Inverse Function
Find f⁻¹(x) for linear, rational, root, log and odd-power functions, with the algebra shown.
f(x) = 2x + 3
- 1Write y = 2x + 3.Start from the definition.
- 2Swap x and y, then solve for y.An inverse undoes the original, so the input and output trade places before you isolate y.
- 3Undo each operation in reverse.x = 2y + 3 becomes y = (x − 3) ÷ 2.
An inverse undoes a function. Swap x and y, then solve for y: from y = 2x + 3 you get x = 2y + 3 and so f⁻¹(x) = (x − 3)/2. Only one-to-one functions have inverses, which is why a parabola needs its domain restricted first.
Swapping is the whole method
If f sends 4 to 11, then f⁻¹ sends 11 back to 4. Inputs and outputs trade places, and that is exactly what swapping x and y does on the page. Everything after the swap is ordinary equation solving: isolate y by undoing each operation in the reverse order it was applied.
Reverse order is the part worth stressing. If f multiplies by 2 and then adds 3, the inverse subtracts 3 and then divides by 2 — not the other way round. It is the same logic as taking off a coat before a jumper.
Not every function has one
An inverse must send each output back to a single input, so the original has to be one-to-one — no two inputs sharing an output. The horizontal line test makes this visible: if any horizontal line crosses the graph twice, the inverse would have to send that height to two places, and no function does that. A parabola fails it, which is why f(x) = x² is normally restricted to x ≥ 0 before anyone writes √x as its inverse.
the definition, and the check worth doing on every answer
- 1 Write the function as y = … Replace f(x) with y, so f(x) = 2x + 3 becomes y = 2x + 3.
- 2 Swap x and y. That gives x = 2y + 3 — the same relationship read in the opposite direction.
- 3 Solve for y, undoing operations in reverse. Subtract 3 to get x − 3 = 2y, then divide by 2 to get y = (x − 3)/2.
- 4 Write it as f⁻¹(x). So f⁻¹(x) = (x − 3)/2.
- 5 Check by composing both ways. f(f⁻¹(x)) = 2·(x − 3)/2 + 3 = x, and the other order gives x too, so the inverse is right.
The families with an exact inverse
Each carries its own restriction, which the tool states alongside the formula.
| Function | Inverse | Restriction |
|---|---|---|
| ax + b (a ≠ 0) | (x − b)/a | None |
| (ax + b)/(cx + d) | (b − dx)/(cx − a) | x ≠ a/c |
| √(ax + b) | (x² − b)/a | x ≥ 0 |
| ln(ax + b) | (eˣ − b)/a | None |
| ax³ + b | ∛((x − b)/a) | None |
| ax² + bx + c | h ± √((x − k)/a) | Domain must be restricted first |
Domain and range trade places too
Because inputs and outputs swap, the domain of f becomes the range of f⁻¹ and the range of f becomes the domain of f⁻¹. That is why the inverse of √(x − 3) carries the restriction x ≥ 0: the original produced only non-negative outputs, so the inverse can only accept them. Dropping that restriction turns a correct formula into a wrong one, because x² + 3 accepts negatives that the true inverse does not.
One more caution: f⁻¹ does not mean 1/f. The superscript −1 marks an inverse function, not a reciprocal — the inverse of 2x + 3 is (x − 3)/2, nothing like 1/(2x + 3). And graphically, f and f⁻¹ are mirror images across the line y = x, which is a quick visual check when an answer looks wrong.