The range of the function is all possible values of #f(x)# it can have. It can also be defined as the domain of #f^-1(x)#.

To find #f^-1(x)#:

#y=1/(x-1)^2#

Switch the variables:

#x=1/(y-1)^2#

Solve for #y#.

#1/x=(y-1)^2#

#y-1=sqrt(1/x)#

#y=sqrt(1/x)+1#

As #sqrt(x)# will be undefined when #x<0#, we can say that this function is undefined when #1/x<0#. But as #n/x#, where #n!=0#, can never equal zero, we cannot use this method. However, remember that, for any #n/x#, when #x=0# the function is undefined.

So the domain of #f^-1(x)# is #(-oo,0)uu(0,oo)#

It so follows that the range of #f(x)# is #(-oo,0)uu(0,oo)#.