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, inverses, composites, and transformations into one coherent toolkit, and learn the exam techniques that turn knowledge into marks.
Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
Imagine you are sitting in an exam and see this question: "The function $f(x) = \sqrt{x - 1}$ has an inverse $f^{-1}(x)$. Find $f^{-1}(x)$, state its domain and range, and sketch both $y = f(x)$ and $y = f^{-1}(x)$ on the same set of axes." Outline the steps you would take and estimate how long you should spend on a 4-mark question like this.
Type your initial response below — you will revisit this at the end of the lesson.
Write your initial response in your book. You will revisit it at the end of the lesson.
Wrong: √(a + b) = √a + √b.
Right: The square root of a sum is not the sum of square roots; √(a+b) cannot be simplified this way.
📚 Core Content
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 |
🧪 Activities
1 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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Answer in your workbook.
2 Find the inverse of $f(x) = \dfrac{2}{x - 1} + 3$. State the domain and range of both $f$ and $f^{-1}$.
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Answer in your workbook.
3 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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Answer in your workbook.
1 You have a 4-mark sketching question but only 2 minutes left. What is the minimum you must include to earn the most marks?
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Answer in your workbook.
2 A question asks you to "show that $f(x)$ is even." What is the exact working you would write?
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Answer in your workbook.
Earlier you were asked: How would you approach a 4-mark inverse + sketch question, and how long should you spend?
For 4 marks, you should allocate about 5 minutes. Here is an efficient approach:
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.
Now revisit your initial response. What did you get right? What has changed in your thinking?
Look back at your initial response in your book. Annotate it with what you now understand differently.
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
5 random questions from a replayable lesson bank — feedback shown immediately
✍️ Short Answer
8. Consider $f(x) = x^2 - 4x + 5$ defined for $x \geq 2$. (a) Show that $f$ is one-to-one on this domain. (b) Find $f^{-1}(x)$. (c) State the domain and range of $f^{-1}$. 4 MARKS
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Answer in your workbook.
9. 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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Answer in your workbook.
10. 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
Type your answer below:
Answer in your workbook.
1. Domain of $f$: $x \geq -1$ (radicand must be non-negative). Domain of $g$: all real $x$. For $f(g(x)) = \sqrt{(2x - 3) + 1} = \sqrt{2x - 2}$, we need $2x - 2 \geq 0$, so $x \geq 1$.
2. $y = \frac{2}{x - 1} + 3$. Swap: $x = \frac{2}{y - 1} + 3 \Rightarrow x - 3 = \frac{2}{y - 1} \Rightarrow y - 1 = \frac{2}{x - 3} \Rightarrow y = \frac{2}{x - 3} + 1$. Domain of $f$: $x \neq 1$. Range of $f$: $y \neq 3$. Domain of $f^{-1}$: $x \neq 3$. Range of $f^{-1}$: $y \neq 1$.
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.
1. With only 2 minutes, label the asymptotes (if any), the vertex/turning point with coordinates, and the $x$- and $y$-intercepts. Even without a perfect curve, correctly labelled features earn most of the marks.
2. Calculate $f(-x)$ and show it equals $f(x)$. Example: $f(-x) = (-x)^2 + 2 = x^2 + 2 = f(x)$. Conclude: since $f(-x) = f(x)$ for all $x$ in the domain, $f$ is even.
1. C — Functions require exactly one output per input.
2. A — $g(2) = 4$, $f(4) = 11$.
3. A — Square root needs $x \geq 2$, denominator needs $x \neq 5$.
4. A — Swap and solve: $y = \frac{x + 6}{3}$.
5. A — Definition of an even function.
Q8 (4 marks): (a) $f(x) = (x - 2)^2 + 1$. Vertex at $(2, 1)$. For $x \geq 2$, the parabola is increasing, so it is one-to-one [1]. (b) $y = (x - 2)^2 + 1 \Rightarrow x - 2 = \sqrt{y - 1} \Rightarrow f^{-1}(x) = 2 + \sqrt{x - 1}$ [1.5]. (c) Domain of $f^{-1}$ = range of $f$ = $[1, \infty)$ [0.5]. Range of $f^{-1}$ = domain of $f$ = $[2, \infty)$ [0.5].
Q9 (5 marks): (a) Maximum revenue is \$800 when $n = 5$, i.e., 500 units [1]. (b) $-20(n - 5)^2 + 800 = 0 \Rightarrow (n - 5)^2 = 40 \Rightarrow n = 5 \pm 2\sqrt{10}$. Since $n \geq 0$, both are valid: approximately $n = 0.68$ and $n = 9.32$ (68 and 932 units) [2]. (c) Parabola opening downward, vertex at $(5, 800)$, $n$-intercepts at $5 \pm 2\sqrt{10}$, $R$-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].
The ultimate Module 1 challenge — use all your functions knowledge to defeat the final boss. Pool: lessons 1–15.
Tick when you've finished all activities and checked your answers.