Answer Key

Functions

Pack A — Answers

#	Question	Answer
1	Given $f(x) = 3x + 2$, find $f(4)$.	$f(4) = 14$
2	Given $f(x) = x^2 + 1$, find $f(3)$.	$f(3) = 10$
3	Given $f(x) = 2x + 5$, find $f(-3)$.	$f(-3) = -1$
4	Given $f(x) = 3x + 2$, find $x$ such that $f(x) = 17$.	$x = 5$
5	A function maps $x$ to $2x + 3$. Find the image of $5$.	13
6	A function maps $x$ to $3x - 1$. What value of $x$ maps to $14$?	$x = 5$
7	A function $f$ has domain $\{1, 2, 3, 4, 5\}$ and rule $f(x) = 3x$. List the range.	$\{3, 6, 9, 12, 15\}$
8	Given $g(x) = x^2 - 4$, find $g(0)$ and $g(3)$.	$g(0) = -4$, $g(3) = 5$
9	A linear function maps $0 \to 5$ and $1 \to 8$. Find $f(x)$.	$f(x) = 3x + 5$
10	A function machine multiplies by $4$ then adds $7$. Write the function $f(x)$.	$f(x) = 4x + 7$
11	State the domain and range of $f(x) = 2x + 3$ where $x \in \mathbb{R}$.	Domain: $\mathbb{R}$; Range: $\mathbb{R}$
12	Find the range of $f(x) = x^2 + 3$ for $x \in \mathbb{R}$.	$f(x) \geq 3$ (i.e. $[3, \infty)$)
13	A function $f(x) = 3x + 1$ has domain $0 \leq x \leq 4$. State the range.	Range: $1 \leq f(x) \leq 13$
14	Given $f(x) = x + 3$ and $g(x) = 2x$, find $(f \circ g)(4)$.	$(f \circ g)(4) = 11$
15	Given $f(x) = 2x + 1$ and $g(x) = x + 3$, find $(f \circ g)(x)$.	$(f \circ g)(x) = 2x + 7$
16	Given $f(x) = x^2$ and $g(x) = x + 1$, find (a) $(f \circ g)(3)$ and (b) $(g \circ f)(3)$.	(a) 16 (b) 10
17	Find the inverse of $f(x) = 3x + 2$.	$f^{-1}(x) = \dfrac{x - 2}{3}$
18	Given $f(x) = 2x + 1$, find $f^{-1}(9)$.	$f^{-1}(9) = 4$
19	A function $f$ is defined by $f: x \mapsto 2x - 6$. Find $f(5)$ and an $x$ with $f(x) = 0$.	$f(5) = 4$; $f(x) = 0$ at $x = 3$
20	A linear function $f$ satisfies $f(1) = 5$ and $f(4) = 14$. Find $f(x)$.	$f(x) = 3x + 2$
21	Given $f(x) = 3x + 1$ and $g(x) = x^2$, find $(f \circ g)(x)$.	$(f \circ g)(x) = 3x^2 + 1$
22	Given $f(x) = 2x + 3$ and $g(x) = x^2$, find $(g \circ f)(x)$.	$(g \circ f)(x) = (2x + 3)^2 = 4x^2 + 12x + 9$
23	State the largest possible domain of $f(x) = \sqrt{x - 4}$.	$x \geq 4$
24	State the largest possible domain of $f(x) = \dfrac{1}{x - 3}$.	$x \in \mathbb{R}, x \neq 3$
25	Given $f(x) = x^2$ and $g(x) = 3x + 4$, solve $f(x) = g(x)$.	$x = 4$ or $x = -1$
26	Find the inverse of $f(x) = \dfrac{x + 3}{2}$.	$f^{-1}(x) = 2x - 3$
27	Show that for $f(x) = 3x - 1$ and $f^{-1}(x) = \dfrac{x + 1}{3}$, we have $(f \circ f^{-1})(7) = 7$.	$(f \circ f^{-1})(7) = 7$ ✓
28	Given $f(x) = x + 3$, $g(x) = 2x$, $h(x) = x^2$, find $(f \circ g \circ h)(2)$.	11
29	Given $f(x) = \sqrt{x}$ and $g(x) = x + 5$, find $(f \circ g)(11)$.	4
30	Find the inverse of $f(x) = (x - 3)^2$ for $x \geq 3$.	$f^{-1}(x) = \sqrt{x} + 3$ (for $x \geq 0$)
31	Given $f(x) = 2x + 1$ and $g(x) = x + 3$, find $(f \circ g)^{-1}(x)$.	$(f \circ g)^{-1}(x) = \dfrac{x - 7}{2}$
32	Find the largest possible domain of $f(x) = \dfrac{1}{\sqrt{x - 4}}$.	$x > 4$
33	A car rental costs a £$25$ fixed fee plus £$12$ per day. (a) Write $C(n)$ for $n$ days. (b) Find $C^{-1}(c)$ and interpret.	(a) $C(n) = 12n + 25$ (b) $C^{-1}(c) = \dfrac{c - 25}{12}$ (days for cost $c$)
34	A function $f$ has graph $y = (x - 3)^2 + 2$. Find the (a) minimum value and (b) range of $f$.	(a) Min = 2 at $x = 3$ (b) Range: $y \geq 2$
35	For $f(x) = 2x - 1$ and $g(x) = x^2 + 3$, solve $(f \circ g)(x) = 13$.	$x = \pm 2$
36	A function $f(x) = ax + b$ satisfies $f(0) = 3$ and $f^{-1}(11) = 2$. Find $a$ and $b$.	$a = 4$, $b = 3$
37	For $f(x) = x + 1$, find $f(f(f(x)))$.	$x + 3$
38	Given $f(x) = 3x + 2$, find the value of $x$ for which $f(x) = f^{-1}(x)$.	$x = -1$
39	A water tank has volume function $V(t) = 8t + 20$ litres after $t$ minutes ($t \geq 0$). After how many minutes does the tank contain $100$ litres?	10 minutes
40	For $f(x) = \sqrt{x}$ (domain $x \geq 0$) and $g(x) = x - 4$, find the domain of $(f \circ g)(x)$.	$x \geq 4$

