AS & A Level · Practice questions · Functions: domain, range and composition

Which one acts first?

Six original Cambridge-style questions on domain, range, one-one functions and composite functions, ending with a shopping problem where the order of the functions matters.

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

Sketch for the range, inside-out for composites.

Range: sketch over the domain; check both ends and any vertex inside it.
Write the range in terms of $f(x)$, with $<$ or $\le$ chosen carefully.
One-one: no horizontal line meets the graph twice; restrict to one side of a vertex.
Composite: $fg(x) = f\big(g(x)\big)$, so $g$ acts first.
Existence: $fg$ needs the range of $g$ inside the domain of $f$.

Recall

[2 marks]

The function f is defined by $f(x) = 5 - 2x$ for $-1 \le x \le 3$. Find the range of f.

$f$ is decreasing: $f(-1) = 7$ and $f(3) = -1$. ✓
Range: $-1 \le f(x) \le 7$. ✓

Calculation

[4 marks]

$f(x) = x^{2} + 2$ and $g(x) = 2x - 3$ for $x \in \mathbb{R}$.

(a) Find $fg(x)$ and $gf(x)$.

(b) Solve $fg(x) = gf(x)$.

(a) $fg(x) = (2x - 3)^{2} + 2 = 4x^{2} - 12x + 11$. ✓
$gf(x) = 2(x^{2} + 2) - 3 = 2x^{2} + 1$. ✓
(b) $4x^{2} - 12x + 11 = 2x^{2} + 1 \Rightarrow 2x^{2} - 12x + 10 = 0 \Rightarrow x^{2} - 6x + 5 = 0$. ✓
$(x - 1)(x - 5) = 0$, so $x = 1$ or $x = 5$. ✓

Calculation

[3 marks]

The function f is defined by $f(x) = 2x^{2} - 8x + 5$ for $x \in \mathbb{R}$. Express $f(x)$ in the form $a(x + b)^{2} + c$ and hence state the range of f.

$2(x^{2} - 4x) + 5 = 2\big[(x - 2)^{2} - 4\big] + 5$. ✓
$= 2(x - 2)^{2} - 3$. ✓
$2(x - 2)^{2} \ge 0$, so the range is $f(x) \ge -3$. ✓

Analysis

[4 marks]

The function f is defined by $f(x) = (x + 1)^{2} - 4$ for $x \ge k$, and g is defined by $g(x) = 2x + 5$ for $x \in \mathbb{R}$.

(a) State the least value of $k$ for which f is one-one, and the range of f for this value of $k$.

(b) For this value of $k$, find $gf(x)$ and state its range.

(a) The vertex is at $x = -1$, so the least $k$ is $-1$. ✓
Range of f: $f(x) \ge -4$. ✓
(b) $gf(x) = 2\big[(x + 1)^{2} - 4\big] + 5 = 2(x + 1)^{2} - 3$. ✓
Since $f(x) \ge -4$, $gf(x) \ge 2(-4) + 5 = -3$. Range: $gf(x) \ge -3$. ✓

Proof

[3 marks]

The functions f and g are defined by $f : x \mapsto \sqrt{x}$ for $x \ge 0$ and $g : x \mapsto x - 5$ for $x \in \mathbb{R}$.

Explain why the composite function $gf$ exists but $fg$ does not.

Range of f is $f(x) \ge 0$, which lies inside the domain of g (all real numbers), so $gf(x) = \sqrt{x} - 5$ exists. ✓
Range of g is all real numbers, including negative values such as $g(0) = -5$. ✓
f cannot take a negative input ($\sqrt{-5}$ is undefined), so the range of g is not within the domain of f and $fg$ does not exist. ✓

Analysis

[5 marks]

A shop sells items priced from $20 to $100. It offers 20% off, given by $f(x) = 0.8x$, and a $5 voucher, given by $g(x) = x - 5$, where $x$ is a price in dollars.

(a) Find $fg(x)$ and $gf(x)$, and describe in words what each represents.

(b) Which order is better for the customer, and by how much?

(a) $fg(x) = 0.8(x - 5) = 0.8x - 4$: voucher first, then 20% off. ✓
$gf(x) = 0.8x - 5$: 20% off first, then the voucher. ✓
(b) $gf(x)$ is always $1 less than $fg(x)$, so taking the discount first saves the customer $1. ✓
(c) $0.8x - 5 = 35 \Rightarrow 0.8x = 40 \Rightarrow x = 50$: the original price was $50. ✓
$gf$ is increasing: $gf(20) = 11$ and $gf(100) = 75$, so the range is $11 \le gf(x) \le 75$. ✓

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.