Domain and Range

The domain of a function is the set of all inputs it accepts, and the range is the set of all outputs it can produce. Knowing them tells you where a graph exists and what values a formula can actually give, which matters for graphing, solving, and real-world problems.

Key ideas

Domain and range

• Domain: all possible $x$-values (inputs).
• Range: all possible $y$-values (outputs).

Writing them in set notation

Meaning	Set notation
all real numbers	$\{x \in \mathbb{R}\}$
$x$ is at least $2$	$\{x \in \mathbb{R} \mid x \ge 2\}$
$x$ is any real number except $-3$	$\{x \in \mathbb{R} \mid x \ne -3\}$
$y$ is from $-1$ up to, but not including, $4$	$\{y \in \mathbb{R} \mid -1 \le y \lt 4\}$
a few separate values	$\{1, 2, 5\}$

Read $\{x \in \mathbb{R} \mid x \ge 2\}$ as “all real numbers $x$ such that $x$ is greater than or equal to $2$”.

From a graph

• Domain: scan from left to right. Which $x$-values does the graph cover?
• Range: scan from bottom to top. Which $y$-values does it cover?
• A closed dot ● means that point is included. An open dot ○ means it isn’t.
• An arrow means the graph keeps going forever in that direction.

From an equation

In this course, watch for two restrictions:

1. You can’t divide by zero. For $\dfrac{1}{x - 3}$, the domain excludes $x = 3$.
2. You can’t take the square root of a negative number (in the real numbers). For $\sqrt{x - 5}$, you need $x - 5 \ge 0$, so $x \ge 5$.

If neither happens, as with lines and parabolas, the domain is all real numbers.

Real-world restrictions

In context, the situation can limit the domain and range: time can’t be negative, a number of people must be a whole number, and a ball’s height can’t go below the ground. The parent functions page lists the domain and range of the four basic graphs.

Worked examples

Example 1: From a graph

Find the domain and range of the function graphed below.

Solution.

• Domain: the graph runs from $x = -3$ (closed dot, included) to $x = 4$ (open dot, not included): $\{x \in \mathbb{R} \mid -3 \le x \lt 4\}$.
• Range: the lowest point is $(1, -2)$ and the highest is $(-3, 6)$. Every height in between is covered: $\{y \in \mathbb{R} \mid -2 \le y \le 6\}$.

Notice that the range does not come from the endpoints alone. The lowest point is in the middle of the graph.

Example 2: From equations

Find the domain and range of each function.

(a) $f(x) = 2x - 7$ $\qquad$ (b) $g(x) = x^2 + 3$ $\qquad$ (c) $h(x) = \sqrt{x - 5}$ $\qquad$ (d) $k(x) = \dfrac{1}{x + 2}$

Solution.

(a) A slanted line goes forever in both directions. Domain $\{x \in \mathbb{R}\}$, range $\{y \in \mathbb{R}\}$.

(b) Any $x$ works. Since $x^2 \ge 0$, the smallest output is $0 + 3 = 3$. Domain $\{x \in \mathbb{R}\}$, range $\{y \in \mathbb{R} \mid y \ge 3\}$.

(c) Need $x - 5 \ge 0$, so $x \ge 5$. A square root is never negative, and it can be $0$ (at $x = 5$). Domain $\{x \in \mathbb{R} \mid x \ge 5\}$, range $\{y \in \mathbb{R} \mid y \ge 0\}$.

(d) Need $x + 2 \ne 0$, so $x \ne -2$. A fraction with numerator $1$ can never equal $0$. Domain $\{x \in \mathbb{R} \mid x \ne -2\}$, range $\{y \in \mathbb{R} \mid y \ne 0\}$.

Example 3: A trickier square root

Find the domain and range of $f(x) = \sqrt{6 - 2x} + 1$.

Solution. The expression under the root must not be negative:

$$\begin{aligned} 6 - 2x &\ge 0 \\ -2x &\ge -6 \\ x &\le 3 && \text{dividing by } -2 \text{ flips the inequality} \end{aligned}$$

The square root itself is at least $0$, so $f(x)$ is at least $0 + 1 = 1$.

Domain $\{x \in \mathbb{R} \mid x \le 3\}$, range $\{y \in \mathbb{R} \mid y \ge 1\}$.

Example 4: A real-world function

A ball is thrown upward from the ground. Its height in metres after $t$ seconds is $h(t) = -5t^2 + 20t$, until it lands. Find the domain and range.