Pack B — Answers

#	Question	Answer
1	Given $f(x) = 5x + -3$, find $f(4)$.	$f(4) = 17$
2	Given $f(x) = x^2 + -4$, find $f(5)$.	$f(5) = 21$
3	Given $f(x) = 4x + -1$, find $f(-2)$.	$f(-2) = -9$
4	Given $f(x) = 4x + -1$, find $x$ such that $f(x) = 19$.	$x = 5$
5	A function maps $x$ to $2x + 3$. Find the image of $7$.	17
6	A function maps $x$ to $3x - 1$. What value of $x$ maps to $20$?	$x = 7$
7	A function $f$ has domain $\{1, 2, 3, 4, 5\}$ and rule $f(x) = 4x$. List the range.	$\{4, 8, 12, 16, 20\}$
8	Given $g(x) = x^2 - 9$, find $g(0)$ and $g(4)$.	$g(0) = -9$, $g(4) = 7$
9	A linear function maps $0 \to 2$ and $1 \to 6$. Find $f(x)$.	$f(x) = 4x + 2$
10	A function machine multiplies by $6$ then adds $1$. Write the function $f(x)$.	$f(x) = 6x + 1$
11	State the domain and range of $f(x) = 5x + -1$ where $x \in \mathbb{R}$.	Domain: $\mathbb{R}$; Range: $\mathbb{R}$
12	Find the range of $f(x) = x^2 + -2$ for $x \in \mathbb{R}$.	$f(x) \geq -2$ (i.e. $[-2, \infty)$)
13	A function $f(x) = 2x + 5$ has domain $0 \leq x \leq 6$. State the range.	Range: $5 \leq f(x) \leq 17$
14	Given $f(x) = x + 5$ and $g(x) = 3x$, find $(f \circ g)(2)$.	$(f \circ g)(2) = 11$
15	Given $f(x) = 3x + 2$ and $g(x) = x + 4$, find $(f \circ g)(x)$.	$(f \circ g)(x) = 3x + 14$
16	Given $f(x) = x^2$ and $g(x) = x + 2$, find (a) $(f \circ g)(4)$ and (b) $(g \circ f)(4)$.	(a) 36 (b) 18
17	Find the inverse of $f(x) = 5x + -4$.	$f^{-1}(x) = \dfrac{x + 4}{5}$
18	Given $f(x) = 3x + -2$, find $f^{-1}(13)$.	$f^{-1}(13) = 5$
19	A function $f$ is defined by $f: x \mapsto 3x - 9$. Find $f(4)$ and an $x$ with $f(x) = 0$.	$f(4) = 3$; $f(x) = 0$ at $x = 3$
20	A linear function $f$ satisfies $f(1) = 7$ and $f(5) = 23$. Find $f(x)$.	$f(x) = 4x + 3$
21	Given $f(x) = 2x + 5$ and $g(x) = x^2$, find $(f \circ g)(x)$.	$(f \circ g)(x) = 2x^2 + 5$
22	Given $f(x) = 3x + 1$ and $g(x) = x^2$, find $(g \circ f)(x)$.	$(g \circ f)(x) = (3x + 1)^2 = 9x^2 + 6x + 1$
23	State the largest possible domain of $f(x) = \sqrt{x - 7}$.	$x \geq 7$
24	State the largest possible domain of $f(x) = \dfrac{1}{x - 5}$.	$x \in \mathbb{R}, x \neq 5$
