Answer
(a) $(-\infty,0)$.
(b) See graph.
(c) range $(-\infty,\infty)$, V.A. $x=0$.
(d) $ f^{-1}(x)=-e^{-x} $
(e) domain $(-\infty,\infty)$, range $(-\infty,0)$.
(f) See graph.
Work Step by Step
(a) Given $f(x)=-ln(-x)$, we can find the domain $-x\gt0$ or $(-\infty,0)$.
(b) See graph.
(c) From the graph, we can determine the range $(-\infty,\infty)$, asymptote(s) V.A. $x=0$.
(d) $f(x)=-ln(-x) \Longrightarrow y=-ln(-x) \Longrightarrow x=-ln(-y) \Longrightarrow y=-e^{-x} \Longrightarrow f^{-1}(x)=-e^{-x} $
(e) For $ f^{-1}(x)$, we can find the domain $(-\infty,\infty)$, range $(-\infty,0)$.
(f) See graph.
