Questions & Answers: "Function - Rational Inequality"

For the function to be defined, the expression under the square root must be non-negative: (x + 5) / (x² - 9) ≥ 0. Factor the denominator: (x + 5) / [(x - 3)(x + 3)] ≥ 0. The critical points are -5, -3, and 3. Testing intervals: 1) x > 3: Positive. 2) -3 < x < 3: Negative. 3) -5 ≤ x < -3: Positive. 4) x < -5: Negative. Since we need the result to be ≥ 0, we select the positive intervals. Note that x = -3 and x = 3 must be excluded because they result in division by zero (hence open brackets), while x = -5 is included because the numerator can be zero (hence a closed bracket). This leads to x ∈ [-5, -3) ∪ (3, ∞). Option 2 represents the intervals where the expression is negative. Option 3 and 4 are mathematically incomplete or ignore the critical points of the numerator.

Do not cross-multiply, as the sign of (x-2) is unknown. Instead, subtract 1 from both sides: [(2x - 3) / (x - 2)] - 1 < 0. Simplify the fraction: (2x - 3 - x + 2) / (x - 2) < 0, which yields (x - 1) / (x - 2) 2, negative for 1 < x < 2, and positive for x 1. Option 3 is wrong because the inequality is strict (<) and x = 2 makes the denominator zero. Option 4 uses incorrect calculation steps.

To solve (x - 1) / (x^2 + x - 6) ≥ 0, first factor the denominator: (x - 1) / [(x + 3)(x - 2)] ≥ 0. The critical points are x = 1 (from numerator) and x = -3, 2 (from denominator). Plotting these on a number line divides it into four intervals: (-∞, -3), (-3, 1], [1, 2), and (2, ∞). Testing the sign in each interval: 1) For x > 2, the expression is positive. 2) For 1 < x < 2, the expression is negative. 3) For -3 < x < 1, the expression is positive. 4) For x < -3, the expression is negative. We need ≥ 0, so we select (-3, 1] ∪ (2, ∞). Option 2 is wrong because -3 and 2 must be excluded to avoid division by zero. Option 3 describes the solution for ≤ 0. Option 4 is incomplete.

To solve (x - 1) / (x + 2) > 0, we identify the critical points where the numerator or denominator is zero, which are x = 1 and x = -2. These points divide the number line into three intervals: (-∞, -2), (-2, 1), and (1, ∞). Testing a value in each: for x = -3, (-4)/(-1) = 4 > 0 (True); for x = 0, (-1)/(2) = -0.5 0 (True). Since the inequality is strict (>), we exclude the endpoints. Thus, the solution is (-∞, -2) ∪ (1, ∞).

First, factor the numerator and denominator: [(x - 2)(x - 3)] / [(x - 1)(x + 1)] ≤ 0. The critical points are -1, 1, 2, and 3. Testing intervals: (-∞, -1) yields (+), (-1, 1) yields (-), (1, 2) yields (+), [2, 3] yields (-), and (3, ∞) yields (+). The inequality is satisfied in (-1, 1) and [2, 3]. Note that x cannot be -1 or 1 as they are in the denominator, while 2 and 3 are included because they make the expression zero.

The factor (x - 1)² is always non-negative for all real x. Therefore, the sign of the expression depends on (x - 2) / (x - 3), except at x = 1 where the expression is 0 (which satisfies ≥ 0). Solving (x - 2) / (x - 3) ≥ 0: critical points are 2 and 3. Testing intervals gives (-∞, 2] ∪ (3, ∞). Since x = 1 is already included in the interval (-∞, 2], the final solution is x ∈ (-∞, 2] ∪ (3, ∞).

Do not cross-multiply as (x-1) can be negative. Rearrange: 1/(x-1) - 2 < 0 → (1 - 2(x-1))/(x-1) < 0 → (1 - 2x + 2)/(x-1) < 0 → (3 - 2x)/(x-1) 0. Critical points are 1 and 1.5. Testing intervals: for x < 1, (+)/( - ) = - (False); for 1 < x 0. If x=0, -3/-1=3>0 (True). If x=1.2, -0.6/0.2=-3 (False). If x=2, 1/1=1>0 (True). So solution is (-∞, 1) ∪ (1.5, ∞).

The critical points are x = 3 (from the numerator) and x = -5 (from the denominator). Note that x cannot be -5 because it would make the denominator zero. Testing intervals: for x 0; for -5 < x < 3 (e.g., 0), (-3)/(5) 3 (e.g., 4), (1)/(9) > 0. The inequality ≤ 0 is satisfied in the interval (-5, 3]. We include 3 because of the 'equal to' part of the inequality, but exclude -5 because the expression is undefined there.

For the quadratic x² + x + 1, the discriminant D = b² - 4ac = 1² - 4(1)(1) = -3. Since D < 0 and the leading coefficient (a=1) is positive, x² + x + 1 is always positive for all real x. Therefore, the sign of the whole expression depends solely on the denominator (x - 2). For the expression to be less than zero, x - 2 must be less than zero. Thus, x < 2, or x ∈ (-∞, 2).

For the square root to be defined, the expression inside must be ≥ 0: (x - 4) / (x - 6) ≥ 0. Critical points are 4 and 6. The denominator cannot be zero, so x ≠ 6. Testing intervals: for x ≤ 4, e.g., 0, (-4)/(-6) = 2/3 ≥ 0 (True). For 4 < x < 6, e.g., 5, (1)/(-1) = -1 6, e.g., 7, (3)/(1) = 3 ≥ 0 (True). Thus, the domain is (-∞, 4] ∪ (6, ∞).

Solve (x + 1) / (x - 1) - 1 ≥ 0 → (x + 1 - (x - 1)) / (x - 1) ≥ 0 → 2 / (x - 1) ≥ 0. For this to be true, the denominator (x - 1) must be positive, i.e., x - 1 > 0, which means x > 1. Among the options, x = 2, 5, and 10 are all greater than 1. However, x = 0 is less than 1, so it does not satisfy the inequality.

Rearrange: (2x - 3) / (3x - 5) - 3 ≥ 0 → [2x - 3 - 3(3x - 5)] / (3x - 5) ≥ 0 → (2x - 3 - 9x + 15) / (3x - 5) ≥ 0 → (-7x + 12) / (3x - 5) ≥ 0. Multiply by -1: (7x - 12) / (3x - 5) ≤ 0. Critical points are 12/7 (≈1.71) and 5/3 (≈1.66). The interval satisfying the 'less than or equal to' sign is between the roots: (5/3, 12/7]. We exclude 5/3 (denominator) and include 12/7.

Critical points are -2, 1, and 3. Powers of (x-1) and (x+2) are odd (3 and 1), so the sign changes at every critical point. Testing intervals: x < -2 (e.g., -3): (-)(-)/(-) = - (True). -2 < x < 1 (e.g., 0): (-)(+)/(-) = + (False). 1 < x 3 (e.g., 4): (+)(+)/(+) = + (False). Combining intervals where the expression is ≤ 0 and excluding x=3 (denominator), we get (-∞, -2] ∪ [1, 3).