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.
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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 confidencethat 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).
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