# How do you find the domain and range of h(x)= 10/(x^2-2x)?

Jul 22, 2015

Analyse behaviour of the denominator and hence $h \left(x\right)$ to find:

domain $= \left(- \infty , 0\right) \cup \left(0 , 2\right) \cup \left(2 , \infty\right)$

and

range $= \left(- \infty , - 10\right] \cup \left(0 , \infty\right)$

#### Explanation:

$h \left(x\right) = \frac{10}{{x}^{2} - 2 x} = \frac{10}{x \left(x - 2\right)} = \frac{10}{{\left(x - 1\right)}^{2} - 1}$

The denominator is zero when $x = 0$ or $x = 2$, so $h \left(x\right)$ is undefined for those values of $x$.

When $x \in \left(0 , 2\right)$, ${x}^{2} - 2 x \in \left[- 1 , 0\right)$,

so $h \left(x\right) \in \left(- \infty , - 10\right]$

When $x \in \left(- \infty , 0\right) \cup \left(2 , \infty\right)$, ${x}^{2} - 2 x \in \left(0 , \infty\right)$,

so $h \left(x\right) \in \left(0 , \infty\right)$