H2 Maths Inverse Functions: Let the Domain Choose the Branch

A student preparing for JC2 asks in a public SGExams discussion how to improve H2 Maths when functions are difficult. One useful place to rebuild understanding is the inverse of a quadratic: the algebra is short, but the domain makes the answer meaningful.

The 2026 SEAB H2 Mathematics 9758 syllabus includes inverse functions, conditions for their existence, domain restriction to obtain an inverse, and the relationship between the graphs of a one-to-one function and its inverse. The following examples are original teaching examples, not questions reproduced from an examination.

Find the information the formula leaves out

Take f(x) = (x − 2)² + 1. If every real input is allowed, f(1) = 2 and f(3) = 2. An inverse would have to send the same input, 2, back to both 1 and 3. That would not give a single output for each input. The unrestricted quadratic is therefore not one-to-one and does not have an inverse function on its full real domain.

Sketch the parabola and draw a horizontal line through height 2. Its two intersections show the same problem visually. The horizontal-line test concerns whether different inputs share an output. It is different from the vertical-line test used to check whether a graph represents a function in the first place.

Restrict the domain before rearranging

Now define f(x) = (x − 2)² + 1 for x ≥ 2. Only the right-hand branch remains. The smallest output is 1, at x = 2, and the outputs increase without bound. Thus the domain is [2, ∞) and the range is [1, ∞).

Write y = (x − 2)² + 1. Rearranging gives (x − 2)² = y − 1. Because the original domain says x ≥ 2, we know x − 2 ≥ 0. Hence x = 2 + √(y − 1). Renaming the input gives f⁻¹(x) = 2 + √(x − 1), for x ≥ 1.

Explain why the plus sign is necessary

The original input must be recovered on the permitted branch. If the output of f is 10, its original input on x ≥ 2 is 5. The inverse gives 2 + √9 = 5. Choosing the minus sign would give −1, which is outside the original domain.

If instead a different function uses the same quadratic formula with domain x ≤ 2, its inverse is 2 − √(x − 1), again with input x ≥ 1. The algebraic expression alone does not choose the branch. The original domain does. These are two different functions because their allowed inputs differ.

Swap the domain and range carefully

The inverse accepts the original outputs and returns the original inputs. For the right-hand branch, its domain is therefore [1, ∞), and its range is [2, ∞). Stating only the inverse formula leaves out part of the definition.

Do not confuse f⁻¹(x) with 1/f(x). For example, f(5) = 10, while f⁻¹(10) = 5. The reciprocal at input 10 would instead be 1/f(10) = 1/65. Inverse notation describes undoing the mapping; it does not instruct you to divide by the function value.

Check the inverse in both directions

For x ≥ 2, substituting into f⁻¹(f(x)) gives 2 + √((x − 2)²) = 2 + |x − 2| = x. The domain is what allows the absolute value to become x − 2. For an inverse input t ≥ 1, substituting f⁻¹(t) into f gives (√(t − 1))² + 1 = t.

A graph check should agree: (5, 10) on the original becomes (10, 5) on the inverse. The endpoint (2, 1) becomes (1, 2). Reflection in y = x exchanges the coordinates, including the endpoints and their restrictions.

Practise changing one condition

Try h(x) = (x + 1)² − 4 with x ≥ −1. Its range is [−4, ∞), and h⁻¹(x) = −1 + √(x + 4) for x ≥ −4. Explain the branch before checking a few values. Then change the original domain to x ≤ −1 and explain why the minus branch is needed.

Use the JC H2 Maths collection for further practice. For each inverse question, record the original domain, original range, branch justification and inverse domain. This keeps the reasoning visible when the formula becomes less familiar.