Since the given functions are
[tex]f(x)=\frac{2}{x},g(x)=3x+12[/tex]To find f(g(x)), substitute x in f by g(x)
[tex](f\circ g)(x)=\frac{2}{3x+12}[/tex]The domain is the values of x that make the function defined
The function is undefined if the denominator = 0
Then equate the denominators by 0 to find the values of x which make the denominator = 0
[tex]\begin{gathered} x=0 \\ 3x+12=0 \\ 3x+12-12=0-12 \\ 3x=-12 \\ \frac{3x}{3}=-\frac{12}{3} \\ x=-4 \end{gathered}[/tex]Then the domain is all values of x except {-4, 0}
[tex]D=\mleft\lbrace x\colon x\in R,x\ne-4,0\mright\rbrace[/tex]To find g(f(x)), substitute x in g by f(x)
[tex]\begin{gathered} (g\circ f)(x)=3(\frac{2}{x})+12 \\ (g\circ x)=\frac{6}{x}+12 \end{gathered}[/tex]The domain of it is all values of x except x = 0
[tex]D=\mleft\lbrace x\colon x\in R,x\ne0\mright\rbrace[/tex]To find f(f(x)), substitute x in f by f(x)
[tex]\begin{gathered} (f\circ f)(x)=\frac{2}{\frac{2}{x}} \\ (f\circ f)(x)=2\times\frac{x}{2} \\ (f\circ f)(x)=x \end{gathered}[/tex]The domain of it is all values of x except x = 0
[tex]D=\mleft\lbrace x\colon x\in R,x\ne0\mright\rbrace[/tex]To find g(g(x)), substitute x in g by g(x)
[tex](g\circ g)(x)=3(3x+12)+12[/tex]Simplify it
[tex]\begin{gathered} (g\circ g)(x)=3(3x)+3(12)+12 \\ (g\circ g)(x)=9x+36+12 \\ (g\circ g)(x)=9x+48 \end{gathered}[/tex]Since there is no denominator in this function, then
The domain is all values of x
[tex]D=\mleft\lbrace x\colon x\in R\mright\rbrace[/tex]