What is the domain of #f(x)=x/(x^3+8) #?

1 Answer
Stefan V.
Sep 6, 2015

Domain: #(-oo, -2) uu (-2, + oo)#

Explanation:

You need to exclude from the function's domain any value of #x# that would make the denominator equal to zero.

This means that you need to exclude any value of #x# for which

#x^3 + 8 = 0#

This is equivalent to

#x^3 + 2""^3 = 0#

You can factor this expression by using the formula

#color(blue)(a^3 + b^3 = (a+b) * (a^2 - ab + b^2))#

to get

#(x+2)(x^2 - 2x + 2^2) = 0#

#(x+2)(x^2 - 2x + 4) = 0#

This equation will have three solutions, but only one will be real.

#x+2 = 0 implies x_1 = -2#

and

#x^2 - 2x + 4 = 0#

#x_(2,3) = (-(2) +- sqrt((-2)^2 - 4 * 1 * 4))/(2 * 1)#

#color(red)(cancel(color(black)(x_(2,3) = (2 +- sqrt(-12))/2))) -># produces two complex roots

Since these two roots will be complex numbers, the only value of #x# that must be excluded from the function's domain is #x=-2#, which means that, in interval notation, the domain of the function will be #(-oo, -2) uu (-2, + oo)#.

Related questions
See all questions in Domain and Range of a Function
Impact of this question
2091 views around the world
You can reuse this answer
Creative Commons License