Given the function;
[tex]y=\frac{1}{x+3}+2[/tex]
First, we find the inverse of the function as;
[tex]\begin{gathered} \operatorname{Re}place\text{ x with y, we have;} \\ x=\frac{1}{y+3}+2 \\ \text{Then, solve for y, we have;} \\ y=\frac{-3x+7}{x-2} \end{gathered}[/tex]
Then, we get the domain of the inverse function, we have;
[tex]x<2\text{ or x>2}[/tex]
Thus, the graph of the function is;
Hence, the range of the function is;
[tex]\mleft\lbrace y\colon y\in\mathfrak{\Re },y\ne2\mright\rbrace[/tex]
CORRECT OPTION: B
