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.

$y=x^2$y = x² 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 $y=x^2$y = x², keep only $x\ge 0$x ≥ 0. The right half of the parabola only rises, so no horizontal line meets it twice.

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

The inverse on that half

On $x\ge 0$x ≥ 0, $y=x^2$y = x² gives $x=\sqrt{y}$x = √y, taking the positive root because x is not negative. Writing x for the input, $f^{-1}\left(x\right)=\sqrt{x}$f⁻¹(x) = √x.

Its graph is the kept half reflected in the line y = x. The two curves meet on that line, where $x^2=x$x² = x: at (0, 0) and (1, 1).

Cutting at the vertex

For $f\left(x\right)={\left(x-2\right)}^2+1$f(x) = (x − 2)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 $x\ge a$x ≥ a is $x\ge 2$x ≥ 2.

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

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

The other half

$g\left(x\right)=x^2-6x+5$g(x) = x² − 6x + 5 completes the square to ${\left(x-3\right)}^2-4$(x − 3)² − 4, with its vertex at (3, −4). Keep $x\le 3$x ≤ 3 this time.

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

The usual slips

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

Keeping $x\ge 0$x ≥ 0 out of habit from $y=x^2$y = x². The cut belongs at this curve’s own vertex.

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

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

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