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For f(x) = (x³ + 4x² - 11x - 30)/(x² - 6x - 7), determine the set of x for which f(x) > 0.

1. x in (-5, -2) U (-1, 3) U (7, inf)
2. x in (-inf, -5) U (-2, -1) U (3, 7)
3. x in (-5, -1) U (3, 7)
4. x in (-inf, -5) U (-1, 7)

Correct answer: x in (-5, -2) U (-1, 3) U (7, inf)

Solution

Numerator x³ + 4x² - 11x - 30 factors as (x+5)(x+2)(x-3) (roots -5, -2, 3). Denominator x² - 6x - 7 = (x-7)(x+1) (roots 7, -1). Critical points in order: -5, -2, -1, 3, 7. A sign chart of (x+5)(x+2)(x-3)/((x-7)(x+1)) gives f(x) > 0 on (-5, -2) U (-1, 3) U (7, inf).

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