Question
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
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