Consider the function \( f(x)=\sqrt{5 x-15} \) for the domain \( [3, \infty) \). 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 function \( f^{-1}(x) \) of \( f(x) = \sqrt{5x - 15} \), we first substitute \( y \) for \( f(x) \): \[ y = \sqrt{5x - 15} \] Now, solve for \( x \): 1. Square both sides: \[ y^2 = 5x - 15 \] 2. Rearrange to solve for \( x \): \[ 5x = y^2 + 15 \] \[ x = \frac{y^2 + 15}{5} \] 3. Switch \( x \) and \( y \) to express the inverse: \[ f^{-1}(x) = \frac{x^2 + 15}{5} \] Next, we need to determine the domain of \( f^{-1} \). The range of the original function \( f(x) \), given that the domain is \( [3, \infty) \), is calculated by evaluating \( f(3) \): \[ f(3) = \sqrt{5 \cdot 3 - 15} = \sqrt{0} = 0 \] As \( x \) approaches infinity, \( f(x) \) also approaches infinity. Thus, the range of \( f(x) \) is \( [0, \infty) \), which means the domain of \( f^{-1}(x) \) is also \( [0, \infty) \). So we can conclude: \[ f^{-1}(x) = \frac{x^2 + 15}{5} \text{ for the domain } [0, \infty) \]