Let f(x)=-1+(x+1)^2, where x>=-1, then the set S=x:f^-1(x)=f(x) equals

Answers (1)

Solution:      The given function is invertible on [-1,infty ).

                      Sloveing for  x , we have   x=-1+sqrty+1.

                     Rightarrow      f^-1(y)=-1+sqrty+1,

                    Rightarrow     f^-1(x)=-1+sqrtx+1.

                             f^-1(x)=f(x)    gives    sqrtx+1[(x+1)^frac32-1]=0

                    Rightarrow      x=0   or x=-1.

                   	herefore                 S=�,-1.

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