IB Maths AA HL Topic 2 — Functions Paper 1 & 2 ~6 min read

Self-Inverse Functions

A self-inverse function is one that equals its own inverse — apply it twice and you get back to where you started. The classic example: f(x) = 1/x. Take any number, take its reciprocal, take the reciprocal again, and you’ve got the original. Most rational functions of the form (ax + b)/(cx + d) are self-inverse if and only if d = −a. Useful to spot, useful to prove, useful for shortcut algebra.

📘 What you need to know

Definition: f is self-inverse if f(x) = f⁻¹(x) — the inverse and the original are the same function.
Equivalent test: applying f twice undoes everything ⟹ f(f(x)) = x.
Common examples: identity f(x) = x, reciprocal f(x) = 1/x, any line of gradient −1 like f(x) = 12 − x.
Rational shortcut: f(x) = (ax + b)/(cx + d) is self-inverse precisely when a + d = 0 (i.e. d = −a).
Graph property: a self-inverse function’s graph is unchanged under reflection in the line y = x.
Two ways to prove it: either show f(f(x)) = x directly, or find f⁻¹(x) and show it equals f(x). Pick whichever is easier.

The two conditions — they say the same thing

Condition 1

f(x) = f⁻¹(x)

function is its own inverse

Condition 2

f(f(x)) = x

applying it twice gets you back

🤔 Why are they the same?

If f = f⁻¹, then f(f(x)) = f(f⁻¹(x)) = x automatically. And conversely, if f(f(x)) = x, then f undoes itself, so f matches the definition of f⁻¹.

Common self-inverse functions

Identity

f(x) = x

does literally nothing

Reciprocal

f(x) = 1/x

flip, then flip again

Lines, gradient −1

f(x) = k − x

e.g. 12 − x, 7 − x

Rational shortcut: the rational function f(x) = (ax + b)/(cx + d) is self-inverse if and only if a + d = 0. Just check: are the coefficients of x on top and the constant on the bottom opposite in sign? If yes, self-inverse.

Graphical view — symmetric in y = x

Since f = f⁻¹, and the graph of f⁻¹ is the reflection of f in y = x, a self-inverse function’s graph is its own reflection. The graph is unchanged when flipped across the line y = x.

A self-inverse curve — its own mirror image in y = x

How to prove a function is self-inverse

Method 1 — composition

show f(f(x)) = x

substitute the function into itself, simplify; if it gives x, you’re done

Method 2 — find inverse

find f⁻¹(x), check it equals f(x)

use the swap-and-rearrange method from the inverse functions note

For rational functions like (ax + b)/(cx + d), Method 1 (composition) is usually faster. The numerator and denominator both simplify with the same factor cancelling out — feels almost magical.

Worked examples

WE 1

Verify the reciprocal function is self-inverse

Show that f(x) = 1/x, x ≠ 0 is self-inverse.

Compute f(f(x)) f(f(x)) = f(1/x) = 1 / (1/x) = x ✓ f(f(x)) = x, so f is self-inverse flip, then flip again — back to the original

WE 2

A linear self-inverse function

Show that f(x) = 12 − x is self-inverse.

Compute f(f(x)) f(f(x)) = f(12 − x) = 12 − (12 − x) = 12 − 12 + x = x ✓ f(x) = 12 − x is self-inverse any line of gradient −1, written as f(x) = k − x, is self-inverse — the graph is its own reflection in y = x

WE 3

Prove a rational function is self-inverse using composition

Show that f(x) = 4x + 3x − 4, x ≠ 4 is self-inverse.

Compute f(f(x)) — substitute f(x) into itself f(f(x)) = [4·f(x) + 3] / [f(x) − 4] Numerator: 4·f(x) + 3 = 4·(4x+3)/(x−4) + 3 = [4(4x+3) + 3(x−4)] / (x−4) = [16x + 12 + 3x − 12] / (x−4) = 19x / (x−4) Denominator: f(x) − 4 = (4x+3)/(x−4) − 4 = [(4x+3) − 4(x−4)] / (x−4) = [4x + 3 − 4x + 16] / (x−4) = 19 / (x−4) Divide numerator by denominator f(f(x)) = [19x/(x−4)] / [19/(x−4)] = 19x/19 = x ✓ f(f(x)) = x, so f is self-inverse a + d = 4 + (−4) = 0 — matches the rational shortcut, so this was expected

WE 4

Prove a rational function is self-inverse by finding its inverse

Show that f(x) = 5x − 1x − 5, x ≠ 5 is self-inverse.

