Answer
$f^{-1}(x)=sin^{-1}(\frac{x-2}{5})$
range of $f(x)$: $[-3,7]$,
domain and range of $f^{-1}(x)$: $[-3,7]$ and $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
Work Step by Step
1. $f(x)=5sin(x)+2 \Longrightarrow y=5sin(x)+2 \Longrightarrow x=5sin(y)+2 \Longrightarrow y=sin^{-1}(\frac{x-2}{5}) \Longrightarrow f^{-1}(x)=sin^{-1}(\frac{x-2}{5})$
2. We can find the range of $f(x)$: $[-3,7]$,
3. We can find the domain and range of $f^{-1}(x)$: $[-3,7]$ and $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
