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