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