Let \( f(x)=(x+1)^{2} \) 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 find a domain on which the function \( f(x) = (x + 1)^2 \) is one-to-one and non-decreasing, we need to analyze the function's behavior. 1. **Determine the nature of the function:** - The function \( f(x) = (x + 1)^2 \) is a quadratic function that opens upwards. It has a vertex at \( x = -1 \), which is the minimum point. 2. **Identify intervals for one-to-one and non-decreasing behavior:** - A function is one-to-one if it is either strictly increasing or strictly decreasing. Since \( f(x) \) is a parabola opening upwards, it is decreasing on the interval \( (-\infty, -1] \) and increasing on the interval \( [-1, \infty) \). - To ensure that \( f \) is non-decreasing and one-to-one, we can restrict the domain to \( [-1, \infty) \). 3. **Find the inverse of \( f \) on the restricted domain:** - To find the inverse, we start by setting \( y = f(x) \): \[ y = (x + 1)^2 \] - Next, we solve for \( x \): \[ \sqrt{y} = x + 1 \quad \text{(taking the positive root since we are in the interval } [-1, \infty)) \] \[ x = \sqrt{y} - 1 \] - Thus, the inverse function is: \[ f^{-1}(x) = \sqrt{x} - 1 \] Now, we can summarize the results: - The domain on which \( f \) is one-to-one and non-decreasing is \( [-1, \infty) \). - The inverse of \( f \) restricted to this domain is: \[ f^{-1}(x) = \sqrt{x} - 1 \]