What is the domain of: $$\left(\frac{5x+4}{x^2+9x+8}\right)^{1/3}$$
I got $(-\infty, -8) \cup (-8,-1) \cup (-1, \infty).$
But according to Wolfram Alpha it is $(-8, -1) \cap [-4/5, \infty)$.
Could someone please tell me why I am wrong? The way I got my answer is by looking at the equation and seeing that a cubed root can be a negative number, but the denominator can not actually equal $0$. Thank you!
Both of you are right. As a real function, $\left(\frac{5x+4}{x^2+9x+8}\right)^{1/3}$ it is defined whenever $\left(\frac{5x+4}{x^2+9x+8}\right)$ is defined, that is whenever $x^2+9x+8 \neq 0$. You already found the roots, $x=-1$ and $x=-8$.
However, as a complex valued function, there is another concept of cubic root, namely the principal root, $x^{1/3} = \exp(\frac{1}{3}\ln(x))$. By default, wolfram alpha computes the domain for the principal root, you can change that by clicking on "use the real-valued root instead".
By default Wolfram Alpha is interpreting ^(1/3) as the principal cube root of 3
What you want to calculate is the real cube root, what you can do this way:
Now even wolfram alpha is saying the domain is $\left\lbrace x : x\neq -8, x\neq -1\right\rbrace$
