Answer
$f^{-1}(x)=\frac{1}{2}sin^{-1}(\frac{x}{3})-\frac{1}{2}$
range of $f(x)$: $[-3,3]$,
domain and range of $f^{-1}(x)$: $[-3,3]$ and $[-\frac{1}{2}-\frac{\pi}{4},-\frac{1}{2}+\frac{\pi}{4}]$.
Work Step by Step
1. $f(x)=3sin(2x+1) \Longrightarrow y=3sin(2x+1) \Longrightarrow x=3sin(2y+1) \Longrightarrow y=\frac{1}{2}sin^{-1}(\frac{x}{3})-\frac{1}{2} \Longrightarrow f^{-1}(x)=\frac{1}{2}sin^{-1}(\frac{x}{3})-\frac{1}{2}$
2. We can find the range of $f(x)$: $[-3,3]$,
3. We can find the domain and range of $f^{-1}(x)$: $[-3,3]$ and $[-\frac{1}{2}-\frac{\pi}{4},-\frac{1}{2}+\frac{\pi}{4}]$.