Let f:R→R be given byf(x)=(x+1)2−1,  x≥−1 . Then the set of values of x for which f(x)=f−1(x) is given by :

Clearly, f is invertible and f−1(x)=−1+x+1 Now, f(x)=f−1(x) ⇒        (x+1)2−1=−1+x+1 ⇒        (x+1)2=x+1 ⇒        x+1[(x+1)3/2−1]=0 ⇒        x+1=0  or  (x+1)3/2=1 ⇒        x+1=0  or  x+1=1 ⇒        x=−1  or  x=0 Therefore, the set of values of x for which f(x)=f−1(x) is {−1,0} .