Question

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

Answer 1

Domain = `RR-{-2}`
Range = `RR`

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-oo)f(x) =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]}