Mathematics · AQA 8300 §A8
Functions
- Evaluate functions at specific inputs using function notation $f(x)$
- Find composite functions $fg(x)$ applying functions in the correct right-to-left order
- Find and verify inverse functions $f^{-1}(x)$ using the swap-and-rearrange method
- Solve equations involving functions including $f(x) = g(x)$
- Find $f^2(x)$ and other iterated and three-way composite functions
🔑 Core Concepts
Function Notation and Evaluation
A function is a mathematical rule that maps each input to exactly one output. The notation $f(x)$ — read "f of x" — denotes the output produced when $x$ is the input to the rule named $f$. Functions can also be written as $f: x \mapsto \text{expression}$, which is read "f maps x to...".
DEFINITION — Function
A function $f$ with domain $D$ assigns to each $x \in D$ exactly one output value $f(x)$. Every input has one — and only one — output. The set of all outputs is the range of $f$.
To evaluate $f$ at a specific value $a$: substitute $a$ for every occurrence of $x$ in the rule and simplify.
Example: if $f(x) = 3x^2 - 2x + 4$, then:
$$f(5) = 3(5)^2 - 2(5) + 4 = 75 - 10 + 4 = 69$$Evaluating a Function
$$f(a) \text{ — replace every } x \text{ in } f(x) \text{ with } a$$
$a$ = the specific input value
$f(a)$ = the resulting output
Always bracket negative or complex substitutions
COMMON MISTAKE — f(x) is not f × x
$f(x)$ is function notation, not multiplication. $f(3)$ means "evaluate function $f$ at input 3", not "$f$ multiplied by 3". This is a fundamental distinction that must be understood before tackling composite functions.
EXAM TIP — Always Bracket Negative Substitutions
For $f(x) = x^2 + 5x$, computing $f(-4)$ requires $(-4)^2 + 5(-4) = 16 - 20 = -4$. Without brackets: $-4^2 = -16$ (wrong). Every negative input must be enclosed in brackets to ensure the squaring acts on the negative sign.
Domain and Range
Understanding domain and range is essential for Grade 9 work, particularly when finding inverses and identifying where composite functions are defined.
DEFINITION — Domain and Range
Domain: the set of all permissible input values $x$ for the function.Range: the set of all output values $f(x)$ produced by inputs from the domain.
Common domain restrictions:
• Division by zero: if $f(x) = \dfrac{1}{x-a}$, then $x \neq a$
• Even roots: if $f(x) = \sqrt{g(x)}$, then $g(x) \geq 0$
EXAM TIP — Stating Domain Restrictions
For $f(x) = \dfrac{3}{2x - 6}$: set $2x - 6 \neq 0 \Rightarrow x \neq 3$. State this explicitly as part of your answer.For $f(x) = \sqrt{4 - x}$: require $4 - x \geq 0 \Rightarrow x \leq 4$.
One-to-One and Many-to-One Functions
Whether a function has an inverse depends critically on its mapping type. This distinction is a key Grade 9 concept.
DEFINITION — Mapping Types
One-to-one (injective): each output value is produced by exactly one input. No two different inputs give the same output. Example: $f(x) = 3x + 1$.Many-to-one: two or more different inputs can produce the same output. Example: $f(x) = x^2$, since $f(3) = f(-3) = 9$.
IMPORTANT — Inverse Functions Require One-to-One
A many-to-one function like $f(x) = x^2$ over all real numbers cannot have an inverse function, because "undoing" it creates ambiguity: the output 9 could map back to $+3$ or $-3$, which is not a function (one input must give one output). The domain must be restricted (e.g., to $x \geq 0$) before an inverse can be defined.
EXAM TIP — Horizontal Line Test
Draw a horizontal line across the graph. If it intersects the graph at most once everywhere, the function is one-to-one and has an inverse. If it intersects twice or more at any height, the function is many-to-one and the domain must be restricted first.
Composite Functions
A composite function chains two (or more) functions: the output of one becomes the input of the next. The crucial rule is the order of application: in $fg(x)$, apply $g$ first, then $f$.
DEFINITION — Composite Function
$$fg(x) = f\bigl(g(x)\bigr)$$
The inner function $g$ is applied to $x$ first. Its output is then used as the input to the outer function $f$. Read the notation right to left.
