Answer
(a) $(0,\infty)$.
(b) See graph.
(c) range $(-\infty,\infty)$, V.A. $x=0$.
(d) $ f^{-1}(x)=\frac{1}{2}e^{x+3}$
(e) domain $(-\infty,\infty)$, range $(0,\infty)$.
(f) See graph.
Work Step by Step
(a) Given $f(x)=ln(2x)-3$, we can find the domain $2x\gt0$ or $(0,\infty)$.
(b) See graph.
(c) From the graph, we can determine the range $(-\infty,\infty)$, asymptote(s) V.A. $x=0$.
(d) $f(x)=ln(2x)-3 \Longrightarrow y=ln(2x)-3 \Longrightarrow x=ln(2y)-3 \Longrightarrow y=\frac{1}{2}e^{x+3} \Longrightarrow f^{-1}(x)=\frac{1}{2}e^{x+3}$
(e) For $ f^{-1}(x)$, we can find the domain $(-\infty,\infty)$, range $(0,\infty)$.
(f) See graph.