Step 1: Set y = f(x) and swap x and y y = (5x − 1)/(x − 5) swap → x = (5y − 1)/(y − 5) Step 2: Multiply through by (y − 5) x(y − 5) = 5y − 1 xy − 5x = 5y − 1 Step 3: Group y terms xy − 5y = 5x − 1 y(x − 5) = 5x − 1 y = (5x − 1)/(x − 5) Step 4: Compare f⁻¹(x) = (5x − 1)/(x − 5) = f(x) ✓ f⁻¹(x) = f(x), so f is self-inverse again: a + d = 5 + (−5) = 0 — fits the rational pattern

WE 5

Find k for which a function is self-inverse

Find the value of k such that f(x) = kx + 12x − 3 is self-inverse.

Use the rational shortcut: a + d = 0 here a = k, d = −3 need k + (−3) = 0 k = 3 k = 3, giving f(x) = (3x + 12)/(x − 3) Quick check by composition f(f(x)) numerator: 3·(3x+12)/(x−3) + 12 = [9x + 36 + 12x − 36]/(x−3) = 21x/(x−3) denominator: (3x+12)/(x−3) − 3 = [3x + 12 − 3x + 9]/(x−3) = 21/(x−3) f(f(x)) = 21x/21 = x ✓ memorise the “a + d = 0” shortcut — saves a lot of algebra in exam questions

WE 6

Use the self-inverse property to find a value

The function f is self-inverse, and f(2) = 11. Find the value of f(11).

Use the property f(f(x)) = x f(f(2)) = 2 f(11) = 2 (since f(2) = 11) f(11) = 2 alternative reasoning: since f = f⁻¹, f(2) = 11 means f⁻¹(11) = 2, which is also f(11)

💡 Top tips

Rational shortcut: for (ax + b)/(cx + d), self-inverse iff a + d = 0. Memorise this.
Method 1 (composition) is usually fastest for rational functions — both numerator and denominator carry a common factor that cancels.
Method 2 (find inverse) is fastest for linear functions — quick to invert, easy to compare.
If you’re stuck, sketch the function on your GDC and check whether it’s symmetric in y = x.
For “find f(k)” questions where f is self-inverse, just look for the value x with f(x) = k — that x is your answer.
Identity, reciprocal, and 𝑘 − 𝑥 lines are the three families you should recognise instantly.
Always state the domain restrictions when working with rational self-inverses (e.g. x ≠ k from the denominator).

⚠ Common mistakes

Confusing self-inverse with even or odd. Self-inverse is about reflection in y = x. Even is reflection in the y-axis. Odd is rotation about the origin. Three different things.
Stopping after computing f(f(x)) without simplifying. The expression looks messy at first — keep going until it cancels to x.
Using a · d = 0 instead of a + d = 0. The condition is on the sum, not the product.
Forgetting to check the domain. A rational self-inverse like (3x + 12)/(x − 3) needs x ≠ 3.
Treating any rational function as self-inverse. Only those satisfying a + d = 0 qualify. The general (ax + b)/(cx + d) is usually NOT self-inverse.
Algebra errors when computing f(f(x)) — especially when handling the common denominator (x − k) carefully.
Reporting “f⁻¹(x) is different” when the algebra produced a different-looking but equivalent expression. Plot both on a GDC to confirm.

Self-inverse, periodic, odd, even — all four are special structural properties that the IB likes to test, because spotting them turns hard problems into easy ones. The next note shifts focus from algebraic properties to graphing functions and their key features — what to label on a sketch, how to use your GDC, and the difference between “sketch” and “draw”.

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