Answer
(a) $(0,\infty)$.
(b) See graph
(c) $(-\infty,\infty)$, $x=0$.
(d) $f^{-1}(x)=\frac{1}{2}e^{x+3}$.
(e) $(-\infty,,\infty)$, $(0,\infty)$.
(f) See graph.
Work Step by Step
(a) Use the domain requirement(s) for the function $2x\gt0$, we have $x\gt0$ or $(0,\infty)$.
(b) See graph for $f(x)=ln(2x)-3$
(c) We can determine the range as $(-\infty,\infty)$ and asymptote(s) as $x=0$.
(d) We can find $f^{-1}(x)=\frac{1}{2}e^{x+3}$. In short $x=ln(2y)-3\longrightarrow y=\frac{1}{2}e^{x+3}$
(e) We can find the domain of $f^{-1}(x)$ as $(-\infty,,\infty)$ and range as $(0,\infty)$.
(f) See graph.
