Answer
(a) $(0,\infty)$.
(b) See graph.
(c) range $(-\infty,\infty)$, V.A. $x=0$.
(d) $f^{-1}(x)=e^{x-2}$
(e) domain $(-\infty,\infty)$, range $(0,\infty)$.
(f) See graph.
Work Step by Step
(a) Given $f(x)=2+ln(x)$, we can find the domain $x\gt0$ or $(0,\infty)$.
(b) See graph.
(c) From the graph, we can determine the range $(-\infty,\infty)$, asymptote(s) V.A. $x=0$.
(d) $f(x)=2+ln(x) \Longrightarrow y=2+ln(x) \Longrightarrow x=2+ln(y) \Longrightarrow y=e^{x-2} \Longrightarrow f^{-1}(x)=e^{x-2}$
(e) For $ f^{-1}(x)$, we can find the domain $(-\infty,\infty)$, range $(0,\infty)$.
(f) See graph.
