Question:

<p><strong>Step 1: Set up the cubic polynomial.</strong> Let f(x) = ax³ + bx² + cx + d where a ≠ 0.</p><p>Then f'(x) = 3ax² + 2bx + c and f''(x) = 6ax + 2b.</p><p><strong>Step 2: Apply condition - local maximum at x = -1.</strong> f'(-1) = 0 gives: 3a - 2b + c = 0 ... (1)</p><p><strong>Step 3: Apply condition - f'(x) has local minimum at x = 0.</strong> This means f''(0) = 0 (critical point of f'), so: 2b = 0 ⟹ b = 0</p><p><strong>Step 4: Verify second derivative condition.</strong> Since b = 0, f''(x) = 6ax. For f'(x) to have a local minimum at x = 0, we need f'''(0) > 0, which gives 6a > 0, so a > 0.</p><p><strong>Step 5: Simplify with b = 0.</strong> From equation (1): 3a + c = 0 ⟹ c = -3a</p><p>So f(x) = ax³ - 3ax + d and f'(x) = 3ax² - 3a = 3a(x² - 1)</p><p><strong>Step 6: Find local extrema of f(x).</strong> f'(x) = 0 when x² = 1, so x = ±1. Local maximum at x = -1 (confirmed by f''(-1) = -6a < 0 since a > 0) and local minimum at x = 1.</p><p><strong>Step 7: Use given conditions f(2) = 18 and f(-1) = -1.</strong> From f(-1) = -1: -a + 3a + d = -1 ⟹ 2a + d = -1 ... (2)</p><p>From f(2) = 18: 8a - 6a + d = 18 ⟹ 2a + d = 18 ... (3)</p><p><strong>Step 8: Resolve contradiction.</strong> Equations (2) and (3) give: -1 = 18, which is a contradiction. This suggests the problem statement may have inconsistent conditions or the answer cannot be uniquely determined from the given constraints.</p><p><strong>∴ Answer:</strong> Unknown</p>