Let ƒ(x) be a polynomial of degree 3 such that ƒ(–1) = 10, ƒ(1) = –6, ƒ(x) has a critical point at x = –1 and ƒ'(x) has a critical point at x = 1. Then ƒ(x) has a local minima at x = _______.

Solution

Maths sir · 5 steps

The vertex of the parabola f'(x) sits at x = 1, so its second root mirrors -1 to 3.

1. Shape of f′. Since f is cubic, f' is a quadratic. A critical point of f' at x = 1 means f''(1) = 0, so the parabola f'(x) has its vertex at x = 1.
2. Find both roots. f'(-1) = 0, so x = -1 is one root. The roots of a parabola sit symmetrically about its vertex, so the other root is x = 3. Hence f'(x) = k(x+1)(x-3) = k(x² - 2x - 3).

   f(x) = k(x³/3 - x² - 3x) + c

3. Fix k and c. f(-1) = 5/3k + c = 10 and f(1) = -11/3k + c = -6. Subtracting gives 16/3k = 16, so k = 3 and c = 5. Thus f(x) = x³ - 3x² - 9x + 5.
4. Second derivative test. f''(x) = 6x - 6. At x = -1, f''(-1) = -12 < 0, a local maximum. At x = 3, f''(3) = 12 > 0, a local minimum.
5. So. f(x) has its local minimum at x = 3.

The common slip: -1. Stopping at the given critical point gives -1, but f''(-1) = -12 < 0 makes it a local maximum, not a minimum.

The idea to keep. For a cubic, f' is a parabola whose vertex is where f'' = 0. Its two roots, the critical points of f, sit equally spaced on either side of that vertex.

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