IB Maths AI HL Functions Toolkit Paper 1 & 2 f⁻¹, reflection, restriction ~9 min read

Inverse Functions

An inverse function f⁻¹ undoes whatever f does: if f(x) = y then f⁻¹(y) = x. Algebraically you swap x and y and rearrange; graphically you reflect in the line y = x. A many-to-one function has no inverse until you restrict its domain.

📘 What you need to know

Inverse undoes the function: (f ∘ f⁻¹)(x) = (f⁻¹ ∘ f)(x) = x.
Algebra steps: swap x and y in y = f(x), then solve for y.
Graph: y = f⁻¹(x) is the reflection of y = f(x) in the line y = x.
Domain ↔ range swap: domain of f = range of f⁻¹, range of f = domain of f⁻¹.
One-to-one required: an inverse exists only if f is one-to-one; restrict the domain of many-to-one functions.
Solve f(x) = b using x = f⁻¹(b).
Caution: f⁻¹(x) is NOT the reciprocal 1/f(x).

Finding an inverse algebraically

Two steps. Step 1: write y = f(x) and swap x and y — you now have x = f(y). Step 2: rearrange to make y the subject. The result y = f⁻¹(x). Always check your work by composing: f(f⁻¹(x)) should simplify back to x. If it doesn’t, you’ve made an algebraic slip.

Reflection in y = x

The graph of y = f⁻¹(x) is the mirror image of y = f(x) across the line y = x. So if f passes through (a, b), then f⁻¹ passes through (b, a). When f intersects the line y = x at a point, f⁻¹ intersects it at the same point — that’s why f(x) = x often gives the simplest way to find where f and f⁻¹ cross. There may also be other crossings not on y = x.

Reflecting (2, 1) on y = f(x) in y = x gives (1, 2) on y = f⁻¹(x). Every point swaps its coordinates.

Inverse function at a glance swap x & y in y = f(x), then make y the subject (f ∘ f⁻¹)(x) = x · reflection in y = x · domain ↔ range

Restricting many-to-one functions

A function only has an inverse when it’s one-to-one — no y-value is produced twice. Quadratics fail this (each output comes from two inputs symmetric about the vertex), and so does any trig function (each output repeats every period). To rescue them, restrict the domain to one side of the vertex (for a quadratic) or to a single half-cycle (for a trig function). The biggest sensible restriction for a(x − h)² + k is x ≥ h or x ≤ h; once restricted, the inverse exists and the ± in ±√… gets resolved by matching the range of the inverse to the (restricted) domain of the original.

Identity check. After finding f⁻¹, evaluate f(f⁻¹(some value)). It must give you that value back. If it doesn’t, recheck the algebra.

🧭 Recipe — finding an inverse

Check one-to-one; restrict the domain if needed (quadratics: choose a side of the vertex; trig: choose one cycle).
Swap x and y in y = f(x).
Make y the subject by rearranging.
Resolve any ± using the rule range of f⁻¹ = domain of f.
State domain and range of f⁻¹ by swapping them with those of f.

Worked examples

WE 1

Inverse of a linear function

Given f(x) = 3x − 7. (a) Find f⁻¹(x). (b) Find f⁻¹(8) and verify.

(a) y = 3x − 7, swap x = 3y − 7 3y = x + 7 f⁻¹(x) = (x + 7)/3 (b) sub x = 8 f⁻¹(8) = (8 + 7)/3 = 15/3 = 5 verify f(5) = 3(5) − 7 = 15 − 7 = 8 ✓ f⁻¹(8) = 5

WE 2

Inverse involving a square root

Given g(x) = √(x − 4) + 3 for x ≥ 4. (a) Find g⁻¹(x). (b) State the domain of g⁻¹.

