Answer
(a) $(-1,\infty)$.
(b) See graph.
(c) range $(-\infty,\infty)$, V.A. $x=-1$.
(d) $ f^{-1}(x)=e^{-x/2}-1$
(e) domain $(-\infty,\infty)$, range $(-1,\infty)$.
(f) See graph.
Work Step by Step
(a) Given $f(x)=-2ln(x+1)$, we can find the domain $x+1\gt0$ or $(-1,\infty)$.
(b) See graph.
(c) From the graph, we can determine the range $(-\infty,\infty)$, asymptote(s) V.A. $x=-1$.
(d) $f(x)=-2ln(x+1) \Longrightarrow y=-2ln(x+1) \Longrightarrow x=-2ln(y+1) \Longrightarrow y=e^{-x/2}-1 \Longrightarrow f^{-1}(x)=e^{-x/2}-1$
(e) For $ f^{-1}(x)$, we can find the domain $(-\infty,\infty)$, range $(-1,\infty)$.
(f) See graph.
