Answer
$f^{-1}(x)=\frac{1}{3}cos^{-1}(-\frac{x}{2})$
range of $f(x)$: $[-2,2]$,
domain and range of $f^{-1}(x)$: $[-2,2]$ and $[0, \frac{\pi}{3}]$.
Work Step by Step
1. $f(x)=-2cos(3x) \Longrightarrow y=-2cos(3x) \Longrightarrow x=-2cos(3y) \Longrightarrow y=\frac{1}{3}cos^{-1}(-\frac{x}{2}) \Longrightarrow f^{-1}(x)=\frac{1}{3}cos^{-1}(-\frac{x}{2})$
2. We can find the range of $f(x)$: $[-2,2]$,
3. We can find the domain and range of $f^{-1}(x)$: $[-2,2]$ and $[0, \frac{\pi}{3}]$.