25	Given $f(x) = x^2$ and $g(x) = 5x + 6$, solve $f(x) = g(x)$.	$x = 6$ or $x = -1$
26	Find the inverse of $f(x) = \dfrac{x + 5}{4}$.	$f^{-1}(x) = 4x - 5$
27	Show that for $f(x) = 2x - 5$ and $f^{-1}(x) = \dfrac{x + 5}{2}$, we have $(f \circ f^{-1})(9) = 9$.	$(f \circ f^{-1})(9) = 9$ ✓
28	Given $f(x) = x + 5$, $g(x) = 3x$, $h(x) = x^2$, find $(f \circ g \circ h)(2)$.	17
29	Given $f(x) = \sqrt{x}$ and $g(x) = x + 7$, find $(f \circ g)(18)$.	5
30	Find the inverse of $f(x) = (x - 5)^2$ for $x \geq 5$.	$f^{-1}(x) = \sqrt{x} + 5$ (for $x \geq 0$)
31	Given $f(x) = 3x + 2$ and $g(x) = x + 5$, find $(f \circ g)^{-1}(x)$.	$(f \circ g)^{-1}(x) = \dfrac{x - 17}{3}$
32	Find the largest possible domain of $f(x) = \dfrac{1}{\sqrt{x - 6}}$.	$x > 6$
33	A car rental costs a £$40$ fixed fee plus £$15$ per day. (a) Write $C(n)$ for $n$ days. (b) Find $C^{-1}(c)$ and interpret.	(a) $C(n) = 15n + 40$ (b) $C^{-1}(c) = \dfrac{c - 40}{15}$
34	A function $f$ has graph $y = (x - 5)^2 + -3$. Find the (a) minimum value and (b) range of $f$.	(a) Min = $-3$ at $x = 5$ (b) Range: $y \geq -3$
35	For $f(x) = 3x - 2$ and $g(x) = x^2 + 1$, solve $(f \circ g)(x) = 28$.	$x = \pm 3$
36	A function $f(x) = ax + b$ satisfies $f(0) = 5$ and $f^{-1}(17) = 3$. Find $a$ and $b$.	$a = 4$, $b = 5$
37	For $f(x) = x + 1$, find $f(f(f(x)))$.	$8x$
38	Given $f(x) = 5x + -4$, find the value of $x$ for which $f(x) = f^{-1}(x)$.	$x = 1$
39	A water tank has volume function $V(t) = 12t + 30$ litres after $t$ minutes ($t \geq 0$). After how many minutes does the tank contain $150$ litres?	10 minutes
40	For $f(x) = \sqrt{x}$ (domain $x \geq 0$) and $g(x) = x - 7$, find the domain of $(f \circ g)(x)$.	$x \geq 7$

Problems — Worked Solutions

**Temperature converter.** A function converts temperature from Celsius to Fahrenheit: $F(c) = \frac{9}{5}c + 32$. (a) Find $F(0)$ and $F(100)$. (b) Find $F^{-1}(F)$, the inverse function (Fahrenheit to Celsius). (c) Find $F^{-1}(212)$. Interpret. (d) Find the temperature where Celsius and Fahrenheit are equal: $F(c) = c$.

