Question Number 161150 by mathlove last updated on 13/Dec/21
Commented by MJS_new last updated on 13/Dec/21
$${y}=\sqrt{\mathrm{2}{x}^{\mathrm{2}} }\geqslant\mathrm{0} \\ $$$${y}=\sqrt{\mathrm{2}}\mid{x}\mid\:\Rightarrow\:{xy}=\sqrt{\mathrm{2}}\mid{x}\mid{x}=\begin{cases}{{x}<\mathrm{0};\:−\sqrt{\mathrm{2}}{x}^{\mathrm{2}} }\\{{x}\geqslant\mathrm{0};\:\sqrt{\mathrm{2}}{x}^{\mathrm{2}} }\end{cases} \\ $$