IGCSE · Practice questions · Inverse functions

Swap inputs and outputs.

Six original Cambridge-style questions on inverse functions: explaining many-one functions, finding inverses of exponential, logarithmic and quadratic functions, stating domains, sketching reflections in y = x and finding where a function meets its inverse.

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

Make \(x\) the subject, then swap.

Check the function is one-one; if not, name two inputs with the same output.
Write \(y = f(x)\) and make \(x\) the subject: complete the square, take logs or raise \(e\) to a power.
Choose the sign of any square root to match the domain of \(f\).
Swap the letters and write \(f^{-1}(x) = \ldots\).
State the domain of \(f^{-1}\): it is the range of \(f\). Sketches reflect in \(y = x\).

Recall

[2 marks]

The function f is defined by \(f(x) = x^{2} - 6x + 1\) for \(x \in \mathbb{R}\).

(a) Explain why f does not have an inverse.

(b) State the largest domain of the form \(x \ge k\) on which f has an inverse.

(a) \(f(0) = 1\) and \(f(6) = 36 - 36 + 1 = 1\): two inputs give the same output, so f is many-one and has no inverse. ✓
(b) \(f(x) = (x - 3)^{2} - 8\), vertex at \(x = 3\), so \(x \ge 3\). ✓

Calculation

[3 marks]

The function f is defined by \(f(x) = 5e^{2x} - 1\) for \(x \in \mathbb{R}\). Find \(f^{-1}(x)\) and state its domain.

\(y + 1 = 5e^{2x} \Rightarrow e^{2x} = \dfrac{y + 1}{5} \Rightarrow 2x = \ln\left(\dfrac{y + 1}{5}\right)\). ✓
\(f^{-1}(x) = \dfrac{1}{2}\ln\left(\dfrac{x + 1}{5}\right)\). ✓
\(5e^{2x} > 0\), so \(f(x) > -1\): the domain of \(f^{-1}\) is \(x > -1\). ✓

Calculation

[4 marks]

The function g is defined by \(g(x) = x^{2} + 4x - 1\) for \(x \ge -2\).

(a) Express \(g(x)\) in the form \((x + a)^{2} + b\).

(b) Find \(g^{-1}(x)\) and state its domain and range.

(a) \(g(x) = (x + 2)^{2} - 4 - 1 = (x + 2)^{2} - 5\). ✓
(b) \((x + 2)^{2} = y + 5\), and \(x + 2 \ge 0\), so \(x = -2 + \sqrt{y + 5}\). ✓
\(g^{-1}(x) = -2 + \sqrt{x + 5}\). ✓
Domain of \(g^{-1}\) = range of g: \(x \ge -5\). Range of \(g^{-1}\) = domain of g: \(g^{-1}(x) \ge -2\). ✓

Analysis

[4 marks]

The function f is defined by \(f(x) = 2 + \ln(x - 3)\) for \(x > 3\).

(a) Find \(f^{-1}(x)\) and state its domain.

(b) State the range of \(f^{-1}\).

(a) \(\ln(x - 3) = y - 2 \Rightarrow x - 3 = e^{y - 2} \Rightarrow x = 3 + e^{y - 2}\), so \(f^{-1}(x) = 3 + e^{x - 2}\). ✓
\(\ln(x - 3)\) takes every real value, so the range of f is all real numbers: the domain of \(f^{-1}\) is \(x \in \mathbb{R}\). ✓
(b) The range of \(f^{-1}\) is the domain of f: \(f^{-1}(x) > 3\). ✓
(c) \(3 + e^{x - 2} = 4 \Rightarrow e^{x - 2} = 1 \Rightarrow x - 2 = 0 \Rightarrow x = 2\). ✓
Check: \(f(4) = 2 + \ln 1 = 2\), so \(f^{-1}(2) = 4\). ✓

Analysis

[5 marks]

The function f is defined by \(f(x) = e^{x} - 2\) for \(x \in \mathbb{R}\).

(a) Find \(f^{-1}(x)\) and state its domain.

(b) On one diagram, sketch \(y = f(x)\), \(y = f^{-1}(x)\) and \(y = x\). Give the coordinates of the points where each curve meets the axes, and the equation of each asymptote. Leave answers in exact form.

(a) \(y + 2 = e^{x} \Rightarrow x = \ln(y + 2)\), so \(f^{-1}(x) = \ln(x + 2)\) for \(x > -2\) (since \(e^{x} > 0\) gives \(f(x) > -2\)). ✓
(b) \(y = f(x)\): \(f(0) = -1\), so \((0, -1)\); \(e^{x} = 2\) gives \((\ln 2, 0)\); asymptote \(y = -2\). ✓
\(y = f^{-1}(x)\): reflect in \(y = x\), so \((-1, 0)\) and \((0, \ln 2)\), asymptote \(x = -2\). ✓
Check: \(\ln(-1 + 2) = \ln 1 = 0\) and \(\ln(0 + 2) = \ln 2\). ✓
The sketch shows \(f\) rising steeply to the right and flattening onto \(y = -2\) to the left; \(f^{-1}\) rising slowly to the right and dropping steeply towards \(x = -2\); the two curves are mirror images in the dashed line \(y = x\). ✓

Analysis

[5 marks]

The function f is defined by \(f(x) = x^{2} - 4x + 6\) for \(x \ge 2\).

(a) Explain why f has an inverse.

(b) Find \(f^{-1}(x)\) and state its domain.

(a) \(f(x) = (x - 2)^{2} + 2\), with vertex \((2, 2)\). On \(x \ge 2\) the graph only rises, so each output comes from exactly one input: f is one-one. ✓
(b) \((x - 2)^{2} = y - 2\), and \(x - 2 \ge 0\), so \(x = 2 + \sqrt{y - 2}\): \(f^{-1}(x) = 2 + \sqrt{x - 2}\) for \(x \ge 2\). ✓
(c) f is increasing, so the graphs meet on \(y = x\): \(x^{2} - 4x + 6 = x \Rightarrow x^{2} - 5x + 6 = 0\). ✓
\((x - 2)(x - 3) = 0\), so \(x = 2\) or \(x = 3\); both lie in the domain \(x \ge 2\). ✓
The graphs meet at \((2, 2)\) and \((3, 3)\). Check: \(f(3) = 9 - 12 + 6 = 3\) and \(f^{-1}(3) = 2 + \sqrt{1} = 3\). ✓

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.