The shape of the lesson
By the end of the lesson, learners can
- Function notation: In the lab's functions tab, type inputs into the function machine and watch the mapping diagram fill in, each arrow joining an input to its output.
- Inverse functions: Open the inverse functions tab and drag the blue point along \(y = f(x)\).
- Composite functions: In the composite functions tab, chain two machines together and swap their order.
- \(f(a)\): replace every \(x\) in the rule with \(a\), using brackets.
- Solve \(f(x) = k\): set the rule equal to \(k\) and solve.
- Inverse: write \(y = f(x)\), make \(x\) the subject, then write \(f^{-1}(x)\).
- Composite: \(gf(x) = g(f(x))\), so the function nearest \(x\) acts first.
- Simplify algebraic composites fully, expanding brackets in denominators.
One picture every idea returns to
\(f(a)\) means substitute \(a\) into the rule. \(f^{-1}\) undoes \(f\): write \(y = f(x)\) and rearrange to make \(x\) the subject. \(gf(x) = g(f(x))\): apply \(f\) FIRST, then \(g\).
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, Input, rule, output.. Extended. 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: Function notation | In the lab's functions tab, type inputs into the function machine and watch the mapping diagram fill in, each arrow joining an input to its output. 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 Evaluating and solving on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Inverse functions | Open the inverse functions tab and drag the blue point along \(y = f(x)\). Project tab 2 of the lab (Functions and differentiation lab, inverse functions) and let a student drive it while the class predicts. Model the worked example A linear inverse on the board, one line at a time. Your turn: \(f(x) = 3x - 5\). Find \(f^{-1}(7)\) without finding \(f^{-1}(x)\). | Pairs |
| 21 to 30 min | Teach and check: Composite functions | In the composite functions tab, chain two machines together and swap their order. 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 Numbers through two functions on the board, one line at a time. Your turn: \(f(x) = 2x + 1\). Find \(ff(x)\) in its simplest form. | Cooperative Concept Mapping |
| 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, 17 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.
Cooperative Concept Mapping
Teams receive the key terms on cards and arrange them on a large sheet, drawing and labelling the links. Compare maps between teams before showing the model map.
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 |
|---|---|
| In \(2x^2\) only the \(x\) is squared: \(h(-2) = 2 \times (-2)^2 + 3 = 2 \times 4 + 3 = 11\), not \((2 \times -2)^2 + 3 = 19\). | Return to Function notation: ask a pair to state the correct version and show it on the lab before moving on. |
| \(f^{-1}(x)\) is NOT \(\dfrac{1}{f(x)}\). | The \(-1\) means 'inverse', not a power: for \(f(x) = 3x - 5\), \(f^{-1}(x) = \dfrac{x + 5}{3}\), not \(\dfrac{1}{3x - 5}\). |
| \(gf(x)\) means \(f\) first, even though \(g\) is written first. | It is also not a product: \(gf(x) \ne g(x) \times f(x)\). |
Functions: notation, inverse and composite: worksheet
6 questions, 17 marks. Written in the style of the papers; no past paper question is reproduced.
\(f(x) = 7 - 2x\). Find (a) \(f(3)\) (b) \(f(-4)\) (c) the value of \(x\) for which \(f(x) = 11\).
\(h(x) = 2x^2 + 3\) with domain \(\{-2, -1, 0, 1, 2\}\). Draw a mapping diagram and state the range.
\(f(x) = \dfrac{2x - 1}{5}\). Find \(f^{-1}(x)\).
\(f(x) = 4x - 1\) and \(g(x) = x^2 + 2\). Find (a) \(fg(x)\) (b) \(gf(x)\), simplifying each.
\(f(x) = \dfrac{3}{x + 2}\) and \(g(x) = (3x + 5)^2\). Find \(fg(x)\). Give your answer as a fraction in its simplest form.
A temperature of \(x\) °C is converted to °F by \(f(x) = 1.8x + 32\). A temperature of \(x\) kelvin is converted to °C by \(g(x) = x - 273\).
(a) Find \(f(25)\).
(b) Find \(f^{-1}(x)\) and use it to convert 98.6 °F to °C.
(c) Find \(fg(300)\) and say what it represents.
Answer key and marking notes
- (a) \(7 - 6 = 1\).
- (b) \(7 + 8 = 15\).
- (c) \(7 - 2x = 11\), so \(-2x = 4\) and \(x = -2\).
- Arrows: \(-2 \to 11\), \(-1 \to 5\), \(0 \to 3\), \(1 \to 5\), \(2 \to 11\).
- Range \(= \{3, 5, 11\}\).
- \(y = \dfrac{2x - 1}{5}\), so \(5y = 2x - 1\) and \(5y + 1 = 2x\).
- \(f^{-1}(x) = \dfrac{5x + 1}{2}\). (Check: \(f(3) = 1\) and \(f^{-1}(1) = 3\).)
- (a) \(fg(x) = f(x^2 + 2) = 4(x^2 + 2) - 1 = 4x^2 + 7\).
- (b) \(gf(x) = g(4x - 1) = (4x - 1)^2 + 2 = 16x^2 - 8x + 1 + 2 = 16x^2 - 8x + 3\).
- \(fg(x) = \dfrac{3}{(3x + 5)^2 + 2}\).