Composite Function Formula
$$fg(x) = f\!\bigl(g(x)\bigr) \qquad \text{apply } g \text{ first, then } f$$
$g(x)$ = inner function — evaluated first
$f(\cdot)$ = outer function — applied to result of $g$
$fg \neq gf$ in general
COMMON MISTAKE — Reversing the Composite Order
$fg(x) \neq gf(x)$. In $fg(x)$, the function written closest to the $x$ (i.e., $g$) is applied first. In $gf(x)$, $f$ is applied first. This is a very common source of dropped marks.
EXAM TIP — Three Composite Functions
For $fgh(x) = f(g(h(x)))$: apply $h$ first, feed the result into $g$, then feed that result into $f$. Always work from the innermost bracket outwards. Write each step clearly as a separate line.
MEMORY TRICK — Socks Before Shoes
Think of $fg(x)$ as getting dressed: $g$ is your socks (applied first, closest to you), $f$ is your shoes (applied second, on the outside). You always put socks on before shoes. Similarly, in $fg(x)$, apply $g$ before $f$.
The Iterated Function $f^2(x)$
$f^2(x)$ uses superscript notation but does not mean squaring the output. It means $f$ applied to itself — a special case of composition.
DEFINITION — Iterated Function
$$f^2(x) = f\!\bigl(f(x)\bigr) = ff(x)$$
Apply $f$ once to get $f(x)$, then apply $f$ again to that result. This is simply $ff(x)$ — the composite of $f$ with itself.
Iterated Function
$$f^2(x) = f\!\bigl(f(x)\bigr) \qquad \text{NOT} \quad \bigl[f(x)\bigr]^2$$
$f^2$ = apply $f$ twice (compose with itself)
$f^3(x) = f(f(f(x)))$ — apply $f$ three times
COMMON MISTAKE — Squaring the Output
If $f(x) = 2x + 1$, then $f^2(x) = f(f(x)) = f(2x+1) = 2(2x+1)+1 = 4x+3$.It does NOT equal $(2x+1)^2 = 4x^2 + 4x + 1$. The superscript 2 on a function name always denotes iteration, not exponentiation of the output.
Inverse Functions $f^{-1}(x)$
The inverse function $f^{-1}$ reverses the effect of $f$. Where $f$ maps $a \to b$, the inverse $f^{-1}$ maps $b \to a$. Finding the inverse algebraically uses the swap-and-rearrange method.
DEFINITION — Inverse Function
$f^{-1}(x)$ is the function satisfying:
$$f\!\bigl(f^{-1}(x)\bigr) = x \quad \text{and} \quad f^{-1}\!\bigl(f(x)\bigr) = x$$
The domain of $f^{-1}$ equals the range of $f$; the range of $f^{-1}$ equals the domain of $f$.
Write $y = f(x)$
→
Swap $x$ and $y$: write $x = f(y)$
→
Rearrange to make $y$ the subject
→
State $f^{-1}(x) =$ expression & domain
Finding the Inverse Function
$$\text{Let } y = f(x) \;\xrightarrow{\text{swap } x \leftrightarrow y}\; x = f(y) \;\xrightarrow{\text{rearrange}}\; f^{-1}(x) = y$$
Step 1: write as $y = f(x)$
Step 2: swap $x$ and $y$ throughout
Step 3: make $y$ the subject
Step 4: state domain of $f^{-1}$
Verification Identity
$$f\!\bigl(f^{-1}(x)\bigr) = x \qquad \text{for all } x \text{ in the domain of } f^{-1}$$
Use this to check your inverse is correct
Also: $f^{-1}(f(x)) = x$ — both must hold
EXAM TIP — Inverse of a Rational Function
For $f(x) = \dfrac{ax+b}{cx+d}$, after swapping and cross-multiplying:$x(cy + d) = ay + b \Rightarrow cxy + dx = ay + b$
Collect $y$ terms: $cxy - ay = b - dx \Rightarrow y(cx - a) = b - dx$
Then divide: $f^{-1}(x) = \dfrac{b-dx}{cx-a}$, domain $x \neq \dfrac{a}{c}$.
EXAM TIP — Domain Swaps with Inverse
If $f$ has domain $x \geq 3$ and range $y \geq 5$, then $f^{-1}$ has domain $x \geq 5$ and range $y \geq 3$. Domain and range swap exactly when taking the inverse.
