Inverse as reflection

The inverse undoes the function: The inverse f⁻¹ reverses f: if f turns a into b, then f⁻¹ turns b back into a.

In symbols, f(a) = b ⟺ f⁻¹(b) = a — inputs and outputs swap roles.

f⁻¹ is NOT 1/f: The notation f⁻¹ means the inverse function, not the reciprocal: f⁻¹(x) ≠ 1/f(x).

The −1 is a label, not a power here.

Reflect the graph in y = x: The graph of f⁻¹ is the mirror image of f in the line y = x.

Every point (a, b) on f becomes (b, a) on f⁻¹ — the coordinates swap.

A point on y = x stays put: If a point already lies on y = x, its reflection is itself.

That's why f and f⁻¹ meet on the line y = x.

Swap, then solve: To find f⁻¹: (1) write y = f(x), (2) swap x and y, (3) solve for y.

The result is f⁻¹(x).

Check with a point: If f(1) = 5, then f⁻¹(5) should give 1 back.

(Or confirm f(f⁻¹(x)) = x.)

Domain and range trade places: Because the reflection swaps coordinates, domain of f⁻¹ = range of f and range of f⁻¹ = domain of f.

So f⁻¹ sometimes needs a restricted domain.

Find where they meet: Since f and f⁻¹ intersect on y = x, you can find their intersection point(s) by solving f(x) = x.

The Section B inverse question: This classic question rolls four skills into one: find f⁻¹, give its domain and range, then find where f and f⁻¹ meet.

We'll do it one part at a time, with f(x) = √(6x − 3), for x ≥ ½.

How do we know they meet on y = x?: f⁻¹ is f reflected in the line y = x.

For a curve like this one that always increases, f and f⁻¹ can only cross on that mirror line — so at any meeting point y = x, which means f(x) = x.

Solving that is far quicker than solving f(x) = f⁻¹(x).

Two easy marks to lose: The domain of f⁻¹ is the range of f (here x ≥ 0 — not all x).

And after squaring, check each answer really fits f(x) = √(6x − 3).