Answer
$f^{-1}(x)=tan^{-1}(-x)-1$
range of $f(x)$: $(-\infty,\infty)$
domain and range of $f^{-1}(x)$: $(-\infty,\infty)$ and $[-1-\frac{\pi}{2}, \frac{\pi}{2}-1]$
Work Step by Step
Step 1. $f(x)=-tan(x+1)-3 \longrightarrow y=-tan(x+1)-3 \longrightarrow x=-tan(y+1)-3\longrightarrow y=tan^{-1}(-x)-1 \longrightarrow f^{-1}(x)=tan^{-1}(-x)-1$
Step 2. We can find the domain and range of $f(x)$: $[-1-\frac{\pi}{2}, \frac{\pi}{2}-1]$ and $(-\infty,\infty)$
Step 3. We can find the domain and range of $f^{-1}(x)$: $(-\infty,\infty)$ and $[-1-\frac{\pi}{2}, \frac{\pi}{2}-1]$
