Functions

Board Coverage

Board	Paper	Notes
AQA	Paper 1, 2	Functions, composition, inverse, transformations
Edexcel	P1, P2	Similar; modulus in P1
OCR (A)	Paper 1, 2	Includes composite functions
CIE (9709)	P1	Functions, domain, range, inverse

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1. Definitions

Definition. A function $f$ from a set $A$ (the domain) to a set $B$ (the codomain) is a rule that assigns to each element $a \in A$ exactly one element $f(a) \in B$. We write $f: A \to B$.

Definition. The range (or image) of $f$ is the set $\{f(a) : a \in A\} \subseteq B$ — the set of all values actually attained.

Definition. The natural domain of a real-valued function defined by an algebraic expression is the largest subset of $\mathbb{R}$ for which the expression is defined. Common restrictions:

• Denominators cannot be zero: $x \neq 0$ in $\frac{1}{x}$.
• Square roots require non-negative arguments: $x \geq 0$ in $\sqrt{x}$.
• Logarithms require positive arguments: $x > 0$ in $\ln x$.

Details

Example

Find the natural domain of $f(x) = \sqrt{x + 2} + \frac{1}{x - 1}$.

We need: $x + 2 \geq 0$ (for the square root) AND $x - 1 \neq 0$ (for the denominator).

So $x \geq -2$ and $x \neq 1$.

Domain: $[-2, 1) \cup (1, \infty)$.

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2. Composition of Functions

Definition. Given functions $f: A \to B$ and $g: B \to C$, the composition $g \circ f: A \to C$ is defined by:

$(g \circ f)(x) = g(f(x))$

Theorem. Function composition is associative: $(h \circ g) \circ f = h \circ (g \circ f)$.

Theorem. Function composition is not commutative in general: $f \circ g \neq g \circ f$.

Details

Example

Given $f(x) = 2x + 1$ and $g(x) = x^2$, find $f \circ g$ and $g \circ f$.

$(f \circ g)(x) = f(g(x)) = f(x^2) = 2x^2 + 1$.

$(g \circ f)(x) = g(f(x)) = g(2x + 1) = (2x + 1)^2 = 4x^2 + 4x + 1$.

Clearly $f \circ g \neq g \circ f$.

warning

warning right to left: $(f \circ g)(x) = f(g(x))$.

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3. Inverse Functions

Definition. A function $f: A \to B$ is injective (one-to-one) if $f(a_1) = f(a_2) \implies a_1 = a_2$ for all $a_1, a_2 \in A$. Equivalently, distinct inputs give distinct outputs.

Definition. A function $f: A \to B$ is surjective (onto) if for every $b \in B$, there exists $a \in A$ such that $f(a) = b$.

Definition. A function that is both injective and surjective is bijective.

Theorem. A function $f: A \to B$ has an inverse function $f^{-1}: B \to A$ if and only if $f$ is bijective.

Proof.

($\Rightarrow$) If $f^{-1}$ exists, then $f$ must be injective (otherwise $f^{-1}$ would be multiply-defined) and surjective (otherwise $f^{-1}$ would be undefined for elements not in the range).

($\Leftarrow$) If $f$ is bijective, then for each $b \in B$ there exists exactly one $a \in A$ with $f(a) = b$. Define $f^{-1}(b) = a$. This is well-defined and satisfies $f^{-1}(f(a)) = a$ and $f(f^{-1}(b)) = b$. $\blacksquare$

Intuition. An inverse function "undoes" the original function. For this to work, the original function must pair each input with a unique output (injectivity) and must cover every element of the codomain (surjectivity).

3.1 Finding Inverse Functions

Method.

1. Write $y = f(x)$.
2. Solve for $x$ in terms of $y$.
3. Replace $y$ with $x$ (and $x$ with $y$) to get $f^{-1}(x)$.
4. State the domain of $f^{-1}$ (which equals the range of $f$).

