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Suppose the cubic x³ - px + q has three distinct real roots where p > 0 and q > 0. Then which one of the following holds?

1. The cubic has minima at √(p/3) and maxima at -√(p/3)
2. The cubic has minima at -√(p/3) and maxima at √(p/3)
3. The cubic has minima at both √(p/3) and -√(p/3)
4. The cubic has maxima at both √(p/3) and -√(p/3)

Correct answer: The cubic has minima at √(p/3) and maxima at -√(p/3)

Solution

f'(x) = 3x^2 - p = 0 gives x = +/-sqrt(p/3). Since f''(x) = 6x, f'' > 0 at x = +sqrt(p/3) (minimum) and f'' < 0 at x = -sqrt(p/3) (maximum).

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