Answer
(a) $(-\infty,\infty)$.
(b) see graph.
(c) $(-2,\infty)$, H.A. $y=-2$.
(d) $f^{-1}(x)=log_4(x+2)-1$.
(e) $(-2,\infty)$, $(-\infty,\infty)$.
(f) see graph.
Work Step by Step
Given $f(x)=4^{x+1}-2$, we have:
(a) the domain of $f(x)$: $(-\infty,\infty)$.
(b) see graph.
(c) From the graph, we can determine the range $(-2,\infty)$, asymptote(s) H.A. $y=-2$.
(d) $f(x)=4^{x+1}-2 \Longrightarrow y=4^{x+1}-2 \Longrightarrow x=4^{y+1}-2 \Longrightarrow y=log_4(x+2)-1 \Longrightarrow f^{-1}(x)=log_4(x+2)-1$.
(e) we can find the domain and the range of $f^{-1}(x)$: $(-2,\infty)$, $(-\infty,\infty)$.
(f) see graph.
