Let f(x) be a non-constant twice-differentiable function on R such that f(2 + x) = f(2 - x) and f'(1/2) = f'(1) = 0. Find the minimum number of roots of the equation f''(x) = 0 in the open interval (0, 4).

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Accepted Answer

4. f(2+x) = f(2-x) means f is symmetric about x = 2. Differentiating: f'(2+x) = -f'(2-x). Setting x = 0: f'(2) = -f'(2), so f'(2) = 0. Setting x = 3/2: f'(7/2) = -f'(1/2) = 0. Setting x = 1: f'(3) = -f'(1) = 0. So f' has zeros at: 1/2, 1, 2, 3, 7/2 in [0, 4]. By Rolle's theorem applied to f' on each sub-interval: (1/2, 1), (1, 2), (2, 3), (3, 7/2) — at least one zero of f'' in each interval = at least 4 zeros in (0, 4).

Let f(x) be a non-constant twice-differentiable function on R such that f(2 + x) = f(2 - x) and f'(1/2) = f'(1) = 0. Find the minimum number of roots of the equation f''(x) = 0 in the open interval (0, 4).

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