what is the domain of x^2-2xy+y^2/x-3
Answers
Answer:
(-∞, -3] U [1, ∞)
Step-by-step explanation:
We know that this (x^2 + 2x - 3) is inside a square root, meaning that it can't be negative: [x^2 + 2x - 3 ≥ 0]. Therefore, what values of x make it so that the value inside the square root is nonnegative? If we set x^2 + 2x - 3 = 0 (just like a quadratic function), we can factor and get (x-1)(x+3) = 0. This means our zeroes (the values that make the equation 0) are 1 and -3, because (x-1) = 0 if x = 1, (x+3) = 0 if x = -3, and 0 x [anything] = 0. From here, we can make three intervals, and we can take any arbitrary number from each interval and plug it into the original equation.
- Interval a (-∞, -3) → -5 lies between -3 and -∞, so let's plug in -5 (note that you could also plug in -4, -6, -7.5, etc, but I chose -5 because I thought it would be easiest). If we plug in 5, we get (5-1)(5+3) = (4)(8) = 32, which is positive. This means that if x equals any value within interval a, then the function [x^2 + 2x - 3] will be positive.
- Interval b (-3, 1) → 0 lies between -3 and 1, so let's plug it in (again, you could choose other values within the interval as well). If we plug in 0, we get (0-1)(0+3) = (-1)(3) = -3, which is negative. This means that if x equals any value within interval b, then the function [x^2 + 2x - 3] will be negative.
- Interval c (1, ∞) → 2 lies between 1 and ∞, so let's plug it in (again, you could choose other values within the interval as well). If we plug in 2, we get (2-1)(2+3) = (1)(5) = 5, which is positive. This means that if x equals any value within interval c, then the function [x^2 + 2x - 3] will be positive.
As we mentioned above, we know that the value of [x^2 + 2x - 3] must be nonnegative because it's inside a square root. Therefore, the only intervals in which this is true would be (-∞, -3] U [1, ∞) → note that i used inclusive brackets for the 1 and the -3 because these two values both make the equation equal to 0, which still works within a square root. Thus, the domain of is (-∞, -3] U [1, ∞).