Details

Example

Find the inverse of $f(x) = \frac{2x + 3}{x - 1}$, $x \neq 1$.

$y = \frac{2x + 3}{x - 1}$

$y(x - 1) = 2x + 3$

$yx - y = 2x + 3$

$yx - 2x = y + 3$

$x(y - 2) = y + 3$

$x = \frac{y + 3}{y - 2}$

So $f^{-1}(x) = \frac{x + 3}{x - 2}$, with domain $x \neq 2$.

The range of $f$ is all real numbers except $f(x) = 2$: $\frac{2x + 3}{x - 1} = 2 \implies 2x + 3 = 2x - 2 \implies 3 = -2$, impossible. So the range is $\mathbb{R} \setminus \{2\}$, which equals the domain of $f^{-1}$. ✓

Theorem. The graph of $y = f^{-1}(x)$ is the reflection of $y = f(x)$ in the line $y = x$.

Proof. If $(a, b)$ lies on $y = f(x)$, then $b = f(a)$, so $a = f^{-1}(b)$, meaning $(b, a)$ lies on $y = f^{-1}(x)$. Swapping coordinates is reflection in $y = x$. $\blacksquare$

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4. The Modulus Function

Definition. The modulus (absolute value) function is defined by:

$|x| = \begin{cases} x & \mathrm{if } x \geq 0 \\ -x & \mathrm{if } x < 0 \end{cases}$

Properties:

$$\begin{aligned} |ab| &= |a| \cdot |b| \\ |a + b| &\leq |a| + |b| \quad \mathrm{(Triangle inequality)} \\ |a|^2 &= a^2 \\ |x| &= \sqrt{x^2} \end{aligned}$$

Theorem (Triangle Inequality). $|a + b| \leq |a| + |b|$ for all $a, b \in \mathbb{R}$.

Proof. We consider cases based on the signs of $a$ and $b$.

Case 1: $a, b \geq 0$. Then $|a + b| = a + b = |a| + |b|$. Equality holds.

Case 2: $a, b \leq 0$. Then $a + b \leq 0$, so $|a + b| = -(a + b) = -a - b = |a| + |b|$. Equality holds.

Case 3: $a \geq 0$, $b \leq 0$. If $a + b \geq 0$: $|a + b| = a + b \leq a + (-b) = |a| + |b|$. If $a + b \leq 0$: $|a + b| = -(a + b) = -a - b \leq a - b = |a| + |b|$ (since $a \geq 0$ implies $a \geq -a$, i.e., $2a \geq 0$). $\blacksquare$

4.1 Solving Modulus Equations

To solve $|f(x)| = g(x)$:

1. Case 1: $f(x) \geq 0$, so $f(x) = g(x)$.
2. Case 2: $f(x) < 0$, so $-f(x) = g(x)$.
3. Check solutions against the case conditions.

Details

Example

Solve $|2x - 3| = x + 2$.

Case 1: $2x - 3 \geq 0$, i.e., $x \geq \frac{3}{2}$.

$2x - 3 = x + 2 \implies x = 5$. Check: $5 \geq \frac{3}{2}$ ✓

Case 2: $2x - 3 < 0$, i.e., $x < \frac{3}{2}$.

$-(2x - 3) = x + 2 \implies -2x + 3 = x + 2 \implies -3x = -1 \implies x = \frac{1}{3}$. Check: $\frac{1}{3} < \frac{3}{2}$ ✓

Solutions: $x = \frac{1}{3}$ and $x = 5$.

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5. Transformations of Graphs

5.1 Translations

Theorem. The graph of $y = f(x - a)$ is the graph of $y = f(x)$ translated by $a$ units in the positive $x$-direction.

Proof. Let $g(x) = f(x - a)$. The point $(x_0, y_0)$ lies on $y = g(x)$ if and only if $y_0 = g(x_0) = f(x_0 - a)$. This means $(x_0 - a, y_0)$ lies on $y = f(x)$. So the point $(x_0, y_0)$ on $g$ corresponds to the point $(x_0 - a, y_0)$ on $f$ — a shift right by $a$. $\blacksquare$

Intuition. Replacing $x$ with $x - a$ means "to get the same output, I need to input $a$ more." The graph shifts right to compensate.

