Question Detail

To classify critical points, we need the first and second derivatives.y = 11 - 12x + 6x² - x³y' = -12 + 12x - 3x²Set y' = 0 to find critical points: -3x² + 12x - 12 = 0 => -3(x² - 4x + 4) = 0 => -3(x - 2)² = 0. This gives x = 2 as the only critical point.Now, find the second derivative: y'' = 12 - 6x.Evaluate y'' at x = 2: y''(2) = 12 - 6(2) = 0.Since y''(2) = 0, it is neither a local maximum nor a local minimum. We need to check the third derivative. y''' = -6. Since y'''(2) = -6 ≠ 0, the point x = 2 is a point of inflexion.