Answer
(a) $(-\infty,\infty)$.
(b) See graph.
(c) range $(2,\infty)$, H.A. $y=2$.
(d) $ f^{-1}(x)=ln(x-2)-ln3$
(e) domain $(2,\infty)$, range $(-\infty,\infty)$.
(f) See graph.
Work Step by Step
(a) Given $f(x)=3e^{x}+2$, we can find the domain $(-\infty,\infty)$.
(b) See graph.
(c) From the graph, we can determine the range $(2,\infty)$, asymptote(s) H.A. $y=2$.
(d) $f(x)=3e^{x}+2\Longrightarrow y=3e^{x}+2\Longrightarrow x=3e^{y}+2 \Longrightarrow y=ln(x-2)-ln3 \Longrightarrow f^{-1}(x)=ln(x-2)-ln3$
(e) For $ f^{-1}(x)$, we can find the domain $(2,\infty)$, range $(-\infty,\infty)$.
(f) See graph.
