A function has an inverse only if each output comes from exactly one input. A many-to-one function breaks that rule, so on its full domain it has no inverse. Restricting the domain to one side of a turning point repairs it.
Additional Mathematics questions often ask “explain why f has no inverse” and then “find the largest domain of the form x ≥ k for which f⁻¹ exists”. This follows finding an inverse and checking its domain, and it connects to the turning-point work in quadratic structure and discriminants.
How do you decide whether an inverse exists?
- Test with two inputs. Look for different x-values that share an output. For a quadratic, pairs equally far either side of the turning point always match.
- Use the picture. A horizontal line cutting the graph more than once means many-to-one.
- Restrict the domain to one side of the turning point, including the turning point itself.
- Find the inverse on that domain and state its domain as the range of f.
Worked example
The function f is defined by f(x) = x² − 6x + 5 for real x. Show that f has no inverse, then find the inverse when the domain is x ≥ 3.
Step 1, complete the square: f(x) = (x − 3)² − 4. The turning point is at x = 3.
Step 2, counterexample: f(0) = 5 and f(6) = 36 − 36 + 5 = 5. Two different inputs give the same output, so f is many-to-one and has no inverse on all real x.
Step 3, restrict to x ≥ 3: on this interval f only increases, so no output repeats. The range is f(x) ≥ −4.
Step 4, find the inverse: y = (x − 3)² − 4, so (x − 3)² = y + 4. Since x ≥ 3, take the positive root: x − 3 = √(y + 4), so x = 3 + √(y + 4).
Answer: f⁻¹(x) = 3 + √(x + 4), for x ≥ −4.
Check: f(5) = 25 − 30 + 5 = 0, and f⁻¹(0) = 3 + 2 = 5. It returns to 5.
The mistake to watch for
A common slip is to choose a restriction that looks natural but does not sit on one side of the turning point, such as x ≥ 0.
Mistaken answer: “Restrict to x ≥ 0 and the inverse exists.”
On x ≥ 0 we still have f(1) = 1 − 6 + 5 = 0 and f(5) = 0. Two inputs, same output, so it is still many-to-one.
The correction is to start from the turning point. Here it is at x = 3, so the restriction must be x ≥ 3 or x ≤ 3. Either side works for an inverse to exist, but the formula for the inverse changes: for x ≤ 3 it becomes 3 − √(x + 4).
Check yourself
Try these, then open each answer.
1. Explain why f(x) = x² − 4 for real x has no inverse.
Show answer
f(1) = −3 and f(−1) = −3. Two different inputs give the same output, so f is many-to-one. An inverse would have to send −3 back to both 1 and −1, which is impossible for a function.
2. f(x) = (x + 1)² is given for real x. Find the largest domain of the form x ≥ k for which f⁻¹ exists, and find f⁻¹(x).
Show answer
The turning point is at x = −1, so k = −1. On x ≥ −1 the range is f(x) ≥ 0. From y = (x + 1)² we get x + 1 = √y, so f⁻¹(x) = √x − 1, for x ≥ 0. Check: f(1) = 4 and f⁻¹(4) = 2 − 1 = 1.
3. Does g(x) = x³ for all real x have an inverse?
Show answer
Yes. Every output comes from exactly one input, for example 8 only from x = 2. So g is one-to-one and g⁻¹(x) = ∛x for all real x, with no restriction needed.
Where this leads next
With the five lessons behind you, work through the functions and restrictions practice set and note which error type returns. The function composition and inverse explorer shows where a graph fails the horizontal line test, and the quadratic structure explorer helps you find turning points.
If explaining a restriction in words is the hard part, a teacher can rehearse it with you in online one-to-one Additional Mathematics tuition.