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