Inverse of a Square

Part of the Radicals & Rational Exponents section of Coddy's Algebra 2 journey. Lesson 37 of 43.

The rule f(x) = x^2 has no inverse as it stands. Both 2 and -2 give 4, so an inverse asked about 4 would need two answers. The fix is to restrict f to inputs x >= 0. Then each output comes from one input only.

Take f(x) = x^2 - 3 for x >= 0. Swap x and y as usual: x = y^2 - 3. Solve for y: moving the -3 across gives x + 3 = y^2, and the square root of both sides gives two answers, y = √(x + 3) and y = -√(x + 3).

The restriction picks one. The inverse's outputs are f's inputs, and those are at least 0. The minus root gives negative outputs, so it goes: f^-1(x) = √(x + 3). Check it: f(2) is 1, and the inverse sends 1 to √(4), which is 2. On a restricted domain, the inverse of a square is a square root.

Challenge

Find the inverse of f(x) = x^2 + 1 for x >= 0. The board holds the swapped equation x = y^2 + 1. Write the lines yourself: first the square alone, then y. Since f only takes inputs x >= 0, its inverse only gives outputs of at least 0, so keep only the plus root.

Try it yourself

x = y^2 + 1

Solve for y