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

Sep 22, 2015

Domain = $\mathbb{R} - \left\{- 2\right\}$
Range = $\mathbb{R}$

#### Explanation:

First factorize the denominator to find possible points of discontinuity.
Factorize it as a sum of 2 cubes, and then as a trinomial to get

x/((x+2)(x^2-2x+4)

The trinomial in the denominator is an irreducible quadratic and has no real roots.

Hence the only vertical asymptote is at $x = - 2$

The horizontal asymptote would usually occur at ${\lim}_{x - \infty} f \left(x\right) = 0$, since the denominator dominates the numerator and increases faster.
However in this case, 0 is in the range of the function since x = 0 in the numerator would set the output value y to 0.

So the domain is all real numbers x except -2, and the range is all real numbers y

Graphically :
graph{x/(x^3+8) [-10, 10, -5, 5]}