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

Mar 7, 2018

$\left(- \infty , 0\right) \cup \left(0 , \infty\right)$

#### Explanation:

The range of the function is all possible values of $f \left(x\right)$ it can have. It can also be defined as the domain of ${f}^{-} 1 \left(x\right)$.

To find ${f}^{-} 1 \left(x\right)$:

$y = \frac{1}{x - 1} ^ 2$

Switch the variables:

$x = \frac{1}{y - 1} ^ 2$

Solve for $y$.

$\frac{1}{x} = {\left(y - 1\right)}^{2}$

$y - 1 = \sqrt{\frac{1}{x}}$

$y = \sqrt{\frac{1}{x}} + 1$

As $\sqrt{x}$ will be undefined when $x < 0$, we can say that this function is undefined when $\frac{1}{x} < 0$. But as $\frac{n}{x}$, where $n \ne 0$, can never equal zero, we cannot use this method. However, remember that, for any $\frac{n}{x}$, when $x = 0$ the function is undefined.

So the domain of ${f}^{-} 1 \left(x\right)$ is $\left(- \infty , 0\right) \cup \left(0 , \infty\right)$

It so follows that the range of $f \left(x\right)$ is $\left(- \infty , 0\right) \cup \left(0 , \infty\right)$.