# What is the domain and range of h(x)= 10/(x^2-2x)?

Jun 19, 2016

Domain is $\left(- \infty , 0\right) \cup \left(0 , 2\right) \cup \left(2 , + \infty\right)$
Range is $\left(- \infty , - \frac{40}{9}\right] \cup \left(0 , + \infty\right)$

#### Explanation:

The domain is obtained by solving:

${x}^{2} - 2 x \ne 0$

$x \left(x - 2\right) \ne 0$

$x \ne 0 \mathmr{and} x \ne 2$

You can find the range by calculating the inverse function

Let y=h(x)

so

$y = \frac{10}{{x}^{2} - 3 x}$

$y {x}^{2} - 3 x y - 10 = 0$

$x = \frac{3 y \pm \sqrt{9 {y}^{2} - 4 y \left(- 10\right)}}{2 y}$

you can find its domain by solving:

$9 {y}^{2} + 40 y \ge 0 \mathmr{and} y \ne 0$

$y \left(9 y + 40\right) \ge 0 \mathmr{and} y \ne 0$

$y \le - \frac{40}{9} \mathmr{and} y > 0$