Answer:
See below
Step-by-step explanation:
When we talk about the function [tex]f(x)=\sqrt{x}[/tex], the domain and codomain are generally defaulted to be subsets of the Real set. Once [tex]$f:[0,\infty)\to [0,\infty)$[/tex] and [tex]$f\subseteq [0,\infty)\times [0,\infty)$[/tex] such that [tex]$(x,y_1)\in f\wedge(x,y_2)\in f \Rightarrow y_1=y_2$[/tex] for [tex]$f(x)=\sqrt x$[/tex]. Therefore,
[tex]\[\sqrt{\cdot}: \mathbb R_{\geq 0} \to \mathbb R_{\geq 0} \][/tex]
[tex]\[x \mapsto \sqrt{x}\][/tex]
But this table just shows the perfect square solutions.
