IGCSE · Practice questions · Functions: notation, inverse and composite

Machines in sequence.

Extended. Six original Cambridge-style questions on functions: notation, domain and range, inverse functions and composite functions.

Original questions All questions on this page are original work, written in the Cambridge IGCSE style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Substitute, rearrange, apply in order.

\(f(a)\): replace every \(x\) in the rule with \(a\), using brackets.
Solve \(f(x) = k\): set the rule equal to \(k\) and solve.
Inverse: write \(y = f(x)\), make \(x\) the subject, then write \(f^{-1}(x)\).
Composite: \(gf(x) = g(f(x))\), so the function nearest \(x\) acts first.
Simplify algebraic composites fully, expanding brackets in denominators.

Recall

[3 marks]

\(f(x) = 7 - 2x\). Find (a) \(f(3)\) (b) \(f(-4)\) (c) the value of \(x\) for which \(f(x) = 11\).

(a) \(7 - 6 = 1\). ✓
(b) \(7 + 8 = 15\). ✓
(c) \(7 - 2x = 11\), so \(-2x = 4\) and \(x = -2\). ✓

Recall

[2 marks]

\(h(x) = 2x^2 + 3\) with domain \(\{-2, -1, 0, 1, 2\}\). Draw a mapping diagram and state the range.

Arrows: \(-2 \to 11\), \(-1 \to 5\), \(0 \to 3\), \(1 \to 5\), \(2 \to 11\). ✓
Range \(= \{3, 5, 11\}\). ✓

Calculation

[2 marks]

\(f(x) = \dfrac{2x - 1}{5}\). Find \(f^{-1}(x)\).

\(y = \dfrac{2x - 1}{5}\), so \(5y = 2x - 1\) and \(5y + 1 = 2x\). ✓
\(f^{-1}(x) = \dfrac{5x + 1}{2}\). (Check: \(f(3) = 1\) and \(f^{-1}(1) = 3\).) ✓

Calculation

[3 marks]

\(f(x) = 4x - 1\) and \(g(x) = x^2 + 2\). Find (a) \(fg(x)\) (b) \(gf(x)\), simplifying each.

(a) \(fg(x) = f(x^2 + 2) = 4(x^2 + 2) - 1 = 4x^2 + 7\). ✓
(b) \(gf(x) = g(4x - 1) = (4x - 1)^2 + 2 = 16x^2 - 8x + 1 + 2 = 16x^2 - 8x + 3\). ✓

Analysis

[3 marks]

\(f(x) = \dfrac{3}{x + 2}\) and \(g(x) = (3x + 5)^2\). Find \(fg(x)\). Give your answer as a fraction in its simplest form.

\(fg(x) = \dfrac{3}{(3x + 5)^2 + 2}\). ✓
\((3x + 5)^2 + 2 = 9x^2 + 30x + 25 + 2 = 9x^2 + 30x + 27\). ✓
Take out the common factor 3: \(9x^2 + 30x + 27 = 3(3x^2 + 10x + 9)\). ✓
\(fg(x) = \dfrac{3}{3(3x^2 + 10x + 9)} = \dfrac{1}{3x^2 + 10x + 9}\). ✓

Analysis

[4 marks]

A temperature of \(x\) °C is converted to °F by \(f(x) = 1.8x + 32\). A temperature of \(x\) kelvin is converted to °C by \(g(x) = x - 273\).

(a) Find \(f(25)\).

(b) Find \(f^{-1}(x)\) and use it to convert 98.6 °F to °C.

(a) \(f(25) = 45 + 32 = 77\) (°F). ✓
(b) \(y = 1.8x + 32\), so \(x = \dfrac{y - 32}{1.8}\): \(f^{-1}(x) = \dfrac{x - 32}{1.8}\). ✓
\(f^{-1}(98.6) = \dfrac{66.6}{1.8} = 37\) °C. ✓
(c) \(g(300) = 27\), then \(f(27) = 48.6 + 32 = 80.6\). It converts 300 K to 80.6 °F. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.