Answer

(a) 32, 212 (b) $C(f) = \frac{5}{9}(f - 32)$ (c) 100°C (boiling point) (d) $c = -40$

(a) $F(0) = 32$; $F(100) = \frac{9}{5}(100) + 32 = 180 + 32 = 212$. (b) Solve $f = \frac{9}{5}c + 32$ for $c$: $f - 32 = \frac{9}{5}c$, so $c = \frac{5}{9}(f - 32)$. Inverse: $F^{-1}(f) = \frac{5}{9}(f - 32)$. (c) $F^{-1}(212) = \frac{5}{9}(180) = 100$. Interpret: 212°F = 100°C (boiling point of water). (d) Set $\frac{9}{5}c + 32 = c$: $\frac{9}{5}c - c = -32$, so $\frac{4}{5}c = -32$, $c = -40$. So **$-40°$C = $-40°$F**.

**Function machines.** Given $f(x) = 3x - 1$ and $g(x) = x^2 + 2$: (a) Find $f(2)$, $g(2)$, $f(g(2))$, and $g(f(2))$. (b) Find a formula for $(f \circ g)(x)$ and $(g \circ f)(x)$. (c) Solve $(f \circ g)(x) = 14$.

Answer

(a) 5, 6, 17, 27 (b) $f \circ g = 3x^2 + 5$; $g \circ f = (3x-1)^2 + 2$ (c) $x = \pm\sqrt{3}$

(a) $f(2) = 5$; $g(2) = 6$; $f(g(2)) = f(6) = 3(6) - 1 = 17$; $g(f(2)) = g(5) = 25 + 2 = 27$. (b) $(f \circ g)(x) = f(x^2 + 2) = 3(x^2 + 2) - 1 = 3x^2 + 5$. $(g \circ f)(x) = g(3x - 1) = (3x-1)^2 + 2 = 9x^2 - 6x + 3$. (c) $3x^2 + 5 = 14 \Rightarrow x^2 = 3 \Rightarrow x = \pm\sqrt{3}$.

**Inverse practice.** Find the inverse of each function. State any domain restrictions needed. (a) $f(x) = 5x - 7$ (b) $g(x) = \dfrac{2x + 3}{4}$ (c) $h(x) = (x - 2)^2$ (for $x \geq 2$)

Answer

(a) $f^{-1}(x) = \frac{x+7}{5}$ (b) $g^{-1}(x) = \frac{4x - 3}{2}$ (c) $h^{-1}(x) = \sqrt{x} + 2$ for $x \geq 0$

(a) $y = 5x - 7 \Rightarrow x = \frac{y + 7}{5}$. So $f^{-1}(x) = \frac{x + 7}{5}$. (b) $y = \frac{2x + 3}{4} \Rightarrow 4y = 2x + 3 \Rightarrow x = \frac{4y - 3}{2}$. So $g^{-1}(x) = \frac{4x - 3}{2}$. (c) $y = (x - 2)^2$. Since $x \geq 2$, $x - 2 \geq 0$, so $\sqrt{y} = x - 2$, giving $x = \sqrt{y} + 2$. So $h^{-1}(x) = \sqrt{x} + 2$, defined for $x \geq 0$.

**Modelling a fence.** A farmer has 60 m of fencing to enclose a rectangular field, one side of which uses an existing wall (so no fencing on that side). If the side perpendicular to the wall has length $x$ metres: (a) Express the side along the wall in terms of $x$. (b) Express the area $A(x)$ as a function of $x$. (c) State the domain of $A$. (d) Find the value of $x$ that maximises the area.