Solution. The ball is on the ground when $h(t) = 0$:

$$-5t^2 + 20t = 0 \quad\Rightarrow\quad -5t(t - 4) = 0 \quad\Rightarrow\quad t = 0 \text{ or } t = 4$$

So the ball is in the air from $t = 0$ to $t = 4$ seconds.

The parabola is symmetric, so its highest point is halfway between the zeros, at $t = 2$: $h(2) = -20 + 40 = 20$ metres.

Domain $\{t \in \mathbb{R} \mid 0 \le t \le 4\}$, range $\{h \in \mathbb{R} \mid 0 \le h \le 20\}$.

Common mistakes

Swapping domain and range. Domain is about $x$ (inputs, left to right). Range is about $y$ (outputs, bottom to top).

Including an open-dot value. In Example 1, $x = 4$ is not in the domain, because the dot there is open.

Finding the range from the endpoints only. A graph can dip or rise in the middle. Look for the lowest and highest points, not just where it starts and stops.

Writing $x \ne 5$ for a square root. For $\sqrt{x - 5}$, values like $x = 2$ don’t work either. The restriction is $x \ge 5$. Save $\ne$ for dividing by zero.

Forgetting to flip the inequality when dividing by a negative number, as in Example 3.

Ignoring the real world. In Example 4, the formula works for any $t$, but negative times and times after the ball lands make no sense.

Practice

1. (Warm-up) State the domain and range of $\{(-2, 5), (0, 3), (1, 5), (4, 8)\}$.

Solution

Domain $\{-2, 0, 1, 4\}$. Range $\{3, 5, 8\}$ (list $5$ only once).

2. (Warm-up) State the domain and range of $f(x) = -4x + 1$.

Solution

It’s a slanted line. Domain $\{x \in \mathbb{R}\}$, range $\{y \in \mathbb{R}\}$.

3. (Warm-up) State the domain and range of $f(x) = x^2 - 6$.

Solution

Domain $\{x \in \mathbb{R}\}$. Since $x^2 \ge 0$, the smallest output is $-6$: range $\{y \in \mathbb{R} \mid y \ge -6\}$.

4. (Core) State the domain and range of $g(x) = \sqrt{x + 7}$.

Solution

Need $x + 7 \ge 0$, so $x \ge -7$. Domain $\{x \in \mathbb{R} \mid x \ge -7\}$, range $\{y \in \mathbb{R} \mid y \ge 0\}$.

5. (Core) State the domain and range of $h(x) = \dfrac{1}{x - 3}$.

Solution

Need $x - 3 \ne 0$, so $x \ne 3$. The fraction can never equal $0$.

Domain $\{x \in \mathbb{R} \mid x \ne 3\}$, range $\{y \in \mathbb{R} \mid y \ne 0\}$.

6. (Core) State the domain and range of $f(x) = \sqrt{10 - 2x}$.

Solution

$$10 - 2x \ge 0 \quad\Rightarrow\quad -2x \ge -10 \quad\Rightarrow\quad x \le 5$$

Domain $\{x \in \mathbb{R} \mid x \le 5\}$, range $\{y \in \mathbb{R} \mid y \ge 0\}$.

7. (Core) Field trip tickets cost $6 each, and a school can buy at most 120. The cost in dollars is $C(n) = 6n$, where $n$ is the number of tickets. State the domain and range.

Solution

You can only buy a whole number of tickets, from $0$ to $120$.

Domain $\{0, 1, 2, \dots, 120\}$. Range $\{0, 6, 12, \dots, 720\}$, the multiples of $6$ from $0$ to $720$.

8. (Challenge) State the domain and range of $f(x) = -2(x - 1)^2 + 8$.

Solution

Any $x$ works: domain $\{x \in \mathbb{R}\}$.

Since $(x - 1)^2 \ge 0$, the term $-2(x - 1)^2$ is at most $0$. So $f(x)$ is at most $8$, and it equals $8$ when $x = 1$.

Range $\{y \in \mathbb{R} \mid y \le 8\}$.

9. (Challenge) Find the domain of $f(x) = \dfrac{\sqrt{x - 2}}{x - 5}$.

Solution

There are two restrictions:

• the square root needs $x - 2 \ge 0$, so $x \ge 2$
• the denominator needs $x - 5 \ne 0$, so $x \ne 5$

Domain $\{x \in \mathbb{R} \mid x \ge 2, \ x \ne 5\}$.