Question:

<p><strong>Step 1: Express f(x) as an even function.</strong> Since f is even and has degree 2018, we write: f(x) = a₀ + a₂x² + a₄x⁴ + ... + a₂₀₁₈x²⁰¹⁸. This is equivalent to f(x) = g(x²) where g(t) is a polynomial of degree 1009.</p><p><strong>Step 2: Use the constraint that f is even.</strong> Since f is even: f(-4) = f(4) = 2. But we're given f(-4) = 2, so f(4) = 2 (consistency check).</p><p><strong>Step 3: Identify critical constraint points.</strong> We have f(1) = 1, f(2) = 0, f(3) = -5, and f(4) = 2. Since f is even, we also know f(-1) = 1, f(-2) = 0, f(-3) = -5, f(-4) = 2.</p><p><strong>Step 4: Apply Rolle's Theorem to f(x).</strong> Between consecutive points where we know the function values, f'(x) must have zeros: On [-4,-3], [-3,-2], [-2,-1], [-1,1], [1,2], [2,3], [3,4], there are at least 7 zeros of f'(x) (one in each interval where f changes sign or by careful application).</p><p><strong>Step 5: Apply Rolle's Theorem to f'(x).</strong> f'(x) is an odd function (derivative of even function). f'(x) has at least 7 zeros. Between consecutive zeros of f'(x), f''(x) must have zeros. With 7 zeros of f'(x), Rolle's theorem guarantees at least 6 zeros of f''(x) in the interior intervals.</p><p><strong>Step 6: Account for zeros at the origin.</strong> Since f'(x) is odd, f'(0) = 0. The zeros of f'(x) come in symmetric pairs ±a₁, ±a₂, ... plus 0. With at least 7 zeros of f'(x) arranged symmetrically, there are at least 6 intervals between consecutive critical points, giving at least 6 zeros of f''(x) by Rolle's theorem.</p><p><strong>Step 7: Verify the minimum.</strong> The degree of f''(x) is 2016. By Rolle's theorem applied systematically to the 7 zeros of f'(x), we guarantee at least 6 zeros of f''(x). Given the constraints and the structure of even polynomials, the minimum is 6.</p><p><strong>∴ Answer: 6</strong></p>