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3 Responses
Swapping x and y gives a true inverse only if the function is one-to-one on the domain you’ve chosen (horizontal-line test); for f(x)=x^2 you must first restrict the domain-x≥0 gives the inverse √x, while x≤0 gives the inverse −√x-so the ± just reflects those two possible choices. Which restriction (x≥0 or x≤0) fits the inputs you actually want to allow?
Swapping x and y works only when the function is one-to-one (passes the horizontal line test), so restrict to a monotone slice first-e.g., for f(x)=x^2 choose x≥0 to get y=√x or x≤0 to get y=−√x; see https://www.khanacademy.org/math/algebra-home/alg-functions/inverse-functions-alg/a/restricting-domains-of-functions-to-find-inverses. Hope this helps!
Great question! The “swap x and y” move really finds the inverse relation; it becomes a true inverse function only when the original f is one-to-one on its domain (passes the horizontal line test). If f isn’t one-to-one, you must first restrict its domain to a region where it’s strictly increasing or strictly decreasing; then the inverse is single-valued. For f(x) = x^2 on all real numbers, it’s not one-to-one, so swapping y = x^2 to x = ±√y gives two branches. To get an inverse function, pick a monotone branch of f. Example: restrict f to x ≥ 0; then y = x^2 with x ≥ 0 implies x = √y (the positive root), so the inverse function is f^{-1}(x) = √x with domain [0, ∞) and range [0, ∞). Alternatively, if you restrict to x ≤ 0, the inverse is f^{-1}(x) = −√x. A quick “always works” case: if f(x) = x^3 + 1 (which is one-to-one on all reals), swapping y = x^3 + 1 to x = ³√(y − 1) gives f^{-1}(x) = ³√(x − 1) immediately. The rule of thumb: ensure injectivity first, then swap-and-solve.