Which of the following intervals are part of the solution set of x 2 − 3 x − 4 x 2 − x − 2 ≥ 0 ? {"version":"1.1","math":"\dfrac{x^2-3x-4}{x^2-x-2}\ge 0?"}

Question

Which of the following intervals are part of the solution set of      x 2  − 3 x − 4    x 2  − x − 2    ≥ 0 ? {"version":"1.1","math":"\dfrac{x^2-3x-4}{x^2-x-2}\ge 0?"}

BlackTom AI · Accepted Answer

We start by simplifying the inequality (x^2 - 3x - 4)/(x^2 - x - 2) ≥ 0, while keeping in mind the domain restrictions where the denominator is zero. First, factor numerator and denominator: - Numerator: x^2 - 3x - 4 = (x - 4)(x + 1) - Denominator: x^2 - x - 2 = (x - 2)(x + 1) Thus, the expression becomes ((x - 4)(x + 1))/((x - 2)(x + 1)). We can cancel the (x + 1) factor, but we must preserve the domain restriction x ≠ -1 (since the original expression is undefined there). After cancellation, the expression behaves like (x - 4)/(x - 2) for x ≠ -1, 2. The point x = 4 makes the numerator zero, so the whole expression equals 0 there (and 4 is allowed since the denominator at 4 is 2, nonzero). Now perform a sign analysis for (x - 4)/(x - 2) with the domain restriction x ≠ -1, 2: - Critical points: -1 (hole), 2 (vertical asymptote/undefined), 4 (numerator zero). - Test intervals: (-∞, -1), (-1, 2), (2, 4), (4, ∞). - Signs: • For x < -1, pick x = -2: ( -6 )/( -4 ) > 0, so positive. • For -1 < x < 2, pick x = 0: ( -4 )/( -2 ) > 0, so positive. • For 2 < x < 4, pick x = 3: ( -1 )/( 1 ) < 0, so negative. • For x > 4, pick x = 5: ( 1 )/( 3 ) > 0, so positive. - Since we want ≥ 0, include intervals where the expression is positive or zero. We must also exclude x = -1 and x = 2 (domain issues). Therefore, the solution set is (-∞, -1) ∪ (-1, 2) ∪ [4, ∞). Now evaluate each provided interval to see if it lies entirely inside the solution set (a interval is "part of the solution set" only if every point in it is in the solution set): - [0, 2]: This interval includes x = 2, but the original expression is undefined at x = 2, so this interval is not entirely contained in the solution set. Additionally, it includes values very close to -1 and 0 where the domain and sign considerations matter; in any case, the endpoint 2 makes it invalid. - (-5, 2): This interval crosses x = -1, which is a hole in the graph (not included in the domain). Because -1 lies inside (-5, 2), this interval cannot be fully contained in the solution set. - (-17, -1): This interval lies entirely to the left of -1 and does not include -1 itself. Since for x < -1 the expression is positive, every point in (-17, -1) is in the solution set. This interval is valid as a subset. - [4, 10]: This interval starts at 4, where the expression equals 0, and continues to 10. However, it includes all points greater than or equal to 4, which are in the solution set except at any points where the domain would fail. Since the domain failures are at x = -1 and x = 2, and this interval does not cross them, [4, 10] is also entirely contained in the solution set. - (-1, 2): This interval straddles the tricky region around -1 and up to 2. It includes values just greater than -1 where the expression is defined and positive, but since -1 is not included and the interval extends up to 2 (which is undefined), the interval as a whole is not entirely contained in the solution set. Summary of each option: - [0, 2] — not contained in the solution set because it includes x = 2, which is undefined. - (-5, 2) — not contained because it includes x = -1 (a hole) within the interval. - (-17, -1) — contained, since every point is < -1 and the expression is positive there. - [4, 10] — contained, since all points x ≥ 4 are in the solution set (the only domain issues are at -1 and 2). - (-1, 2) — not contained, because while many points in (-1, 2) are in the solution set, the interval includes points arbitrarily close to -1 from the right where the function is defined; however, the issue is that the interval is not entirely in the domain due to the exclusion at x = -1 and the endpoint 2 being undefined, so it cannot be considered fully contained. Therefore, the intervals that are part of the solution set, meaning entirely contained within the solution set, are (-17, -1) and [4, 10].

Which of the following intervals are part of the solution set of x 2 − 3 x − 4 x 2 − x − 2 ≥ 0 ? {"version":"1.1","math":"\dfrac{x^2-3x-4}{x^2-x-2}\ge 0?"}

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