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Let f(x) = cos(pi*x) + 10x + 3x² + x³ for x in [-2, 3]. What is the absolute minimum value of f(x) on this interval?

1. 0
2. 3 - 2*pi
3. -2
4. -15

Correct answer: -15

Solution

Differentiating: f'(x) = -pi*sin(pi*x) + 10 + 6x + 3x². The polynomial part 3x² + 6x + 10 = 3(x+1)² + 7 >= 7 for all x, while |pi*sin(pi*x)| <= pi < 7. Therefore f'(x) > 0 on [-2, 3], meaning f is strictly increasing. The absolute minimum occurs at x = -2: f(-2) = cos(-2*pi) + 10(-2) + 3(4) + (-8) = 1 - 20 + 12 - 8 = -15.

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