Functions · worked example

Inverse functions: choose the branch using the domain

The algebra often gives two possible expressions. The original domain tells you which one belongs to the inverse. Work through that decision here, then check it on a fresh question.

By Sterling Education Studio · 3 October 2026 · Original practice examples

The key idea

To find an inverse function, first check that the original function is one-to-one on its stated domain. Rearrange y = f(x), choose the branch that fits that domain, then swap the variables. The inverse’s domain is the original range.

One example, with the decision made explicit

Let f(x) = (x − 2)² + 1, for x ≤ 2. Find f⁻¹, including its domain.

Check whether the inverse exists.
On x ≤ 2, the graph is the left-hand part of a parabola. Each output occurs once on this restricted domain, so the function is one-to-one.
Find the original range.
The minimum occurs at x = 2, where f(2) = 1. Values grow without bound as x moves left. The range is y ≥ 1.
Rearrange, keeping both branches for now.
y − 1 = (x − 2)²
x − 2 = ±√(y − 1)
Use the domain before choosing the sign.
Because x ≤ 2, we know x − 2 ≤ 0. Choose the negative branch: x = 2 − √(y − 1).
Swap the variable names and state the domain.
f⁻¹(x) = 2 − √(x − 1), for x ≥ 1
The inverse’s range is x ≤ 2, matching the original domain.

Why the positive square root fails here

The expression 2 + √(x − 1) would return values above 2 for inputs above 1. Those values lie outside the original domain. Rearranging the equation correctly is only part of the work: the inverse must send each output back to an allowed original input.

For example, f(0) = 5. The correct inverse returns f⁻¹(5) = 2 − √4 = 0. The positive branch returns 4, which is not in the stated domain.

A useful self-check

Choose a simple allowed input, find its output, and apply your inverse. You should return to the same input. One successful check can expose a wrong branch, but it is not a proof that the function is one-to-one everywhere.

Keep the domain and range attached to the formula

For this example, the original function maps (−∞, 2] to [1, ∞), and its inverse maps [1, ∞) back to (−∞, 2]. Writing these sets alongside the formulas helps prevent a correct expression being paired with an incorrect domain.

For a composite function g(f(x)), check that every output produced by f on the proposed domain is an allowed input to g. If it is not, the composite is not defined for all of that proposed domain; a suitable restriction may be needed. This is the same input-output habit used when checking an inverse.

Try the decision on a different function

Let g(x) = 3 − (x + 1)², for x ≥ −1. Find g⁻¹ and its domain. Before expanding anything, write down the sign of x + 1.

Show the worked answer

Since x ≥ −1, we have x + 1 ≥ 0. From y = 3 − (x + 1)², obtain (x + 1)² = 3 − y and choose x + 1 = √(3 − y).

g⁻¹(x) = −1 + √(3 − x), for x ≤ 3

The original range is y ≤ 3, so this becomes the inverse’s domain. Its range is x ≥ −1. As a check, g(1) = −1 and g⁻¹(−1) = −1 + √4 = 1.

Four questions to ask before finishing

Is the function one-to-one on the stated domain?
What is its full range, including any endpoints?
Does the chosen branch return only allowed original inputs?
Have I stated the inverse’s domain and checked one simple pair?

If the branch choice went wrong, record the specific missed decision: “I did not use the sign forced by the original domain.” On the next question, make that sign check before taking a square root.

Keep learning

These are original teaching examples, not past-paper questions or an official mark scheme. Use the 2026 H2 Mathematics 9758 syllabus guide to check the wider topic requirements for that examination year.

Planning what comes next? Read the promos recovery plan or JC2 preparation guide.