Answer
(a) $(-\infty,\infty)$.
(b) See graph
(c) $(0,\infty)$, $y=0$.
(d) $ f^{-1}(x)=ln\frac{x}{3}+2$.
(e) $(0,\infty)$, $(-\infty,\infty)$.
(f) See graph.
Work Step by Step
(a) We can find the domain of $f$ as $(-\infty,\infty)$.
(b) See graph for $f(x)=3e^{x-2}$
(c) We can determine the range of $f$ as $(0,\infty)$, asymptote(s) as $y=0$.
(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 of $f^{-1}$ as $(0,\infty)$, range as $(-\infty,\infty)$.
(f) See graph.