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

Get ready for your Honors Mathematics 3 exam with our engaging quizzes. Use flashcards and multiple choice questions with explanations to enhance your study. Prepare effectively for the test!

Multiple Choice

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

Explanation:
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.

Subscribe

Get the latest from Passetra

You can unsubscribe at any time. Read our privacy policy