Answer
(a) $(-\infty,\infty)$.
(b) see graph.
(c) $(0,\infty)$, $y=0$.
(d) $ f^{-1}(x)=ln(\frac{x}{3})+2$.
(e) $(0,\infty)$ and $(-\infty,\infty)$.
(f) see graph.
Work Step by Step
Given $f(x)=3e^{x-2}$, we have:
(a) the domain of $f(x)$: $(-\infty,\infty)$.
(b) see graph.
(c) From the graph, we can determine the range of $f(x)$: $(0,\infty)$, asymptote(s) of $f(x)$: H.A. $y=0$.
(d) $f(x)=3e^{x-2} \Longrightarrow y=3e^{x-2} \Longrightarrow x=3e^{y-2} \Longrightarrow y=ln(\frac{x}{3})+2 \Longrightarrow f^{-1}(x)=ln(\frac{x}{3})+2$.
(e) we can find the domain and the range of $f^{-1}(x)$: $(0,\infty)$ and $(-\infty,\infty)$.
(f) see graph.