The function f(x) = x^3 - 6x^2 + 9x + 1 has a local maximum at which point?

Core Concept

First, find the derivative: f'(x) = 3x^2 - 12x + 9. Set it to zero: 3x^2 - 12x + 9 = 0, which simplifies to x^2 - 4x + 3 = 0. Factoring gives (x - 1)(x - 3) = 0, so critical points are at x = 1 and x = 3. Using the second derivative test: f''(x) = 6x - 12. f''(1) = -6 < 0, so x = 1 is a local maximum. f''(3) = 6 > 0, so x = 3 is a local minimum.