Functions in IGCSE Maths
Function notation, domain and range, composite functions, and inverse functions.
0580 Extended · E2.13Function notation
A function is a rule that takes an input and produces exactly one output. In IGCSE Maths you'll see two ways of writing the same function:
\[ \begin{aligned} f(x) = 2x + 3 \quad &\text{or} \\[6pt] f: x \mapsto 2x + 3 \end{aligned} \]Both mean the same thing: take \(x\), double it, and add 3. The notation \(f(x)\) is used most often, but the mapping notation \(f: x \mapsto\) does appear on exam papers.
To find \(f(4)\), replace every \(x\) in the function with 4:
\[ f(4) = 2(4) + 3 = 11 \]This works with expressions too. If you're asked to find \(f(2a)\), substitute \(2a\) in place of \(x\):
\[ f(2a) = 2(2a) + 3 = 4a + 3 \]\(g(x) = x^2 - 3x + 1\).
(a) Find \(g(5)\).
(b) Find \(g(-2)\).
Part (a)
\[ \begin{aligned} g(5) &= (5)^2 - 3(5) + 1 \\[6pt] &= 25 - 15 + 1 \\[6pt] &= 11 \end{aligned} \]Part (b)
\[ \begin{aligned} g(-2) &= (-2)^2 - 3(-2) + 1 \\[6pt] &= 4 + 6 + 1 \\[6pt] &= 11 \end{aligned} \]Domain & range New from 2025
Domain and range were added to the IGCSE 0580 Extended syllabus from 2025. This content doesn't appear on UK GCSE papers, so you won't find it in most GCSE revision guides.
The domain is the set of all allowed input values (\(x\)-values). Some functions have restrictions:
\(f(x) = \dfrac{1}{x-2}\) is undefined when \(x = 2\), so the domain is all real numbers except 2.
\(g(x) = \sqrt{x}\) requires \(x \geqslant 0\).
The range is the set of all possible output values (\(y\)-values or \(f(x)\)-values).
For \(f(x) = x^2\), the output is always \(\geqslant 0\), so the range is \(f(x) \geqslant 0\).
For \(g(x) = 5 - x\), any real number can be an output, so the range is all real numbers.
\(f(x) = \dfrac{3}{x + 1}\). State the domain and range of \(f\).
The function is undefined when the denominator is zero: \(x + 1 = 0 \Rightarrow x = -1\).
Domain: all real numbers, \(x \neq -1\).
The output \(\frac{3}{x+1}\) can take any value except 0 (since the numerator is always 3, the fraction can never equal zero).
Range: all real numbers, \(f(x) \neq 0\).
Composite functions Extended only
A composite function applies one function, then feeds the result into another. The notation \(fg(x)\) means apply \(g\) first, then apply \(f\) to the result:
\[ fg(x) = f\big(g(x)\big) \]Read it right to left: \(g\) first, then \(f\). The function closest to \(x\) acts first.
The notation \(f^2(x)\) means apply \(f\) twice — it does not mean squaring the function:
\[ f^2(x) = f\big(f(x)\big) \]If \(f(x) = 2x + 1\), then:
\[ \begin{aligned} f^2(x) &= f(2x+1) \\[6pt] &= 2(2x+1) + 1 \\[6pt] &= 4x + 3 \end{aligned} \]\(f(x) = 2x + 3\) and \(g(x) = x^2\).
(a) Find \(fg(x)\).
(b) Find \(gf(x)\).
Part (a)
\(fg(x) = f\big(g(x)\big)\). Apply \(g\) first, then \(f\):
\[ \begin{aligned} fg(x) &= f(x^2) \\[6pt] &= 2(x^2) + 3 \\[6pt] &= 2x^2 + 3 \end{aligned} \]Part (b)
\(gf(x) = g\big(f(x)\big)\). Apply \(f\) first, then \(g\):
\[ \begin{aligned} gf(x) &= g(2x+3) \\[6pt] &= (2x+3)^2 \end{aligned} \]\(fg(x) \neq gf(x)\) in general. The order matters. In the example above, \(fg(x) = 2x^2 + 3\) but \(gf(x) = (2x+3)^2 = 4x^2 + 12x + 9\). Very different results.
Inverse functions Extended only
The inverse function \(f^{-1}(x)\) reverses what \(f\) does. If \(f\) takes an input and produces an output, \(f^{-1}\) takes that output and returns the original input:
\[ \begin{aligned} f\big(f^{-1}(x)\big) &= x \\[6pt] f^{-1}\big(f(x)\big) &= x \end{aligned} \]1. Write \(y = f(x)\).