Similarly, $y = f(x) + b$ translates up by $b$.

5.2 Reflections

Transformation	Effect
$y = f(-x)$	Reflection in the $y$-axis
$y = -f(x)$	Reflection in the $x$-axis

Proof for $y = f(-x)$ as reflection in the $y$-axis. If $(x_0, y_0)$ is on $y = f(x)$, then $y_0 = f(x_0)$. The point $(-x_0, y_0)$ satisfies $y_0 = f(-(-x_0)) = f(x_0)$, so it lies on $y = f(-x)$. Reflecting $(x_0, y_0)$ in the $y$-axis gives $(-x_0, y_0)$. $\blacksquare$

5.3 Stretches

Transformation	Effect
$y = f(ax)$	Horizontal stretch, scale factor $\frac{1}{a}$
$y = af(x)$	Vertical stretch, scale factor $a$

Proof for $y = f(ax)$ as horizontal stretch. If $(x_0, y_0)$ is on $y = f(x)$, then on $y = f(ax)$, the same $y$-value occurs when $ax = x_0$, i.e., $x = \frac{x_0}{a}$. So $(\frac{x_0}{a}, y_0)$ is on the new graph — a horizontal stretch by factor $\frac{1}{a}$. $\blacksquare$

warning

Horizontal transformations are "backwards": $f(x - a)$ shifts right (not left), and $f(ax)$ stretches by $\frac{1}{a}$ (not $a$). This is the single most common error in this topic.

5.4 Combined Transformations

When multiple transformations are applied, the order matters. The convention is:

$y = af(x - p) + q$

represents: horizontal translation by $p$ (right), vertical stretch by factor $a$, vertical translation by $q$ (up).

tip

tip (outside). The order inside-out matters.

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6. Even and Odd Functions

Definition. A function $f$ is even if $f(-x) = f(x)$ for all $x$ in its domain. A function $f$ is odd if $f(-x) = -f(x)$ for all $x$ in its domain.

Theorem. The graph of an even function is symmetric about the $y$-axis. The graph of an odd function has rotational symmetry of order 2 about the origin.

Proof. For even $f$: the point $(-x, f(-x)) = (-x, f(x))$ is on the graph whenever $(x, f(x))$ is. These two points are reflections of each other across the $y$-axis. $\blacksquare$

For odd $f$: the point $(-x, f(-x)) = (-x, -f(x))$ is on the graph whenever $(x, f(x))$ is. Rotating $(x, f(x))$ by $180^\circ$ about the origin gives $(-x, -f(x))$. $\blacksquare$

Algebraic Properties.

Operation	Even $\times$ Even	Odd $\times$ Odd	Even $\times$ Odd
Result	Even	Even	Odd

Proof (for $f$ odd, $g$ odd $\implies fg$ even). $(fg)(-x) = f(-x)g(-x) = (-f(x))(-g(x)) = f(x)g(x) = (fg)(x)$. $\blacksquare$

Proof (for $f$ even, $g$ odd $\implies fg$ odd). $(fg)(-x) = f(-x)g(-x) = f(x)(-g(x)) = -f(x)g(x) = -(fg)(x)$. $\blacksquare$

Theorem. The only function that is both even and odd is $f(x) = 0$ (the zero function on a symmetric domain).

Proof. $f$ even: $f(-x) = f(x)$. $f$ odd: $f(-x) = -f(x)$. Therefore $f(x) = -f(x)$, so $2f(x) = 0$, hence $f(x) = 0$ for all $x$. $\blacksquare$

Details

Example

Classify $f(x) = x^3 - x$ and $g(x) = \cos(x^2)$.