🗺️ Visual Notes
Functions
Notation & Evaluation
- $f(x)$ = rule mapping input to output
- Evaluate: substitute $x = a$ with brackets
- $f: x \mapsto$ expression notation
- $f(a)$ = output when input is $a$
Composite Functions
- $fg(x) = f(g(x))$: $g$ applied first
- Read right to left — inner first
- $fg \neq gf$ in general
- $fgh(x)$: apply $h$, then $g$, then $f$
Inverse Functions
- $f^{-1}$ undoes $f$ completely
- Swap $x$ and $y$, then rearrange
- $f(f^{-1}(x)) = x$ always
- State domain restriction explicitly
Domain & Range
- Domain: valid input values
- Range: all possible outputs
- Restrict: division by zero, $\sqrt{\text{negative}}$
- Domain of $f^{-1}$ = Range of $f$
Iterated Functions
- $f^2(x) = f(f(x))$ — apply twice
- NOT $[f(x)]^2$ — do not square output
- $f^3(x) = f(f(f(x)))$
- Expand inner result, then substitute again
Solving Equations
- $f(x) = k$: substitute and solve
- $f(x) = g(x)$: equate, form quadratic
- Use $f^{-1}$ to undo: $f(x) = k \Rightarrow x = f^{-1}(k)$
- Check solutions lie in the domain
One-to-One vs Many-to-One: Comparison
| Property | One-to-One (Injective) | Many-to-One |
|---|---|---|
| Definition | Each output comes from exactly one input | Two or more inputs share the same output |
| Example | $f(x) = 3x + 1$, $f(x) = e^x$ | $f(x) = x^2$, $f(x) = x^4 - 2$ |
| Has inverse? | Yes — over its full domain | Only with domain restriction |
| Horizontal line test | Any horizontal line crosses graph at most once | Some horizontal line crosses graph twice or more |
| Graph shape | Strictly increasing or decreasing | Has a turning point (e.g., a minimum) |
| Domain restriction? | Not needed for inverse | Needed — e.g., restrict $x \geq 0$ for $f(x) = x^2$ |
Process: Finding an Inverse Function (Step by Step)
Example: $f(x) = \dfrac{2x+3}{x-1},\ x \neq 1$
→
Write as equation: $y = \dfrac{2x+3}{x-1}$
→
Swap $x$ and $y$: $x = \dfrac{2y+3}{y-1}$
→
Cross-multiply: $x(y-1) = 2y+3$
→
Expand & collect $y$: $xy - 2y = x + 3$
→
Factorise: $y(x-2) = x+3$
→
$f^{-1}(x) = \dfrac{x+3}{x-2},\ x \neq 2$
Process: Finding a Composite Function
Identify inner and outer functions in $fg(x)$
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Write $g(x)$ fully — this is your new input
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Substitute $g(x)$ into $f$, replacing every $x$
→
Expand all brackets (use $(a+b)^2 = a^2+2ab+b^2$)
→
Collect like terms and simplify fully
✏️ Worked Examples
Grade 4–5 — Evaluating and Basic Composite
Given $f(x) = 4x - 3$ and $g(x) = x^2 + 1$:
(a) Find $f(5)$ (b) Find $g(-2)$ (c) Find $fg(3)$
(a) Find $f(5)$ (b) Find $g(-2)$ (c) Find $fg(3)$
1
Part (a) — Evaluate f(5)
Substitute $x = 5$ into $f(x) = 4x - 3$:
$$f(5) = 4(5) - 3 = 20 - 3 = 17$$
2
Part (b) — Evaluate g(−2) — use brackets for negative input
Substitute $x = -2$ into $g(x) = x^2 + 1$, enclosing in brackets:
$$g(-2) = (-2)^2 + 1 = 4 + 1 = 5$$
Note: $(-2)^2 = +4$, not $-4$. Brackets are essential.
3
Part (c) — Find fg(3): apply g first, then f
Step 1 — compute $g(3)$: $g(3) = (3)^2 + 1 = 10$Step 2 — apply $f$ to the result: $f(10) = 4(10) - 3 = 37$
So $fg(3) = f(g(3)) = f(10) = 37$
$f(5) = 17$ | $g(-2) = 5$ | $fg(3) = 37$
Grade 6–7 — Composite Expressions and Inverse
Given $f(x) = 2x + 5$ and $g(x) = x^2 - 2$:
(a) Find $fg(x)$ as a simplified expression
(b) Find $gf(x)$ as a simplified expression
(c) Find $f^{-1}(x)$
(d) Verify that $f(f^{-1}(x)) = x$
(a) Find $fg(x)$ as a simplified expression
(b) Find $gf(x)$ as a simplified expression
(c) Find $f^{-1}(x)$
(d) Verify that $f(f^{-1}(x)) = x$
1
Part (a) — fg(x): apply g first, then f
$fg(x) = f(g(x)) = f(x^2 - 2)$Replace every $x$ in $f(x) = 2x + 5$ with $(x^2 - 2)$: $$fg(x) = 2(x^2 - 2) + 5 = 2x^2 - 4 + 5 = 2x^2 + 1$$
2
Part (b) — gf(x): apply f first, then g
$gf(x) = g(f(x)) = g(2x + 5)$Replace every $x$ in $g(x) = x^2 - 2$ with $(2x + 5)$: $$gf(x) = (2x + 5)^2 - 2 = 4x^2 + 20x + 25 - 2 = 4x^2 + 20x + 23$$ Note: $fg(x) \neq gf(x)$, confirming order matters.
