Answer
(a) $(-\infty,\infty)$.
(b) See graph.
(c) range $(-\infty,0)$, H.A. $y=0$.
(d) $ f^{-1}(x)=log_3(-x)-1$
(e) domain $(-\infty,0)$, range $(-\infty,\infty)$.
(f) See graph.
Work Step by Step
(a) Given $f(x)=-3^{x+1}$, we can find the domain $(-\infty,\infty)$.
(b) See graph.
(c) From the graph, we can determine the range $(-\infty,0)$, asymptote(s) H.A. $y=0$.
(d) $f(x)=-3^{x+1}\Longrightarrow y=-3^{x+1}\Longrightarrow x=-3^{y+1} \Longrightarrow y=log_3(-x)-1 \Longrightarrow f^{-1}(x)=log_3(-x)-1$
(e) For $ f^{-1}(x)$, we can find the domain $(-\infty,0)$, range $(-\infty,\infty)$.
(f) See graph.