$f(-x) = (-x)^3 - (-x) = -x^3 + x = -(x^3 - x) = -f(x)$. So $f$ is odd.

$g(-x) = \cos((-x)^2) = \cos(x^2) = g(x)$. So $g$ is even.

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7. Composite Function Domain and Range

Theorem. The domain of $g \circ f$ is $\{x \in \mathrm{dom}(f) : f(x) \in \mathrm{dom}(g)\}$.

Proof. For $(g \circ f)(x) = g(f(x))$ to be defined, we need $x \in \mathrm{dom}(f)$ (so $f(x)$ exists) AND $f(x) \in \mathrm{dom}(g)$ (so $g$ can accept $f(x)$ as input). $\blacksquare$

The range of $g \circ f$ is the image under $g$ of the set $\{f(x) : x \in \mathrm{dom}(g \circ f)\}$, which is a subset of the range of $g$.

Details

Example

Given $f(x) = x^2$ with domain $\mathbb{R}$ and $g(x) = \sqrt{x}$ with domain $[0, \infty)$, find the domain and range of $g \circ f$.

Domain: We need $x \in \mathbb{R}$ (always true) and $f(x) = x^2 \in [0, \infty)$ (always true). So $\mathrm{dom}(g \circ f) = \mathbb{R}$.

Range: $g(f(x)) = \sqrt{x^2} = |x|$. The range of $|x|$ over $\mathbb{R}$ is $[0, \infty)$.

Details

Example

Given $f(x) = \frac{1}{x - 1}$ with domain $\mathbb{R} \setminus \{1\}$ and $g(x) = \ln x$ with domain $(0, \infty)$, find the domain of $g \circ f$.

We need $\frac{1}{x-1} > 0$, so $x - 1 > 0$, giving $x > 1$.

$\mathrm{dom}(g \circ f) = (1, \infty)$.

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8. Self-Inverse Functions

Definition. A function $f$ is self-inverse if $f^{-1} = f$, i.e., $f(f(x)) = x$ for all $x$ in the domain of $f$.

Theorem. If $f$ is self-inverse, then $f$ is bijective and $f = f^{-1}$.

Proof. If $f(f(x)) = x$ for all $x$, then $f$ has an inverse (namely $f$ itself), so $f$ is bijective. And by definition of inverse, $f^{-1} = f$. $\blacksquare$

Common self-inverse functions:

• $f(x) = \frac{a}{x}$ (reciprocal, $x \neq 0$): $f(f(x)) = a/(a/x) = x$. ✓
• $f(x) = a - x$ (reflection): $f(f(x)) = a - (a - x) = x$. ✓
• $f(x) = \frac{ax + b}{cx - a}$ (general fractional linear, $c \neq 0$): verify $f(f(x)) = x$.

Proof for $f(x) = (ax+b)/(cx-a)$. Let $f(x) = \frac{ax+b}{cx-a}$. Then:

$f(f(x)) = \frac◆LB◆a \cdot \frac{ax+b}{cx-a} + b◆RB◆◆LB◆c \cdot \frac{ax+b}{cx-a} - a◆RB◆ = \frac◆LB◆\frac{a(ax+b) + b(cx-a)}{cx-a}◆RB◆◆LB◆\frac{c(ax+b) - a(cx-a)}{cx-a}◆RB◆ = \frac{a^2 x + ab + bcx - ab}{acx + bc - acx + a^2} = \frac{(a^2 + bc)x}{a^2 + bc} = x$

$\blacksquare$ (provided $a^2 + bc \neq 0$).

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9. Modulus Inequalities with Functions

To solve inequalities involving the modulus function:

Type 1: $|f(x)| \lt g(x)$ where $g(x) > 0$.

This is equivalent to the compound inequality $-g(x) \lt f(x) \lt g(x)$.

Proof. $|f(x)| \lt g(x) \iff -g(x) \lt f(x) \lt g(x)$ follows directly from the definition of modulus. $\blacksquare$

Type 2: $|f(x)| \gt g(x)$.

