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. 0 is one of them,I don't know the other ones.Thank you!
Assuming you meant
f(x) = 1/(1+(1+1/x))
excluded values occur when
1+1+1/x = 0
2+1/x = 0
x = -1/2
Well, that's probably wrong. How about
f(x) = 1/(1+1/(1+1/x))
If 1+1/x = 0, then
x = -1
1+1/(1+1/x)) = 0
if x = -1/2
So, now we have three values:
0,-1,-1/2 whose sum is -3/2
To determine the real numbers that are not in the domain of the function f(x), we need to find the values of x for which the denominator becomes zero. Let's solve for x in the equation (1+1/x) = -1.
1 + 1/x = -1
Subtracting 1 from both sides, we have:
1/x = -2
To get rid of the fraction, we can multiply both sides by x:
x * (1/x) = x * (-2)
This simplifies to:
1 = -2x
Now we isolate x:
x = 1/(-2)
Simplifying the expression, we have:
x = -1/2
Therefore, the two other real numbers not in the domain of f(x) are -1/2 and 0. The sum of these three numbers is:
Sum = 0 + (-1/2) + (-1/2) = -1.