Let y=f(x)=(1/(2x+2))2,
(2x+2))2=1/y, 2x+2=1/√y, x=½(1/√y-2)=1/(2√y)-1. f(x) has two parts to its curve which reflect one another in the line x=-1. The inverse can only represent one of these parts.
x=f-1(y)=1/(2√y)-1, so f-1(x)=1/(2√x)-1 which is not a true function because the square root of x can be positive or negative. x can only map via a function to a unique point. This function only maps the positive root. The domain of the inverse is (0,∞). The range is (-1,∞). -1 is a horizontal asymptote and the vertical axis is the vertical asymptote.