This is equivalent to: $f(x) > g(x)$ OR $f(x) < -g(x)$.

Details

Example

Solve $|x^2 - 3x| \lt 4$.

This gives $-4 \lt x^2 - 3x \lt 4$, i.e., two separate inequalities:

$x^2 - 3x - 4 \lt 0 \implies (x-4)(x+1) \lt 0 \implies -1 \lt x \lt 4$.

$x^2 - 3x + 4 \gt 0$: discriminant $\Delta = 9 - 16 = -7 \lt 0$, and the coefficient of $x^2$ is positive, so this is always true.

Solution: $-1 \lt x \lt 4$.

Details

Example

Solve $|2x - 1| \geq x + 2$.

Case 1: $2x - 1 \geq 0$ (i.e., $x \geq 1/2$): $2x - 1 \geq x + 2 \implies x \geq 3$. Combined with $x \geq 1/2$: $x \geq 3$.

Case 2: $2x - 1 < 0$ (i.e., $x < 1/2$): $-(2x-1) \geq x + 2 \implies -2x + 1 \geq x + 2 \implies -1 \geq 3x \implies x \leq -1/3$. Combined with $x < 1/2$: $x \leq -1/3$.

Solution: $x \leq -1/3$ or $x \geq 3$, i.e., $x \in (-\infty, -1/3] \cup [3, \infty)$.

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10. Transformation Order — Why It Matters

When multiple transformations are applied to $y = f(x)$, the order matters because horizontal and vertical transformations interact differently.

Rule. For $y = af(bx + c) + d$:

1. Apply horizontal transformations first: the argument is $bx + c$, which is a horizontal stretch by factor $1/b$ then a horizontal shift of $-c/b$.
2. Apply vertical transformations second: vertical stretch by $|a|$, reflection if $a < 0$, then vertical shift $d$.

warning

Common Pitfall The horizontal shift in $f(bx + c)$ is $-c/b$, NOT $-c$. The stretch "absorbs" part of the shift. This is the single most common error in transformation problems.

Details

Example

Describe the transformations mapping $y = x^2$ to $y = 3(2x - 4)^2 + 5$.

Reading from inside out:

1. $x^2 \to (2x)^2$: horizontal stretch, factor $1/2$.
2. $(2x)^2 \to (2x - 4)^2 = [2(x - 2)]^2 = 4(x-2)^2$: horizontal shift right by 2.
3. $4(x-2)^2 \to 3 \cdot 4(x-2)^2 = 12(x-2)^2$: vertical stretch, factor 3.
4. $12(x-2)^2 \to 12(x-2)^2 + 5$: vertical shift up by 5.

Note: the horizontal shift is 2 (not 4), because $2x - 4 = 2(x - 2)$.

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11. Problem Set

Problem 1. Given $f(x) = \frac{1}{x - 3} + 2$, find the domain, range, and inverse function.

Details

Solution

Domain: $x \neq 3$, i.e., $\mathbb{R} \setminus \{3\}$.

Range: As $x \to 3^+$, $f(x) \to +\infty$; as $x \to 3^-$, $f(x) \to -\infty$. As $x \to \pm\infty$, $f(x) \to 2$. So $f(x) \neq 2$.

Range: $\mathbb{R} \setminus \{2\}$.

Inverse: $y = \frac{1}{x - 3} + 2 \implies y - 2 = \frac{1}{x - 3} \implies x - 3 = \frac{1}{y - 2} \implies x = \frac{1}{y - 2} + 3$.

$f^{-1}(x) = \frac{1}{x - 2} + 3$, domain $x \neq 2$.

If you get this wrong, revise: Inverse functions

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Problem 2. Given $f(x) = x^2 - 4x + 9$ for $x \geq 2$, find $f^{-1}(x)$.

Details

Solution

Completing the square: $f(x) = (x - 2)^2 + 5$.

