For $\backslash ($f$($x$,$ y$)=3+\backslash$sqrt$\{2$x $-$ y $+1\}\backslash ),$ the domain is determined by the requirement that the expression under the square root, $\backslash (2$x $-$ y $+1\backslash ),$ must be non$-$negative. Therefore, $\backslash (2$x $-$ y $+1\backslash$geq $0\backslash ).$ The range of $\backslash ($f$\backslash ),$ given its definition, is $\backslash ($f$($x$,$ y$)>3\backslash )$ since the square root function outputs values $\backslash (\backslash$geq $0\backslash ),$ and adding $3$ shifts these values upwards.For $\backslash ($g$($x$,$ y$)=\backslash$frac$\{\backslash$sin$($x$^2+$ y$^2-1)\}\{$x$^2+$ y$^2-1\}\backslash ),$ the domain excludes points where $\backslash ($x$^2+$ y$^2-1=0\backslash )$ to prevent division by zero, thus $\backslash ($x$^2+$ y$^2$

eq $1\backslash ).$ The range of $\backslash ($g$\backslash )$ is all real numbers because the function $\backslash (\backslash$frac$\{\backslash$sin z$\}\{$z$\}\backslash )($with $\backslash ($z $=$ x$^2+$ y$^2-1\backslash ))$ is bounded and continuous except where $\backslash ($z $=0\backslash ),$ and it approaches $1$ as $\backslash ($z$\backslash )$ approaches $0$ from either side.