Restricting the Domain for an Inverse

When an inverse cannot exist

An inverse takes an output of f and returns the input it came from. That only works if each output came from exactly one input.

$y = x^{2}$ fails this: 4 comes from both x = −2 and x = 2, so an inverse given 4 would have two answers. On the graph, the horizontal line y = 4 meets the curve twice.

One-to-one

A function is one-to-one when every output comes from a single input. The graph shows it directly: every horizontal line meets the curve at most once. This is the horizontal-line test.

A function that is not one-to-one can be made so by restricting its domain. For $y = x^{2}$, keep only $x \ge 0$. The right half of the parabola only rises, so no horizontal line meets it twice.

Keeping $x \le 0$ would work too. A domain is a choice, and the inverse depends on which half was kept.

The inverse on that half

On $x \ge 0$, $y = x^{2}$ gives $x = \sqrt{y}$, taking the positive root because x is not negative. Writing x for the input, $f^{-1}\left(x\right) = \sqrt{x}$.

Its graph is the kept half reflected in the line y = x. The two curves meet on that line, where $x^{2} = x$: at (0, 0) and (1, 1).

Cutting at the vertex

For $f\left(x\right) = \left(x - 2\right)^{2} + 1$ the vertex is at (2, 1). Either side of x = 2 the curve only rises or only falls, so the largest domain of the form $x \ge a$ is $x \ge 2$.

Solve $y = \left(x - 2\right)^{2} + 1$ for x: $y - 1 = \left(x - 2\right)^{2}$, so $x - 2 = \sqrt{y - 1}$, positive because $x \ge 2$. Then $x = 2 + \sqrt{y - 1}$, and $f^{-1}\left(x\right) = 2 + \sqrt{x - 1}$.

The inverse takes the outputs of f, so its domain is the range of f. On $x \ge 2$ the lowest output is f(2) = 1, so $f^{-1}$ is defined for $x \ge 1$. A check: f(3) = 2, and $f^{-1}\left(2\right) = 2 + \sqrt{1} = 3$.

The other half

$g\left(x\right) = x^{2} - 6x + 5$ completes the square to $\left(x - 3\right)^{2} - 4$, with its vertex at (3, −4). Keep $x \le 3$ this time.

Then $x - 3 = -\sqrt{y + 4}$, negative because $x \le 3$, so $g^{-1}\left(x\right) = 3 - \sqrt{x + 4}$ for $x \ge -4$. A check: g(1) = 0, and $g^{-1}\left(0\right) = 3 - \sqrt{4} = 1$.

The usual slips

Cutting at the height of the vertex instead of its position: for $\left(x - h\right)^{2} + k$ the cut is at x = h, and k is where the range starts.

Keeping $x \ge 0$ out of habit from $y = x^{2}$. The cut belongs at this curve’s own vertex.

Taking the wrong root: on $x \ge h$ the inverse adds the square root; on $x \le h$ it subtracts it.

Undoing the steps in their original order. f subtracts h, squares, then adds k, so $f^{-1}$ subtracts k, takes the root, then adds h.

Giving the domain of f as the domain of $f^{-1}$. The inverse’s domain is f’s range.