For $x \geq 2$, $f$ is injective (strictly increasing).

$y = (x - 2)^2 + 5 \implies (x - 2)^2 = y - 5 \implies x - 2 = \sqrt{y - 5}$

(taking the positive root since $x \geq 2$).

$x = \sqrt{y - 5} + 2$

$f^{-1}(x) = \sqrt{x - 5} + 2$, domain $x \geq 5$.

If you get this wrong, revise: Inverse functions

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Problem 3. Solve $|3x + 1| = 2x + 5$.

Details

Solution

Case 1: $3x + 1 \geq 0$, i.e., $x \geq -\frac{1}{3}$.

$3x + 1 = 2x + 5 \implies x = 4$. Check: $4 \geq -\frac{1}{3}$ ✓

Case 2: $3x + 1 < 0$, i.e., $x < -\frac{1}{3}$.

$-(3x + 1) = 2x + 5 \implies -3x - 1 = 2x + 5 \implies -5x = 6 \implies x = -\frac{6}{5}$.

Check: $-\frac{6}{5} < -\frac{1}{3}$ ✓

Solutions: $x = -\frac{6}{5}$ and $x = 4$.

If you get this wrong, revise: Modulus equations

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Problem 4. Given $f(x) = 2x - 1$ and $g(x) = x^2 + 3$, find $(g \circ f)(x)$ and solve $(g \circ f)(x) = 12$.

Details

Solution

$(g \circ f)(x) = g(f(x)) = g(2x - 1) = (2x - 1)^2 + 3 = 4x^2 - 4x + 4$.

$4x^2 - 4x + 4 = 12 \implies 4x^2 - 4x - 8 = 0 \implies x^2 - x - 2 = 0 \implies (x - 2)(x + 1) = 0$.

$x = 2$ or $x = -1$.

If you get this wrong, revise: Composition

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Problem 5. The graph of $y = f(x)$ passes through $(0, 1)$ and $(3, -2)$. State the coordinates of the corresponding points on: (a) $y = f(x + 2)$ (b) $y = -f(x)$ (c) $y = f(2x)$ (d) $y = 3f(x) - 1$

Details

Solution

(a) $y = f(x + 2)$: shift left by 2. Points: $(-2, 1)$ and $(1, -2)$.

(b) $y = -f(x)$: reflect in $x$-axis. Points: $(0, -1)$ and $(3, 2)$.

(c) $y = f(2x)$: horizontal stretch factor $\frac{1}{2}$. Points: $(0, 1)$ and $\left(\frac{3}{2}, -2\right)$.

(d) $y = 3f(x) - 1$: vertical stretch factor 3, then shift down 1. Points: $(0, 2)$ and $(3, -7)$.

If you get this wrong, revise: Transformations

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Problem 6. The function $f$ is defined by $f(x) = x^3 - 3x + 1$. Show that $f$ is not injective on $\mathbb{R}$, and find the largest interval containing $x = 0$ on which $f$ is injective.

Details

Solution

$f'(x) = 3x^2 - 3 = 3(x - 1)(x + 1)$.

$f'(x) = 0$ at $x = \pm 1$. $f'(x) < 0$ for $-1 < x < 1$ (decreasing), and $f'(x) > 0$ for $x < -1$ or $x > 1$ (increasing).

Since $f$ is decreasing on $(-1, 1)$ and increasing on $(-\infty, -1)$ and $(1, \infty)$, it is not injective on all of $\mathbb{R}$. For example, $f(-2) = -8 + 6 + 1 = -1$ and $f(0) = 1$ and $f(1) = -1$. So $f(-2) = f(1) = -1$ with $-2 \neq 1$.

The largest interval containing $0$ on which $f$ is strictly monotonic (hence injective) is $[-1, 1]$.

If you get this wrong, revise: Injectivity and Differentiation

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Problem 7. Solve the inequality $|x - 3| > |2x + 1|$.

