So if f(x)=√(x-3), I know the original domain would be [3, ∞) and the range would be [0,∞), but what would the domain and the range be for the inverse be, which is f

^{-1}=x^{2}+3, and why would it be that way? My teacher told us the domain of the original one to one would be the range of the inverse and the range of the original would the the domain of the inverse, but wouldn't the domain be (-∞, ∞) because any integer can be squared and the result be real? And if you drew the graph, the parabola would extend to both sides of the x axis forever, right? So why would the domain be [0, ∞)?I may have follow up questions about this topic too! Thanks for your help!

## Comments

^{-1 }and y^{-1}, both x and x^{-1}must follow the first rule. That is not a mirror.^{-1}(x) has a mirror domain/range, is impossible.^{2}, the exponent being an even number.^{2}^{2}.Comment