The shape of the lesson
By the end of the lesson, learners can
- Domain and range: In the lab, choose a function and follow the mapping diagram as each input in the domain is sent to its output: the set of outputs is the range.
- One-one or many-one?: A function is one-one if every output comes from exactly one input.
- Composite functions: In the lab, type in \(f(x)\) and \(g(x)\) and compare \(fg(x)\), where \(x\) goes into \(g\) first, with \(gf(x)\).
- Range: sketch over the domain; check both ends and any vertex inside it.
- Write the range in terms of \(f(x)\), with \(<\) or \(\le\) chosen carefully.
- One-one: no horizontal line meets the graph twice; restrict to one side of a vertex.
- Composite: \(fg(x) = f\big(g(x)\big)\), so \(g\) acts first.
- Existence: \(fg\) needs the range of \(g\) inside the domain of \(f\).
One picture every idea returns to
The domain is the set of inputs; the range is the set of outputs it produces. \(fg(x) = f\big(g(x)\big)\): apply \(g\) FIRST. The composite \(fg\) exists only when the range of \(g\) lies within the domain of \(f\).
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Inputs, outputs and chains.. A function is a rule together with the set of inputs it accepts. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: Domain and range | In the lab, choose a function and follow the mapping diagram as each input in the domain is sent to its output: the set of outputs is the range. Project tab 1 of the lab (Functions and differentiation lab, functions) and let a student drive it while the class predicts. Model the worked example Linear and reciprocal on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: One-one or many-one? | A function is one-one if every output comes from exactly one input. Model the worked example Restricting a domain on the board, one line at a time. | Pairs |
| 21 to 30 min | Teach and check: Composite functions | In the lab, type in \(f(x)\) and \(g(x)\) and compare \(fg(x)\), where \(x\) goes into \(g\) first, with \(gf(x)\). Project tab 3 of the lab (Functions and differentiation lab, composite functions) and let a student drive it while the class predicts. Model the worked example Building fg and gf on the board, one line at a time. Your turn: \(f(x) = x^{2} - 1\) and \(g(x) = 3x + 2\) for \(x \in \mathbb{R}\). Find \(gf(2)\) and \(fg(x)\). | Round Table |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 21 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Composite functions, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Round Table
One sheet travels round the team. Each student adds one line of working or one idea and passes it on, so the solution is built four hands at a time and every contribution is visible.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| Do not just substitute the two ends of the domain. | For Fig. 1 that gives \(7 \le f(x) \le 12\), missing every output between 3 and 7. Always look for a vertex inside the domain. |
| The least \(k\) is the \(x\)-coordinate of the vertex, not its \(y\)-coordinate. | For \((x - 3)^{2} - 8\) the answer is \(k = 3\), not \(-8\). |
| \(fg(x)\) is NOT \(f(x) \times g(x)\). | With \(f(x) = 2x + 1\) and \(g(x) = x^{2}\), the product is \(2x^{3} + x^{2}\), but \(fg(x) = 2x^{2} + 1\). |
Functions: domain, range and composition: worksheet
6 questions, 21 marks. Written in the style of the papers; no past paper question is reproduced.
The function f is defined by \(f(x) = 5 - 2x\) for \(-1 \le x \le 3\). Find the range of f.
\(f(x) = x^{2} + 2\) and \(g(x) = 2x - 3\) for \(x \in \mathbb{R}\).
(a) Find \(fg(x)\) and \(gf(x)\).
(b) Solve \(fg(x) = gf(x)\).
The function f is defined by \(f(x) = 2x^{2} - 8x + 5\) for \(x \in \mathbb{R}\). Express \(f(x)\) in the form \(a(x + b)^{2} + c\) and hence state the range of f.
The function f is defined by \(f(x) = (x + 1)^{2} - 4\) for \(x \ge k\), and g is defined by \(g(x) = 2x + 5\) for \(x \in \mathbb{R}\).
(a) State the least value of \(k\) for which f is one-one, and the range of f for this value of \(k\).
(b) For this value of \(k\), find \(gf(x)\) and state its range.
The functions f and g are defined by \(f : x \mapsto \sqrt{x}\) for \(x \ge 0\) and \(g : x \mapsto x - 5\) for \(x \in \mathbb{R}\).
Explain why the composite function \(gf\) exists but \(fg\) does not.
A shop sells items priced from $20 to $100. It offers 20% off, given by \(f(x) = 0.8x\), and a $5 voucher, given by \(g(x) = x - 5\), where \(x\) is a price in dollars.
(a) Find \(fg(x)\) and \(gf(x)\), and describe in words what each represents.
(b) Which order is better for the customer, and by how much?
