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?

sandwich Jan 21, 2024

#1**+1 **

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__.

proyaop Feb 19, 2024