Details

Solution

Square both sides (both sides are non-negative):

$$\begin{aligned} (x - 3)^2 &> (2x + 1)^2 \\ x^2 - 6x + 9 &> 4x^2 + 4x + 1 \\ -3x^2 - 10x + 8 &> 0 \\ 3x^2 + 10x - 8 &< 0 \end{aligned}$$

$(3x - 2)(x + 4) < 0$

$-4 < x < \frac{2}{3}$

If you get this wrong, revise: Modulus function

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Problem 8. Given $f(x) = e^{2x}$ and $g(x) = \ln(x + 1)$, find $f^{-1}$, $g^{-1}$, and the domain of $f \circ g$.

Details

Solution

$f^{-1}(x)$: $y = e^{2x} \implies \ln y = 2x \implies x = \frac◆LB◆\ln y◆RB◆◆LB◆2◆RB◆$.

$f^{-1}(x) = \frac{1}{2}\ln x$, domain $x > 0$.

$g^{-1}(x)$: $y = \ln(x + 1) \implies x + 1 = e^y \implies x = e^y - 1$.

$g^{-1}(x) = e^x - 1$, domain all $\mathbb{R}$.

$(f \circ g)(x) = f(g(x)) = f(\ln(x + 1)) = e^{2\ln(x+1)} = (x + 1)^2$.

Domain of $f \circ g$: we need $x + 1 > 0$ (for $g$), so $x > -1$.

If you get this wrong, revise: Composition and Inverse functions

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Problem 9. Sketch the graph of $y = |x^2 - 4|$, showing the coordinates of all points where the graph meets the axes.

Details

Solution

$y = |x^2 - 4| = |(x - 2)(x + 2)|$.

When $x^2 - 4 \geq 0$ (i.e., $x \leq -2$ or $x \geq 2$): $y = x^2 - 4$ (parabola opening up).

When $x^2 - 4 < 0$ (i.e., $-2 < x < 2$): $y = -(x^2 - 4) = 4 - x^2$ (parabola opening down).

$y$-intercept: $x = 0 \implies y = 4$. Point: $(0, 4)$.

$x$-intercepts: $x^2 - 4 = 0 \implies x = \pm 2$. Points: $(-2, 0)$ and $(2, 0)$.

The graph is the standard parabola $y = x^2 - 4$ with the part between $x = -2$ and $x = 2$ reflected above the $x$-axis.

If you get this wrong, revise: Modulus function and Transformations

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Problem 10. Prove that $f: [0, \infty) \to [0, \infty)$ defined by $f(x) = x^2 + 4x$ is bijective, and find $f^{-1}$.

Details

Solution

Injective: $f(x) = x^2 + 4x = (x+2)^2 - 4$. For $x \geq 0$, $x + 2 > 0$, so $(x+2)^2$ is strictly increasing, hence $f$ is strictly increasing, hence injective.

Surjective: For any $y \geq 0$: $x^2 + 4x - y = 0$. By the quadratic formula: $x = \frac◆LB◆-4 + \sqrt{16 + 4y}◆RB◆◆LB◆2◆RB◆ = -2 + \sqrt{4 + y}$. Since $y \geq 0$: $\sqrt{4 + y} \geq 2$, so $x \geq 0$. Thus every $y \geq 0$ has a preimage.

Inverse: $y = x^2 + 4x \implies x^2 + 4x - y = 0 \implies x = \frac◆LB◆-4 + \sqrt{16 + 4y}◆RB◆◆LB◆2◆RB◆$ (taking the positive root since $x \geq 0$).

$f^{-1}(x) = -2 + \sqrt{4 + x} = \sqrt{x + 4} - 2$, domain $x \geq 0$.

If you get this wrong, revise: Inverse functions

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Problem 11. Classify each function as even, odd, or neither: (a) $f(x) = x^4 - x^2$ (b) $g(x) = x^3 + x$ (c) $h(x) = x + 1$ (d) $k(x) = |x|$

Details

Solution

(a) $f(-x) = (-x)^4 - (-x)^2 = x^4 - x^2 = f(x)$. Even.

