The shape of the lesson
By the end of the lesson, learners can
- Undo, in the reverse order: In the lab, pick a function and watch its inverse appear as the reflection in \(y = x\): every point \((a, b)\) on \(y = f(x)\) becomes \((b, a)\).
- Inverses of quadratics: A quadratic is only one-one on one side of its vertex, so the domain is always restricted.
- Mirror in y = x: Because the inverse swaps every \((a, b)\) to \((b, a)\), the graphs of \(f\) and \(f^{-1}\) are reflections of each other in \(y = x\).
- Check the function is one-one on its domain.
- Write \(y = f(x)\) and make \(x\) the subject (complete the square for a quadratic).
- Choose the sign of any square root to match the domain of \(f\).
- Swap letters to write \(f^{-1}(x)\).
- State the domain of \(f^{-1}\): it is the range of \(f\).
One picture every idea returns to
To find \(f^{-1}\): write \(y = f(x)\), make \(x\) the subject, then swap the letters. The domain of \(f^{-1}\) is the range of \(f\), the range of \(f^{-1}\) is the domain of \(f\), and the graph of \(y = f^{-1}(x)\) is the reflection of \(y = f(x)\) in the line \(y = x\).
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, Run the function backwards.. An inverse function undoes what the function did. 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: Undo, in the reverse order | In the lab, pick a function and watch its inverse appear as the reflection in \(y = x\): every point \((a, b)\) on \(y = f(x)\) becomes \((b, a)\). 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 Linear on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Inverses of quadratics | A quadratic is only one-one on one side of its vertex, so the domain is always restricted. Model the worked example Positive root on the board, one line at a time. Your turn: \(f(x) = x^{2} + 6x + 2\) for \(x \ge -3\). Find \(f^{-1}(x)\). | Pairs |
| 21 to 30 min | Teach and check: Mirror in y = x | Because the inverse swaps every \((a, b)\) to \((b, a)\), the graphs of \(f\) and \(f^{-1}\) are reflections of each other in \(y = x\). Model the worked example Where a function meets its inverse on the board, one line at a time. | Rally Coach |
| 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 Mirror in y = x, 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.
Rally Coach
Pairs share one worksheet. One partner solves the first question aloud while the other coaches, checking each line and offering a prompt rather than the answer; they swap roles for the next question.
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 |
|---|---|
| \(f^{-1}(x)\) is NOT \(\dfrac{1}{f(x)}\). | For \(f(x) = 3x - 5\), \(\dfrac{1}{3x - 5}\) is a reciprocal; the inverse is \(\dfrac{x + 5}{3}\). |
| Writing \(\pm\sqrt{\ }\) in the final answer is wrong: an inverse function gives ONE output. | Test your choice: \(g^{-1}\) must output values \(\le \frac{1}{3}\), and \(\frac{1 - \sqrt{x + 5}}{3}\) does. |
| Sketch \(f^{-1}\) only over the RANGE of \(f\). | In Fig. 1 the inverse starts at \((-2, 0)\); drawing \(\sqrt{x + 2}\) to the left of \(-2\), or reflecting the whole parabola, earns no marks. |
Inverse functions: worksheet
6 questions, 21 marks. Written in the style of the papers; no past paper question is reproduced.
The function f is defined by \(f(x) = \dfrac{5x - 2}{3}\) for \(x \in \mathbb{R}\). Find \(f^{-1}(x)\).
The function f is defined by \(f(x) = 4 - (x - 1)^{2}\) for \(x \ge 1\). Find \(f^{-1}(x)\) and state the domain of \(f^{-1}\).
The function f is defined by \(f(x) = 2x^{2} + 12x + 11\) for \(x \ge -3\).
(a) Express \(f(x)\) in the form \(a(x + b)^{2} + c\).
(b) Find \(f^{-1}(x)\), stating its domain.
The function f is defined by \(f(x) = \dfrac{3}{2x - 1}\) for \(x > \dfrac{1}{2}\).
