First, find the derivative: f'(x) = 3x² - 3. Set f'(x) = 0 to find critical points: 3(x² - 1) = 0, so x = 1 or x = -1. Now use the second derivative test: f''(x) = 6x. For x = 1, f''(1) = 6 > 0, indicating a local minimum. The value at x = 1 is f(1) = 1³ - 3(1) + 2 = 0. For x = -1, f''(-1) = -6 < 0, indicating a local maximum. The value at x = -1 is f(-1) = (-1)³ - 3(-1) + 2 = -1 + 3 + 2 = 4. Therefore, the local minimum value is 0. Option B represents the local maximum. Options C and D are not values at the critical points.