What is the range of the inverse of the function \( f(x)=\sqrt{2-x} \) ? \( (-\infty, 2] \) \( (-\infty, 2) \) Cannot be determined from this limited information. \( (-\infty, \infty) \)

To determine the range of the inverse of the function \( f(x) = \sqrt{2 - x} \), let's follow these steps: 1. **Find the Domain and Range of \( f(x) \):** - **Domain of \( f(x) \):** The expression inside the square root must be non-negative. \[ 2 - x \geq 0 \implies x \leq 2 \] So, the domain of \( f(x) \) is \( (-\infty, 2] \). - **Range of \( f(x) \):** Since \( \sqrt{2 - x} \) is always non-negative and can take any value from 0 to \( +\infty \) as \( x \) approaches \( -\infty \). \[ \text{Range of } f(x) = [0, +\infty) \] 2. **Determine the Inverse Function \( f^{-1}(y) \):** - Start with \( y = \sqrt{2 - x} \). - Solve for \( x \): \[ y^2 = 2 - x \implies x = 2 - y^2 \] So, \( f^{-1}(y) = 2 - y^2 \). 3. **Find the Range of \( f^{-1}(y) \):** - The domain of \( f^{-1}(y) \) is the range of \( f(x) \), which is \( [0, +\infty) \). - As \( y \) increases from 0 to \( +\infty \): \[ f^{-1}(y) = 2 - y^2 \quad \text{decreases from } 2 \text{ to } -\infty \] - Therefore, the range of \( f^{-1}(y) \) is \( (-\infty, 2] \). **Answer:** \( (-\infty, 2] \)