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