(b) $g(-x) = (-x)^3 + (-x) = -x^3 - x = -(x^3 + x) = -g(x)$. Odd.

(c) $h(-x) = -x + 1 \neq h(x)$ and $h(-x) \neq -h(x)$. Neither.

(d) $k(-x) = |-x| = |x| = k(x)$. Even.

If you get this wrong, revise: Even and odd functions

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Problem 12. Given $f(x) = 2x + 3$ with domain $\mathbb{R}$ and $g(x) = \sqrt{x - 1}$ with domain $[1, \infty)$, find the domain of $f \circ g$ and $g \circ f$.

Details

Solution

$f \circ g$: domain is $\{x \geq 1 : g(x) \in \mathbb{R}\} = [1, \infty)$ (since $f$ accepts all reals).

$g \circ f$: domain is $\{x \in \mathbb{R} : f(x) \geq 1\} = \{x : 2x + 3 \geq 1\} = \{x : x \geq -1\} = [-1, \infty)$.

If you get this wrong, revise: Composite function domain

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Problem 13. Verify that $f(x) = \frac{3x + 2}{x - 3}$ is self-inverse.

Details

Solution

$f(f(x)) = \frac◆LB◆3 \cdot \frac{3x+2}{x-3} + 2◆RB◆◆LB◆\frac{3x+2}{x-3} - 3◆RB◆ = \frac◆LB◆\frac{3(3x+2) + 2(x-3)}{x-3}◆RB◆◆LB◆\frac{3x+2 - 3(x-3)}{x-3}◆RB◆ = \frac{9x + 6 + 2x - 6}{3x + 2 - 3x + 9} = \frac{11x}{11} = x$

Since $f(f(x)) = x$, $f$ is self-inverse. ✓

If you get this wrong, revise: Self-inverse functions

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Problem 14. Solve the inequality $|x^2 - 5x + 6| \geq |x - 2|$.

Details

Solution

Factor: $x^2 - 5x + 6 = (x-2)(x-3)$. So $|(x-2)(x-3)| \geq |x-2|$.

If $x = 2$: both sides are $0$, so equality holds. $x = 2$ is a solution.

If $x \neq 2$: divide both sides by $|x-2| > 0$:

$|x - 3| \geq 1$

This gives $x - 3 \geq 1$ or $x - 3 \leq -1$, i.e., $x \geq 4$ or $x \leq 2$.

Combined with $x \neq 2$: $x \leq 2$ or $x \geq 4$.

Solution: $x \in (-\infty, 2] \cup [4, \infty)$.

If you get this wrong, revise: Modulus inequalities

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Problem 15. The graph of $y = f(x)$ passes through $(1, 3)$ and $(-2, 5)$. State the coordinates of the corresponding points on the graph of $y = 2f(3x - 1) + 4$.

Details

Solution

A point $(x_0, y_0)$ on $y = f(x)$ corresponds to a point on the new graph where $f(3x - 1) = y_0$, i.e., $3x - 1 = x_0$, so $x = (x_0 + 1)/3$. The new $y$-value is $2y_0 + 4$.

For $(1, 3)$: new point is $\left(\frac{1+1}{3}, 2 \times 3 + 4\right) = \left(\frac{2}{3}, 10\right)$.

For $(-2, 5)$: new point is $\left(\frac{-2+1}{3}, 2 \times 5 + 4\right) = \left(-\frac{1}{3}, 14\right)$.

If you get this wrong, revise: Transformation order

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tip

tip Ready to test your understanding of Functions? The diagnostic test contains the hardest questions within the A-Level specification for this topic, each with a full worked solution.

Unit tests probe edge cases and common misconceptions. Integration tests combine Functions with other pure mathematics topics to test synthesis under exam conditions.

See Diagnostic Guide for instructions on self-marking and building a personal test matrix.