If f (x) = x * 3 + 3ax * 2 + BX + A * 2 has extreme value 0 when x = - 1, find the value of constant a and B

If f (x) = x * 3 + 3ax * 2 + BX + A * 2 has extreme value 0 when x = - 1, find the value of constant a and B

First, we derive f (x) = x * 3 + 3ax * 2 + BX + A * 2 to get f '(x) = 3x * 2 + 6AX + B
Because when x = - 1, there is an extreme value, which means f '(- 1) = 0 and f (- 1) = 0
So we get 3-6a + B = 0 and - 1 + 3a-b + A * 2 = 0
The results show that B = 6a-3 and a * 2 + 3a-1 = B
Substitute B into a * 2-3a + 2 = 0
The solution is a = 1 or a = 2
The corresponding solution is b = 3 or B = 9
So when a = 1, B = 3
Or when a = 2, B = 9

If f (x) = x ^ 3-3ax ^ 2 + 1 obtains the extremum at x = 1, find the value of a and all extremum of F (x)

Answer: answer: F (x) (x = x (x) = x (179); - 3ax (178; + 1) derivation: F '(x) = 3x (178; - 6AX further derivation: F' (x) (x) = 6x-6a at x = 1: F '(1) = 3-6a = 0, f (1) = 3-6a = 0, f (1) = 6-6a (6-6a) is not equal to 0, the solution is: a = 1 / 2, so: F (x) (x) = x (x (x) (3x) (3x) (3x) (3x) (3x) so: F (x (x) as: F (x (x) = 3x (x) (3x) (3x) (3x) in this paper: let: F (x (x (x) as: F (x (x) as: F (x (x (x) we: F (x (x) this: F (x (x (x) this: F (x (x) our: F (x (x) such: F (x) we it's not easy

Given the function f (x) = 2x ^ 3 + 3ax ^ 2 + 3bx + C, we obtain the extremum at x = 1 and x = 2. (1) find the value of a and B; (2) if the equation f (x) = 0 has three roots, find the extremum of C
Value range

(1) F '(x) = 6x ^ 2 + 6AX + 3BF' (1) = 6 + 6A + 3B = 0 f '(2) = 24 + 12a + 3B = 0. A = - 3B = 4 (2) f (x) = 2x ^ 3-9x ^ 2 + 12x + C f' (x) = 6x ^ 2-18x + 12 = 6 (x-1) (X-2). F (1) is the maximum and f (2) is the minimum. The equation f (x) = 0 has three roots, then f (1) = 5 + C > 0 and f (2) = 4 + C