Let f(x) be a cubic polynomial with f(1) = -10, f(-1) = 6, and has a local minima at x = 1, and f'(x) has a local minima at x = -1. Then f(3) is equal to ____________.

Solution

Maths sir · 4 steps

Turn both conditions into facts about f': it is zero at x = 1, and its own minimum sits at x = -1.

1. Shape of f′. Since f is cubic, f' is a quadratic. A quadratic with its minimum at x = -1 has the form f'(x) = A(x+1)² + B with A > 0. Its minimum point is where f''(-1) = 0, not where f'(-1) = 0.
2. Use x = 1. A local minimum of f at x = 1 means f'(1) = 0, so 4A + B = 0 and B = -4A. That gives f'(x) = A(x² + 2x - 3) = A(x+3)(x-1).

   f(x) = A(x³/3 + x² - 3x) + C

3. Fix A and C. f(1) = -5/3A + C = -10 and f(-1) = 11/3A + C = 6. Subtracting gives 16/3A = 16, so A = 3 and C = -5. Then f(x) = x³ + 3x² - 9x - 5, and f''(1) = 12 > 0 confirms the minimum at x = 1.
4. So. f(3) = 27 + 27 - 27 - 5 = 22.

The common slip: 70. Taking f'(-1) = 0 instead of f''(-1) = 0 gives f(x) = 4x³ - 12x - 2 and f(3) = 70.

The idea to keep. An extremum of f' is a condition on f'', not on f'. Write the derivative in a form that already builds in its turning point, then integrate and fit the constants.

Answer from the NTA final answer key. Solution by PrepWiser.