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