(c) Using \(gf\), find the original price of an item that costs $35, and the range of \(gf\) for the shop's prices.
Answer key and marking notes
- \(f\) is decreasing: \(f(-1) = 7\) and \(f(3) = -1\).
- Range: \(-1 \le f(x) \le 7\).
- (a) \(fg(x) = (2x - 3)^{2} + 2 = 4x^{2} - 12x + 11\).
- \(gf(x) = 2(x^{2} + 2) - 3 = 2x^{2} + 1\).
- (b) \(4x^{2} - 12x + 11 = 2x^{2} + 1 \Rightarrow 2x^{2} - 12x + 10 = 0 \Rightarrow x^{2} - 6x + 5 = 0\).
- \((x - 1)(x - 5) = 0\), so \(x = 1\) or \(x = 5\).
- \(2(x^{2} - 4x) + 5 = 2\big[(x - 2)^{2} - 4\big] + 5\).
- \(= 2(x - 2)^{2} - 3\).
- \(2(x - 2)^{2} \ge 0\), so the range is \(f(x) \ge -3\).
- (a) The vertex is at \(x = -1\), so the least \(k\) is \(-1\).
- Range of f: \(f(x) \ge -4\).
- (b) \(gf(x) = 2\big[(x + 1)^{2} - 4\big] + 5 = 2(x + 1)^{2} - 3\).
- Since \(f(x) \ge -4\), \(gf(x) \ge 2(-4) + 5 = -3\). Range: \(gf(x) \ge -3\).
- Range of f is \(f(x) \ge 0\), which lies inside the domain of g (all real numbers), so \(gf(x) = \sqrt{x} - 5\) exists.
- Range of g is all real numbers, including negative values such as \(g(0) = -5\).
- f cannot take a negative input (\(\sqrt{-5}\) is undefined), so the range of g is not within the domain of f and \(fg\) does not exist.
- (a) \(fg(x) = 0.8(x - 5) = 0.8x - 4\): voucher first, then 20% off.
- \(gf(x) = 0.8x - 5\): 20% off first, then the voucher.
- (b) \(gf(x)\) is always $1 less than \(fg(x)\), so taking the discount first saves the customer $1.
- (c) \(0.8x - 5 = 35 \Rightarrow 0.8x = 40 \Rightarrow x = 50\): the original price was $50.
- \(gf\) is increasing: \(gf(20) = 11\) and \(gf(100) = 75\), so the range is \(11 \le gf(x) \le 75\).
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
The function f is defined by \(\text{f}(x) = 5 - 2x\) for \(-1 \le x \le 3\). Find the range of f.
- A\(-1 \le \text{f}(x) \le 7\)
- B\(-1 < \text{f}(x) < 7\)
- C\(3 \le \text{f}(x) \le 11\)
- D\(-1 \le \text{f}(x) \le 3\)
\(\text{f}(x) = x^{2} - 3\) and \(\text{g}(x) = 2x + 1\), both for \(x \in \mathbb{R}\). Find \(\text{fg}(2)\).
- A\(22\)
- B\(3\)
- C\(5\)
- D\(25\)
\(\text{f}(x) = 3x - 1\) and \(\text{g}(x) = x^{2} + 2\), both for \(x \in \mathbb{R}\). Find \(\text{fg}(x)\).
- A\(3x^{2} + 5\)
- B\(9x^{2} - 6x + 3\)
- C\(3x^{3} - x^{2} + 6x - 2\)
- D\(3x^{2} + 1\)
Which of these functions is one-one?
- A\(\text{f}(x) = x^{2} + 1\) for \(x \ge 0\)
- B\(\text{f}(x) = x^{2} + 1\) for \(x \in \mathbb{R}\)
- C\(\text{f}(x) = (x - 2)^{2}\) for \(x \ge 0\)
- D\(\text{f}(x) = x^{2} - 4x\) for \(-1 \le x \le 5\)
The function f is defined by \(\text{f}(x) = x^{2} - 6x + 11\) for \(1 \le x \le 6\). Find the range of f.
- A\(2 \le \text{f}(x) \le 11\)
- B\(6 \le \text{f}(x) \le 11\)
- C\(2 \le \text{f}(x) \le 6\)
- D\(\text{f}(x) \ge 2\)
The function f is defined by \(\text{f}(x) = 2x^{2} + 8x - 3\) for \(x \ge k\). Find the least value of \(k\) for which f is one-one.
- A\(-2\)
- B\(-11\)
- C\(2\)
- D\(-4\)
\(\text{f}(x) = 3x - 2\) and \(\text{g}(x) = x^{2} + 1\), both for \(x \in \mathbb{R}\). Solve the equation \(\text{gf}(x) = 17\).
