Swap x and y, then solve: To find f⁻¹: write y = f(x), swap x and y, then solve for y. (Geometrically this reflects the graph in y = x — see 2.2.3.)
IB-style question — a quick inverse
Find the inverse of f(x) = 5x − 2.
Step by step
- y = 5x − 2, then swap.
- Solve for y.
Final answer
f⁻¹(x) = (x + 2)/5.
Same three steps every time: Write y =, swap x ↔ y, make y the subject. It works for any one-to-one function.
Clear the fraction, then isolate y: For a rational function, swap, then multiply up to clear the denominator and gather the y-terms before isolating y.
IB-style question — a rational inverse
Find the inverse of f(x) = (2x + 1)/(x − 3).
Step by step
- Swap x and y.
- Multiply up and expand.
- Gather y-terms and factor.
- Divide.
Final answer
f⁻¹(x) = (3x + 1)/(x − 2).
Collect the y-terms on one side: After multiplying up, move every term containing y to one side, factor out y, then divide.
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f and f⁻¹ undo each other: A correct inverse satisfies f(f⁻¹(x)) = x and f⁻¹(f(x)) = x — composing a function with its inverse gives back x. This is a quick way to check your answer.
IB-style question — verify by composing
Check that f⁻¹(x) = (x + 2)/5 is the inverse of f(x) = 5x − 2.
Step by step
- Compose f with f⁻¹.
- Simplify.
Final answer
f(f⁻¹(x)) = x, so the inverse is correct.
A fast exam check: If a composite of your two functions doesn't simplify to x, you've made a slip — re-check the algebra.
Domain and range swap: The domain of f⁻¹ is the range of f, and the range of f⁻¹ is the domain of f — they trade places. Sometimes you must restrict the domain so the inverse is a function.
IB-style question — a restricted inverse
f(x) = x² for x ≥ 0. Find f⁻¹ and state its domain.
Step by step
- Swap and solve (take the positive root, since x ≥ 0).
- Domain of f⁻¹ = range of f.
Final answer
f⁻¹(x) = √x, with domain x ≥ 0.
Why restrict?: Without x ≥ 0, x² isn't one-to-one (it fails the horizontal-line test), so it wouldn't have a single inverse. The restriction fixes that.