The square root function y = sqrt(x) is the inverse of which function (with appropriate domain restriction)?

The square root function y = sqrt(x) is the inverse of which function (with appropriate domain restriction)?

• A. y = sqrt(x)
• B. The inverse is not a function
• C. y = x^2 for x ≥ 0
• D. y = x^2 for x ≤ 0

Answer: (D)

When a function has the square root as its inverse, the original function must square its input but only on nonnegative values to keep the mapping one-to-one. Squaring on all real numbers isn’t one-to-one (since x and -x give the same square), so we restrict the domain to nonnegative x. On that domain, f(x) = x^2 sends [0, ∞) to [0, ∞) and has the inverse f^{-1}(x) = sqrt(x) with domain [0, ∞). Therefore the function whose inverse is sqrt(x) is y = x^2 with x ≥ 0. If we restricted to x ≤ 0, the inverse would be -sqrt(x), not sqrt(x). So the correct original function is y = x^2 for x ≥ 0.

When a function has the square root as its inverse, the original function must square its input but only on nonnegative values to keep the mapping one-to-one. Squaring on all real numbers isn’t one-to-one (since x and -x give the same square), so we restrict the domain to nonnegative x. On that domain, f(x) = x^2 sends [0, ∞) to [0, ∞) and has the inverse f^{-1}(x) = sqrt(x) with domain [0, ∞). Therefore the function whose inverse is sqrt(x) is y = x^2 with x ≥ 0. If we restricted to x ≤ 0, the inverse would be -sqrt(x), not sqrt(x). So the correct original function is y = x^2 for x ≥ 0.