Analysis only. It needs the inverse-as-a-reflection idea from SL 2.2 and feeds the transformations at SL 2.11, where composition is what the order of transformations actually is.
Set the slider to x = −1 first and ask whether order matters. Both composites read 4, so the class says no. Then move to x = 2 and the two read 25 and 7.
That sequence is the lesson, not the two numbers. One test value proved the wrong thing, and the crossing is marked on the figure so they can see the trap was a single point rather than bad luck.
Then write both formulas out: (x + 3)² against x² + 3. The gap is 6x + 6, which is zero once. A formula settles in one line what sampling cannot settle at all.
| Question | Answer |
|---|---|
| 1. (f ∘ g)(2) | f(5) = 25. |
| 2. (g ∘ f)(2) | g(4) = 7. |
| 3. h⁻¹(11) for h(x) = 2x + 3 | (11 − 3)/2 = 4. |
| 4. Where the composites agree | 6x + 6 = 0, so x = −1, both giving 4. |
| 5. Why x² has no inverse unrestricted | B. 3 and −3 share an image, so an inverse would have to send 9 to two places. |
Question 4 carries the teaching. A student who can find the crossing has understood that the two composites are functions to be compared, not procedures to be run.
1 markThe composite written as f(g(x)), with the order right.
1 markThe substitution, or the simplified formula.
1 markFor an inverse, the rearrangement, and the domain restriction where one is needed.
The domain is the mark most often left blank. On any inverse built from a square or a square root, insist on a stated domain even when the question does not ask in those words.
| They wrote | What happened |
|---|---|
| 7 on question 1 | The order read off the letters rather than the brackets. In f ∘ g the g touches the x, so it goes first. |
| 13 | 2² + 3². They have squared both numbers separately instead of composing. |
| 4 on question 1 | f(2) alone. The g never got applied. |
| 25 on question 2 | The other order again, one question later. Worth asking which function is next to the x. |
| 25 on question 3 | h(11) rather than h⁻¹(11). They have run the function forwards. |
| 2.5 | 11/2 − 3: undone in the same order as the doing. An inverse reverses the order, so subtract the 3 first. |
| 8 | 11 − 3 and stopped. Halfway. |
| 0 on question 4 | Tested x = 0, where the two are 9 and 3, and reported the test value rather than solving. Push them to 6x + 6 = 0. |
| 4 on question 4 | The shared VALUE at the crossing, not where it happens. Both are worth saying aloud: at x = −1 both composites give 4. |
"Why is it f(g(x)) and not g(f(x))?" Because of the brackets. Write f ∘ g and f(g(x)) one above the other and read the second from the inside out. It is the same convention as function notation everywhere else, and it is worth connecting to the matrix order in a parallel course if any of the class take Applications.
"Is the inverse the same as 1/f(x)?" No, and the notation invites it. f⁻¹ is a function that undoes f; (f(x))⁻¹ is a reciprocal. For f(x) = 2x + 3 the inverse is (x − 3)/2 and the reciprocal is 1/(2x + 3), and they are nothing alike. Say it once, early.
"How do I find the domain of an inverse?" It is the range of the original, and the range of the inverse is the domain of the original. They swap. That one sentence answers most of the questions in this sub-topic, and it is worth putting on the board rather than deriving each time.
"Can a function be its own inverse?" Yes, and they meet one at 2.8, 1/x, and a second at AHL 2.14, (x + 1)/(x − 1). Graphically those are the curves symmetric about y = x. Mentioning it here gives 2.8 somewhere to land.
| Demonstrate | Define f and g once, then put f(g(2)) and g(f(2)) on consecutive lines: 25 and 7. Then type f(g(x)) with the x, and the Nspire returns the composite as a formula in x. Do not promise the class it will be tidied up: the CX II is non-CAS and may hand back (x+3)² as it stands. That second step is still the one that matters, because it replaces a sample with a proof. |
|---|---|
| Where they stick | On the Casio, composing means referring to Y1 from inside Y3, and the Y symbols come from VARS → GRAPH rather than from the alpha keys. Typing the letter Y and the digit 1 gives Y times 1. They then need F1 to deselect Y1 and Y2 so only the composites draw, or four curves appear and nobody can tell which is which. |
| The check | Run the inverse back through the function. If h⁻¹(11) = 4, then h(4) must be 11. One line, and it catches every rearrangement slip including the 2.5. |
Analysis Paper 1 has no calculator, and this sub-topic is mostly Paper 1 work. Use the machine to check an algebraic answer, not to produce one.
| Step | What |
|---|---|
| 1 | Two machines on the board: a squarer and an adder. Feed 2 through in both orders. 25 and 7. |
| 2 | Notation: f ∘ g, then f(g(x)) underneath. Read inside out. |
| 3 | Figure at x = −1. Ask whether order matters. Collect the no. |
| 4 | Move to x = 2. Then solve 6x + 6 = 0 to show the agreement was one point. |
| 5 | The identity function, and an inverse as the thing that produces it. |
| 6 | Find an inverse in three steps, then check it by running it back. |
| 7 | Why x² needs a restriction, with 3 and −3 on the board. |
| 8 | Domains and ranges swapping, as one sentence. |
Do not say "composition is just doing one then the other". It is, and the sentence leaves out the only hard part, which is which one is first. Say "the one next to the x goes first", which is both true and usable.
Do not say "the inverse swaps x and y". It describes the algebraic trick and it makes students think the inverse is a relabelling. It is a different function, with its own domain, and on a graph it is the reflection in y = x. The swap is a method for finding it, not what it is.