IB Maths AA SL Topic 2 — Functions Paper 1 & 2 🎯 Skill ~3 min practice

AA SL Inverse Functions skills

An inverse function undoes the original. The recipe is three steps: write y = f(x), swap x and y, then solve for y. The new equation is f⁻¹(x). Master the swap-and-solve method and inverses become mechanical.

The Method

f(f⁻¹(x)) = x and f⁻¹(f(x)) = x f and f⁻¹ cancel each other out

Write f(x) as y = … (so the function looks like an equation in x and y)
Swap every x and y (this is the move that creates the inverse)
Solve for y, then rename y as f⁻¹(x) — done.

Swap and solve — three moves

Example: find the inverse of f(x) = 2x + 6

Step 1 — write as y y = 2x + 6

→

Step 2 — swap x & y x = 2y + 6

→

Step 3 — solve for y y = (x − 6) / 2

Final answer: f⁻¹(x) = (x − 6) / 2

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The graph is a reflection in y = x

The inverse swaps inputs and outputs, so geometrically y = f⁻¹(x) is just y = f(x) reflected in the line y = x. Every point (a, b) on f becomes (b, a) on f⁻¹.

Worked examples

WE 1 EASY

Find the inverse of f(x) = 3x − 4.

step 1 — write as y y = 3x − 4step 2 — swap x and y x = 3y − 4step 3 — solve for y x + 4 = 3y y = (x + 4) / 3f⁻¹(x) = (x + 4) / 3 linear inverses are the easiest — just three moves and you’re done!

WE 2 MEDIUM

Find the inverse of f(x) = x + 1x − 2.

step 1 — write as y y = (x + 1) / (x − 2)step 2 — swap x and y x = (y + 1) / (y − 2)step 3 — solve for y x(y − 2) = y + 1 xy − 2x = y + 1 collect y terms on one side xy − y = 2x + 1 y(x − 1) = 2x + 1 y = (2x + 1) / (x − 1)f⁻¹(x) = (2x + 1) / (x − 1) cross-multiply first, then group all y terms — the rest is just algebra!

WE 3 HARD

Find the inverse of f(x) = √(x − 3) + 5, given x ≥ 3.

step 1 — write as y y = √(x − 3) + 5step 2 — swap x and y x = √(y − 3) + 5step 3 — solve for y isolate the root first x − 5 = √(y − 3) square both sides (x − 5)² = y − 3 y = (x − 5)² + 3f⁻¹(x) = (x − 5)² + 3, x ≥ 5 always state the domain of f⁻¹ — it’s the range of f!

Practice questions

Try each one yourself first, then click the question to reveal the worked answer. Always do all three steps in order — don’t try to shortcut.

Q1 EASY Find the inverse of f(x) = x + 7. Show answer ▼Hide answer ▲

y = x + 7 → x = y + 7 y = x − 7 f⁻¹(x) = x − 7

Q2 EASY Find the inverse of f(x) = 5x + 2. Show answer ▼Hide answer ▲

y = 5x + 2 → x = 5y + 2 x − 2 = 5y → y = (x − 2) / 5 f⁻¹(x) = (x − 2) / 5

Q3 MEDIUM Find the inverse of f(x) = 4x + 1. Show answer ▼Hide answer ▲

y = 4 / (x + 1) → x = 4 / (y + 1) x(y + 1) = 4 y + 1 = 4/x y = 4/x − 1 f⁻¹(x) = 4/x − 1

Q4 MEDIUM Find f⁻¹(x) for f(x) = 2x² − 3, x ≥ 0. Show answer ▼Hide answer ▲

y = 2x² − 3 → x = 2y² − 3 x + 3 = 2y² → y² = (x + 3) / 2 x ≥ 0 → take positive root only y = √((x + 3) / 2) f⁻¹(x) = √((x + 3) / 2) domain restriction tells you which root to keep!

Q5 HARD Find the inverse of f(x) = 3x − 1x + 4. Show answer ▼Hide answer ▲

swap x and y x = (3y − 1) / (y + 4)cross-multiply x(y + 4) = 3y − 1 xy + 4x = 3y − 1collect y terms xy − 3y = −1 − 4x y(x − 3) = −(4x + 1) y = −(4x + 1) / (x − 3) = (4x + 1) / (3 − x)f⁻¹(x) = (4x + 1) / (3 − x) flip the sign of denominator instead of leaving the minus on top — looks cleaner.

⚠ Common mistakes

Treating f⁻¹(x) as 1/f(x). The “−1” superscript means inverse, not reciprocal. f⁻¹(x) ≠ 1f(x).
Forgetting to swap before solving. If you solve for x without swapping first, you just rearrange — you don’t get the inverse.
Sign errors when collecting y terms. When y appears on both sides, move them carefully — write each step.
Ignoring domain restrictions. If f has x ≥ 0, the inverse needs the positive root only — not ±.
Not checking the inverse exists. A function only has an inverse if it’s one-to-one (passes the horizontal line test). Quadratics need a domain restriction first.
Forgetting the domain swap. Domain of f⁻¹ = range of f. Range of f⁻¹ = domain of f. They trade.

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Want the theory?

Read the full Inverse Functions notes for the link to one-to-one functions, the horizontal line test, and how the domain and range swap between f and f⁻¹.

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