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Module 1 · L15 of 15 ~45 min ⚡ +50 XP in Learn · +25 to complete

Module Synthesis & Exam Technique

A systems engineer does not think about resistors, capacitors, and inductors as separate ideas, they combine them into circuits that solve real problems. In this final lesson, you will do the same with Module 1: synthesise functions, domains, composites and graph features into one coherent toolkit, and learn the exam techniques that turn knowledge into marks.

Today's hook, You see in an exam: "Let $f(x) = \sqrt{x - 1}$ and $g(x) = 2x$. Find $f(g(x))$, state its domain and range, and sketch it." What steps would you take, and how long should 4 marks take you?

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Three printable worksheets that build from foundations to mastery, or build your own from any module’s questions.

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Imagine you are sitting in an exam and see this question: "Let $f(x) = \sqrt{x - 1}$ and $g(x) = 2x$. Find $f(g(x))$, state its domain and range, and sketch $y = f(g(x))$." Outline the steps you would take and estimate how long you should spend on a 4-mark question like this.

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Formula reference · Module 1 summary

complete toolkit

Module 1 Core Toolkit

Function: one input $\to$ exactly one output

Piecewise: a different rule on each part of the domain

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

Even: $f(-x) = f(x)$ Odd: $f(-x) = -f(x)$

Transformations: $y = af(b(x - h)) + k$

$|f(x)|$: fold up $\frac{1}{f(x)}$: reciprocal

Key insight: Most exam questions in this module combine 2–3 of these ideas. Read carefully to identify which tools are needed.

What you'll master

Know

Key facts

The complete scope of Module 1: IQ1, IQ2, IQ3
Standard question types and mark allocations
Common traps that appear in exam questions

Understand

Concepts

How domain, range, composites and graph features interconnect
Why exam technique is as important as content knowledge
How to break down multi-step questions efficiently

Can do

Skills

Solve mixed problems combining multiple topics
Manage time effectively in an exam setting
Show working that maximises partial credit

Key terms

Function

A relation where each input has exactly one output.

Domain

The set of all possible input values for a function.

Range

The set of all possible output values for a function.

Composite Function

A function formed by applying one function to the output of another: f(g(x)).

Combined Transformation

A sequence of translations, reflections and dilations applied to a parent function.

Domain Restriction

Limiting the domain so the rule is actually defined. Like this: $f(x) = \sqrt{x - 3}$ needs $x \geq 3$, because a square root of a negative number is not a real number.

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Module 1 at a glance

Work through the core explanation before applying it.

Module 1 at a glance

synthesis · +3 XP at end

Inquiry Question 1, Working with Functions

Functions & relations: vertical line test, function notation, evaluating functions
Domain & range: interval notation, restrictions (denominator, radical, logarithm)
Piecewise & absolute value: evaluating piecewise rules, solving $|ax + b| = c$
Odd & even: symmetry tests, algebraic verification

Inquiry Question 2, Composite Functions

Piecewise functions: which rule applies on which part of the domain, and whether the pieces join
Composite functions: $f(g(x))$, order matters, finding domains of composites
Working with functions: combining the above in multi-step problems

Inquiry Question 3, Graph Transformations

Translations: horizontal and vertical shifts
Reflections: in $x$-axis ($-f(x)$) and $y$-axis ($f(-x)$)
Dilations: vertical ($af(x)$) and horizontal ($f(bx)$)
Combined transformations: general form $y = af(b(x - h)) + k$
Sketching & modelling: tracking key features, choosing parent functions
Further transformations: $|f(x)|$, $f(|x|)$, $\frac{1}{f(x)}$

The misconception checklist. Two extremely common errors: (1) $f(g(x))$ does NOT mean $f(x) \times g(x)$, a composite substitutes $g$ into $f$; (2) $f(2x - 4)$ is NOT a shift left 4, factor first to get $f(2(x - 2))$, then the horizontal shift is right 2.

IQ1: Function = one output per input; domain restrictions: $\neq 0$, $\geq 0$, $> 0$; even $\leftrightarrow$ $y$-axis symmetry; IQ2: composite = work from the inside out, and $f(g(x))$ is usually not $g(f(x))$

Pause, copy the Module 1 quick-reference: function definition + three domain restrictions + even/odd tests + composite evaluation order + the two-step domain check into your book.

Quick check: For $f(x) = x^2$ and $g(x) = x - 5$, which expression is $(f \circ g)(x)$?

Exam technique for Module 1

exam skills

We just saw all the key facts, functions, domain restrictions, composites and graph features, condensed into a reference card. That raises a question: knowing the content is one thing, but how do you manage time and avoid common traps under exam conditions? This card answers it → the 1–1.5 min-per-mark time rule and the must-label sketch checklist.