3
Part (c) — Find f⁻¹(x) using swap-and-rearrange
Write: $y = 2x + 5$Swap $x$ and $y$: $x = 2y + 5$
Rearrange: $x - 5 = 2y \implies y = \dfrac{x-5}{2}$ $$f^{-1}(x) = \frac{x-5}{2}$$
4
Part (d) — Verify f(f⁻¹(x)) = x algebraically
Substitute $f^{-1}(x) = \dfrac{x-5}{2}$ into $f$:
$$f\!\left(\frac{x-5}{2}\right) = 2 \cdot \frac{x-5}{2} + 5 = (x-5) + 5 = x \checkmark$$
$fg(x) = 2x^2 + 1$ | $gf(x) = 4x^2 + 20x + 23$ | $f^{-1}(x) = \dfrac{x-5}{2}$ | Verified $f(f^{-1}(x)) = x$
Grade 9 — Rational Function: Iterated, Inverse, and Proof
Given $f(x) = \dfrac{x + 2}{x - 1}$, $x \neq 1$, and $g(x) = 3x - 1$:
(a) Find $f^2(x) = f(f(x))$, simplifying your answer fully.
(b) Find $f^{-1}(x)$, stating its domain.
(c) Prove algebraically that $f(f^{-1}(x)) = x$.
(d) Find $fg(x)$ and state the value of $x$ excluded from its domain.
(a) Find $f^2(x) = f(f(x))$, simplifying your answer fully.
(b) Find $f^{-1}(x)$, stating its domain.
(c) Prove algebraically that $f(f^{-1}(x)) = x$.
(d) Find $fg(x)$ and state the value of $x$ excluded from its domain.
1
Part (a) — Find f²(x): substitute f(x) back into f
We need $f\!\left(\dfrac{x+2}{x-1}\right)$ — replace every $x$ in $f(x) = \dfrac{x+2}{x-1}$ with $\dfrac{x+2}{x-1}$:
$$f^2(x) = \frac{\dfrac{x+2}{x-1} + 2}{\dfrac{x+2}{x-1} - 1}$$
Multiply numerator and denominator by $(x-1)$ to clear fractions:
$$= \frac{(x+2) + 2(x-1)}{(x+2) - 1(x-1)} = \frac{x+2+2x-2}{x+2-x+1} = \frac{3x}{3} = x$$
So $f^2(x) = x$ — this function is its own inverse (it is self-inverse)!
2
Part (b) — Find f⁻¹(x)
Write: $y = \dfrac{x+2}{x-1}$Swap: $x = \dfrac{y+2}{y-1}$
Multiply both sides by $(y-1)$: $x(y-1) = y + 2$
Expand: $xy - x = y + 2$
Collect $y$ terms: $xy - y = x + 2$
Factorise: $y(x-1) = x+2$
$$f^{-1}(x) = \frac{x+2}{x-1}$$ Domain: $x \neq 1$ (since $x - 1 \neq 0$). This confirms $f$ is self-inverse: $f^{-1}(x) = f(x)$.
3
Part (c) — Prove f(f⁻¹(x)) = x algebraically
Substitute $f^{-1}(x) = \dfrac{x+2}{x-1}$ into $f$:
$$f\!\left(\frac{x+2}{x-1}\right) = \frac{\dfrac{x+2}{x-1}+2}{\dfrac{x+2}{x-1}-1}$$
Multiply top and bottom by $(x-1)$:
$$= \frac{(x+2)+2(x-1)}{(x+2)-1\cdot(x-1)} = \frac{3x}{3} = x \checkmark$$
Since the result equals $x$, the proof is complete.
