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

Question

Eli Assouline · Accepted Answer

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

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 the same as

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

This expression can be factored using the formula

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

to obtain

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

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

There are three possible answers to this equation, but only one of them is true.

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

additionally

#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)#.

Ethan Adkins · Answer

undefined The domain of $ f(x) = \frac{x}{x^3 + 8} $ is all real numbers except for the values that make the denominator equal to zero. Therefore, the domain is $ (-\infty, -2) \cup (-2, \infty) $.