Answer
(a) $(3,\infty)$.
(b) See graph.
(c) range $(-\infty,\infty)$, V.A. $x=3$.
(d) $ f^{-1}(x)=e^x+3$
(e) domain $(-\infty,\infty)$, range $(3,\infty)$.
(f) See graph.
Work Step by Step
(a) Given $f(x)=ln(x-3)$, we can find the domain $x-3\gt0$ or $(3,\infty)$.
(b) See graph.
(c) From the graph, we can determine the range $(-\infty,\infty)$, asymptote(s) V.A. $x=3$.
(d) $f(x)=ln(x-3) \Longrightarrow y=ln(x-3) \Longrightarrow x=ln(y-3) \Longrightarrow y=e^x+3 \Longrightarrow f^{-1}(x)=e^x+3$
(e) For $ f^{-1}(x)$, we can find the domain $(-\infty,\infty)$, range $(3,\infty)$.
(f) See graph.
