# How do you find the domain of h(x) = (x-1) / ( [x^3] - 9x) ?

Apr 1, 2018

$\left(- \infty , - 3\right) U \left(- 3 , 0\right) U \left(0 , 3\right) U \left(3 , \infty\right)$

#### Explanation:

For this function, we can plug in any value of $x$ and get an output, except for any values of $x$ which cause the denominator ${x}^{3} - 9 x$ to equal zero.

So, to determine the domain, solve the denominator for zero:

${x}^{3} - 9 x = 0$

$x \left({x}^{2} - 9\right) = 0$

$x = 0$

${x}^{2} - 9 = 0$

${x}^{2} = 9$

$x = \pm 3$

So, the domain does not include $x = - 3 , 0 , 3$. In interval notation, the domain is

$\left(- \infty , - 3\right) U \left(- 3 , 0\right) U \left(0 , 3\right) U \left(3 , \infty\right)$

Apr 1, 2018

See below

#### Explanation:

Well let's factor the function first:
$h \left(x\right) = \frac{x - 1}{{x}^{3} - 9 x}$

$h \left(x\right) = \frac{x - 1}{x \left({x}^{2} - 9\right)}$

$h \left(x\right) = \frac{x - 1}{x \left(x + 3\right) \left(x - 3\right)}$

No removable discontinuities, so x cannot equal $0 , - 3 , 3$ as they would have the denominator equal $0$ and these will be the vertical asymptotes:

So Domain in interval notation can be said to be:
(-∞, -3)∪(-3, 0)∪(0, 3)∪(3,∞)

Or in Set builder:
{x|x∈ℝ, x ≠-3, 0, 3}

Graph:
graph{(x-1)/(x^3-9x) [-8.54, 11.46, 11, 21]}