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. I think one number is 0. Thank you!
To find the three real numbers that are not in the domain of the function f(x) = 1 / (1 + (1 + 1/x)), we need to identify the values of x that would result in an undefined operation.
The function f(x) is defined for all real numbers except for those that make the denominator equal to zero. In this case, the denominator is 1 + (1 + 1/x), so we need to find values of x that make this expression equal to zero.
Simplifying the expression, we have:
1 + (1 + 1/x) = 0
Combining like terms:
1 + 1 + 1/x = 0
Adding the numbers:
2 + 1/x = 0
Subtracting 2 from both sides:
1/x = -2
Taking the reciprocal of both sides, we get:
x = -1/2
So, one of the numbers that are not in the domain of f(x) is x = -1/2.
To determine the other two numbers, we need to find additional values of x that can make the denominator equal to zero.
Setting 1 + (1 + 1/x) equal to zero, we get:
1 + 1 + 1/x = 0
Combining like terms:
2 + 1/x = 0
Subtracting 2 from both sides:
1/x = -2
Taking the reciprocal of both sides:
x = -1/2
So, x = -1/2 is another number that is not in the domain of f(x).
To find the third number, we can substitute any other value that makes the denominator zero into the equation. For example, let's try x = 0.
Setting 1 + (1 + 1/x) equal to zero, we have:
1 + (1 + 1/0) = 0
1/0 is undefined, so this expression is not well-defined for x = 0.
Therefore, the three numbers that are not in the domain of f(x) are x = -1/2, x = -1/2, and x = 0.
To find the sum of these three numbers, we add them together:
(-1/2) + (-1/2) + 0 = -1
So, the sum of the three numbers is -1.