(a) Find \(f^{-1}(x)\) and state the domain of \(f^{-1}\).
(b) Solve \(f^{-1}(x) = 2\).
The function g is defined by \(g(x) = \dfrac{x + 4}{x - 1}\) for \(x > 1\). Show that g is self-inverse, that is, \(g^{-1}(x) = g(x)\).
A workshop's cost, in dollars, of making \(x\) chairs is modelled by \(c(x) = 0.5x^{2} + 20\) for \(x \ge 0\).
(a) Explain why c has an inverse.
(b) Find \(c^{-1}(x)\) and state its domain.
(c) Find \(c^{-1}(92)\) and explain what it means.
Answer key and marking notes
- \(y = \dfrac{5x - 2}{3} \Rightarrow 3y = 5x - 2 \Rightarrow x = \dfrac{3y + 2}{5}\).
- \(f^{-1}(x) = \dfrac{3x + 2}{5}\).
- \((x - 1)^{2} = 4 - y\), and \(x - 1 \ge 0\), so \(x - 1 = \sqrt{4 - y}\).
- \(f^{-1}(x) = 1 + \sqrt{4 - x}\).
- Range of f is \(f(x) \le 4\), so the domain of \(f^{-1}\) is \(x \le 4\).
- (a) \(2(x^{2} + 6x) + 11 = 2\big[(x + 3)^{2} - 9\big] + 11 = 2(x + 3)^{2} - 7\).
- (b) \(\dfrac{y + 7}{2} = (x + 3)^{2}\), and \(x + 3 \ge 0\), so \(x = -3 + \sqrt{\dfrac{y + 7}{2}}\).
- \(f^{-1}(x) = -3 + \sqrt{\dfrac{x + 7}{2}}\).
- Range of f is \(f(x) \ge -7\), so the domain of \(f^{-1}\) is \(x \ge -7\).
- (a) \(2x - 1 = \dfrac{3}{y} \Rightarrow 2x = \dfrac{3 + y}{y} \Rightarrow x = \dfrac{y + 3}{2y}\).
- \(f^{-1}(x) = \dfrac{x + 3}{2x}\). For \(x > \frac{1}{2}\), \(2x - 1 > 0\) takes every positive value, so \(f(x) > 0\): domain of \(f^{-1}\) is \(x > 0\).
- (b) \(\dfrac{x + 3}{2x} = 2 \Rightarrow x + 3 = 4x \Rightarrow x = 1\).
- Check: \(f(2) = \dfrac{3}{3} = 1\), so \(f^{-1}(1) = 2\).
- \(y(x - 1) = x + 4 \Rightarrow xy - y = x + 4 \Rightarrow x(y - 1) = y + 4\).
- \(x = \dfrac{y + 4}{y - 1}\), so \(g^{-1}(x) = \dfrac{x + 4}{x - 1}\), the same rule as g.
- Domains match too: \(g(x) = 1 + \dfrac{5}{x - 1} > 1\) for \(x > 1\), so \(g^{-1}\) has domain \(x > 1\), the same as g.
- (a) On \(x \ge 0\) the graph is the right half of a parabola with vertex \((0, 20)\), which is increasing, so c is one-one.
- (b) \(y - 20 = 0.5x^{2} \Rightarrow x^{2} = 2(y - 20)\), and \(x \ge 0\), so \(x = \sqrt{2(y - 20)}\).
- \(c^{-1}(x) = \sqrt{2(x - 20)}\) for \(x \ge 20\) (the range of c).
- (c) \(c^{-1}(92) = \sqrt{2 \times 72} = \sqrt{144} = 12\).
- A budget of 92 dollars pays for exactly 12 chairs.
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 \(\text{f}(x) = \dfrac{2x + 7}{5}\) for \(x \in \mathbb{R}\). Find \(\text{f}^{-1}(x)\).
