Answer
(a) $(2,\infty)$.
(b) see graph.
(c) $(-\infty,\infty)$, V.A. $x=2$.
(d) $ f^{-1}(x)=5^{1-x}+2 $.
(e) $(-\infty,\infty)$, $(2,\infty)$.
(f) see graph.
Work Step by Step
Given $f(x)=1-log_5(x-2)$, we have:
(a) the domain of $f(x)$: $(2,\infty)$.
(b) see graph.
(c) From the graph, we can determine the range $(-\infty,\infty)$, asymptote(s) V.A. $x=2$.
(d) $f(x)=1-log_5(x-2) \Longrightarrow y=1-log_5(x-2) \Longrightarrow x=1-log_5(y-2) \Longrightarrow y=5^{1-x}+2 \Longrightarrow f^{-1}(x)=5^{1-x}+2 $.
(e) we can find the domain and the range of $f^{-1}(x)$: $(-\infty,\infty)$, $(2,\infty)$.
(f) see graph.