4
Part (d) — Find fg(x) and its domain exclusion
$fg(x) = f(g(x)) = f(3x-1)$. Replace every $x$ in $f$ with $(3x-1)$:
$$fg(x) = \frac{(3x-1)+2}{(3x-1)-1} = \frac{3x+1}{3x-2}$$
Domain exclusion: $(3x - 2) \neq 0 \Rightarrow x \neq \dfrac{2}{3}$
$f^2(x) = x$ (self-inverse) | $f^{-1}(x) = \dfrac{x+2}{x-1},\ x\neq 1$ | Proved $f(f^{-1}(x)) = x$ | $fg(x) = \dfrac{3x+1}{3x-2},\ x \neq \dfrac{2}{3}$
❓ Exam Questions
Question 1
1 mark
$f(x) = 7 - 3x$. Find $f(-4)$.
Answer:
$f(-4) = 7 - 3(-4) = 7 + 12 = 19$
Mark scheme: 1 mark for correct evaluation. Common error: writing $7 - 3 \times -4 = 7 + 12$ without brackets on the $-4$ — award mark if correct answer reached, but advise use of brackets.
$f(-4) = 7 - 3(-4) = 7 + 12 = 19$
Mark scheme: 1 mark for correct evaluation. Common error: writing $7 - 3 \times -4 = 7 + 12$ without brackets on the $-4$ — award mark if correct answer reached, but advise use of brackets.
Question 2
2 marks
$f(x) = 6x - 1$. Find $f^{-1}(x)$.
Method:
Let $y = 6x - 1$ [M1 — correct method: swap and rearrange]
Swap: $x = 6y - 1$
$x + 1 = 6y$
$y = \dfrac{x + 1}{6}$
Answer: $f^{-1}(x) = \dfrac{x+1}{6}$ [A1]
2 marks
Let $y = 6x - 1$ [M1 — correct method: swap and rearrange]
Swap: $x = 6y - 1$
$x + 1 = 6y$
$y = \dfrac{x + 1}{6}$
Answer: $f^{-1}(x) = \dfrac{x+1}{6}$ [A1]
2 marks
Question 3
3 marks
$f(x) = x^2 - 3$ and $g(x) = 4x + 1$.
(a) Find $gf(x)$ as a simplified expression. [2]
(b) Find $gf(2)$. [1]
(a) $gf(x) = g(f(x)) = g(x^2 - 3) = 4(x^2 - 3) + 1 = 4x^2 - 12 + 1 = 4x^2 - 11$ [M1 correct method, A1 simplified]
(b) $gf(2) = 4(2)^2 - 11 = 16 - 11 = 5$ [A1 — or substitute $x=2$ directly into the expression]
3 marks total
(b) $gf(2) = 4(2)^2 - 11 = 16 - 11 = 5$ [A1 — or substitute $x=2$ directly into the expression]
3 marks total
Question 4
4 marks
$f(x) = 2x + 3$ and $g(x) = \dfrac{1}{x}$, $x \neq 0$.
(a) Find $f^{-1}(x)$. [2]
(b) Solve $f(x) = g(x)$, giving your answers to 3 significant figures. [2]
(a) $y = 2x+3 \to x = 2y+3 \to y = \dfrac{x-3}{2}$, so $f^{-1}(x) = \dfrac{x-3}{2}$ [M1 A1]
(b) Set $f(x) = g(x)$: $\quad 2x + 3 = \dfrac{1}{x}$
Multiply by $x$: $2x^2 + 3x = 1$
$2x^2 + 3x - 1 = 0$ [M1 — correct quadratic formed]
Using the quadratic formula: $x = \dfrac{-3 \pm \sqrt{9 + 8}}{4} = \dfrac{-3 \pm \sqrt{17}}{4}$
$x = 0.281$ or $x = -1.78$ (3 s.f.) [A1 — both values]
4 marks total
(b) Set $f(x) = g(x)$: $\quad 2x + 3 = \dfrac{1}{x}$
Multiply by $x$: $2x^2 + 3x = 1$
$2x^2 + 3x - 1 = 0$ [M1 — correct quadratic formed]
Using the quadratic formula: $x = \dfrac{-3 \pm \sqrt{9 + 8}}{4} = \dfrac{-3 \pm \sqrt{17}}{4}$
$x = 0.281$ or $x = -1.78$ (3 s.f.) [A1 — both values]
4 marks total
Question 5
6 marks
$f(x) = \dfrac{3x - 2}{x + 4}$, $x \neq -4$.
