+1347 Let
$$f(x) = \frac{1}{1+\frac{2}{1+\frac 3x}}.$$
There are three real numbers $x$ that are not in the domain of $f(x)$. What is the sum of those three numbers?
+1622 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.
