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?

learnmgcat Jun 29, 2024

#1**+1 **

In this case, note that the only way x is possibily not in the domain is if the denominator of a fraction is 0.

First off, the smallest fraction 3/x. x can't be 0 since 3/0 is undefined.

Next, we have

\(1+3/x \neq 0\\ 3/x\neq-1\\ -x \neq 3\\ x \neq -3\)

Thus, x cannot be negative 3. Lastly, let's focus on the last fraction. We have

\(1+\frac{2}{1+3/x} \neq 0\\ \frac{2}{1+3/x}\neq-1\\ 2\neq-1-3/x\\ 3\neq -3/x\\ x \neq -1\)

So x also can't possibly be -1.

Thus, we have 3 following solutions. We have x cannot be 0, -1 and -3.

So our answer is 0, -1, -3.

Thanks! :)

NotThatSmart Jun 29, 2024