Functions · step 4 of 13

Inverse functions

f has an inverse function exactly when it is bijective, and then f⁻¹ undoes it: f⁻¹(f(x)) = x and f(f⁻¹(y)) = y. Finding f⁻¹ means solving y = f(x) for x, and the domain and codomain swap in the process. When f is not injective you can often restrict the domain to make it so — which is exactly how √ and arcsin are defined.

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Method: how to approach it

The order below is what examiners expect to see, and each step carries its own marks.

Confirm bijectivity firstNo inverse exists otherwise. If f is not injective, restrict the domain and say so explicitly.
Set y = f(x) and solve for xRearrange to express x in terms of y. Every step must be reversible.
Swap the rolesWrite f⁻¹(y) = your expression. The domain of f⁻¹ is the codomain (image) of f and vice versa.
Verify both compositionsCheck f⁻¹(f(x)) = x and f(f⁻¹(y)) = y. One direction alone only proves a one-sided inverse.

Worked example

Find the inverse of f : [0,∞) → [0,∞), f(x) = x² + 1 — first fixing its codomain.

On [0,∞) the function is injective (x² is increasing there), and its image is [1, ∞), so take the codomain to be [1,∞).
Set y = x² + 1, so x² = y − 1.
On the domain x ≥ 0 we take the positive root: x = √(y − 1).
Check: f⁻¹(f(x)) = √(x² + 1 − 1) = √(x²) = x for x ≥ 0 ✓.

Answer. f⁻¹ : [1,∞) → [0,∞), f⁻¹(y) = √(y − 1).

Where marks get dropped

These are the specific errors that cost credit on inverse functions questions — QED's rubric penalises each of them separately.

Writing ±√ in the inverse. A function has one output, so the domain restriction must decide the sign.
Forgetting to change the domain and codomain. f⁻¹ has domain equal to the image of f, and stating this is usually a mark.
Confusing f⁻¹ with 1/f. They are unrelated: the inverse of f(x) = 2x is x/2, not 1/(2x).

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Inverse functions — frequently asked questions

Why must f be bijective?

Injectivity ensures each output has at most one origin (so f⁻¹ is single-valued), and surjectivity ensures each element of the codomain has at least one (so f⁻¹ is total).

What is a left inverse?

A g with g∘f = id, which exists exactly when f is injective. A right inverse satisfies f∘g = id and exists when f is surjective. A two-sided inverse needs both.

How does the graph relate?

The graph of f⁻¹ is the reflection of the graph of f in the line y = x, which is why a horizontal-line test on f is a vertical-line test on f⁻¹.

The rest of Functions

Injective, surjective, bijective, composition, inverse. Each subtopic below has its own method, worked example and mark-losing traps.

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