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Consider two continuous functions f and g defined on the interval [-1, 2], which are also twice differentiable on (-1, 2). The values of f and g at x = -1, 0, and 2 are provided in the table below: x = -1 x = 0 x = 2 f(x) = 3 f(x) = 6 f(x) = 0 g(x) = 0 g(x) = 1 g(x) = -1 In the intervals (-1, 0) and (0, 2), the second derivative of (f - 3g), denoted as (f - 3g)'', does not become zero at any point. Which of the following statements is true?

1. The equation f'(x) - 3g'(x) = 0 has exactly three solutions in the union of (-1, 0) and (0, 2).
2. The equation f'(x) - 3g'(x) = 0 has exactly one solution in the interval (-1, 0).
3. The equation f'(x) - 3g'(x) = 0 has exactly one solution in the interval (0, 2).
4. The equation f'(x) - 3g'(x) = 0 has exactly two solutions in (-1, 0) and exactly two solutions in (0, 2).

Correct answer: The equation f'(x) - 3g'(x) = 0 has exactly one solution in the interval (-1, 0).

Solution

The second derivative of (f - 3g) does not become zero, indicating a single turning point in each interval. By Rolle's theorem, f'(x) - 3g'(x) = 0 has exactly one solution in (-1, 0).

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