Define a function and find its domain and range
Distinguish permitted inputs from actual outputs, then test whether different inputs can share an output.
Pure Mathematics 1 (Paper 1) · AS & A Level Mathematics 9709, 2026–2027. Written practice is self-checked; show your reasoning and domain restrictions.
Before you begin
- Substitute values into a rule and complete a quadratic square.
- Understand non-negative square roots and non-zero denominators.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which input is excluded by the rule 1/(x − 1)?
What you will learn
- Explain function, domain, range and one-one using a rule and its stated inputs.
- Find ranges for simple quadratic and reciprocal functions, respecting endpoints.
A rule needs its input set
A function assigns exactly one output to each input in its domain. The domain is the set of permitted inputs; the range is the set of outputs actually obtained from those inputs. Writing f: x ↦ x² describes the rule, while a domain such as x ≥ 0 supplies the permitted inputs. The same rule with a different domain can have a different range or different inverse behaviour.
Different inputs may share an output and still form a function: x² sends both 2 and −2 to 4. A one-one function has different outputs for every pair of different permitted inputs. Graphically, a vertical line meets a function graph at most once; a horizontal line meets a one-one function graph at most once. Test one-one on the stated domain rather than on an imagined wider graph.
Find actual outputs, not just a convenient bound
For f(x) = (x − 2)² + 1 on x ≥ 2, the square takes every value from zero upwards, so the range is y ≥ 1. The endpoint 1 is attained at x = 2. On all real x the same range holds, but the function is no longer one-one: inputs equally spaced on opposite sides of 2 share an output.
For g(x) = 3/(x + 2) on x > −2, the denominator takes every positive value, so the range is y > 0. Zero is approached for large x but never attained at any finite input. For a restricted interval, evaluate permitted endpoints and any turning point inside it. An open endpoint or an approached limiting value may be excluded from the range.
Pause and explain
Why is x ↦ x² on all real x a function but not one-one?
Worked example
Find the range of g(x) = 3/(x + 2) for x > −2 and decide whether g is one-one.
Show the worked solution
- The denominator x + 2 is positive and can take any positive value.
- Dividing 3 by those values produces every positive output, but never zero.
- If two outputs are equal, 3/(x₁ + 2) = 3/(x₂ + 2) gives x₁ = x₂, so g is one-one.
Answer Range y > 0; g is one-one on x > −2.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
For f(x) = 4 − x² on all real x, find the range and decide whether f is one-one.
Give me a hint
Use x² ≥ 0 and compare inputs on opposite sides of zero.
Compare my reasoning
- Since x² ≥ 0, f(x) ≤ 4, with equality at x = 0.
- Every y ≤ 4 is attained by a real x satisfying x² = 4 − y.
- Inputs 1 and −1 both give 3, so the function is not one-one.
Range y ≤ 4; f is not one-one.
Look for these in your work
- Justify that every output in the proposed range is attainable.
- Give two distinct inputs with the same output.
Find the range of h(x) = 1/(x − 1) for x ≥ 3.
Give me a hint
The denominator starts at 2 and can grow without bound.
Compare my reasoning
- For x ≥ 3, x − 1 ≥ 2, so 0 < h(x) ≤ 1/2.
- The largest output 1/2 is attained at x = 3.
- Zero is never attained, while every output between zero and 1/2 has an allowed reciprocal denominator.
0 < y ≤ 1/2.
Look for these in your work
- Include the attained maximum.
- Exclude the limiting output zero.
A supplied sensor model is S(t) = (t − 3)² + 2 for 0 ≤ t ≤ 5. Find its range and decide whether different allowed times can give the same reading.
Give me a hint
Check the vertex and both interval endpoints.
Compare my reasoning
- The vertex t = 3 is permitted and gives the minimum S = 2.
- The endpoint readings are S(0) = 11 and S(5) = 6; the continuous quadratic attains all readings between 2 and 11 on this interval.
- Times 1 and 5 both give 6, so the model is not one-one over the full stated interval.
2 ≤ S ≤ 11; readings can repeat, for example S(1) = S(5) = 6.
Look for these in your work
- Check interior turning points as well as endpoints.
- Use permitted times when demonstrating a repeated reading.
Common mistakes
- Domain describes inputs; range describes actual outputs.
- A limiting value is not automatically an included endpoint.
- A function can be many-to-one; one-one is an additional condition.
What can you explain now?
Find the range of f(x) = (x + 1)² for x ≤ −1. Is f one-one on this domain?
Compare with the explanation
Range y ≥ 0; f is one-one on x ≤ −1.
The allowed inputs use only the left branch of the quadratic, with output zero at −1 and all positive outputs attained once.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Explain domain, range and one-one to someone using one quadratic and one reciprocal example. Explain why a value approached forever may still be outside the range.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.