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

2 Answers
VNVDVI
Apr 1, 2018

(-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)

Hriman
Apr 1, 2018

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]}

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