The range of a function on a restricted domain is the set of output values produced by the allowed inputs only. Change the interval and you can change the range, even though the rule for f(x) stays the same.
Additional Mathematics questions ask for this directly (“find the range of f”), and they set it up for forming composite functions and finding inverses, where the range of one function becomes the domain of another.
How do you read the range from a domain?
Think of the graph between two vertical lines at the ends of the domain. The range is how low and how high the curve goes between them.
- Identify the domain and mark its endpoints on a rough sketch.
- Find any turning point of the curve inside the interval. For a quadratic, complete the square to find it.
- Evaluate f at both endpoints.
- Compare all the values. The smallest is the lower end of the range and the largest is the upper end.
- Write it with inequalities on f(x).
For a straight line there is no turning point, so steps 3 and 4 are enough.
Worked example
The function f is defined by f(x) = x² − 4x + 3 for 0 ≤ x ≤ 5. Find the range of f.
Step 1, complete the square: x² − 4x + 3 = (x − 2)² − 1.
Step 2, turning point: the minimum is −1 at x = 2. The value x = 2 lies inside 0 ≤ x ≤ 5, so it counts.
Step 3, endpoints: f(0) = 3. f(5) = 25 − 20 + 3 = 8. Check with the completed square: (5 − 2)² − 1 = 9 − 1 = 8.
Step 4, compare: the values are −1, 3 and 8. The smallest is −1 and the largest is 8.
Answer: the range is −1 ≤ f(x) ≤ 8.
The mistake to watch for
A common slip is to evaluate only the two endpoints and treat them as the smallest and largest outputs.
Mistaken answer: f(0) = 3 and f(5) = 8, so 3 ≤ f(x) ≤ 8.
The curve dips to −1 at x = 2, which lies between the endpoints. The lowest output is missed.
The correction is to ask whether the turning point sits inside the domain. Here it does, so it decides the minimum. If the domain were 3 ≤ x ≤ 5, the turning point would lie outside it, and the endpoints alone would give the range: f(3) = 0 and f(5) = 8, so 0 ≤ f(x) ≤ 8.
Check yourself
Find the range in each case, then open the answer.
1. f(x) = 2x − 1 for 1 ≤ x ≤ 4
Show answer
It is a straight line that increases. f(1) = 1 and f(4) = 7. Range: 1 ≤ f(x) ≤ 7.
2. g(x) = x² + 2 for −3 ≤ x ≤ 1
Show answer
The minimum is 2 at x = 0, which lies inside the interval. g(−3) = 9 + 2 = 11 and g(1) = 3. The largest is 11. Range: 2 ≤ g(x) ≤ 11.
3. h(x) = 6 − (x − 1)² for 0 ≤ x ≤ 4
Show answer
This curve opens downwards, so the turning point at x = 1 is a maximum of 6, and it lies inside the interval. h(0) = 5 and h(4) = 6 − 9 = −3. Range: −3 ≤ h(x) ≤ 6.
Where this leads next
The next step is forming a composite function in the correct order, which needs you to know what each function can output. If completing the square is slow, revise it in quadratic structure and discriminants. You can also test a rule over an interval in the function composition and inverse explorer.
If you understand ranges but hesitate over which value to read off, a teacher can look at your sketches with you in online one-to-one Additional Mathematics tuition.