(a) Find $f^{-1}(x)$, stating its domain. [4]
(b) Find $f^2(x)$, simplifying your answer fully. [2]
(a)
$y = \dfrac{3x-2}{x+4}$, swap: $x = \dfrac{3y-2}{y+4}$ [M1 — swap method shown]
$x(y+4) = 3y - 2 \implies xy + 4x = 3y - 2$ [M1 — multiply and expand]
$xy - 3y = -4x - 2 \implies y(x-3) = -4x - 2$ [M1 — collect $y$ and factorise]
$f^{-1}(x) = \dfrac{-4x-2}{x-3} = \dfrac{-(4x+2)}{x-3}$, domain $x \neq 3$ [A1 — expression and domain]
(b)
$f^2(x) = f\!\left(\dfrac{3x-2}{x+4}\right) = \dfrac{3\cdot\dfrac{3x-2}{x+4} - 2}{\dfrac{3x-2}{x+4} + 4}$
Multiply top and bottom by $(x+4)$: [M1]
Numerator: $3(3x-2) - 2(x+4) = 9x - 6 - 2x - 8 = 7x - 14$
Denominator: $(3x-2) + 4(x+4) = 3x - 2 + 4x + 16 = 7x + 14$
$f^2(x) = \dfrac{7x-14}{7x+14} = \dfrac{7(x-2)}{7(x+2)} = \dfrac{x-2}{x+2}$ [A1 — fully simplified]
6 marks total
$y = \dfrac{3x-2}{x+4}$, swap: $x = \dfrac{3y-2}{y+4}$ [M1 — swap method shown]
$x(y+4) = 3y - 2 \implies xy + 4x = 3y - 2$ [M1 — multiply and expand]
$xy - 3y = -4x - 2 \implies y(x-3) = -4x - 2$ [M1 — collect $y$ and factorise]
$f^{-1}(x) = \dfrac{-4x-2}{x-3} = \dfrac{-(4x+2)}{x-3}$, domain $x \neq 3$ [A1 — expression and domain]
(b)
$f^2(x) = f\!\left(\dfrac{3x-2}{x+4}\right) = \dfrac{3\cdot\dfrac{3x-2}{x+4} - 2}{\dfrac{3x-2}{x+4} + 4}$
Multiply top and bottom by $(x+4)$: [M1]
Numerator: $3(3x-2) - 2(x+4) = 9x - 6 - 2x - 8 = 7x - 14$
Denominator: $(3x-2) + 4(x+4) = 3x - 2 + 4x + 16 = 7x + 14$
$f^2(x) = \dfrac{7x-14}{7x+14} = \dfrac{7(x-2)}{7(x+2)} = \dfrac{x-2}{x+2}$ [A1 — fully simplified]
6 marks total
Question 6
3 marks
$h(x) = x^2 + 3$, $x \geq 0$.
(a) State why the domain restriction $x \geq 0$ is necessary for $h^{-1}(x)$ to exist. [1]
(b) Find $h^{-1}(x)$. [2]
(a) Without the restriction, $h(x) = x^2 + 3$ is many-to-one (e.g., $h(2) = h(-2) = 7$), so it does not have an inverse. The restriction $x \geq 0$ makes it one-to-one. [B1]
(b)
$y = x^2 + 3$, swap: $x = y^2 + 3$
$x - 3 = y^2 \implies y = \sqrt{x-3}$ (positive root only, since range of $h^{-1}$ = domain of $h \geq 0$)
$h^{-1}(x) = \sqrt{x-3}$, domain $x \geq 3$ [M1 method, A1 answer with domain]
3 marks total
(b)
$y = x^2 + 3$, swap: $x = y^2 + 3$
$x - 3 = y^2 \implies y = \sqrt{x-3}$ (positive root only, since range of $h^{-1}$ = domain of $h \geq 0$)
$h^{-1}(x) = \sqrt{x-3}$, domain $x \geq 3$ [M1 method, A1 answer with domain]
3 marks total
⭐ Grade 9 Model Answers
Full Annotated Answer: Question 5
This question targets the highest-order skills in the Functions topic: rational function inverse with domain, and $f^2(x)$ requiring compound fraction simplification. These are distinguishing skills between Grade 8 and Grade 9.
ANNOTATED ANSWER — Part (a): Inverse of a Rational Function (4 marks)
Setup (M1): $y = \dfrac{3x-2}{x+4}$ — write as equation, then swap $x$ and $y$:
$$x = \frac{3y-2}{y+4}$$
This step earns M1. The examiner requires the swap to be explicit — do not skip it.Cross-multiply (M1):
$x(y + 4) = 3y - 2$
$xy + 4x = 3y - 2$
Expanding correctly earns M1. A common error is sign mistakes here.
Collect and factorise (M1):
$xy - 3y = -4x - 2$
$y(x - 3) = -4x - 2$
This is the critical algebraic step. Moving all $y$ terms to one side and factorising earns M1. Students who forget to factorise cannot complete the problem.
