Let $f:[1, \infty) \to \mathbb{R}, x \mapsto f(x) = x^2-2x+3$
How can one find the inverse function $f^{-1}:W(f) \to [1,\infty)?$
Without the interval I know that for $y$ we get $y=1+\sqrt{x-2}, y = 1 - \sqrt{x-2}$ for the inverse function, but I don't know how it's done when an interval is given.
After graphing the function one knows that the codomain $W(f)$ is given by $W(f) = f([1,\infty)) = [2,\infty)$. But how can one show that?
We first let $f(x)=y$ and solve for $x$ ignoring the range at first: $$f(x)=x^2-2x+3=(x-1)^2+2=y$$ $$x=\pm\sqrt{y-2}+1$$ We reject the negative sign because if $y=3$ we get $x=0$ in this case, outside the (known) range. Hence $x=\sqrt{y-2}+1$ and the inverse is $$f^{-1}:[2,\infty)\to[1,\infty),f(x)=1+\sqrt{x-2}$$ where the domain of $f^{-1}$ may be worked out considering that the domain of $\sqrt x$ is $[0,\infty)$.