Let the minimum value of F (x) = ax ^ 3 + BX ^ 2 + CX be - 8, and the image whose derivative y = f '(x) passes through the point (- 2,0) (2 / 3,0), The minimum value of F (x) is - 8. Can (- 8,0) be brought in? Isn't the minimum a point with zero derivative?

Let the minimum value of F (x) = ax ^ 3 + BX ^ 2 + CX be - 8, and the image whose derivative y = f '(x) passes through the point (- 2,0) (2 / 3,0), The minimum value of F (x) is - 8. Can (- 8,0) be brought in? Isn't the minimum a point with zero derivative?

If x = - 2 is the minimum of F (x), then f (- 2) = - 8A + 4b-2c = - 8F '(- 2) = 12a-4b + C = 0f' (2 / 3) = 4A / 3 + 4B / 3 + C = 0, the solution is a = - 1, B = - 2, C = 4, and the analytic expression of the function is f (x) = - X & # 179; - 2x & # 178; + 4x if x = 2 / 3