Time Allocation

A good rule of thumb is 1 to 1.5 minutes per mark. In a 2-hour exam worth 80 marks, that gives you about 90–120 minutes for writing, leaving 10–20 minutes for checking.

Marks	Suggested time	What to show
1	1 min	Final answer
2	2–3 min	One clear line of working
3	3–4 min	Two–three steps with reasoning
4+	5–6 min	Full working, labelled steps, conclusion

Common Question Types

"Find the domain and range" State restrictions clearly. Use interval notation.
"Find the composite" Substitute in the stated order, then state the domain of the composite explicitly.
"Sketch the graph" Label intercepts, turning points, and asymptotes. Dashed lines for asymptotes earn marks.
"Describe the transformations" Be specific: axis, direction, factor. Do not just say "dilation."
"Evaluate $f(g(x))$" Substitute carefully. Show the inner function first.

Maximising Partial Credit

Even if you cannot finish a question, write down what you know. A correct formula or method statement can earn a mark.
If you get stuck on algebra, define your variables and set up the equation, examiners can award marks for correct setup.
Always check that your answer makes sense: negative time, impossible domains, or graphs crossing asymptotes are red flags.

The "silly mistake" audit. Before you submit, run a 2-minute check on every answer: (1) Did I answer the exact question asked? (2) Are my signs correct? (3) Does the domain/range make sense? (4) Did I label the sketch? This alone can recover 5–10 marks in an exam.

Time rule: 1–1.5 min per mark; 4-mark question = 5–6 min; Every sketch must have: intercepts, turning points, asymptotes labelled with coordinates

Pause, copy the time-per-mark rule (1–1.5 min/mark) and the mandatory sketch checklist (intercepts, turning points, asymptotes, all with coordinates) into your book.

True or false: For a 4-mark question you should aim to spend about 5–6 minutes.

Fill the blanks: drag each token into the correct blank.

substitute restrict factor partial

To evaluate $f(g(x))$, first ___ the inner rule into the outer one. Before reading a horizontal shift, ___ out the coefficient of $x$. If a square root or a denominator appears, you must ___ the domain. Writing a correct method earns ___ credit even if the algebra is wrong.

Common mistakes · the 4 traps that cost marks

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Dodge the traps, then apply

Meet the mistakes that cost marks, then do it yourself.

Confusing $f(g(x))$ with $f(x) \times g(x)$

A composite substitutes one rule into the other; a product multiplies them. With $f(x) = x^2$ and $g(x) = x + 1$, the composite is $(x + 1)^2$ but the product is $x^2(x + 1)$.

✓ Fix: Composite = substitute. Product = multiply. Read the brackets carefully.

Forgetting the inner function when finding a composite's domain

For $f(x) = \sqrt{x}$ and $g(x) = x - 4$, the composite $f(g(x)) = \sqrt{x - 4}$ needs $x \geq 4$, not $x \geq 0$. The inner rule moves the restriction.

✓ Fix: Ask what the inner function outputs, then require that output to lie inside the outer function's domain.

Not factoring out $b$ in combined transformations

$f(2x - 4)$ is NOT a dilation by 2 and a shift left 4. It is $f(2(x - 2))$, so the shift is right 2.

✓ Fix: Always factorise the coefficient of $x$ before reading the horizontal translation.

Sketching without labelled features

A beautifully drawn parabola without a labelled vertex or intercepts will not earn full marks. Labels are essential.

✓ Fix: Every sketch should have at least intercepts and turning points clearly marked with coordinates.

Activity 1, Mixed Topic Problems

Each question below draws on multiple topics from Module 1. Write out your full working.

Let $f(x) = \sqrt{x + 1}$ and $g(x) = 2x - 3$. Find the domain of $f$, the domain of $g$, and the domain of $f(g(x))$.

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Let $f(x) = \dfrac{2}{x - 1} + 3$ and $g(x) = x + 2$. Find $f(g(x))$ and state its domain.

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The graph of $y = f(x)$ passes through $(0, 2)$, $(2, 0)$, and has a maximum at $(1, 3)$. Sketch $y = -2f(x - 1) + 4$ and label the images of these three points.

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Quick check: For $f(g(x))$ where $f(x) = \sqrt{x + 1}$ and $g(x) = 2x - 3$, the domain requires $x \geq$ what value?

Quick-fire practice · 5 reps +2 XP per reveal

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Drill it, then lock it in

Run the quick drill and copy the summary into your book.

What is the domain of $f(x) = \dfrac{1}{\sqrt{x - 2}}$?

Find $f(g(x))$ if $f(x) = 3x + 1$ and $g(x) = x^2$.

Describe all transformations applied to $y = f(x)$ to get $y = -f(2(x + 3)) + 5$.

Is $f(x) = x^3 - x$ odd, even, or neither?

