Answer
(a) $(-1,\infty)$.
(b) See graph
(c) $(-\infty,\infty)$, $x=-1$.
(d) $f^{-1}(x)=e^{-x/2}-1$.
(e) $(-\infty,\infty)$, $(-1,\infty)$.
(f) See graph.
Work Step by Step
(a) Use the domain requirement(s) for the function $x+1\gt0$, we have $x\gt-1$ or $(-1,\infty)$.
(b) See graph for $f(x)=-2ln(x+1)$
(c) We can determine the range as $(-\infty,\infty)$ and asymptote(s) as $x=-1$.
(d) We can find $f^{-1}(x)=e^{-x/2}-1$. In short $x=-2ln(y+1)\longrightarrow y=e^{-x/2}-1$
(e) We can find the domain of $f^{-1}(x)$ as $(-\infty,\infty)$ and range as $(-1,\infty)$.
(f) See graph.
