Find the inverse function $f^{-1}$ of the function $f$. Find the range of $f$ and the domain and range of $\mathrm{f}^{-1}$. \[ f(x)=4 \sin x+7 ;-\frac{\pi}{2} \leq x \leq \frac{\pi}{2} \] Find the inverse, $\mathrm{f}^{-1}$, of $\mathrm{f}$. \[ f^{-1}(x)=\square \] (Use integers or fractions for any numbers in the expression.)

Step 1 ：Given the function \(f(x)=4 \sin x+7 ; -\frac{\pi}{2} \leq x \leq \frac{\pi}{2}\), we are asked to find the inverse function \(f^{-1}\), the range of \(f\), and the domain and range of \(f^{-1}\).

Step 2 ：The inverse of a function is found by swapping the x and y (or f(x)) values, and solving for y. In this case, we need to solve for x in the equation y = 4sin(x) + 7.

Step 3 ：Let's solve for x: \(y = 4\sin(x) + 7\) becomes \(x = \sin^{-1}(\frac{y-7}{4})\). So, the inverse function of f is \(f^{-1}(x) = \sin^{-1}(\frac{x-7}{4})\).

Step 4 ：The range of f is the set of all possible output values (y-values), which for a sine function with amplitude 4 and vertical shift 7, is between 3 and 11. Therefore, the range of f is [3, 11].

Step 5 ：The domain of \(f^{-1}\) is the same as the range of f, and the range of \(f^{-1}\) is the same as the domain of f, which is between -\(\frac{\pi}{2}\) and \(\frac{\pi}{2}\). Therefore, the domain of \(f^{-1}\) is [3, 11] and the range of \(f^{-1}\) is \([-\frac{\pi}{2}, \frac{\pi}{2}]\).

Step 6 ：\(\boxed{f^{-1}(x)=\sin^{-1}\left(\frac{x-7}{4}\right)}\)

Step 7 ：\(\boxed{\text{Range of } f: [3, 11]}\)

Step 8 ：\(\boxed{\text{Domain of } f^{-1}: [3, 11]}\)

Step 9 ：\(\boxed{\text{Range of } f^{-1}: \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]}\)