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?
\(f(x) = \frac{1}{1+\frac{2}{1+\frac 3x}}. \)
x cannot = 0
1 + 3/x = [ x + 3] / x
2 / [ ( x + 3) /x ] = 2x / [x + 3]
x cannot = -3
1 + 2x / [ x + 3] = [ x + 3 + 2x ] / [ x + 3] = [ 3x + 3] / [x + 3]
1/ ([3x + 3] /[x + 3]) = [x + 3] / [ 3x + 3]
x cannot = -1
+829 
