Consider the function $y(x) = \tfrac{1}{2}x^2 -2x + 1$, $(x\geq 2)$. Determine the inverse function $x(y)$ and its domain.
The inverse function can be found by rewriting the function $y(x)$. Here we use the discriminant criterion (at $(*)$):
$$
\begin{align}
y &= \tfrac{1}{2}x^2 -2x + 1\\
0 &= \tfrac{1}{2}x^2 -2x + (1-y)\\
x &\stackrel{(*)}{=} \dfrac{-(-2) \pm \sqrt{(-2)^2- 4\cdot \tfrac{1}{2}\cdot(1-y)}}{2\cdot \tfrac{1}{2}} = \dfrac{2 \pm \sqrt{4- 2(1-y)}}{1} = 2 \pm \sqrt{4 - 2 +2y} = 2\pm \sqrt{2+2y}.
\end{align}
$$
We obtain two possible inverse functions: $ x(y) = 2 + \sqrt{2 + 2y}$ or $x(y) = 2 - \sqrt{2+2y}$. Since it is given that $x\geq 2$, the second one is deleted. Hence, the inverse function is
$$x(y) = 2 + \sqrt{2+2y}.$$
Since you cannot take the square root of a negative number, the domain of this function consists of all $y$ such that
$$2 + 2y \geq 0 \qquad \rightarrow \qquad 2y \geq -2 \qquad \rightarrow \qquad y \geq -1.$$