2. Swap \(x\) and \(y\).
3. Rearrange to make \(y\) the subject.
4. Replace \(y\) with \(f^{-1}(x)\).
\(f(x) = 5x - 7\).
Find \(f^{-1}(x)\).
Write \(y = 5x - 7\). Swap: \(x = 5y - 7\). Rearrange:
\[ x + 7 = 5y \quad \Rightarrow \quad y = \frac{x + 7}{5} \] \[ \boxed{f^{-1}(x) = \frac{x + 7}{5}} \]\(f(x) = \dfrac{3}{x + 2}\), \(x \neq -2\).
Find \(f^{-1}(x)\).
Write \(y = \dfrac{3}{x+2}\). Swap: \(x = \dfrac{3}{y+2}\). Rearrange:
\[ \begin{aligned} x(y + 2) &= 3 \\[6pt] xy + 2x &= 3 \\[6pt] xy &= 3 - 2x \\[6pt] y &= \frac{3 - 2x}{x} \end{aligned} \] \[ \boxed{f^{-1}(x) = \frac{3 - 2x}{x}, \quad x \neq 0} \]A quick way to verify an inverse: compute \(ff^{-1}(x)\) and check you get \(x\). If it simplifies to \(x\), you're correct. This takes 30 seconds and can save you marks.
Exam-style question
\(f(x) = 3x + 1\) and \(g(x) = \dfrac{x}{x - 2}\), \(x \neq 2\).
(a) Find \(f(-4)\). [1]
(b) Find \(gf(x)\). Simplify your answer. [2]
(c) Find \(f^{-1}(x)\). [2]
(d) Solve \(fg(x) = 10\). [3]
(e) State the value of \(x\) that is not in the domain of \(g(x)\). [1]
Part (a)
\[ \begin{aligned} f(-4) &= 3(-4) + 1 \\[6pt] &= -12 + 1 \\[6pt] &= -11 \end{aligned} \]Part (b)
\(gf(x) = g\big(f(x)\big) = g(3x+1)\). Replace \(x\) with \(3x+1\) in \(g\):
\[ \begin{aligned} gf(x) &= \frac{3x + 1}{(3x + 1) - 2} \\[6pt] &= \frac{3x + 1}{3x - 1} \end{aligned} \]Part (c)
Write \(y = 3x + 1\). Swap: \(x = 3y + 1\). Rearrange:
\[ \begin{aligned} x - 1 &= 3y \\[6pt] f^{-1}(x) &= \frac{x - 1}{3} \end{aligned} \]Part (d)
\(fg(x) = f\big(g(x)\big)\). Substituting:
\[ \begin{aligned} fg(x) &= 3\!\left(\frac{x}{x-2}\right) + 1 \\[6pt] &= \frac{3x}{x-2} + 1 \end{aligned} \]Set equal to 10:
\[ \begin{aligned} \frac{3x}{x - 2} + 1 &= 10 \\[6pt] \frac{3x}{x-2} &= 9 \\[6pt] 3x &= 9(x - 2) \\[6pt] 3x &= 9x - 18 \\[6pt] -6x &= -18 \\[6pt] x &= 3 \end{aligned} \]Part (e)
\(g(x) = \dfrac{x}{x-2}\) is undefined when \(x - 2 = 0\), so \(x = 2\) is not in the domain.
This is a typical 9-mark functions question. It tests evaluation (a), composite functions in both orders (b and d), inverse functions (c), and domain (e). Part (d) is the hardest — it combines forming a composite, simplifying a fraction, and solving an equation. Practise that chain of steps until it feels routine.
Common mistakes
\(fg(x)\) means apply \(g\) first, then \(f\). The function nearest to \(x\) goes first. Getting this backwards is the single most common error on functions questions.
\(f^2(x)\) means \(f(f(x))\) — apply \(f\) twice. It does not mean square the output. If \(f(x) = x + 3\), then \(f^2(x) = f(x+3) = x + 6\), not \((x+3)^2\).
When finding an inverse, you must swap \(x\) and \(y\) before rearranging. Simply rearranging \(y = 3x + 1\) to get \(x = \frac{y-1}{3}\) is not the inverse — you need to express it in terms of \(x\).
If the original function has a restricted range, that becomes the domain of the inverse. For \(f(x) = \frac{3}{x+2}\) where \(f(x) \neq 0\), the inverse has domain \(x \neq 0\). State this alongside your answer.
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