Functions

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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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 $\ln x$.

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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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 : $f \circ g \neq g \circ f$.

Example

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

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

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

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

Caution: Warning Right to left: $(f \circ g)(x) = f(g(x))$.

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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).

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 $f$).

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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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., $2a \geq 0$). $\blacksquare$

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.

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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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 $b$.

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 $(-x_0, y_0)$. $\blacksquare$

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 $\frac{1}{a}$. $\blacksquare$

### 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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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 $x$. $\blacksquare$

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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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 $g$.

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 $[0, \infty)$.

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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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:

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

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

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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 $f(x) < -g(x)$.

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

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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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 $d$.

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

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](#3-inverse-functions)

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

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](#3-inverse-functions)

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

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](#41-solving-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 $(g \circ f)(x) = 12$.

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](#2-composition-of-functions)

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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) $y = 3f(x) - 1$

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](#5-transformations-of-graphs)

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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.

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](#3-inverse-functions) and [Differentiation](10-differentiation.mdx)

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

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](#4-the-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 $f \circ g$.

Solution

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

$f^{-1}(x) = \frac{1}{2}\ln x$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](#2-composition-of-functions) and [Inverse functions](#3-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.

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](#4-the-modulus-function) and [Transformations](#5-transformations-of-graphs)

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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 $f^{-1}$.

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: $x = \frac{-4 + \sqrt{16 + 4y}}{2} = -2 + \sqrt{4 + y}$. Since $y \geq 0$: $\sqrt{4 + y} \geq 2$So $x \geq 0$. Thus every $y \geq 0$ has a preimage.

Inverse: $y = x^2 + 4x \implies x^2 + 4x - y = 0 \implies x = \frac{-4 + \sqrt{16 + 4y}}{2}$ (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](#3-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) $k(x) = |x|$

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](#6-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 $g \circ f$.

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](#7-composite-function-domain-and-range)

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

Solution

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

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

If you get this wrong, revise: [Self-inverse functions](#8-self-inverse-functions)

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

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](#9-modulus-inequalities-with-functions)

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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 $y = 2f(3x - 1) + 4$.

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](#10-transformation-order--why-it-matters)

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• Differentiation — Derivatives describe the rate of change of functions and are used to find stationary points and sketch graphs.
• Exponentials and Logarithms — The exponential and logarithmic functions are inverses of each other, a key example of function inversion.
• Integration — Integration finds areas under function curves and reverses the process of differentiation.
• Sequences and Series — Sequences are functions defined on the natural numbers, linking function behaviour to series convergence.

1. Forgetting the $+c$ constant of integration in indefinite integrals, or misusing boundary conditions in definite integrals.

2. Losing marks by not showing sufficient working. Always write out each step, especially in proof questions.

3. Confusing the domain and range of functions, or not considering restrictions (e.g., denominator cannot be zero).

4. Forgetting to check that solutions satisfy the original equation (especially with squaring both sides or dividing by variables).

The key principles covered in this topic are linked in the sub-pages above. Focus on understanding the definitions, applying the formulas or frameworks, and evaluating strengths and limitations of each approach.

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.