How to Solve Functions, Composite Functions and Inverse Functions in IGCSE Additional Mathematics (0606)

Functions can feel difficult in Cambridge IGCSE Additional Mathematics (0606) because several ideas are tested at once: notation, domain and range, composition, one-to-one behaviour and inverse functions. The algebra is often manageable; the harder part is knowing what each piece of notation is asking you to do. Under the current 2025–2027 syllabus, candidates are expected to use function notation, find domains and ranges, form composite functions, explain when an inverse does not exist, find inverses of one-to-one functions and understand the graphical relationship between a function and its inverse. The safest way to approach the topic is therefore not to memorise isolated rules. Treat every question as a sequence of input, operation, output and allowable values.

Start with the idea of input and output

A function is a rule that assigns each allowed input exactly one output. If f(x) = 3x - 5, then f(4) means: put 4 into the rule, so f(4) = 3(4) - 5 = 7. This looks simple, but the same idea continues through harder questions. The domain tells you which inputs are allowed. The range is the set of outputs the function can produce. A one-to-one function gives different outputs for different inputs, while a many-to-one function can send two or more inputs to the same output. That distinction matters because an inverse function can only exist as a function when the original function is one-to-one over the stated domain. For example, f(x) = x² over all real numbers is many-to-one because f(2) = 4 and f(-2) = 4. If you tried to reverse it, the output 4 would have to return to both 2 and -2, so the reverse rule would not be a function. Restricting the domain, for example to x ≥ 0, makes the function one-to-one and allows an inverse to be defined.

Read composite notation from the inside out

For the current syllabus, fg(x) means f(g(x)). The function nearest to x acts first. If f(x) = 2x + 3 and g(x) = x², then fg(x) = f(g(x)) = f(x²) = 2x² + 3. In the opposite order, gf(x) = g(f(x)) = g(2x + 3) = (2x + 3)². These are not the same expression, which is why Cambridge explicitly expects candidates to understand that the order of functions matters. A reliable habit is to write the middle line instead of trying to jump directly to the answer. Write fg(x) = f(g(x)), then substitute the complete expression for g(x) into every x in f. Treat the inner function as one object. If g(x) = x² - 1 and f(x) = 3x + 4, then fg(x) = 3(x² - 1) + 4, not 3x² - 1 + 4. The brackets protect the structure of the inner function.

Domain restrictions are part of the composite function

A composite expression is not complete until its domain makes sense. The output of the inner function must be an allowable input for the outer function. Suppose f(x) = √x and g(x) = 2x - 6. Then fg(x) = √(2x - 6). Because the quantity under the square root must be non-negative, 2x - 6 ≥ 0, so x ≥ 3. The algebraic expression and the domain belong together. The same issue appears with logarithms, denominators and restricted domains. If an outer function contains ln x or lg x, the inner output must be positive. If a denominator contains the inner function, values that make it zero must be excluded. The current syllabus specifically notes that the domain may need to be restricted for an inverse or a composite function to exist. When a question gives a stated domain, use it; when the composition creates a new restriction, identify it rather than assuming every real value is allowed.

Find an inverse by reversing the rule

An inverse function undoes the original function. If f(x) = 3x - 5, then start with y = 3x - 5 and solve for x: x = (y + 5) / 3. Replacing y with x gives f⁻¹(x) = (x + 5) / 3. You can check the result by composing the functions: f(f⁻¹(x)) should simplify to x, and f⁻¹(f(x)) should also simplify to x wherever both compositions are defined. Do not confuse f⁻¹(x) with 1 / f(x). The superscript -1 means inverse function, not reciprocal. For f(x) = 3x - 5, the reciprocal would be 1 / (3x - 5), which is a completely different function. This notation mistake can destroy an otherwise correct solution, so write the inverse notation carefully.

