Question

What is the range of the function `f(x)= 1/(x-1)^2`?

Answer 1

`(-oo,0)uu(0,oo)`

Explanation:

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)`.