AS & A Level · Practice questions · Inverse functions

Swap the inputs and outputs.

Six original Cambridge-style questions on inverse functions: finding them, stating their domains, choosing the right square root and using an inverse in context.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level 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 on its domain.
Write \(y = f(x)\) and make \(x\) the subject (complete the square for a quadratic).
Choose the sign of any square root to match the domain of \(f\).
Swap letters to write \(f^{-1}(x)\).
State the domain of \(f^{-1}\): it is the range of \(f\).

Recall

[2 marks]

The function f is defined by \(f(x) = \dfrac{5x - 2}{3}\) for \(x \in \mathbb{R}\). Find \(f^{-1}(x)\).

\(y = \dfrac{5x - 2}{3} \Rightarrow 3y = 5x - 2 \Rightarrow x = \dfrac{3y + 2}{5}\). ✓
\(f^{-1}(x) = \dfrac{3x + 2}{5}\). ✓

Calculation

[3 marks]

The function f is defined by \(f(x) = 4 - (x - 1)^{2}\) for \(x \ge 1\). Find \(f^{-1}(x)\) and state the domain of \(f^{-1}\).

\((x - 1)^{2} = 4 - y\), and \(x - 1 \ge 0\), so \(x - 1 = \sqrt{4 - y}\). ✓
\(f^{-1}(x) = 1 + \sqrt{4 - x}\). ✓
Range of f is \(f(x) \le 4\), so the domain of \(f^{-1}\) is \(x \le 4\). ✓

Calculation

[4 marks]

The function f is defined by \(f(x) = 2x^{2} + 12x + 11\) for \(x \ge -3\).

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

(b) Find \(f^{-1}(x)\), stating its domain.

(a) \(2(x^{2} + 6x) + 11 = 2\big[(x + 3)^{2} - 9\big] + 11 = 2(x + 3)^{2} - 7\). ✓
(b) \(\dfrac{y + 7}{2} = (x + 3)^{2}\), and \(x + 3 \ge 0\), so \(x = -3 + \sqrt{\dfrac{y + 7}{2}}\). ✓
\(f^{-1}(x) = -3 + \sqrt{\dfrac{x + 7}{2}}\). ✓
Range of f is \(f(x) \ge -7\), so the domain of \(f^{-1}\) is \(x \ge -7\). ✓

Analysis

[4 marks]

The function f is defined by \(f(x) = \dfrac{3}{2x - 1}\) for \(x > \dfrac{1}{2}\).

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

(b) Solve \(f^{-1}(x) = 2\).

(a) \(2x - 1 = \dfrac{3}{y} \Rightarrow 2x = \dfrac{3 + y}{y} \Rightarrow x = \dfrac{y + 3}{2y}\). ✓
\(f^{-1}(x) = \dfrac{x + 3}{2x}\). For \(x > \frac{1}{2}\), \(2x - 1 > 0\) takes every positive value, so \(f(x) > 0\): domain of \(f^{-1}\) is \(x > 0\). ✓
(b) \(\dfrac{x + 3}{2x} = 2 \Rightarrow x + 3 = 4x \Rightarrow x = 1\). ✓
Check: \(f(2) = \dfrac{3}{3} = 1\), so \(f^{-1}(1) = 2\). ✓

Proof

[3 marks]

The function g is defined by \(g(x) = \dfrac{x + 4}{x - 1}\) for \(x > 1\). Show that g is self-inverse, that is, \(g^{-1}(x) = g(x)\).

\(y(x - 1) = x + 4 \Rightarrow xy - y = x + 4 \Rightarrow x(y - 1) = y + 4\). ✓
\(x = \dfrac{y + 4}{y - 1}\), so \(g^{-1}(x) = \dfrac{x + 4}{x - 1}\), the same rule as g. ✓
Domains match too: \(g(x) = 1 + \dfrac{5}{x - 1} > 1\) for \(x > 1\), so \(g^{-1}\) has domain \(x > 1\), the same as g. ✓

Analysis

[5 marks]

A workshop's cost, in dollars, of making \(x\) chairs is modelled by \(c(x) = 0.5x^{2} + 20\) for \(x \ge 0\).

(a) Explain why c has an inverse.

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

(a) On \(x \ge 0\) the graph is the right half of a parabola with vertex \((0, 20)\), which is increasing, so c is one-one. ✓
(b) \(y - 20 = 0.5x^{2} \Rightarrow x^{2} = 2(y - 20)\), and \(x \ge 0\), so \(x = \sqrt{2(y - 20)}\). ✓
\(c^{-1}(x) = \sqrt{2(x - 20)}\) for \(x \ge 20\) (the range of c). ✓
(c) \(c^{-1}(92) = \sqrt{2 \times 72} = \sqrt{144} = 12\). ✓
A budget of 92 dollars pays for exactly 12 chairs. ✓

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.