Functions, Composite and Inverse
A function is a rule that turns an input into an output. The mathematics is straightforward substitution — the marks are lost in notation: what fg(x) means, and which way round to work.
1. Function notation
f(x) = 2x + 5 means: take the input, double it, add 5.
f(3) means substitute x = 3: f(3) = 2(3) + 5 = 11
Two different instructions that look similar:
| Question | Meaning | Method |
|---|---|---|
| Find f(3) | x is given | Substitute |
| Solve f(x) = 11 | the output is given | Form an equation and solve |
For f(x) = 11: 2x + 5 = 11 → 2x = 6 → x = 3
If a number is given for x, substitute. If the whole function equals a number, form an equation. Confusing these two is the most common early error.
Negative inputs need brackets:
For f(x) = x² − 8 with x = −1: f(−1) = (−1)² − 8 = 1 − 8 = −7
Always bracket a negative input. Writing −1² gives −1 on most calculators, which is why an answer like “97” can appear where −7 was correct. The bracket is not optional.
f(a) means substitute a — it is not f × a. For f(x) = 2x + 5, f(a) = 2a + 5, not 20a + 5 or 2 × f × a. Reading the notation as multiplication was a specific recorded error.
2. Composite functions
fg(x) means: do g FIRST, then apply f to the result.
Work from the inside out — the function closest to the x acts first.
Example: f(x) = 2x + 5 and g(x) = x²
fg(3):
- g(3) = 9
- f(9) = 2(9) + 5 = 23
gf(3):
- f(3) = 11
- g(11) = 121
fg(x) and gf(x) are different. 23 and 121 above. Order matters, and questions frequently ask for both to test exactly this.
Finding the composite as an expression:
fg(x) = f(x²) = 2(x²) + 5 = 2x² + 5
gf(x) = g(2x + 5) = (2x + 5)² = 4x² + 20x + 25
Substitute the whole inner function into the outer one, in brackets. (2x + 5)² must then be expanded properly — as x² + … , not as 4x² + 25. This expansion error came up repeatedly.
fg(x) is not f(x) × g(x). Composite functions are applied in sequence, not multiplied. Confusing composition with multiplication was a documented error.
3. Inverse functions
f⁻¹(x) reverses f — it takes the output back to the input.
Method: swap and solve.
Find the inverse of f(x) = 2x + 5:
- Write y = 2x + 5
- Swap x and y: x = 2y + 5
- Solve for y: x − 5 = 2y → y = (x − 5)/2
- f⁻¹(x) = (x − 5)/2
Check: f(3) = 11, and f⁻¹(11) = (11 − 5)/2 = 3 ✓
Always check with a number. Put a value through f, then through f⁻¹, and you should return to where you started. It takes seconds and confirms the whole answer.
f⁻¹(x) does not mean 1/f(x). The −1 is notation for “inverse”, not a power.
Applying a function and then its inverse gets you back to x: ff⁻¹(x) = x.
A harder example: f(x) = (x + 1)/(x − 3)
- y = (x + 1)/(x − 3)
- Swap: x = (y + 1)/(y − 3)
- Multiply out: x(y − 3) = y + 1 → xy − 3x = y + 1
- Collect y terms: xy − y = 3x + 1 → y(x − 1) = 3x + 1
- f⁻¹(x) = (3x + 1)/(x − 1)
Get all the y terms on one side and factorise. That is the step that makes these solvable.
4. Domain and range
The domain is the set of inputs (x-values) the function accepts. The range is the set of outputs (y-values) it produces.
Domain = input = x. Range = output = y. Swapping these was one of the most frequent errors here. A memory hook: d comes before r in the alphabet, just as x comes before y.
Finding the range — think about what values the expression can actually take:
- f(x) = x² → a square is never negative → range: f(x) ≥ 0
- f(x) = x² + 2 → the smallest x² can be is 0 → range: f(x) ≥ 2
- f(x) = 2x + 5 (linear) → range: all real numbers
x² + 2 can never be less than 2. Students assumed it could; testing x = 0, a positive value and a negative value quickly shows the minimum is at x = 0.
Test positive, negative and zero inputs to see how a function behaves before stating the range — this was the practical advice given in lessons.
Domain restrictions arise in two situations:
A denominator cannot be zero. For f(x) = 1/(x − 3), the domain excludes x = 3. You cannot square-root a negative. For f(x) = √(x − 2), the domain is x ≥ 2.
5. Scope note
Some lessons used logarithms to invert exponential functions, e.g. solving 2ˣ = 1/32.
Logarithms are not on the 0580 syllabus. A question like 2ˣ = 1/32 is solved by writing both sides with the same base: 1/32 = 2⁻⁵, so x = −5. If you are sitting 0580 only, you do not need logs — check your own syllabus document.
6. Mistakes that cost marks
Substituting when you should solve, or the reverse.
Not bracketing negative inputs.
Reading f(a) as multiplication.
Doing fg(x) in the wrong order — g acts first.
Treating fg(x) as f(x) × g(x).
Expanding a squared bracket wrongly in a composite.
Reading f⁻¹(x) as 1/f(x).
Forgetting to swap x and y when finding an inverse.
Not collecting y terms in a harder inverse.
Confusing domain with range.
Missing a domain restriction from a denominator or square root.
Assuming a range is all real numbers without testing.
Frequently asked questions
What does f(3) mean? Substitute x = 3 into the function.
What is the difference between finding f(3) and solving f(x) = 3? For f(3) you substitute. For f(x) = 3 you form an equation and solve for x.
What does fg(x) mean? Apply g first, then f to the result.
Is fg(x) the same as gf(x)? No — the order changes the answer.
Is fg(x) the same as f(x) × g(x)? No — composition applies the functions in sequence.
What does f⁻¹(x) mean? The inverse function, which reverses f. It does not mean 1/f(x).
How do I find an inverse function? Write y = f(x), swap x and y, then solve for y.
How do I check my inverse? Put a number through f, then through f⁻¹ — you should get back the original number.
What is the domain? The set of inputs (x-values).
What is the range? The set of outputs (y-values).
When is the domain restricted? When a denominator would be zero, or a square root would be negative.
Quick revision checklist
- I can evaluate f(x) for a given value, bracketing negatives
- I can solve f(x) = k by forming an equation
- I know f(a) means substitution, not multiplication
- I know fg(x) means g first
- I can find fg(x) and gf(x) as expressions
- I expand squared brackets correctly inside composites
- I know composition is not multiplication
- I can find an inverse by swapping x and y
- I can handle inverses with x in the denominator
- I check my inverse with a number
- I know f⁻¹(x) ≠ 1/f(x)
- I know domain = input and range = output
- I can find a range by testing values
- I can spot domain restrictions
- I know logarithms are not required on 0580
These notes cover functions, composite and inverse, in the Cambridge IGCSE Mathematics (0580) syllabus, and are written for Grade 9–11 / Year 10–11 students. They are based on teaching patterns observed across a large set of one-to-one IGCSE Maths lessons, with particular attention to the errors students make most often and the wording examiners reward. Always check the current syllabus and formula list for your own exam series.
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