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

Sep 6, 2015

Domain: $\left(- \infty , - 2\right) \cup \left(- 2 , + \infty\right)$

#### 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

$\textcolor{b l u e}{{a}^{3} + {b}^{3} = \left(a + b\right) \cdot \left({a}^{2} - a b + {b}^{2}\right)}$

to get

$\left(x + 2\right) \left({x}^{2} - 2 x + {2}^{2}\right) = 0$

$\left(x + 2\right) \left({x}^{2} - 2 x + 4\right) = 0$

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

$x + 2 = 0 \implies {x}_{1} = - 2$

and

${x}^{2} - 2 x + 4 = 0$

${x}_{2 , 3} = \frac{- \left(2\right) \pm \sqrt{{\left(- 2\right)}^{2} - 4 \cdot 1 \cdot 4}}{2 \cdot 1}$

$\textcolor{red}{\cancel{\textcolor{b l a c k}{{x}_{2 , 3} = \frac{2 \pm \sqrt{- 12}}{2}}}} \to$ 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 $\left(- \infty , - 2\right) \cup \left(- 2 , + \infty\right)$.