The range of the function is all possible values of f(x)f(x) it can have. It can also be defined as the domain of f^-1(x)f−1(x).
To find f^-1(x)f−1(x):
y=1/(x-1)^2y=1(x−1)2
Switch the variables:
x=1/(y-1)^2x=1(y−1)2
Solve for yy.
1/x=(y-1)^21x=(y−1)2
y-1=sqrt(1/x)y−1=√1x
y=sqrt(1/x)+1y=√1x+1
As sqrt(x)√x will be undefined when x<0x<0, we can say that this function is undefined when 1/x<01x<0. But as n/xnx, where n!=0n≠0, can never equal zero, we cannot use this method. However, remember that, for any n/xnx, when x=0x=0 the function is undefined.
So the domain of f^-1(x)f−1(x) is (-oo,0)uu(0,oo)(−∞,0)∪(0,∞)
It so follows that the range of f(x)f(x) is (-oo,0)uu(0,oo)(−∞,0)∪(0,∞).
