What's the domain of √((x-2)/(x^2+x-6))
(2,infinity] ?
The proposed answer is good for √(x-2).
However, since the denominator becomes negative between -2 and 1, the expression in the square-root radical is therefore positive on this interval.
So the domain can be revised to [2,∞)∪(-2,1).
Note that +2 is within the domain, because the denominator becomes zero. However, -2 and 1 are excluded because the denominator becomes zero.
When I factor out the bottom its x=-3 and x=2, how did you get -2 and 1? Why is +2 included? Doesn't it become undefined when x=2?
To find the domain of the function √((x-2)/(x²+x-6)), we need to consider two things: the square root and the denominator.
First, let's consider the square root (√). The square root of a number (or expression) is only defined for non-negative values. This means that the expression inside the square root, (x-2)/(x²+x-6), must be greater than or equal to zero.
Next, let's consider the denominator (x²+x-6). We can't have division by zero, so we need to exclude any values of x that would make the denominator equal to zero.
To find the value of x that makes the denominator equal to zero, we set the denominator equal to zero and solve for x:
x² + x - 6 = 0
This quadratic equation factors as (x - 2)(x + 3) = 0. Setting each factor equal to zero, we get x - 2 = 0 or x + 3 = 0. Solving these equations, we find x = 2 or x = -3.
Now, let's consider the inequality (x - 2)/(x²+x-6) ≥ 0. We can determine the signs of the expression by examining the signs of the numerator and the denominator.
For x > 2, both the numerator and denominator are positive, so the inequality is satisfied.
For -3 < x < 2, the numerator (x - 2) is negative, while the denominator (x²+x-6) is positive. Since a negative divided by a positive is negative, the inequality is not satisfied.
For x < -3, both the numerator and denominator are negative, so the inequality is satisfied.
Therefore, the solution to the inequality is x ∈ (-∞, -3) ∪ (2, ∞).
Hence, the domain of the function √((x-2)/(x²+x-6)) is (-∞, -3) ∪ (2, ∞).