- A\(x = 2\) or \(x = -\dfrac{2}{3}\)
- B\(x = 2\) only
- C\(x = \pm\dfrac{4}{\sqrt{3}}\)
- D\(x = 2\) or \(x = \dfrac{2}{3}\)
The functions f and g are defined by \(\text{f}(x) = \sqrt{x - 5}\) for \(x \ge 5\) and \(\text{g}(x) = x^{2} - 2x + c\) for \(x \in \mathbb{R}\), where \(c\) is a constant. Find the set of values of \(c\) for which the composite function \(\text{fg}\) exists.
- A\(c \ge 6\)
- B\(c \ge 5\)
- C\(c > 6\)
- D\(c \ge 4\)
Quiz key
\(\text{f}(-1) = 5 + 2 = 7\) and \(\text{f}(3) = 5 - 6 = -1\). The function is decreasing, so every value from \(-1\) to 7 is taken: \(-1 \le \text{f}(x) \le 7\). The tempting strict inequalities exclude the end values, which ARE reached because the domain includes its end points. \(3 \le \text{f}(x) \le 11\) uses \(5 + 2x\), and \(-1 \le \text{f}(x) \le 3\) mixes up the range with the domain.
\(\text{g}(2) = 5\), then \(\text{f}(5) = 25 - 3 = 22\). The tempting 3 is \(\text{gf}(2) = \text{g}(1)\), applying f first. 5 is \(\text{f}(2) \times \text{g}(2) = 1 \times 5\), a product, not a composite, and 25 forgets the \(-3\) in f.
\(\text{fg}(x) = 3(x^{2} + 2) - 1 = 3x^{2} + 6 - 1 = 3x^{2} + 5\). The tempting \(9x^{2} - 6x + 3\) is \(\text{gf}(x) = (3x - 1)^{2} + 2\), the other order. The cubic is the product \(\text{f}(x)\,\text{g}(x)\), and \(3x^{2} + 1\) forgets to multiply the 2 by 3.
\(x^{2} + 1\) is increasing for \(x \ge 0\), so each output comes from exactly one input: one-one. On \(\mathbb{R}\) it is many-one, since \(\text{f}(-1) = \text{f}(1) = 2\). \((x - 2)^{2}\) for \(x \ge 0\) has its vertex at \(x = 2\) inside the domain, so \(\text{f}(0) = \text{f}(4) = 4\). \(x^{2} - 4x\) on \(-1 \le x \le 5\) also has its vertex \(x = 2\) inside, so \(\text{f}(0) = \text{f}(4) = 0\).
\(\text{f}(x) = (x - 3)^{2} + 2\), with vertex \((3,\ 2)\) inside the domain, so the least value is 2. At the ends, \(\text{f}(1) = 6\) and \(\text{f}(6) = 11\); the greater is 11. Range: \(2 \le \text{f}(x) \le 11\). The tempting \(6 \le \text{f}(x) \le 11\) only substitutes the ends and misses the vertex. \(2 \le \text{f}(x) \le 6\) uses the wrong end, and \(\text{f}(x) \ge 2\) is the range for all real \(x\).
\(2x^{2} + 8x - 3 = 2(x^{2} + 4x) - 3 = 2(x + 2)^{2} - 11\). The vertex is at \(x = -2\), and to the right of it f is increasing, so the least \(k\) is \(-2\). The tempting \(-11\) is the \(y\)-coordinate of the vertex, not the \(x\)-coordinate. \(2\) has the wrong sign, and \(-4\) halves 8 without first taking out the factor 2.
\((3x - 2)^{2} + 1 = 17\) gives \((3x - 2)^{2} = 16\), so \(3x - 2 = \pm 4\): \(x = 2\) or \(x = -\dfrac{2}{3}\). The tempting \(x = 2\) only forgets the negative square root. \(\pm\dfrac{4}{\sqrt{3}}\) solves \(\text{fg}(x) = 3x^{2} + 1 = 17\), the wrong order, and \(\dfrac{2}{3}\) comes from \(3x - 2 = -4\) with a sign slip.
\(\text{g}(x) = (x - 1)^{2} + c - 1\), so the range of g is \(\text{g}(x) \ge c - 1\). fg exists when this range lies within the domain of f, \(x \ge 5\): \(c - 1 \ge 5\), so \(c \ge 6\). The tempting \(c \ge 5\) compares the constant \(c\) itself with 5, ignoring the \(-1\) from completing the square. \(c > 6\) wrongly excludes \(c = 6\), where the least output 5 is still allowed (\(\sqrt{0} = 0\)), and \(c \ge 4\) adds 1 instead of subtracting.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.