Let \( f(x)=(x+4)^{2} \). Find a domain on which \( f \) is one-to-one and non-decreasing. Find the inverse of \( f \) restricted to this domain. \( f^{-1}(x)=\square \)

To determine a domain on which the function \( f(x) = (x + 4)^2 \) is one-to-one and non-decreasing, we first analyze the function. 1. **Finding the vertex**: The function \( f(x) = (x + 4)^2 \) is a parabola that opens upwards. The vertex of this parabola is at \( x = -4 \). 2. **Determining intervals**: - The function is non-decreasing for \( x \geq -4 \) because the derivative \( f'(x) = 2(x + 4) \) is non-negative when \( x + 4 \geq 0 \) (i.e., \( x \geq -4 \)). - The function is decreasing for \( x < -4 \). Thus, we can restrict the domain to \( x \geq -4 \) to ensure that \( f \) is one-to-one and non-decreasing. 3. **Finding the inverse**: - We restrict the domain to \( [-4, \infty) \). - To find the inverse, we set \( y = f(x) \): \[ y = (x + 4)^2 \] - To solve for \( x \), we take the square root of both sides: \[ \sqrt{y} = x + 4 \quad \text{(since we are considering the non-negative root)} \] - Rearranging gives: \[ x = \sqrt{y} - 4 \] - Therefore, the inverse function is: \[ f^{-1}(x) = \sqrt{x} - 4 \] In conclusion, the domain on which \( f \) is one-to-one and non-decreasing is \( [-4, \infty) \), and the inverse of \( f \) restricted to this domain is: \[ f^{-1}(x) = \sqrt{x} - 4 \]