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