Inverse of a Square

Part of the Radicals & Rational Exponents section of Coddy's Algebra 2 journey. Lesson 34 of 40.

The rule $f\left(x\right) = {x}^{2}$ 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 $x \ge 0$ . Then each output comes from one input only.

Take $f\left(x\right) = {x}^{2}-3$ for $x \ge 0$ . 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, $y = \sqrt{x+3}$ and $y = -\sqrt{x+3}$ .

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: ${f}^{-1}\left(x\right) = \sqrt{x+3}$ . Check it: f(2) is 1, and the inverse sends 1 to $\sqrt{4}$ , which is 2. On a restricted domain, the inverse of a square is a square root.

Challenge

Find the inverse of $f\left(x\right) = {x}^{2}+1$ for $x \ge 0$ . 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 $x \ge 0$ , its inverse only gives outputs of at least 0, so keep only the plus root.