(a) y = √(x − 4) + 3, swap x = √(y − 4) + 3 x − 3 = √(y − 4) square: (x − 3)² = y − 4 g⁻¹(x) = (x − 3)² + 4 (b) domain of g⁻¹ = range of g on x ≥ 4: √ ≥ 0, so g ≥ 3 domain of g⁻¹: x ≥ 3 verify g(g⁻¹(7)) = g(4² + 4) = g(20) = √16 + 3 = 7 ✓

WE 3

Domain and range of an inverse

Given f(x) = 4x + 1 with domain 2 ≤ x ≤ 8. (a) Find the range of f. (b) Find f⁻¹(x) and state its domain and range.

(a) endpoints f(2) = 8 + 1 = 9; f(8) = 32 + 1 = 33 range of f: 9 ≤ f(x) ≤ 33 (b) swap & solve x = 4y + 1 ⇒ y = (x − 1)/4 f⁻¹(x) = (x − 1)/4 domain ↔ range swap domain f⁻¹: 9 ≤ x ≤ 33 · range f⁻¹: 2 ≤ f⁻¹(x) ≤ 8

WE 4

Restrict a quadratic to invert it

The function f(x) = (x + 1)² − 3 has domain x ≥ m. (a) Find the smallest value of m for which f⁻¹ exists. (b) Find f⁻¹(x). (c) State the domain of f⁻¹.

(a) vertex at x = −1 need one-to-one ⇒ restrict to x ≥ −1 m = −1 (b) swap & solve x = (y + 1)² − 3 (y + 1)² = x + 3 y + 1 = ±√(x + 3) range of f⁻¹ = domain of f = x ≥ −1 so y ≥ −1 ⇒ choose +√ f⁻¹(x) = −1 + √(x + 3) (c) domain f⁻¹ = range of f min of f at vertex: −3 domain f⁻¹: x ≥ −3

WE 5

Inverse of a rational function

Given f(x) = x − 32x + 1. (a) Find f⁻¹(x). (b) State the value of x for which each function is undefined.

(a) swap x = (y − 3)/(2y + 1) cross-multiply x(2y + 1) = y − 3 2xy + x = y − 3 collect y on one side 2xy − y = −x − 3 y(2x − 1) = −(x + 3) f⁻¹(x) = (x + 3)/(1 − 2x) (b) denominator zero f: 2x + 1 = 0 ⇒ x = −½ f⁻¹: 1 − 2x = 0 ⇒ x = ½ f undef at x = −½; f⁻¹ undef at x = ½

WE 6

Intersection of f and f⁻¹

Given f(x) = √(2x − 1) + 1 for x ≥ ½. The graphs of y = f(x) and y = f⁻¹(x) intersect on the line y = x. Find the coordinates of the intersection (exact form).

set f(x) = x √(2x − 1) + 1 = x √(2x − 1) = x − 1 (need x ≥ 1) square both sides 2x − 1 = (x − 1)² 2x − 1 = x² − 2x + 1 x² − 4x + 2 = 0 quadratic formula x = (4 ± √8)/2 = 2 ± √2 check x ≥ 1 2 + √2 ≈ 3.41 ✓ 2 − √2 ≈ 0.59 < 1 ✗ (reject) intersection: (2 + √2, 2 + √2) squaring can create extraneous roots — always check.

💡 Top tips

Always verify: composing f with your f⁻¹ should give back x.
Swap domain & range: the most common mistake is forgetting these flip when you invert.
For quadratics, the vertex is the natural restriction point — choose x ≥ h or x ≤ h.
The ± sign after taking a square root is resolved by the rule “range of f⁻¹ = domain of f“.
To solve f(x) = b exactly, compute x = f⁻¹(b).

⚠ Common mistakes

Confusing f⁻¹(x) with 1/f(x) — the −1 is a function exponent, not a power.
Forgetting to swap x and y at the start — you end up rearranging f(x) instead of finding its inverse.
Picking the wrong sign from ±√… without thinking about the range of f⁻¹.
Trying to invert a many-to-one function without restricting its domain first.
Keeping extraneous roots introduced by squaring — always check each candidate satisfies the original equation.

Chapter complete — you now have the Functions Toolkit (Composite + Inverse) ready for any AI HL Paper 1 or 2 question. Next chapter on the syllabus: Geometry & Trigonometry.

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