- A\(\dfrac{5x - 7}{2}\)
- B\(\dfrac{5}{2x + 7}\)
- C\(\dfrac{5x + 7}{2}\)
- D\(\dfrac{5(x - 7)}{2}\)
The function f is defined by \(\text{f}(x) = x^{2} + 3\) for \(x \ge 1\). State the domain of \(\text{f}^{-1}\).
- A\(x \ge 4\)
- B\(x \ge 1\)
- C\(x \ge 3\)
- D\(x \in \mathbb{R}\)
The function f is defined by \(\text{f}(x) = (x - 1)^{2} + 4\) for \(x \ge 1\). Find an expression for \(\text{f}^{-1}(x)\).
- A\(1 + \sqrt{x - 4}\)
- B\(1 - \sqrt{x - 4}\)
- C\(1 + \sqrt{x + 4}\)
- D\(\dfrac{1}{(x - 1)^{2} + 4}\)
The point \((3,\ 8)\) lies on the graph of the one-one function \(y = \text{f}(x)\). Which point must lie on the graph of \(y = \text{f}^{-1}(x)\)?
- A\((8,\ 3)\)
- B\(\left(3,\ \dfrac{1}{8}\right)\)
- C\((-3,\ -8)\)
- D\((-8,\ -3)\)
The function f is defined by \(\text{f}(x) = \dfrac{3x + 2}{x - 1}\) for \(x > 1\). Find \(\text{f}^{-1}(x)\) and state the domain of \(\text{f}^{-1}\).
- A\(\text{f}^{-1}(x) = \dfrac{x + 2}{x - 3}\) for \(x > 3\)
- B\(\text{f}^{-1}(x) = \dfrac{x + 2}{x - 3}\) for \(x > 1\)
- C\(\text{f}^{-1}(x) = \dfrac{x - 1}{3x + 2}\) for \(x > 3\)
- D\(\text{f}^{-1}(x) = \dfrac{x - 2}{x - 3}\) for \(x > 3\)
The function g is defined by \(\text{g}(x) = (2x + 1)^{2} - 3\) for \(x \le -\dfrac{1}{2}\). Find \(\text{g}^{-1}(x)\).
- A\(\dfrac{-1 - \sqrt{x + 3}}{2}\)
- B\(\dfrac{-1 + \sqrt{x + 3}}{2}\)
- C\(\dfrac{1 - \sqrt{x + 3}}{2}\)
- D\(\dfrac{-1 - \sqrt{x - 3}}{2}\)
The function f is defined by \(\text{f}(x) = x^{2} - 3x + 3\) for \(x \ge \dfrac{3}{2}\). Find the \(x\)-coordinate of the point where the graphs of \(y = \text{f}(x)\) and \(y = \text{f}^{-1}(x)\) meet.
- A\(3\)
- B\(1\) or \(3\)
- C\(1\)
- D\(\dfrac{3}{2}\)
The function f is defined by \(\text{f}(x) = 4x - 9\) for \(x \in \mathbb{R}\). Solve the equation \(\text{f}(x) = \text{f}^{-1}(x)\).
- A\(x = 3\)
- B\(x = \dfrac{9}{4}\)
- C\(x = \dfrac{9}{5}\)
- D\(x = -3\)
Quiz key
\(5y = 2x + 7\), so \(2x = 5y - 7\) and \(x = \dfrac{5y - 7}{2}\). Swapping: \(\text{f}^{-1}(x) = \dfrac{5x - 7}{2}\). Check: \(\text{f}(1) = \dfrac{9}{5}\) and \(\dfrac{5 \times \frac{9}{5} - 7}{2} = 1\). The tempting \(\dfrac{5}{2x + 7}\) is the reciprocal of f, not its inverse. \(5x + 7\) undoes the \(+7\) with another \(+7\), and \(\dfrac{5(x - 7)}{2}\) undoes the steps in the wrong order.
f is increasing for \(x \ge 1\), and \(\text{f}(1) = 4\), so the range of f is \(\text{f}(x) \ge 4\). The domain of \(\text{f}^{-1}\) is the range of f: \(x \ge 4\). The tempting \(x \ge 1\) is the domain of f, which becomes the RANGE of \(\text{f}^{-1}\). \(x \ge 3\) is the range of \(x^{2} + 3\) over all real \(x\), ignoring the restriction \(x \ge 1\).
