Not every function has an inverse. For $f(x) = x^2$, both $2$ and $-2$ go to 4, so an inverse would have to send 4 back to two places at once, and a function can't do that. The fix is to keep only part of the graph. This builds on Checking Inverses by Composition.
Key idea
When a function is not one-to-one, restrict its domain to a piece where it is (for a parabola, one side of the vertex). The inverse then has domain equal to the range of that piece, and its outputs must lie in the restricted domain. That fact tells you which sign of $\pm\sqrt{\phantom{x}}$ to keep.
Why it works
A function has an inverse exactly when each output comes from only one input: the horizontal line test. A parabola turns around at its vertex, so a horizontal line above the vertex meets it twice. On one side of the vertex the parabola only rises (or only falls), so every horizontal line meets that piece at most once. Choosing a side is choosing which input each output goes back to.
Worked examples
Worked example: Keeping the right half
Find the inverse of $f(x) = (x - 3)^2 + 1$ with $x \ge 3$, and give its domain.
- Swap: $x = (y - 3)^2 + 1$, so $(y - 3)^2 = x - 1$.
- Square roots: $y - 3 = \pm\sqrt{x - 1}$.
- The outputs of $f^{-1}$ are the inputs of $f$, which satisfy $y \ge 3$. So keep the plus sign: $y = 3 + \sqrt{x - 1}$.
- The range of $f$ on $x \ge 3$ is $[1, \infty)$, so that is the domain of $f^{-1}$.
Answer: $f^{-1}(x) = 3 + \sqrt{x - 1}$ on $[1, \infty)$
Worked example: Keeping the left half
Same function with $x \le 3$ instead.
- The algebra is identical up to $y - 3 = \pm\sqrt{x - 1}$.
- Now outputs must satisfy $y \le 3$, so keep the minus sign.
Answer: $f^{-1}(x) = 3 - \sqrt{x - 1}$ on $[1, \infty)$
Worked example: Absolute value
Find the inverse of $f(x) = |x + 2|$ with $x \ge -2$.
- For $x \ge -2$ the inside is not negative, so $f(x) = x + 2$ there.
- Its inverse is $x - 2$, and the domain is the range of $f$: $[0, \infty)$.
Answer: $f^{-1}(x) = x - 2$ for $x \ge 0$
Common mistake
Writing the inverse of $x^2$ as $\pm\sqrt{x}$. That rule gives two outputs for one input, so it is not a function. The restriction decides the sign: with $x \ge 0$ the inverse is $\sqrt{x}$; with $x \le 0$ it is $-\sqrt{x}$. Check: $f^{-1}(f(-3))$ must give back $-3$.