Final answer (A1): $$f^{-1}(x) = \frac{-4x-2}{x-3} = \frac{-(4x+2)}{x-3}, \quad x \neq 3$$ The A1 mark requires both the simplified expression AND the domain restriction $x \neq 3$. Omitting the domain loses this mark.
ANNOTATED ANSWER — Part (b): Finding f²(x) for a Rational Function (2 marks)
Set up the substitution (M1):$f^2(x) = f\!\left(\dfrac{3x-2}{x+4}\right)$ — substitute $f(x)$ as the input to $f$: $$= \frac{3 \cdot \dfrac{3x-2}{x+4} - 2}{\dfrac{3x-2}{x+4} + 4}$$ The M1 is for correctly setting up the compound fraction. The error of computing $(f(x))^2$ instead earns no marks.
Multiply top and bottom by $(x+4)$ to clear denominators:
Numerator: $3(3x-2) - 2(x+4) = 9x - 6 - 2x - 8 = 7x - 14$
Denominator: $(3x-2) + 4(x+4) = 3x - 2 + 4x + 16 = 7x + 14$
Simplify (A1): $$f^2(x) = \frac{7x-14}{7x+14} = \frac{7(x-2)}{7(x+2)} = \frac{x-2}{x+2}$$ The A1 requires full simplification — cancelling the factor of 7. An unsimplified $\dfrac{7x-14}{7x+14}$ does not earn A1.
Why These Steps Earn Full Marks at Grade 9
In any 4+ mark functions question, examiners award marks at distinct algebraic milestones: (1) the swap, (2) the expansion/cross-multiplication, (3) the factorisation of $y$, (4) the correct expression plus domain. Similarly, for $f^2(x)$ with a rational function: (1) setting up the compound fraction correctly, (2) simplifying fully including cancellation. Working clearly through each stage, with every line of algebra shown, guarantees method marks even if a numerical slip occurs.
📋 Revision Sheet
Key Definitions
| Term | Meaning |
|---|---|
| $f(x)$ | Output of function $f$ when input is $x$ |
| Domain | Set of all valid input values |
| Range | Set of all possible output values |
| $fg(x)$ | Composite: apply $g$ first, then $f$ |
| $f^{-1}(x)$ | Inverse: reverses the effect of $f$ |
| $f^2(x)$ | $f$ applied twice: $f(f(x))$ |
| One-to-one | Each output has a unique input; inverse exists |
| Many-to-one | Multiple inputs share an output; needs domain restriction for inverse |
Essential Formulae
Composite: $fg(x) = f(g(x))$ — apply $g$ first
Three-way: $fgh(x) = f(g(h(x)))$ — apply $h$, then $g$, then $f$
Iterated: $f^2(x) = f(f(x)) \neq [f(x)]^2$
Inverse steps: $y = f(x) \to$ swap $x,y \to$ make $y$ the subject $\to f^{-1}(x)$
Verification: $f(f^{-1}(x)) = x$ always
Rational inverse: cross-multiply $\to$ expand $\to$ collect $y$ $\to$ factorise $\to$ divide
Domain swap: domain of $f^{-1}$ = range of $f$
Memory Hooks
- Socks before shoes: in $fg(x)$, $g$ goes on first (inner), $f$ second (outer)
- Swap and rearrange: the two words that describe finding any inverse
- $f^2 \neq [f]^2$: apply twice — never square the output
- MECFD: Multiply, Expand, Collect, Factorise, Divide — for rational inverse
- Domain swaps: domain $\leftrightarrow$ range when taking inverse
- $ff^{-1}(x) = x$: the ultimate check for any inverse
Exam Tips
- Always use brackets: $f(-3)$ means substitute $(-3)$, not $-3$
- Show ALL algebraic steps — method marks are awarded for process
- State domain restriction every time you find an inverse
- Label each stage of a composite calculation: "inner function $g(x) = ...$, then $f(...) = ...$"
- For $f^2(x)$ with a rational function, multiply numerator and denominator by the original denominator
- Verify your inverse by checking $f(f^{-1}(x)) = x$ — if it doesn't simplify to $x$, you have an error
- When solving $f(x) = g(x)$, multiply through to clear any fractions, then form a quadratic
🔄 Flashcards
Click each card to flip it and reveal the answer.
✗ Common Mistakes
MISTAKE 1 — Reversing the Composite Order
What students do: For $fg(x)$, they compute $f(x)$ first, then substitute into $g$ — the exact opposite of what is required.Why marks are lost: The entire answer is wrong. Even if the algebra is correct, applying functions in the wrong order earns zero marks.
