# How do you find the domain and range of f(x)=x/(x^3+8) ?

Jan 23, 2018

The domain is $x \in \mathbb{R} - \left\{- 2\right\}$. The range is $y \in \mathbb{R} - \left\{0\right\}$

#### Explanation:

As you cannot divide by $0$, the denominator is $\ne 0$

Therefore,

${x}^{3} + 8 \ne 0$

${x}^{3} \ne - 8$

$x \ne - 2$

The domain is $x \in \mathbb{R} - \left\{- 2\right\}$

To calculate the range, proceed as follows

Let $y = \frac{x}{{x}^{3} + 8}$

$y \left({x}^{3} + 8\right) = x$

${x}^{3} y + 8 y - x = 0$

${x}^{3} y - x + 8 y = 0$

This is a cubic equation in $x$

The solution of this equation is

x=(-4+(((-1/(3y))^2)+(-1/(3y))^3)^(1/2))^(1/3)+(-4-((-1/(3y))^2)+(-1/(3y))^3)^(1/2)))^(1/3)

The only restriction is

$\left({\left(- \frac{1}{3 y}\right)}^{2} + {\left(- \frac{1}{3 y}\right)}^{3}\right) \ge 0$

$\frac{1}{9 {y}^{2}} - \frac{1}{27 {y}^{3}} \ge 0$

$\frac{3 y - 1}{27 {y}^{3}} \ge 0$

So $y \ne 0$

The range is $y \in \mathbb{R} - \left\{0\right\}$

graph{x/(x^3+8) [-7.9, 7.904, -3.95, 3.95]}