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Domain and Range

MATH161
Reference: Stewart 1.1  •  Chapter: 1  •  Section: 1

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Domain and Range

Where Functions Live

Before you can evaluate a function, you need to know: where is it defined? Plug $x = -5$ into $f(x) = \sqrt{x}$ and your calculator protests. Divide by zero and mathematics breaks. The domain tells you which inputs are "legal," and the range tells you what outputs are possible.

Mastering domain and range is your first line of defense against illegal operations—and it builds intuition for later topics like limits and continuity.

Prerequisite Map

Prerequisites
Function RepresentationsSolving Inequalities
→
This skill
Domain and Range
→
Unlocks
ContinuityLimits at InfinityComposite Functions

Quick Reference

Property Value
Chapter 1 - Functions and Limits
Section 1.1
Difficulty Beginner
Time ~20 minutes

Key Concepts

Definitions

Domain: The set of all input values $x$ for which $f(x)$ is defined.

Range: The set of all output values $f(x)$ that the function actually produces.

Finding Domain: The Two Main Restrictions

Restriction Rule Example
Square roots Radicand $\geq 0$ $\sqrt{x-3}$ requires $x \geq 3$
Denominators Denominator $\neq 0$ $\frac{1}{x-2}$ requires $x \neq 2$

Process for finding domain:

  1. Start with all real numbers ($-\infty, \infty$)
  2. Remove values that cause problems:
  3. Express in interval notation

Interval Notation Quick Reference

Notation Meaning Number Line
$(a, b)$ All $x$ with $a < x < b$ Open circles at both ends
$[a, b]$ All $x$ with $a \leq x \leq b$ Closed circles at both ends
$[a, b)$ All $x$ with $a \leq x < b$ Closed at $a$, open at $b$
$(a, \infty)$ All $x > a$ Open at $a$, extends right
$(-\infty, b]$ All $x \leq b$ Extends left, closed at $b$

Union: Use $\cup$ to combine disjoint intervals. Example: $(-\infty, 2) \cup (2, \infty)$ means "all real numbers except 2."

Finding Range

From a graph: Look at the vertical extent—what $y$-values are covered?

From a formula: Analyze the function's behavior:

Practice Problems

Level 1 Simple Square Root Domain

Find the domain of $f(x) = \sqrt{x + 5}$.

Thought Process

For square roots, the expression under the radical must be non-negative. Set up the inequality and solve.

Show Answer

The radicand must be non-negative: $$x + 5 \geq 0$$ $$x \geq -5$$

Domain: $[-5, \infty)$

Level 2 Rational Function Domain

Find the domain of $g(x) = \frac{3x + 1}{x^2 - 9}$.

Thought Process

For rational functions, find where the denominator equals zero and exclude those values. Factor the denominator to find the problematic $x$-values.

Show Answer

The denominator cannot equal zero: $$x^2 - 9 \neq 0$$ $$(x-3)(x+3) \neq 0$$ $$x \neq 3 \text{ and } x \neq -3$$

Domain: $(-\infty, -3) \cup (-3, 3) \cup (3, \infty)$

Or equivalently: "All real numbers except $x = -3$ and $x = 3$."

Level 3 Combined Restrictions

Find the domain of $h(x) = \frac{\sqrt{x - 2}}{x - 5}$.

Thought Process

This function has two restrictions:

  1. Square root: $x - 2 \geq 0$
  2. Denominator: $x - 5 \neq 0$

Find each restriction separately, then combine them (intersection).

Show Answer

Restriction 1 (square root): $$x - 2 \geq 0 \implies x \geq 2$$

Restriction 2 (denominator): $$x - 5 \neq 0 \implies x \neq 5$$

Combine: We need $x \geq 2$ AND $x \neq 5$.

Domain: $[2, 5) \cup (5, \infty)$

Level 4 Finding Domain and Range Together

For $f(x) = \sqrt{16 - x^2}$:

  1. Find the domain.
  2. Sketch the graph by recognizing the shape.
  3. Find the range.
Thought Process

For part (a): Set $16 - x^2 \geq 0$ and solve.

For part (b): The equation $y = \sqrt{16 - x^2}$ can be rewritten. Square both sides: $y^2 = 16 - x^2$, which gives $x^2 + y^2 = 16$. This is a circle! But since $y = \sqrt{\cdot} \geq 0$, we only get the top half.

For part (c): Use the graph to find vertical extent.

Show Answer

(a) Domain: $$16 - x^2 \geq 0$$ $$x^2 \leq 16$$ $$-4 \leq x \leq 4$$

Domain: $[-4, 4]$

(b) Graph:

Squaring both sides of $y = \sqrt{16 - x^2}$ gives $y^2 = 16 - x^2$, or $x^2 + y^2 = 16$.

This is a circle of radius 4 centered at the origin. Since $y = \sqrt{\cdot} \geq 0$, we only see the upper semicircle.

(c) Range:

From the graph, $y$ starts at $0$ (when $x = \pm 4$) and reaches maximum $y = 4$ (when $x = 0$).

Range: $[0, 4]$

Level 5 Constructing a Function with Given Domain
  1. Find a formula for a function whose domain is $(-\infty, 3) \cup (3, \infty)$.
  2. Find a formula for a function whose domain is $[2, 7]$.
  3. Find a formula for a function whose domain is $(-\infty, -1) \cup [4, \infty)$.
  4. Is it possible to have a function with domain $(0, 1) \cup (2, 3)$ using only square roots and rational expressions? Justify your answer.
Thought Process

Work backwards from the domain to construct the function:

  • To exclude a single point, use a denominator that's zero there.
  • To create an interval $[a, b]$, use $\sqrt{(x-a)(b-x)}$ (product is non-negative between roots).
  • To exclude an interval, think about what restrictions create that pattern.

Part (d) requires thinking about what shapes of domains are achievable with our basic restrictions.

Show Answer

(a) Domain excludes only $x = 3$: $$f(x) = \frac{1}{x - 3}$$

(b) Domain is exactly $[2, 7]$. We need $x \geq 2$ AND $x \leq 7$: $$g(x) = \sqrt{(x-2)(7-x)}$$

Check: $(x-2)(7-x) \geq 0$ when $2 \leq x \leq 7$ ✓

(c) Domain is $x < -1$ OR $x \geq 4$. One approach: $$h(x) = \sqrt{\frac{x - 4}{x + 1}}$$

When is $\frac{x-4}{x+1} \geq 0$?

  • Sign chart shows this holds when $x < -1$ or $x \geq 4$ ✓

(d) This is challenging. Standard restrictions give:

  • Denominators: exclude isolated points or nothing
  • Square roots: include or exclude intervals

To get $(0,1) \cup (2,3)$ exactly seems to require excluding both $[1,2]$ and the regions outside $(0,3)$.

One construction using nested expressions: $$k(x) = \sqrt{x(1-x)} + \sqrt{(x-2)(3-x)}$$

For each term to be defined, we need $x \in [0,1]$ for the first and $x \in [2,3]$ for the second. But the SUM is defined only where BOTH are defined... which is never (the intervals don't overlap).

Better approach: $$k(x) = \frac{1}{\sqrt{x(1-x)(x-2)(3-x)}}$$

This requires the radicand to be positive (strictly, for the denominator). Checking signs: $x(1-x)(x-2)(3-x) > 0$ on $(0,1) \cup (2,3)$. ✓

Mastery Checklist

Mental Model

The Bouncer Analogy:

Think of domain restrictions as bouncers at a club:

To find the domain, figure out who gets turned away, and everyone else can enter.

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Last updated: 2026-01-22