Domain and Range

Every function has a domain and a range. The domain is the set of all inputs the function can accept, and the range is the set of all outputs the function actually produces. Understanding these two sets tells you where a function “lives” — what you are allowed to plug in and what you can expect to get out. This matters in pure algebra and in every applied field, where physical constraints naturally limit the inputs and outputs of a formula.

Definitions

• Domain: The set of all $x$-values (inputs) for which the function is defined.
• Range: The set of all $y$-values (outputs) that the function produces.

Think of the domain as “what can I plug in?” and the range as “what comes out?”

Finding Domain and Range from a Table

When a function is given as a set of ordered pairs or a table, the domain is the collection of all $x$-values and the range is the collection of all $y$-values.

Example 1: Find the domain and range

Domain: $\{-3,\; 0,\; 2,\; 5\}$

Range: $\{0,\; 4,\; 9,\; 25\}$

List each value once, even if it appears multiple times in the table.

Finding Domain and Range from a Graph

On a graph, the domain is the set of $x$-values the graph covers (read left to right), and the range is the set of $y$-values the graph covers (read bottom to top).

Example 2: A line segment from $(-2, 1)$ to $(4, 7)$

The graph starts at $x = -2$ and ends at $x = 4$, so the domain is $[-2, 4]$. The lowest $y$-value is 1 and the highest is 7, so the range is $[1, 7]$.

Example 3: The graph of $y = x^2$

The parabola extends infinitely left and right, so the domain is $(-\infty, \infty)$. The lowest point is at $y = 0$ (the vertex), and the graph opens upward forever, so the range is $[0, \infty)$.

Interval Notation

Interval notation is a compact way to describe continuous sets of numbers:

Key rule: Use a square bracket $[$ or $]$ when the endpoint is included, and a parenthesis $($ or $)$ when it is not. Infinity always gets a parenthesis because it is not a number you can reach.

Set-Builder Notation

Set-builder notation describes a set by stating the condition its members must satisfy:

$\{x \mid x \geq 0\} \quad \text{reads as "the set of all } x \text{ such that } x \text{ is greater than or equal to 0"}$

The vertical bar $\mid$ means “such that.” Some textbooks use a colon instead: $\{x : x \geq 0\}$.

Comparing the Two Notations

Both notations say the same thing — use whichever your course requires.

Finding Domain from Equations

For most algebraic functions, the domain is “all real numbers” unless the equation contains one of these restrictions:

Restriction 1: Division by Zero

You cannot divide by zero. If the function has a variable in the denominator, set the denominator not equal to zero.

Example 4: Find the domain of $f(x) = \dfrac{1}{x - 3}$

The denominator is zero when $x - 3 = 0$, i.e., $x = 3$.

Domain: All real numbers except 3

• Interval notation: $(-\infty, 3) \cup (3, \infty)$
• Set-builder notation: $\{x \mid x \neq 3\}$

Example 5: Find the domain of $g(x) = \dfrac{x + 2}{x^2 - 9}$

Factor the denominator: $x^2 - 9 = (x - 3)(x + 3)$

Set each factor not equal to zero: $x \neq 3$ and $x \neq -3$.

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

Restriction 2: Square Roots of Negative Numbers

In real-number math, you cannot take the square root of a negative number. The expression under the radical must be greater than or equal to zero.

Example 6: Find the domain of $h(x) = \sqrt{x - 5}$

Set the radicand greater than or equal to zero:

$x - 5 \geq 0 \implies x \geq 5$

Domain: $[5, \infty)$

Example 7: Find the domain of $f(x) = \sqrt{12 - 4x}$

$12 - 4x \geq 0$

$-4x \geq -12$

$x \leq 3 \quad \text{(flip the inequality when dividing by a negative number)}$

Domain: $(-\infty, 3]$

Combined Restrictions

Example 8: Find the domain of $f(x) = \dfrac{\sqrt{x + 1}}{x - 4}$

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

Restriction 2 (denominator): $x - 4 \neq 0 \implies x \neq 4$

Combine: $x \geq -1$ and $x \neq 4$.

