## Precalculus (6th Edition)

The domain is $(-\infty, 0) \cup (0, \infty)$. The range is $(0, \infty)$.
The denominator is not allowed to be positive since dividing $0$ to any number gives an undefined expression. This means the value of $x$ cannot be zero. Thus, the domain is any real number except zero. In interval notation, the domain is $(-\infty, 0) \cup (0, \infty)$. The range is the set that contains all the possible values of $f(x)$ (or $y$). Note that the value of $\dfrac{1}{x^2}$ will never be equal to zero since when any non-zero number is divided to $1$, the quotient is always a non-zero number. Note further that, since the value of $x^2$ is always non-negative, then the quotient $\dfrac{1}{x^2}$ will always be non-negative as well. Thus, the range is the set of positive real numbers. In interval notation, the range is $(0, \infty)$.