Domain & Range
Try typing $\sqrt{-1}$ into a basic calculator. It throws an error. That is not a bug — it is a domain restriction. Every function has boundaries, and knowing where they are is what separates a correct answer from a careless mistake.
Practise this lesson
Three printable worksheets that build from foundations to mastery — or build your own from any module’s questions.
A basic calculator refuses to calculate $\sqrt{-1}$ and shows an error. Why do you think this happens? What does this tell you about the inputs that the square root function will accept?
Key insight: interval notation uses round brackets for excluded endpoints and square brackets for included endpoints. $\infty$ always takes a round bracket.
Key facts
- the definitions of domain and range
- how to write domains and ranges in interval notation
- the three main types of domain restrictions
Concepts
- why division by zero is undefined
- why square roots of negative numbers are not real
- how the shape of a graph determines its range
Skills
- find the domain of linear, quadratic, rational, and root functions
- find the range of basic functions from their graphs or equations
- express domains and ranges using interval notation
The domain of a function is the complete set of input values ($x$-values) for which the function is defined. The range is the complete set of output values ($y$-values) that the function can produce.
We usually express domains and ranges using interval notation:
- $[a, b]$ means all numbers from $a$ to $b$, including both endpoints
- $(a, b)$ means all numbers from $a$ to $b$, excluding both endpoints
- $[a, b)$ includes $a$ but excludes $b$
- $\infty$ always uses a round bracket because infinity is not a real number we can reach
The three main domain restrictions
Most domain problems in Year 11 involve one of three restrictions:
- Denominator cannot be zero. For rational functions like $f(x) = \dfrac{1}{x-2}$, solve $x-2 \neq 0$ to find the restriction.
- Radicand must be non-negative. For functions with square roots like $f(x) = \sqrt{x+3}$, solve $x+3 \geq 0$.
- Logarithm argument must be positive. For $f(x) = \log_e(x)$, the domain is $x > 0$.
Domain = all allowable input ($x$) values; Range = all possible output ($y$) values; Interval notation: $[$ $]$ = included endpoint; $( )$ = excluded endpoint; $\infty$ always uses $($
Pause — copy the interval notation rules (square bracket = included, round bracket = excluded), the infinity convention, and the three domain restrictions into your book.
Did you get this? True or false: $[2, \infty]$ is correct interval notation for all real numbers greater than or equal to 2.
Quick check: What is the domain of $f(x) = \sqrt{x - 2}$?
We just saw that domain uses interval notation to list the valid $x$-values — including the three main restrictions (denominator, radicand, logarithm). That raises a question: the domain tells us what goes in, but what values can actually come out? This card answers it → finding the range by using the vertex for quadratics and the non-negativity rule for square roots.
The range depends on the shape of the function. For many basic functions, we can determine the range by inspection:
- Linear functions: $y = mx + b$ has range $(-\infty, \infty)$ unless the domain is restricted.
- Quadratic functions: If the parabola opens upward, the range starts at the $y$-coordinate of the vertex. If it opens downward, the range ends at the vertex.
- Square root functions: The principal square root always produces a non-negative result, so $\sqrt{\dots} \geq 0$.
- Absolute value functions: $|x| \geq 0$, so the range depends on any vertical shifts.
Linear: range $(-\infty, \infty)$ unless domain is restricted; Quadratic $a(x-h)^2+k$: range $[k,\infty)$ if $a>0$; range $(-\infty,k]$ if $a<0$
Pause — copy the range rules: linear $(-\infty,\infty)$; quadratic $a(x-h)^2+k$ gives range $[k,\infty)$ if $a>0$ or $(-\infty,k]$ if $a<0$; square root always $\geq 0$.
Fill the blanks: drag each token into the matching blank.
For a quadratic that opens ___, the range starts at the ___ $y$-value. This boundary point is the ___ of the function and is written with a ___ in interval notation.
Worked examples · reveal as you go
Find the domain of $\displaystyle f(x) = \frac{5}{x - 3}$. Write your answer in interval notation.
Find the domain of $f(x) = \sqrt{2x + 6}$.
Find the domain and range of $f(x) = x^2 - 4x + 5$.
Common mistakes · the 4 traps that cost marks
Using square brackets with infinity
Infinity is not a real number, so it can never be "included." Writing $[2, \infty]$ instead of $[2, \infty)$ is incorrect.
✓ Fix: Always use a round bracket next to $\infty$ or $-\infty$.
Confusing domain and range
Domain refers to allowed $x$-values; range refers to possible $y$-values. Some students give the range when asked for the domain, or vice versa.
