Answer
(a) $(0,\infty)$.
(b) See graph
(c) $(-\infty,\infty)$, $x=0$.
(d) $f^{-1}(x)=\frac{10^{2x}}{2}$.
(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 $(0,\infty)$.
(b) See graph for $f(x)=\frac{1}{2}log(2x)$
(c) We can determine the range as $(-\infty,\infty)$ and asymptote(s) as $x=0$.
(d) We can find $f^{-1}(x)=\frac{10^{2x}}{2}$. In short $x=\frac{1}{2}log(2y)\longrightarrow y=\frac{10^{2x}}{2}$
(e) We can find the domain of $f^{-1}(x)$ as $(-\infty,\infty)$ and range as $(0,\infty)$.
(f) See graph.