- \((3x + 5)^2 + 2 = 9x^2 + 30x + 25 + 2 = 9x^2 + 30x + 27\).
- Take out the common factor 3: \(9x^2 + 30x + 27 = 3(3x^2 + 10x + 9)\).
- \(fg(x) = \dfrac{3}{3(3x^2 + 10x + 9)} = \dfrac{1}{3x^2 + 10x + 9}\).
- (a) \(f(25) = 45 + 32 = 77\) (°F).
- (b) \(y = 1.8x + 32\), so \(x = \dfrac{y - 32}{1.8}\): \(f^{-1}(x) = \dfrac{x - 32}{1.8}\).
- \(f^{-1}(98.6) = \dfrac{66.6}{1.8} = 37\) °C.
- (c) \(g(300) = 27\), then \(f(27) = 48.6 + 32 = 80.6\). It converts 300 K to 80.6 °F.
6 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.
\(f(x) = 5 - 2x\). Find \(f(-3)\).
- A\(11\)
- B\(-1\)
- C\(-9\)
- D\(0\)
\(f(x) = \dfrac{2x + 1}{3}\). Find \(f^{-1}(x)\).
- A\(\dfrac{3x - 1}{2}\)
- B\(\dfrac{3}{2x + 1}\)
- C\(\dfrac{3x + 1}{2}\)
- D\(\dfrac{3(x - 1)}{2}\)
\(f(x) = 2x + 3\) and \(g(x) = x^2 - 1\). Find \(gf(2)\).
- A\(48\)
- B\(9\)
- C\(21\)
- D\(49\)
\(f(x) = 2x + 3\) and \(g(x) = x^2 - 1\). Find \(gf(x)\) in its simplest form.
- A\(4x^2 + 12x + 8\)
- B\(2x^2 + 1\)
- C\(4x^2 + 8\)
- D\(2x^3 + 3x^2 - 2x - 3\)
\(f(x) = 10 - 2x\) for the domain \(1 \le x \le 4\). Find the range of \(f\).
- A\(2 \le f(x) \le 8\)
- B\(1 \le f(x) \le 4\)
- C\(f(x) \le 8\)
- D\(12 \le f(x) \le 18\)
\(f(x) = 3x - 4\). Solve \(ff(x) = 20\).
- A\(x = 4\)
- B\(x = \dfrac{8}{3}\)
- C\(x = \dfrac{14}{3}\)
- D\(x = 8\)
Quiz key
\(f(-3) = 5 - 2(-3) = 5 + 6 = 11\). The answer \(-1\) is \(5 - 6\), losing the sign of the product. \(-9\) works out \(5 - 2\) first and then multiplies by \(-3\), ignoring the order of operations. 0 treats \(2x\) as \(2 + x\), giving \(5 - 2 - 3\).
\(y = \dfrac{2x + 1}{3}\), so \(3y = 2x + 1\), \(2x = 3y - 1\) and \(x = \dfrac{3y - 1}{2}\). So \(f^{-1}(x) = \dfrac{3x - 1}{2}\). Check: \(f(1) = 1\) and \(f^{-1}(1) = 1\); \(f(4) = 3\) and \(f^{-1}(3) = 4\). \(\dfrac{3}{2x + 1}\) treats the inverse as a reciprocal. \(\dfrac{3(x - 1)}{2}\) undoes the steps in the wrong order (subtract 1 before multiplying by 3), and \(\dfrac{3x + 1}{2}\) adds 1 instead of subtracting it.
\(f\) first: \(f(2) = 7\). Then \(g(7) = 49 - 1 = 48\). The answer 9 is \(fg(2) = f(3)\), the wrong order. 21 multiplies \(f(2) \times g(2) = 7 \times 3\), but \(gf\) is not a product. 49 forgets the \(-1\) in \(g\).
\(gf(x) = (2x + 3)^2 - 1 = 4x^2 + 12x + 9 - 1 = 4x^2 + 12x + 8\). Check with \(x = 2\): \(16 + 24 + 8 = 48 = gf(2)\). \(2x^2 + 1\) is \(fg(x) = 2(x^2 - 1) + 3\), the other order. \(4x^2 + 8\) squares the bracket as \(4x^2 + 9\), missing the middle term \(12x\). The cubic is the product \(f(x) \times g(x)\).
\(f(1) = 10 - 2 = 8\) and \(f(4) = 10 - 8 = 2\). As \(x\) increases, \(f(x)\) decreases steadily, so the outputs fill \(2 \le f(x) \le 8\). \(1 \le f(x) \le 4\) copies the domain (inputs), not the range. \(f(x) \le 8\) forgets the lower limit set by \(x \le 4\). \(12 \le f(x) \le 18\) uses \(10 + 2x\).
\(ff(x) = 3(3x - 4) - 4 = 9x - 12 - 4 = 9x - 16\). So \(9x - 16 = 20\), \(9x = 36\), \(x = 4\). Check: \(f(4) = 8\), \(f(8) = 20\). \(\dfrac{8}{3}\) comes from \(9x - 4 = 20\), forgetting to multiply the \(-4\) inside the bracket by 3. \(\dfrac{14}{3}\) uses \(2f(x) = 6x - 8\). \(x = 8\) solves \(f(x) = 20\), applying \(f\) only once.
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.