### Instruction: There are three real numbers $x$ that are not in the domain of $$f(x) = \frac{1}{1+\frac{1}{1+\frac 1x}}.$$ What is the sum of those three numbers?### Response: The first thing I notice is that the function $f(x)$ has fractions inside fractions, which means there might be some issues with dividing by zero.
To avoid that, I need to make sure that none of the denominators in the expression are zero.
So I start by looking at the innermost fraction, $\frac 1x$.
That fraction is undefined when $x=0$, so that's one of the numbers I need to exclude from the domain.
Next, I look at the fraction $\frac{1}{1+\frac 1x}$.
That fraction is undefined when the denominator, $1+\frac 1x$, is zero.
To find out when that happens, I set $1+\frac 1x$ equal to zero and solve for $x$.
Subtracting $1$ from both sides, I get $\frac 1x = -1$.
Multiplying both sides by $x$, I get $1 = -x$.
Dividing both sides by $-1$, I get $x = -1$.
So that's another number I need to exclude from the domain.
Finally, I look at the fraction $\frac{1}{1+\frac{1}{1+\frac 1x}}$.
That fraction is undefined when the denominator, $1+\frac{1}{1+\frac 1x}$, is zero.
To find out when that happens, I set $1+\frac{1}{1+\frac 1x}$ equal to zero and solve for $x$.
Subtracting $1$ from both sides, I get $\frac{1}{1+\frac 1x} = -1$.
Multiplying both sides by $1+\frac 1x$, I get $1 = -(1+\frac 1x)$.
Expanding the brackets, I get $1 = -1 - \frac 1x$.
Adding $1$ to both sides, I get $2 = -\frac 1x$.
Multiplying both sides by $x$, I get $2x = -1$.
Dividing both sides by $2$, I get $x = -\frac 12$.
So that's the third number I need to exclude from the domain.
Now I have all three numbers that are not in the domain of $f(x)$: $0, -1$, and $-\frac 12$.
To find their sum, I just add them up: $$0 + (-1) + \left(-\frac 12\right) = -\frac 32$$