✓ Fix: Before answering, underline the word "domain" or "range" in the question. Domain = $x$, Range = $y$.
Forgetting to include the endpoint for square roots
The radicand must be $\geq 0$, not just $> 0$. If $\sqrt{x-2}$ is defined when $x = 2$, the interval should start with $[2$, not $(2$.
✓ Fix: Radicand $\geq 0$ means the boundary value IS in the domain. Use a square bracket.
Writing the range of a quadratic without finding the vertex
Guessing the range of $y = x^2 - 6x + 10$ without calculating the vertex often leads to wrong answers.
✓ Fix: Always find the vertex $y$-coordinate for quadratics. That value is the boundary of the range.
Activity 1 — Find the domain
Find the domain of each function below. Write your answer in interval notation and state the restriction you used.
$\displaystyle f(x) = \frac{4}{x + 2}$
$f(x) = \sqrt{3x - 6}$
$f(x) = x^2 - 2x + 7$
$\displaystyle f(x) = \frac{1}{x^2 - 9}$
Odd one out: Which function below does NOT have the domain $(-\infty, \infty)$?
Quick-fire practice · 5 reps +2 XP per reveal
State the domain of $f(x) = \dfrac{5}{x - 3}$ in interval notation.
State the domain of $g(x) = \sqrt{x + 4}$ in interval notation.
Find the range of $f(x) = x^2 - 6x + 10$.
Write $(-3, 5]$ in words.
State the domain and range of $f(x) = |x| + 2$.
Earlier you were asked: Why does a basic calculator refuse to calculate $\sqrt{-1}$? What does this tell you about the inputs the square root function will accept?
A basic calculator works only with real numbers. The square root of a negative number is not a real number — it belongs to the complex number system, which basic calculators are not designed to handle. This tells us that the square root function has a restricted domain: the expression under the square root must be greater than or equal to zero. In other words, only non-negative inputs are allowed. Domain restrictions like this exist to keep functions well-defined and predictable.
Pick your answer, then rate your confidence — that tells the system what to drill next.
Q8. Explain why the value $x = 2$ is not in the domain of $\displaystyle f(x) = \frac{3}{x - 2}$. In your answer, refer to the mathematical reason why this value must be excluded. (2 marks)
Q9. Find the domain and range of $f(x) = x^2 - 6x + 10$. Show your working for finding the vertex. (3 marks)
Q10. A student claims that the domain of every function they have ever seen is $(-\infty, \infty)$. Evaluate this claim with reference to at least two different types of functions that have restricted domains. Include a specific example and the correct domain for each. (4 marks)
📖 Comprehensive answers (click to reveal)
Multiple choice — drill bank
MC answers and feedback are shown inline as you complete each question. Use the retry button to attempt a fresh set.
Activity 1 — Find the domain model answers
1. Domain: $(-\infty, -2) \cup (-2, \infty)$. Restriction: denominator cannot be zero, so $x \neq -2$.
2. Domain: $[2, \infty)$. Restriction: radicand must be $\geq 0$, so $3x - 6 \geq 0 \Rightarrow x \geq 2$.
3. Domain: $(-\infty, \infty)$. No restrictions — it is a polynomial.
4. Domain: $(-\infty, -3) \cup (-3, 3) \cup (3, \infty)$. Restriction: $x^2 - 9 \neq 0 \Rightarrow x \neq \pm 3$.
Short answer model answers
Q8 (2 marks): When $x = 2$, the denominator $x - 2$ equals zero [1]. Division by zero is undefined in the real number system, so $x = 2$ cannot be an input for this function [1].
Q9 (3 marks): Domain: $(-\infty, \infty)$ because it is a polynomial [1]. Vertex: $x = \dfrac{6}{2} = 3$, so $f(3) = 9 - 18 + 10 = 1$ [1]. Range: $[1, \infty)$ because the parabola opens upward [1].
Q10 (4 marks): The student's claim is false [1]. Many functions have restricted domains. For example, $\displaystyle f(x) = \frac{1}{x}$ has domain $(-\infty, 0) \cup (0, \infty)$ because division by zero is undefined [1–2]. Also, $f(x) = \sqrt{x}$ has domain $[0, \infty)$ because the square root of a negative number is not real [1–2]. Polynomials do have domain $(-\infty, \infty)$, but this does not apply to all functions.
Five timed questions on domain and range. Beat the boss to bank a tier — gold (perfect + fast), silver (80%+), or bronze (cleared).
⚔ Enter the arenaClimb platforms, hit checkpoints, and answer domain and range questions. Quick recall, lighter than the boss.
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