Given the function f (x) = X3 + AX2 + BX + C, the tangent l of the curve y = f (x) at the point x = 1 is only the fourth quadrant, and the slope is 3, and the distance from the origin of the coordinate to the tangent L is 1010, if x = 23, y = f (x) has extremum. (1) find the value of a, B, C; (2) find the maximum and minimum value of y = f (x) on [- 3, 1]

Given the function f (x) = X3 + AX2 + BX + C, the tangent l of the curve y = f (x) at the point x = 1 is only the fourth quadrant, and the slope is 3, and the distance from the origin of the coordinate to the tangent L is 1010, if x = 23, y = f (x) has extremum. (1) find the value of a, B, C; (2) find the maximum and minimum value of y = f (x) on [- 3, 1]

(1) From F (x) = X3 + AX2 + BX + C, we can get f ′ (x) = 3x2 + 2aX + B. when x = 1, the slope of tangent L is 3, we can get 2A + B = 0. When x = 23, y = f (x) has extremum, then f ′ (23) = 0, that is, 4A + 3B + 4 = 0