Consider the function \( f(x)=\sqrt{x+2}-9 \) for the domain \( [-2, \infty) \). Find \( f^{-1}(x) \), where \( f^{-1} \) is the inverse of \( f \). Also state the domain of \( f^{-1} \) in interval notation. \( f^{-1}(x)=\square \) for the domain \( \square \)

To find the inverse of the function \( f(x)=\sqrt{x+2}-9 \), we start by setting \( y = f(x) \): \[ y = \sqrt{x+2} - 9 \] Next, we solve for \( x \) in terms of \( y \): 1. Add 9 to both sides: \[ y + 9 = \sqrt{x+2} \] 2. Square both sides to eliminate the square root: \[ (y + 9)^2 = x + 2 \] 3. Now, isolate \( x \): \[ x = (y + 9)^2 - 2 \] Thus, the inverse function is: \[ f^{-1}(x) = (x + 9)^2 - 2 \] Next, we find the domain of \( f^{-1} \). The range of the original function \( f(x) \) for the domain \( [-2, \infty) \) is found by evaluating: 1. At \( x = -2 \): \[ f(-2) = \sqrt{-2 + 2} - 9 = 0 - 9 = -9 \] 2. As \( x \) approaches \( \infty \), \( f(x) \) approaches \( \infty \). Thus, the range of \( f \) is \( [-9, \infty) \), which specifies the domain of the inverse function. So, we conclude with: \[ f^{-1}(x) = (x + 9)^2 - 2 \quad \text{for the domain } [-9, \infty) \]