#1**0 **

Make sure that x^3 + 4 is not 0. Dividing by 0 is undefined. This means that \(x^3 \ne -4 \Rightarrow x \ne ^3\sqrt{-4} \Rightarrow x \ne 2i\).

This value is not real, so we do not need to consider it. The domain of \(f(x) = \dfrac{6x}{x^3+4}\) is \(x \in (-\infty, \infty)\).

- PM

PartialMathematician Dec 17, 2018

#4**+1 **

Actually, I think \(^3\sqrt-4\) might actually be a real number, so the domain is \(x \in (-\infty, ^3\sqrt-4) \cup (^3\sqrt-4, \infty)\).

PartialMathematician
Dec 17, 2018

#3**+4 **

y = 6x / [ x^3 + 4]

The domain will only be affected by any x that makes the denominator = 0

So....set the denominator to 0 to find this value

x^3 + 4 = 0

x^3 = -4 take the cube root of both sides

x = ∛-4 = - ∛4

So....the domain is

(-infinity, -∛ 4 ) U ( - ∛4, infinity )

Here's a graph : https://www.desmos.com/calculator/qih0ucpatr

CPhill Dec 17, 2018