How to avoid: Remember the rule: the function written closest to $x$ is applied first. In $fg(x)$, $g$ is nearest to $x$, so $g$ goes first. Say aloud: "$fg(x)$ means apply $g$ first, then $f$."
MISTAKE 2 — Confusing $f^2(x)$ with $[f(x)]^2$
What students do: When asked for $f^2(x)$, they square the expression for $f(x)$ instead of applying $f$ twice.Why marks are lost: The notation $f^2(x) = f(f(x))$ is standard — squaring gives a completely different (and incorrect) answer. No marks are awarded.
How to avoid: The superscript on a function name always means iteration, not exponentiation. $f^2(x)$ = "apply $f$ twice". $(f(x))^2$ would be written with explicit brackets around $f(x)$.
MISTAKE 3 — Omitting the Domain Restriction for the Inverse
What students do: They correctly derive $f^{-1}(x)$ algebraically but fail to state the excluded value (e.g., $x \neq 3$ for a rational inverse with denominator $x - 3$).Why marks are lost: The final A1 mark in a 4-mark inverse question almost always requires the domain to be stated. Leaving it out drops a mark even though the expression is correct.
How to avoid: After finding $f^{-1}(x)$, always ask: "Does the denominator equal zero for any value of $x$?" State that value as excluded. Write "domain: $x \neq ...$" explicitly.
MISTAKE 4 — Bracket Error When Substituting Negative Values
What students do: For $f(x) = x^2 - 4x$, they write $f(-3) = -3^2 - 4(-3) = -9 + 12 = 3$ (wrong, because $-3^2 = -9$ without brackets).Why marks are lost: The correct answer is $(-3)^2 - 4(-3) = 9 + 12 = 21$. The omission of brackets on the negative input changes the sign and produces an entirely wrong answer.
How to avoid: Form a habit: whenever substituting a negative value, write it inside brackets before simplifying. Every single time — no exceptions.
MISTAKE 5 — Failing to Collect $y$ Terms When Finding a Rational Inverse
What students do: After expanding $x(y+a) = by + c$ to $xy + ax = by + c$, they try to divide both sides immediately, not realising that $y$ appears on both sides.Why marks are lost: You cannot isolate $y$ by dividing if $y$ appears in both the numerator and denominator of the result. The solution breaks down and no further marks are available after this error.
How to avoid: After expanding, always move all $y$ terms to one side: $xy - by = c - ax$. Then factorise: $y(x - b) = c - ax$. Only then divide: $y = \dfrac{c - ax}{x - b}$.
MISTAKE 6 — Claiming a Many-to-One Function Has an Inverse Without Restricting the Domain
What students do: For $f(x) = x^2$, they write $f^{-1}(x) = \sqrt{x}$ without any qualification or domain restriction.Why marks are lost: Without the restriction $x \geq 0$ (or $x \leq 0$) on $f$, it is many-to-one and $\sqrt{x}$ is not the inverse of $f$ over all reals. The mark requiring explicit acknowledgement of the restriction is lost.
How to avoid: When the function has a turning point or symmetry (quadratic, modulus, even power), state the required domain restriction and explain why it is necessary before giving the inverse.
✅ Final Checklist
Click each item when you are confident with it. Your progress is tracked automatically.
- I can evaluate $f(a)$ by substituting $a$ for $x$ in $f(x)$
- I always enclose negative substitutions in brackets: $f(-3) = (-3)^2$, not $-3^2$
- I understand that $f(x)$ is function notation, not $f$ multiplied by $x$
- I can identify and state the domain of a function (excluding zero-denominators and square roots of negatives)
- I can find the range of a function given its domain
- I can find $fg(x)$ by applying $g$ first and substituting into $f$
- I know that $fg(x) \neq gf(x)$ in general, and I can find both correctly
- I can find three-way composite functions $fgh(x)$ working from right to left
- I know $f^2(x) = f(f(x))$, not $[f(x)]^2$, and can compute it for any function type
- I can find $f^{-1}(x)$ for linear and rational functions using the swap-and-rearrange method
- I can find the inverse of a rational function (multiply, collect $y$, factorise, divide)
- I always state the domain restriction when giving an inverse function
- I can prove $f(f^{-1}(x)) = x$ algebraically with full working
- I can explain why a many-to-one function requires domain restriction before its inverse exists
- I can solve $f(x) = g(x)$ by forming and solving a quadratic equation