It is known that f (x) = 2x ^ 3 + 3ax ^ 2 + 3bx + 8C has extremum at x = 1 and x = 2 1) Find the value of A. B (2) for any x belonging to the closed interval 0 to 3, f (x)

It is known that f (x) = 2x ^ 3 + 3ax ^ 2 + 3bx + 8C has extremum at x = 1 and x = 2 1) Find the value of A. B (2) for any x belonging to the closed interval 0 to 3, f (x)

If f (x) = 2x ^ 3 + 3ax ^ 2 + 3bx + 8C, we can get f (x) '= 6x ^ 2 + 6AX + 3B by derivation. If we get the extremum in x = 1 and x = 2, then f (x)' = K (x-1) (X-2) = 6x ^ 2 + 6AX + 3b, then we can get KX ^ 2-3kx + 2K = 6x ^ 2 + 6AX + 3B. The comparison coefficient is k = 6, - 3K = 6a, 2K = 3b, so a = - 3, B = 4. So f (x) = 2x ^ 3-9x ^ 2 + 12x + 8C

Given that f (x) = 12x2 + 4lnx-5x, f ′ (x) is the derivative of F (x). (I) find the extremum of y = f (x); (II) find the interval where f ′ (x) and f (x) have the same monotonicity

(I) ∵ f (x) = 12x2 + 4lnx-5x, ∵ f '(x) = x + 4x-5 = (x-1) (x-4) x (x > 0), from F' (x) > 0, 0 < x < 1 or x > 4, from F '(x) < 0, 1 < x < 4. When x changes, f' (x), f (x) changes as follows: ⊙ ⊙ x ⊙ x ⊙ 1 ⊙ (1,4) ⊙ (4, + ∞) ⊙ f '(x) ⊙ + ⊙ 0 ⊙ + ⊙ f (x) ⊙ ↗ ⊙ maximum value ⊙ ↘ ⊙ minimum ⊙ ↗ The maximum of F (x) = f (1) = - 92, the minimum of F (x) = f (4) = 8ln2-12 6 points (Ⅱ) Let G (x) = x + 4x-5 (x ＞ 0), G '(x) = (x + 2) (X-2) x, from G' (x) ＞ 0, X ＞ 2, G (x) is an increasing function, from G '(x) ＜ 0, 0 ＜ x ＜ 2, G (x) is a decreasing function 12 points

Finding the extreme value of y = 2x ^ 3-3x ^ 2
It's a process, thank you

y'=6x²-6x=0
x=0,x=1
X1, y '> 0, increasing function
0

Find the extreme value of y = 2x * 2-3x + 5 on [- 2,2]

Y=2X^2-3X+5
=2(x^2-3x/2+9/16)+31/8
=2(x-3/4)^2+31/8
When x = 3 / 4, the minimum is 31 / 8
The extreme value is only 31 / 8
If it's the maximum, the minimum is 31 / 8
The maximum value is 31 / 8 at x = - 2
The maximum is 8 + 6 + 5 = 19