Let f(x)=(x+1)2−1,x≥−1 . Then the set x:f(x)=f−1(x) is
0,−1,−3+i32,−3−i32
{0,1,−1}
{0,−1}
empty
since fx=x+12-1 is continuous function, solution of fx=f-1x lies on the line y=x. Therefore fx=f-1x=x
x+12-1=x ⇒x2+x=0⇒ x=0 or -1 Therefore , the required set is 0,-1