Given f(x) = x^2 - 1 with domain [0, ∞), what is f^{-1}(y)?

Given f(x) = x^2 - 1 with domain [0, ∞), what is f^{-1}(y)?

• A. -sqrt(y+1)
• B. sqrt(y-1)
• C. sqrt(y+1)
• D. (y+1)/2

Answer: (C)

When you restrict the domain to nonnegative x, the function y = x^2 - 1 becomes one-to-one, so it has a unique inverse. To find it, solve y = x^2 - 1 for x in terms of y: x^2 = y + 1, so x = ±√(y + 1). The domain x ≥ 0 chooses the nonnegative root, giving f^{-1}(y) = √(y + 1). This inverse is defined for y ≥ -1, which matches the range of the original function on the restricted domain. The other forms would either pick the wrong sign or correspond to a different relationship, so they don’t satisfy f(f^{-1}(y)) = y with the given domain.

When you restrict the domain to nonnegative x, the function y = x^2 - 1 becomes one-to-one, so it has a unique inverse. To find it, solve y = x^2 - 1 for x in terms of y: x^2 = y + 1, so x = ±√(y + 1). The domain x ≥ 0 chooses the nonnegative root, giving f^{-1}(y) = √(y + 1). This inverse is defined for y ≥ -1, which matches the range of the original function on the restricted domain. The other forms would either pick the wrong sign or correspond to a different relationship, so they don’t satisfy f(f^{-1}(y)) = y with the given domain.