Proficient (Level 5):

I can read a domain and a range off a graph.
I can justify a range by describing how the output behaves as the input moves across the domain.
I can explain why a particular endpoint is included or excluded.

Resources

Textbook: OpenStax Precalculus 2e, Section 1.2 Domain and Range. Page numbers pending faculty verification.
OpenStax Precalculus 2e 1.2 Domain and Range
OpenStax Precalculus 2e Ch 1 Key Concepts

Connections

Looking back:

Functions and Function Notation establishes the input-output process that domain and range describe.
Solving Inequalities supplies the technique for solving the radicand inequality that produces a domain.

Looking ahead:

Rates of Change measures how output changes over an interval inside the domain.
Function Composition is defined only where the output of the inner function lands inside the domain of the outer function.
Inverse Functions swap the roles: the domain of a function becomes the range of its inverse.

Real-world connections:

A ticket price as a function of age has a domain of nonnegative ages, not all real numbers.
A medication dose as a function of body weight has a domain bounded by realistic weights.
The height of a thrown ball as a function of time has a domain from launch to landing and a range from the ground up to the peak.

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Functions and Function Notation	Skills Index	Rates of Change

Last updated: 2026-06-16