question below: answers need 3 decimal places for part 3)

Consider the 01 function f : R2 > R2 defined by f(a:,y) = ($2+yayz+$)' Notice that f(0, 1) = (1,1) = f(1,0), hence f is not injective. 1. The Inverse Function Theorem asserts that f is locally invertible near all points (x, y) in its domain except those satisfying the equation = 0. Note that (0,1) and (1,0) do not satisfy this equation. 2. The system f(:c, y) = (1, 1) has four solutions. Find the two that we do not already know. Order the solutions so that 1:1 =