Answer:
[tex]\lbrace x|x\text{ is a real number and x }\ne\text{ -11},3\rbrace[/tex]Explanation:
Here, we want to get the domain of the given function
We start by dividing the two as follows:
[tex](\frac{f}{g})(x)\text{ = }\frac{f(x)}{g(x)}[/tex][tex]\begin{gathered} So,\text{ we have it that:} \\ \frac{3}{3-x}\times\frac{1}{11+x}\text{ = }\frac{3}{33+3x-11x-x^2}\text{ = }\frac{3}{33-8x-x^2} \end{gathered}[/tex]The domain refers to the possible x-values
To get that, we need to solve the quadratic equation in the denominator
We have that as:
[tex]\begin{gathered} 33-8x-x^2=0 \\ 33-11x+3x-x^2=0 \\ 11(3-x)+x(3-x)=\text{ 0} \\ (11+x)(3-x)\text{ = 0} \\ x\text{ = -11 or 3} \end{gathered}[/tex]So, we have the domain as:
[tex]\mleft\lbrace x|x\text{ is a real number and x }\ne\text{ -11},3\mright\rbrace[/tex]