Let y=n(x)=√(2-x)-x.
x+y=√(2-x), (x+y)2=2-x, x2+2xy+y2=2-x, x2+2xy+x+y2=2,
x2+x(2y+1)=2-y2,
x2+x(2y+1)+(2y+1)2/4=2-y2+(2y+1)2/4,
(x+(2y+1)/2)2=[8-4y2+(4y2+4y+1)]/4=(4y+9)/4,
x+(2y+1)/2=±√(4y+9)/2,
x=[(-2y-1)±√(4y+9)]/2. There appear to be two solutions for x which would lead to two possible inverses. However, both solutions may not satisfy the original equation.
n(1)=0, so let's substitute y=0 in this:
x=(-1±√9)/2=(-1±3)/2⇒x=1 or -2. But n(1)=0 so we must reject the negative square root.
x=[(-2y-1)+√(4y+9)]/2. Let x=m(y)=[(-2y-1)+√(4y+9)]/2.
Therefore m(x)=n-1(x)=[(-2x-1)+√(4x+9)]/2. This is the inverse.