When the function is not one-to-one, restrict the domain first

Consider f(x) = (x - 2)² + 1. Over all real x, this quadratic is many-to-one and does not have an inverse function. If the domain is restricted to x ≥ 2, the right-hand branch is one-to-one. Starting with y = (x - 2)² + 1 gives y - 1 = (x - 2)². Because the original domain requires x ≥ 2, take the positive square root: x - 2 = √(y - 1), so f⁻¹(x) = 2 + √(x - 1). The domain of the inverse is x ≥ 1 because the domain of the inverse comes from the range of the original function. This is a good example of why domain information is not decorative. Without the restriction, there are two possible square-root branches and no single inverse function. With the restriction, the sign choice is determined. Whenever you invert a quadratic or another many-to-one expression, ask what domain has been given or what domain would make the function one-to-one before choosing a branch.

Use the graph to understand what an inverse is doing

The graph of y = f⁻¹(x) is the reflection of y = f(x) in the line y = x. A point (a, b) on the original function becomes (b, a) on the inverse. This is the graphical version of swapping inputs and outputs. It also explains why the domain and range exchange roles: the x-values used by the original become y-values on the inverse, while the original outputs become the inputs of the inverse. The reflection idea is also a quick check. If an inverse graph has been sketched on the same axes but does not look like a mirror image across y = x, something is probably wrong. For a function that is not one-to-one, the original graph fails the horizontal-line test, which is another way of seeing why an unrestricted inverse would not be a function.

Do not mix up f²(x), [f(x)]² and f⁻¹(x)

The notation in this topic is compact, so small reading errors matter. In the current 0606 syllabus, f²(x) means f(f(x)), a function composed with itself. It does not mean [f(x)]². And f⁻¹(x) means the inverse function, not a reciprocal. If f(x) = x + 2, then f²(x) = f(x + 2) = x + 4, while [f(x)]² = (x + 2)² and f⁻¹(x) = x - 2. Three similar-looking expressions represent three different operations.

A quick decision table for function questions

Worked practice example

Let f(x) = 2x - 1 and g(x) = 1 / (x + 3). First, fg(x) = f(g(x)) = 2 / (x + 3) - 1, with x ≠ -3. In the other order, gf(x) = g(f(x)) = 1 / ((2x - 1) + 3) = 1 / (2x + 2), with x ≠ -1. The two domain restrictions are different because the inner function is different in each composition. Now find f⁻¹(x). Since y = 2x - 1, rearranging gives x = (y + 1) / 2, so f⁻¹(x) = (x + 1) / 2. Checking f(f⁻¹(x)) gives 2[(x + 1) / 2] - 1 = x. The check confirms that the inverse undoes f. Notice how the three parts of the question use the same central idea: track what enters the function, what leaves it, and which values are permitted.

Common mistakes to avoid

Most function errors are not caused by advanced algebra. They come from losing track of the structure. Common problems include doing composite functions in the wrong order, forgetting brackets around the inner function, treating f⁻¹ as a reciprocal, ignoring a domain restriction, giving an inverse for a function that is not one-to-one, or taking the wrong square-root branch after restricting a quadratic. Another frequent issue is finding a correct algebraic expression but not stating the excluded or restricted values that make the function valid. Cambridge mark schemes allow mathematically valid alternative methods unless a method is specified, so the aim is not to imitate one fixed layout. The aim is to make the mathematical chain easy to inspect. Write the composition explicitly, show the rearrangement used to find the inverse and keep domain conditions beside the relevant expression. Clear structure reduces both algebra mistakes and notation mistakes.

A practical revision routine for functions

Instead of completing ten nearly identical substitution questions, cycle through the full set of skills:

1. Take two simple functions and calculate fg(x), gf(x), f²(x) and several numerical values such as f(3).
2. Add domain restrictions, using roots, logarithms or denominators, and decide which inputs are actually allowed.
3. Practise identifying whether a function is one-to-one before attempting to invert it.
4. Find inverses algebraically and verify them by composition.
5. Sketch a function and its inverse together, checking the reflection in y = x.
6. Finish with mixed exam questions where the notation is not labelled by topic, because recognition is part of the difficulty.

Put it into practice

Choose one pair of functions from your current Additional Mathematics practice and do more than the first thing the question asks. Find both composite orders, state any domain restrictions, decide whether either function has an inverse, and check one inverse by composition. If you can explain why the order matters and why a domain restriction is needed, you understand the structure rather than only the procedure. That is the level of control this topic needs in Cambridge IGCSE Additional Mathematics (0606).