Asked by Trish Goal
There are three real numbers x that are not in the domain of f(x)=1/ 1+ (1+1/x). What is the sum of those three numbers. Thank you!?
Answers
Answered by
Reiny
looks like you meant to type a partial continued fraction
1/( 1+ (1+1/x) ) , so x≠0
= 1/(1 +(x+1)/x) --> (x+1)/x ≠ -1 , or x ≠ -2
= 1/(x + x+1)/x
= 1/(2x+1)/x
= x/(2x+1) , x ≠ -1/2
so x ≠ 0, -1/2 , -2
the sum of those 3 restricted values = -2 1/2 or -5/2
1/( 1+ (1+1/x) ) , so x≠0
= 1/(1 +(x+1)/x) --> (x+1)/x ≠ -1 , or x ≠ -2
= 1/(x + x+1)/x
= 1/(2x+1)/x
= x/(2x+1) , x ≠ -1/2
so x ≠ 0, -1/2 , -2
the sum of those 3 restricted values = -2 1/2 or -5/2
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.