Answer

(a) $60 - 2x$ m (b) $A(x) = 60x - 2x^2$ (c) $0 < x < 30$ (d) $x = 15$ m (max area = 450 m²)

(a) Total fencing used: $2x + (\text{side along wall}) = 60$. So side along wall = $60 - 2x$. (b) Area = $x \cdot (60 - 2x) = 60x - 2x^2$. (c) For a valid rectangle: $x > 0$ and $60 - 2x > 0 \Rightarrow x < 30$. So domain $0 < x < 30$. (d) $A(x) = -2x^2 + 60x$. Vertex of downward parabola at $x = -\frac{60}{2(-2)} = 15$ m. Max area: $A(15) = 60(15) - 2(225) = 900 - 450 = 450$ m².

**Composition puzzle.** $f$ and $g$ are linear functions with $f(x) = 2x + 1$. If $(f \circ g)(x) = 4x + 7$, find $g(x)$.

Answer

$g(x) = 2x + 3$

Let $g(x) = ax + b$. Then $(f \circ g)(x) = f(ax + b) = 2(ax + b) + 1 = 2ax + 2b + 1$. Set equal to $4x + 7$: - $2a = 4 \Rightarrow a = 2$ - $2b + 1 = 7 \Rightarrow b = 3$ So $g(x) = 2x + 3$. Check: $(f \circ g)(x) = 2(2x + 3) + 1 = 4x + 7$ ✓.

**Identifying functions.** For each relation, decide whether it represents a function. Justify. (a) Each Year 10 student maps to their unique school ID. (b) Each town maps to all citizens who live there. (c) Each $x \in \mathbb{R}$ maps to its square $x^2$. (d) Each positive number $y$ maps to all $x$ with $x^2 = y$.

Answer

(a) Yes (b) No (one-to-many) (c) Yes (d) No (one-to-many)

A function must assign exactly one output to each input. (a) **Function.** Each student → exactly one ID. (b) **Not a function.** A town has many citizens — one input maps to multiple outputs. (It's a relation but not a function.) (c) **Function.** For each $x$, $x^2$ is a single number. (d) **Not a function.** Each positive $y$ has *two* preimages: $\pm\sqrt{y}$. So one input maps to two outputs.

**Domain and range.** A function is defined by $f(x) = \dfrac{1}{x - 3}$. (a) State the largest possible domain. (b) Find the range. (c) Find $f^{-1}(x)$ and its domain.

Answer

(a) $x \neq 3$ (b) $f(x) \neq 0$ (c) $f^{-1}(x) = \frac{1}{x} + 3$, domain $x \neq 0$

(a) Denominator cannot be zero: $x \neq 3$. Domain: $\mathbb{R} \setminus \{3\}$. (b) $\frac{1}{x - 3}$ takes every non-zero real value (as $x \to \pm\infty$, $f \to 0$ but never reaches it; as $x \to 3$, $f \to \pm\infty$). Range: $\mathbb{R} \setminus \{0\}$. (c) Solve $y = \frac{1}{x - 3}$: $y(x - 3) = 1$, so $x = \frac{1}{y} + 3$. Inverse: $f^{-1}(x) = \frac{1}{x} + 3$. Domain: $x \neq 0$ (matches range of $f$, as expected).

**Sketch and interpret.** A function has graph passing through $(0, 4)$, $(2, 0)$, $(3, -1)$, $(5, 1)$, $(6, 4)$. (a) Estimate the range from the data. (b) Is the function one-to-one over $[0, 6]$? Justify. (c) Why does this matter for finding an inverse?

Answer

(a) Approx $[-1, 4]$ (b) No (c) Inverse not a function on this domain

(a) From the given values, $y$ ranges from $-1$ to $4$ (taking values $4, 0, -1, 1, 4$). Estimated range: $[-1, 4]$. (b) The value $y = 4$ appears at both $x = 0$ and $x = 6$ — so two different inputs give the same output. **Not one-to-one**. (c) An inverse function must assign a unique input to each output. If two $x$ values share an output, you cannot decide which $x$ to map back to. So a non-one-to-one function does **not have an inverse** unless you restrict its domain.

**Currency converter.** £1 = €1.18 (May 2026 rate). (a) Write a function $E(p)$ converting £$p$ to euros. (b) Find $E^{-1}$ and interpret. (c) Tomás is travelling from Geneva to London with €500. What does he have in pounds (2 d.p.)?

