Question

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

Answer 1

`(-oo, -3) U (-3,0) U (0,3) U (3,oo)`

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^3-9x` to equal zero.

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

`x^3-9x=0`

`x(x^2-9)=0`

`x=0`

`x^2-9=0`

`x^2=9`

`x=+-3`

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

`(-oo, -3) U (-3,0) U (0,3) U (3,oo)`

Answer 2

See below

Explanation:

Well let's factor the function first:
`h(x)=(x-1)/(x^3-9x)`

`h(x)=(x-1)/(x(x^2-9))`

`h(x)=(x-1)/(x(x+3)(x-3))`

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]}