\(y - 4 = (x - 1)^{2}\), so \(x - 1 = \pm\sqrt{y - 4}\). The domain of f is \(x \ge 1\), so \(\text{f}^{-1}\) must output values \(\ge 1\): take the positive root. \(\text{f}^{-1}(x) = 1 + \sqrt{x - 4}\). The tempting \(1 - \sqrt{x - 4}\) gives outputs \(\le 1\), the inverse of the OTHER half of the parabola. \(\sqrt{x + 4}\) moves the 4 with the wrong sign, and the last option is a reciprocal.
\(\text{f}(3) = 8\) means \(\text{f}^{-1}(8) = 3\), so \((8,\ 3)\) is on \(y = \text{f}^{-1}(x)\): the graph is reflected in the line \(y = x\), which swaps the coordinates. The tempting \(\left(3,\ \frac{1}{8}\right)\) treats \(\text{f}^{-1}\) as \(\dfrac{1}{\text{f}}\). \((-3,\ -8)\) is a rotation about the origin, and \((-8,\ -3)\) is a reflection in \(y = -x\).
\(yx - y = 3x + 2\), so \(x(y - 3) = y + 2\) and \(x = \dfrac{y + 2}{y - 3}\). Swapping: \(\text{f}^{-1}(x) = \dfrac{x + 2}{x - 3}\). Since \(\text{f}(x) = 3 + \dfrac{5}{x - 1}\) and \(\dfrac{5}{x - 1} > 0\) for \(x > 1\), the range of f is \(\text{f}(x) > 3\), which is the domain of \(\text{f}^{-1}\). The tempting \(x > 1\) copies the domain of f. \(\dfrac{x - 1}{3x + 2}\) is the reciprocal, and \(x - 2\) comes from a sign slip when moving \(-y\) across.
\((2x + 1)^{2} = y + 3\), so \(2x + 1 = \pm\sqrt{y + 3}\). For \(x \le -\frac{1}{2}\), \(2x + 1 \le 0\), so take the NEGATIVE root: \(2x + 1 = -\sqrt{y + 3}\), \(x = \dfrac{-1 - \sqrt{y + 3}}{2}\). The tempting \(+\sqrt{\ }\) version gives outputs \(\ge -\frac{1}{2}\), outside the domain of g. \(\dfrac{1 - \sqrt{x + 3}}{2}\) moves the 1 with the wrong sign, and \(\sqrt{x - 3}\) moves the 3 with the wrong sign.
The graphs are reflections of each other in \(y = x\), and for this increasing f they meet on that line. \(x^{2} - 3x + 3 = x\) gives \(x^{2} - 4x + 3 = 0\), so \(x = 1\) or \(x = 3\). Only \(x = 3\) is in the domain \(x \ge \frac{3}{2}\), giving the point \((3,\ 3)\). The tempting "1 or 3" forgets the domain: \(x = 1\) is not an input of f. \(\frac{3}{2}\) is the vertex, where the domain starts, not a meeting point (\(\text{f}(\frac{3}{2}) = \frac{3}{4} \ne \frac{3}{2}\)).
\(\text{f}^{-1}(x) = \dfrac{x + 9}{4}\). \(4x - 9 = \dfrac{x + 9}{4}\) gives \(16x - 36 = x + 9\), so \(15x = 45\) and \(x = 3\). (Equivalently \(4x - 9 = x\).) The tempting \(\frac{9}{4}\) solves \(\text{f}(x) = 0\). \(\frac{9}{5}\) solves \(4x - 9 = -x\), as if the mirror line were \(y = -x\), and \(-3\) is a sign slip.
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.