Answer

(a) $E(p) = 1.18p$ (b) $E^{-1}(e) = e/1.18$ (euros to pounds) (c) £423.73

(a) $E(p) = 1.18p$. (b) $e = 1.18p \Rightarrow p = \frac{e}{1.18}$. So $E^{-1}(e) = \frac{e}{1.18}$. This converts euros back to pounds. (c) $E^{-1}(500) = \frac{500}{1.18} \approx 423.7288 \approx \mathbf{\pounds 423.73}$.

**Restricted-domain inverses.** Consider $f(x) = x^2$ on the full domain $\mathbb{R}$. (a) Why does $f$ not have an inverse on this domain? (b) Restrict to $x \geq 0$. Now what is $f^{-1}$? (c) Restrict to $x \leq 0$. Now what is $f^{-1}$? (d) Verify (b): compute $(f \circ f^{-1})(9)$ and $(f^{-1} \circ f)(3)$.

Answer

(a) Not 1-1 (b) $\sqrt{x}$ (c) $-\sqrt{x}$ (d) Both equal what they should

(a) $f(2) = f(-2) = 4$: two inputs give the same output, so $f$ is not one-to-one and has no inverse on $\mathbb{R}$. (b) Restricted to $x \geq 0$: $f$ is one-to-one and onto $[0, \infty)$. Inverse: $f^{-1}(x) = \sqrt{x}$. (c) Restricted to $x \leq 0$: outputs again cover $[0, \infty)$ but inputs are negative. Inverse takes positive $x$ to negative root: $f^{-1}(x) = -\sqrt{x}$. (d) For (b): - $(f \circ f^{-1})(9) = f(\sqrt{9}) = f(3) = 9$ ✓. - $(f^{-1} \circ f)(3) = f^{-1}(9) = \sqrt{9} = 3$ ✓.

**Self-inverse functions.** A function $f$ is *self-inverse* if $f^{-1}(x) = f(x)$ for all $x$. (a) Show that $f(x) = -x$ is self-inverse. (b) Show that $f(x) = \dfrac{1}{x}$ (for $x \neq 0$) is self-inverse. (c) Find all linear functions of the form $f(x) = ax + b$ that are self-inverse.

Answer

(a) See working (b) See working (c) $f(x) = b$ (constant — not invertible) or $f(x) = -x + b$ for any $b$

(a) $f(f(x)) = f(-x) = -(-x) = x$. So applying $f$ twice gives back $x$ → self-inverse ✓. (b) $f(f(x)) = f(\frac{1}{x}) = \frac{1}{1/x} = x$ ✓. (c) Require $f(f(x)) = x$. Compute: $f(f(x)) = a(ax + b) + b = a^2 x + ab + b$. Set $= x$: $a^2 = 1$ and $ab + b = 0$, i.e. $b(a + 1) = 0$. Case 1: $a = 1$, then $b(2) = 0 \Rightarrow b = 0$. So $f(x) = x$ (the identity, trivially self-inverse). Case 2: $a = -1$, then $b(0) = 0$ — any $b$. So $f(x) = -x + b$ for any $b$. **Family of self-inverse linear functions: $f(x) = -x + b$ (plus the identity).** Geometrically, these are reflections across the line $y = \frac{b}{2}$.

**A composition mystery.** Given $f(x) = 2x + 5$ and $h(x) = 4x + 13$, find a function $g$ such that $(f \circ g)(x) = h(x)$. Also state whether $g$ is unique. Justify.

Answer

$g(x) = 2x + 4$; unique because $f$ is one-to-one

Let $g(x) = ax + b$. Then $f(g(x)) = 2(ax + b) + 5 = 2ax + 2b + 5$. Set equal to $h(x) = 4x + 13$: - $2a = 4 \Rightarrow a = 2$ - $2b + 5 = 13 \Rightarrow b = 4$ So $g(x) = 2x + 4$. **Uniqueness.** $f$ is linear with non-zero gradient → one-to-one → invertible. Given $h$, the function $g$ is forced to equal $f^{-1} \circ h$, which is unique: $g(x) = \frac{h(x) - 5}{2} = \frac{4x + 8}{2} = 2x + 4$ ✓. So **$g$ is unique**.