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.
Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
Why do you think a basic calculator refuses to calculate the square root of a negative number? And what does this tell you about the inputs that the square root function will accept?
Type your initial response below — you will revisit this at the end of the lesson.
Write your initial response in your book. You will revisit it at the end of the lesson.
Wrong: f⁻¹(x) means 1/f(x).
Right: f⁻¹(x) denotes the inverse function, not the reciprocal. The reciprocal would be written [f(x)]⁻¹ or 1/f(x).
📚 Core Content
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:
Most domain problems in Year 11 involve one of three restrictions:
The range depends on the shape of the function. For many basic functions, we can determine the range by inspection:
🧮 Worked Examples
🧪 Activities
1 $\displaystyle f(x) = \frac{4}{x + 2}$
Type your answer:
Answer in your workbook.
2 $f(x) = \sqrt{3x - 6}$
Type your answer:
Answer in your workbook.
3 $f(x) = x^2 - 2x + 7$
Type your answer:
Answer in your workbook.
4 $\displaystyle f(x) = \frac{1}{x^2 - 9}$
Type your answer:
Answer in your workbook.
1 $f(x) = (x - 2)^2 + 3$
Type your answer:
Answer in your workbook.
2 $f(x) = |x| + 1$
Type your answer:
Answer in your workbook.
3 $\displaystyle f(x) = \sqrt{x + 4} - 1$
Type your answer:
Answer in your workbook.
Earlier you were asked: Why do you think a basic calculator refuses to calculate the square root of a negative number? And what does this tell you about the inputs that 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.
Now revisit your initial response. What did you get right? What has changed in your thinking?
Look back at your initial response in your book. Annotate it with what you now understand differently.
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
5 random questions from a replayable lesson bank — feedback shown immediately
✍️ Short Answer
8. 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
Type your answer below:
Answer in your workbook.
9. Find the domain and range of $f(x) = x^2 - 6x + 10$. Show your working for finding the vertex. 3 MARKS
Type your answer below:
Answer in your workbook.
10. 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
Type your answer below:
Answer in your workbook.
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: $x^2 - 9 \neq 0 \Rightarrow x \neq \pm 3$.
1. Domain: $(-\infty, \infty)$; Range: $[3, \infty)$. Vertex at $(2, 3)$, opens upward.
2. Domain: $(-\infty, \infty)$; Range: $[1, \infty)$. Minimum value of $|x|$ is 0, so minimum of $f(x)$ is $0 + 1 = 1$.
3. Domain: $[-4, \infty)$ (since $x + 4 \geq 0$); Range: $[-1, \infty)$ (since $\sqrt{x+4} \geq 0$, so $f(x) \geq -1$).
1. B — $x - 2 \geq 0 \Rightarrow x \geq 2$, so $[2, \infty)$.
2. B — $x - 4 = 0 \Rightarrow x = 4$, which is excluded.
3. C — Round bracket at $-3$, square bracket at $5$.
4. C — Minimum value is $0 + 3 = 3$, so $[3, \infty)$.
5. C — Polynomials are defined for all real $x$.
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: $x = \frac{6}{2} = 3$, 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.
Defend your ship by blasting the correct answers for Domain & Range. Scores count toward the Asteroid Blaster leaderboard.
Play Asteroid Blaster →Domain & Range
Tick when you've finished all activities and checked your answers.