We will have the following:
*First: We have the expression:
[tex]\frac{x^2+2x}{3x^2-1}[/tex]
Now, we solve for the values in which g(x) = 0, that is:
[tex]3x^2-1=0\Rightarrow3x^2=1\Rightarrow x^2=\frac{1}{3}[/tex][tex]\Rightarrow x=\pm\sqrt[]{\frac{1}{3}}\Rightarrow x=\pm\frac{1}{\sqrt[]{3}}[/tex]
Thus, the domain of (f / g) (x) is:
[tex]\mleft\lbrace x\in R\colon x\ne\mright?\frac{1}{\sqrt[]{3}}\&x\ne-\frac{1}{\sqrt[]{3}}\rbrace[/tex]
In interval notation:
[tex](-\infty,-\frac{1}{\sqrt[]{3}})\cup(-\frac{1}{\sqrt[]{3}},\frac{1}{\sqrt[]{3}})\cup(\frac{1}{\sqrt[]{3}},\infty)[/tex]