Domain: $[-1, 4) \cup (4, \infty)$

Finding Range from Equations

Finding the range algebraically is often harder than finding the domain. Here are strategies:

1. Linear functions $f(x) = mx + b$ (with $m \neq 0$): Range is $(-\infty, \infty)$.
2. Quadratic functions $f(x) = ax^2 + bx + c$: Find the vertex. If $a > 0$ (opens up), range is $[y_{\text{vertex}}, \infty)$. If $a < 0$ (opens down), range is $(-\infty, y_{\text{vertex}}]$.
3. Square root functions $f(x) = \sqrt{x}$: Outputs are always non-negative. Range is $[0, \infty)$.

Example 9: Find the range of $f(x) = x^2 - 6x + 5$

Complete the square or use the vertex formula. The vertex is at $x = \dfrac{-(-6)}{2(1)} = 3$:

$f(3) = 9 - 18 + 5 = -4$

Since $a = 1 > 0$, the parabola opens upward.

Range: $[-4, \infty)$

Example 10: Find the range of $f(x) = \sqrt{x - 2} + 1$

The smallest value of $\sqrt{x - 2}$ is $0$ (when $x = 2$), so the smallest value of $f(x)$ is $0 + 1 = 1$. As $x$ increases, $\sqrt{x - 2}$ grows without bound.

Range: $[1, \infty)$

Real-World Application: Retail — Revenue Function

A retail store sells a product at $25 each. The revenue function is:

$R(n) = 25n$

where $n$ is the number of units sold.

Domain in context: You cannot sell a negative number of units, and the store has a stock limit of, say, 500 units. So the domain is $\{n \mid 0 \leq n \leq 500\}$ or in interval notation $[0, 500]$.

Range in context: When $n = 0$, revenue is $0. When $n = 500$, revenue is $25(500) =$ $12,500. So the range is $[0, 12500]$.

Notice how the real-world context restricts both the domain and range far more than the algebraic expression alone would suggest. Without context, $R(n) = 25n$ has domain and range of all real numbers.

Real-World Application: HVAC — Temperature Model

An HVAC technician models indoor temperature (degrees Fahrenheit) as a function of time $t$ (hours after the system turns on):

$T(t) = -0.5t^2 + 8t + 60, \quad 0 \leq t \leq 12$

Domain: The system runs for 12 hours, so $t$ ranges from 0 to 12: $[0, 12]$.

Range: Find the vertex to determine the maximum temperature. The vertex occurs at $t = \dfrac{-8}{2(-0.5)} = 8$:

$T(8) = -0.5(64) + 8(8) + 60 = -32 + 64 + 60 = 92$

Check the endpoints: $T(0) = 60$ and $T(12) = -0.5(144) + 96 + 60 = -72 + 96 + 60 = 84$.

The minimum output is 60 (at $t = 0$) and the maximum is 92 (at $t = 8$).

Range: $[60, 92]$

Common Mistakes to Avoid

1. Forgetting to flip the inequality sign. When you multiply or divide an inequality by a negative number, the direction reverses. Missing this is the most common error in domain problems involving square roots.
2. Using a square bracket with infinity. Infinity is not a number. Always write $\infty)$ or $(-\infty$, never $[\infty$ or $\infty]$.
3. Assuming the range is always all real numbers. Many functions have restricted ranges. Quadratics, absolute values, and square root functions all have limited output sets.
4. Ignoring context restrictions. In applied problems, the domain is often restricted by physical constraints (non-negative quantities, maximum capacity, time bounds) even when the algebra allows all real numbers.
5. Confusing domain with range. Domain is inputs ($x$-values); range is outputs ($y$-values). On a graph: domain reads left-right, range reads bottom-top.

Practice Problems

Test your understanding with these problems. Click to reveal each answer.

Key Takeaways

• Domain = all valid inputs; Range = all resulting outputs
• Division by zero and square roots of negatives are the two main restrictions that limit domain in algebra
• Interval notation uses brackets $[ \; ]$ for included endpoints and parentheses $( \; )$ for excluded endpoints; infinity always gets a parenthesis
• Set-builder notation describes a set by its defining condition: $\{x \mid \text{condition}\}$
• For quadratics, the vertex determines the minimum or maximum, which defines the range boundary
• In applied problems, context often restricts the domain and range beyond what the algebra alone requires

Return to Algebra for more topics in this section.