SOLUTION: Let f(x) = 1/(1 + 2/(1 + 3/x). There are three real numbers x that are not in the domain of f(x). What is the sum of those three numbers?

Algebra ->  Percentage-and-ratio-word-problems -> SOLUTION: Let f(x) = 1/(1 + 2/(1 + 3/x). There are three real numbers x that are not in the domain of f(x). What is the sum of those three numbers?      Log On


   

Question 1204249: Let f(x) = 1/(1 + 2/(1 + 3/x). There are three real numbers x that are not in the domain of f(x). What is the sum of those three numbers?
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
f%28x%29+=+1%2F%281+%2B+2%2F%281+%2B+3%2Fx%29%29.....x is in denominator, so x%3C%3E0

f%28x%29+=+1%2F%281+%2B+2%2F%28%28x+%2B+3%29%2Fx%29%29

f%28x%29+=+1%2F%281+%2B+1%2F%281+%2B+2x%2F%28x+%2B+3%29%29%29.....x%2B3 is in denominator, so x%3C%3E-3

f%28x%29+=+1%2F%28%283%28x+%2B+1%29%29%2F%28x+%2B+3%29%29

f%28x%29+=+%28x+%2B+3%29%2F%283+%28x+%2B+1%29%29....x%2B1 is in denominator, so x%3C%3E-1

domain:
{ x element R+: x%3C%3E-3 and x%3C%3E-1 and x%3C%3E0 }

interval:
(-infinity,-3) U (-3,-1) U (-1,0) U (0,infinity)

the sum of those three numbers is:
-3%2B%28-1%29%2B0=-4