Function

Observing, we have 3 overarching fractions in which their denominators cannot be 0, and thus if we set their denominators equal to 0, we can find the 3 values of x that are not in the domain of f(x) (domain means all the values of x that "work/are allowed" for the function):

1) $1 + {2\over{1 + {3\over{x}}}}=0$

Solve for x: $-1 = {2\over{1 + {3\over{x}}}}$; $-1 - {3\over{x}}=2$; x = -1

2) However, what if $1 + {3\over{x}} = 0$? Solving for x, you would get x = -3

3) Lastly, x = 0 is quite obvious, because then 3/0 would be undefined.

Their sum is -1 + (-3) + 0 = -4.