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 \( f^{-1}(x)=\square \)

We begin with the function f(x) = (x + 1)². To have f be one-to-one and non-decreasing, we need to choose a domain on which the function is monotonic (either always non-decreasing or always non-increasing) and does not repeat values. 1. Observe that f(x) is a parabola that has its vertex where the derivative equals zero. Alternatively, note that the expression is a perfect square and reaches its minimum when x + 1 = 0, i.e. when x = -1. 2. For x ≥ -1, the function is non-decreasing (in fact, it is increasing) because as x increases from -1 onward, (x + 1) increases, and squaring a nonnegative number preserves the order. Thus, the domain on which f is one-to-one and non-decreasing is:   x ∈ [−1, ∞) Now, we find the inverse function f⁻¹(x) for x ≥ 0 (since f(x) = (x+1)² gives outputs in [0, ∞) when x ≥ -1). Start with:   y = (x + 1)² Solve for x:   √y = x + 1         (Since x + 1 ≥ 0 for x ≥ −1)   x = √y − 1 Thus, the inverse function is:   f⁻¹(y) = √y − 1 If we use x as the variable for the inverse function, then:   f⁻¹(x) = √x − 1 This gives the final answer:   f⁻¹(x) = √x − 1, for x ∈ [0, ∞).