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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?

 Jun 29, 2024
 #1
avatar+1230 
+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! :)

 Jun 29, 2024

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