Answer
$f^{-1}(x)=e^x-2$, domain $(-\infty,\infty)$, rang $(-2,\infty)$
Work Step by Step
Step 1. Find the inverse: $y=ln{x+2} \longrightarrow x=e^y-2 \longrightarrow f^{-1}(x)=e^x-2$
Steo 2. For $f^{-1}(x)=e^x-2$, we can find domain $(-\infty,\infty)$, rang $(-2,\infty)$