### Instruction: What is the domain of the function $$k(y) = \frac{1}{2y+1}~?$$ Express your answer in interval notation.### Response: I need to find the set of all possible values of y that make the function k(y) well-defined.
The only thing that could make k(y) undefined is if the denominator, 2y+1, is equal to zero, because then I would have a division by zero, which is not allowed.
So I want to avoid the value of y that makes 2y+1 = 0, and that value is y = -1/2.
This means that y = -1/2 is not in the domain of k(y).
Every other value of y is fine, because it will make the denominator non-zero and the function k(y) valid.
So the domain of k(y) is the set of all real numbers except -1/2.
In interval notation, this means that the domain is (-infinity, -1/2) union (-1/2, infinity), where the parentheses indicate that the endpoints are not included.
I can write this more compactly as (-infinity, -1/2) U (-1/2, infinity).