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.
IB-style question — undo a value
Given that f(2) = 7, write down f⁻¹(7).
Step by step
- f sends 2 → 7, so f⁻¹ sends 7 back to 2.
Final answer
f⁻¹(7) = 2.
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.
[Diagram: math-inverse-reflection] - Available in full study mode
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.
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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).
IB-style question — a linear inverse
Find the inverse of f(x) = 2x + 3.
Step by step
- Write y = f(x).
- Swap x and y.
- Solve for y.
Final answer
f⁻¹(x) = (x − 3)/2.
IB-style question — a fraction
Find the inverse of f(x) = (x − 4)/3.
Step by step
- Write y, then swap x and y.
- Multiply up and solve for y.
Final answer
f⁻¹(x) = 3x + 4.
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.
IB-style question — the swap in action
f(x) = √x has domain x ≥ 0 and range y ≥ 0. Describe its inverse.
Step by step
- Swap x and y: x = √y, so y = x².
- The domain of f⁻¹ is the range of f.
Final answer
f⁻¹(x) = x², restricted to x ≥ 0 (inherited from f's range).
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.