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

2 Answers
Apr 1, 2018

Answer:

#(-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)#

Apr 1, 2018

Answer:

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]}