If $f(x) = 2x + 1$ and $g(x) = x^2$, find $f(g(3))$.

Revisit your thinking

Earlier you were asked: How would you approach a 4-mark composite + domain question, and how long should you spend?

For 4 marks, you should allocate about 5 minutes. Here is an efficient approach:

Form the composite (1.5 min): substitute the inner rule into the outer one, in the stated order, and simplify
State domain and range (1 min): start from the inner function's domain, then require its output to lie in the outer function's domain
Sketch the result (2 min): draw $y = f(g(x))$, label intercepts and any endpoint, and mark the excluded region on the $x$-axis

The key is to show each step clearly. Even if your algebra has a small error, you can still earn marks for correct method, correct domain/range reasoning, and a well-labelled sketch.

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Multiple choice

Answer the drill bank and rate your confidence.

Multiple choice

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Pick your answer, then rate your confidence, that tells the system what to drill next.

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Short answer

Write full responses, then check them against the model answers.

Short answer

ApplyBand 44 marks

Q8. Let $f(x) = \sqrt{x - 1}$ and $g(x) = x^2$. (a) Find $f(g(x))$. (b) State its domain. (c) Explain why $g(f(x))$ has a different domain. (4 marks)

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ApplyBand 45 marks

Q9. A company's daily revenue $R$ (in dollars) from selling $n$ hundred units is modelled by $R(n) = -20(n - 5)^2 + 800$. (a) What is the maximum daily revenue and how many units produce it? (b) Find the break-even quantities (where $R = 0$). (c) Sketch the graph of $R(n)$ for $n \geq 0$, labelling key features. (5 marks)

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EvaluateBand 53 marks

Q10. A student writes: "If $f(x) = x^2$ and $g(x) = \sqrt{x}$, then $f(g(x)) = x$ for all real $x$." Evaluate this statement, identifying any errors and stating the correct domain for which the statement is true. (3 marks)

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Comprehensive answers (click to reveal)

Multiple choice, drill bank

MC answers and feedback are shown inline as you complete each question. Use the retry button to attempt a fresh set.

The one-output rule. Functions require exactly one output per input.

Evaluating a composite. $g(2) = 4$, $f(4) = 11$.

Combined domain restrictions. Square root needs $x \geq 2$, denominator needs $x \neq 5$.

Finding an inverse. Swap and solve: $y = \frac{x + 6}{3}$.

Definition of an even function. Definition of an even function.

Activity 1, model answers

1. Domain of $f$: $x \geq -1$. Domain of $g$: all real $x$. For $f(g(x)) = \sqrt{(2x - 3) + 1} = \sqrt{2x - 2}$, need $2x - 2 \geq 0$, so $x \geq 1$.

2. $y = \frac{2}{x - 1} + 3$. $f(g(x)) = \frac{2}{(x + 2) - 1} + 3 = \frac{2}{x + 1} + 3$. The denominator cannot be zero, so the domain is $x \neq -1$.

3. $(0, 2) \rightarrow (1, 0)$; $(2, 0) \rightarrow (3, 4)$; $(1, 3) \rightarrow (2, -2)$. The graph is reflected in the $x$-axis, dilated vertically by 2, and shifted right 1 and up 4.

Short answer model answers

Q8 (4 marks): (a) $f(x) = \sqrt{x^2 - 1}$ [1]. (b) Need $x^2 - 1 \geq 0$, so $x \leq -1$ or $x \geq 1$ [1.5]. (c) $g(f(x)) = \left(\sqrt{x - 1}\right)^2 = x - 1$, but the inner square root already forces $x \geq 1$, so its domain is $[1, \infty)$ [1]. The restriction comes from whichever function is applied first [0.5].

Q9 (5 marks): (a) Maximum revenue is $\$800$ when $n = 5$, i.e., 500 units [1]. (b) $-20(n - 5)^2 + 800 = 0 \Rightarrow (n - 5)^2 = 40 \Rightarrow n = 5 \pm 2\sqrt{10}$. so $n \approx 11.32$ or $n \approx -1.32$. Since $n \geq 0$, only $n \approx 11.32$ is valid, about 1132 units [2]. (c) Parabola opening downward, vertex at $(5, 800)$, the only $n$-intercept in the domain at $n = 5 + 2\sqrt{10} \approx 11.32$, $R$-intercept at $(0, 300)$ [2 marks for correctly labelled sketch].

Q10 (3 marks): The student's statement is incorrect [0.5]. The error is ignoring the domain of $g(x) = \sqrt{x}$, which requires $x \geq 0$ [1]. Also, $f(g(x)) = (\sqrt{x})^2 = x$ only for $x \geq 0$; for negative $x$, $g(x)$ is undefined [1]. The correct domain is $x \geq 0$ [0.5].

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