Inverse functions: how to find them and check your answer

Inverse functions come up across the Australian senior secondary curriculum right through to first-year university calculus, yet many students find the topic surprisingly slippery. The idea is straightforward enough in concept: an inverse function undoes what the original function does, taking an output back to the input that produced it. The challenge is that not every function has an inverse, and the algebraic steps to find one are easy to muddle when you are working under timed conditions.

Students sitting the VCE in Melbourne or the HSC in Sydney will recognise inverse functions from past papers where examiners love to combine them with logarithms, quadratics, or domain restrictions. The skill also shows up in specialist mathematics, where inverse trigonometric functions need their own care.

For learners who prefer a more visual approach, working through the relationship between a graph and its reflection across the line y = x can be a great way to build intuition before tackling symbolic manipulation. Once the graphical picture clicks, the algebra usually follows.

The good news is that inverse functions follow a small, reliable set of rules, and with a bit of practice you can both find them and verify them with confidence.

What makes a function invertible

The first question to ask before you try to find an inverse is whether one actually exists. A function is only invertible if it is one-to-one, meaning each output comes from exactly one input. If two different x-values produce the same y-value, no inverse function can reverse that relationship cleanly.

You can check this graphically using the horizontal line test: if any horizontal line crosses the graph more than once, the function fails the test and has no inverse over its full domain. Many functions can be salvaged, however, by restricting their domain to a portion where they are one-to-one.

Take f(x) = x². As written, this is not one-to-one because both x = 3 and x = -3 give f(x) = 9. But if we restrict the domain to x ≥ 0, then f becomes invertible and we can write f⁻¹(x) = √x. This trick of restricting the domain shows up constantly in textbooks, particularly when introducing inverse trigonometric functions like arcsin and arccos, which only make sense on a carefully chosen interval.

The horizontal line test in practice

Drawing quick sketch graphs is often faster than trying to reason about injectivity symbolically, especially for less familiar functions. Sketch f(x) = 2x + 1 in your head and it is obviously one-to-one, since any straight line of positive gradient satisfies the horizontal line test without effort.

More interesting cases include f(x) = x³ - x, which has a local maximum and minimum and therefore fails the horizontal line test over its full domain. But restrict x to (-∞, -0.5] or [-0.5, 0.5] or [0.5, ∞) and the function becomes invertible in pieces. Working through examples like this is a staple of senior mathematics revision.

If you want a structured set of practice items to work through, a well-organised study checklist can help you tick off each variation of the test before exams.

Algebraic steps for finding an inverse

Once you know an inverse exists, the algebraic procedure is short enough to memorise. Write the function as y = f(x), then swap x and y to get x = f(y). Solve the resulting equation for y, and what you end up with is f⁻¹(x).

A worked example: let f(x) = (3x + 2) / (x - 1), with x ≠ 1. Start by swapping to get x = (3y + 2) / (y - 1). Multiply both sides by (y - 1) to obtain x(y - 1) = 3y + 2, which expands to xy - x = 3y + 2. Group the y-terms on one side: xy - 3y = x + 2, so y(x - 3) = x + 2, giving y = (x + 2) / (x - 3). So f⁻¹(x) = (x + 2) / (x - 3), valid for x ≠ 3.

The domain and range of the original function become the range and domain of the inverse, respectively. In this example, the excluded x = 1 of f becomes an excluded y = 1 of f, and the vertical asymptote of the inverse at x = 3 corresponds to the horizontal asymptote of the original.

Domain and range considerations

Paying attention to domain and range statements is what separates a complete answer from a careless one in ATAR-style marking. If f has domain [0, 10] and range [0, 100], the inverse has domain [0, 100] and range [0, 10].

These sets flip every time, and examiners in Queensland's QCE system regularly test this detail explicitly. Getting the swaps right is worth easy marks.

Functions with restricted domains, such as the inverse trigonometric functions defined only on [-π/2, π/2], also need explicit domain notes written into your final answer.

Verifying the inverse by composition

The defining property of an inverse is that composing it with the original function returns the input unchanged. In symbols, f(f⁻¹(x)) = x and f⁻¹(f(x)) = x for every x in the relevant domain.

To verify, take your candidate inverse and substitute it back into f. With f(x) = 3x + 1, the inverse is f⁻¹(x) = (x - 1) / 3. Now compute f(f⁻¹(x)) = 3 · ((x - 1) / 3) + 1 = (x - 1) + 1 = x. Check.

The other direction gives f⁻¹(f(x)) = (3x + 1 - 1) / 3 = 3x / 3 = x. Both compositions return x, so the inverse is correct. This is the most reliable verification method available, and it catches sign errors, swapped operations, and forgotten domain restrictions that are easy to overlook when working quickly through NAPLAN-style problems.

Common mistakes worth flagging

A frequent error is forgetting that y = x² does not have an inverse as written, and writing f⁻¹(x) = √x without mentioning the ± sign or domain restriction. Another is mixing up the order when swapping x and y, then solving and getting something close to right but with the original variables left in place.

Students also sometimes forget to flip domain and range statements when describing the inverse, which costs marks even when the algebra itself is correct.

Finally, watch out for functions that look invertible but contain subtle symmetries, such as f(x) = x³ - 3x, which has two turning points and fails the horizontal line test. Restricting the domain is the only way forward.

Working with inverse function tools

For students tackling tougher functions, online resources such as curryplum.com offer step-by-step solutions and calculators that walk through each manipulation. These are useful for checking your own algebra rather than skipping the work.

If you are revising for the HSC, VCE, or QCE, building a personal routine of sketching first, solving algebraically, then verifying by composition tends to produce reliable answers under exam pressure.

Pair this with targeted past-paper questions, and inverse functions shift from being a stumbling block to a dependable source of marks.

When you next meet a problem asking you to find f⁻¹(x), pause for a moment before diving into algebra. Sketch the graph, check the horizontal line test, restrict the domain if needed, then swap, solve, and verify. That five-second pause is what separates a clean solution from a messy one, and it is the single habit that lifts most students from guesswork to genuine understanding.