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