# What is the domain and range of h(x)= 1 /x^2?

Jul 22, 2016

Domain of $h , {D}_{h}$, is $\mathbb{R} - \left\{0\right\}$.

Range of $h , {R}_{h} = \left(0 , \infty\right)$.

#### Explanation:

We know that Division by $0$ is not permissible , so, we can not take x=0. Except this, x can take any value from $\mathbb{R}$, which means that the Domain of $h , {D}_{h}$, is $\mathbb{R} - \left\{0\right\}$.

Next, $\forall x \in {D}_{h} , {x}^{2} > 0$, and so is, $\frac{1}{x} ^ 2$.

Also let us note that, for each $x \in {D}_{h} , h \left(x\right) \ne 0$, because, if for some $x \in {D}_{h} , h \left(x\right) = \frac{1}{x} ^ 2 = 0 \rightarrow 1 = 0$, an impossible result.

Thus, Range of $h , {R}_{h} = \left(0 , \infty\right)$.