Given the function f (x) = - 1 / 3x ^ 3 + BX ^ 2 + CX + BC, its derivative is f '(x). Let g (x) = | f' (x) |, note that the maximum value of function g (x) in the interval [- 1,1] is m When C = 1, if M ≥ K holds for any B. try to find the maximum value of K

Given the function f (x) = - 1 / 3x ^ 3 + BX ^ 2 + CX + BC, its derivative is f '(x). Let g (x) = | f' (x) |, note that the maximum value of function g (x) in the interval [- 1,1] is m When C = 1, if M ≥ K holds for any B. try to find the maximum value of K

one
When C is 1. F '(x) = - x ^ 2 + 2bx + 1 g (1) = | - 2b | g (- 1) = | 2B | so when G (1) = g (- 1) g (x) takes the maximum value, X is between - 1 and 1, and when x = B, it is the maximum. If G (x) is brought in, we can get / b ^ 2 + 1 / > = k, so the maximum value of K is 1

Solving inequality 2loga (x-1) ≥ loga (AX + 1) (a > 1)

2log(x-1)≥log(ax+1)log(x-1)^2≥log(ax+1)x≠1 and ax+1 >0 and (x-1)^2 ≥ ax+1x≠1 and x> 1/a and x^2 - (2+a)x ≥0x≠1 and x> 1/a and "x≤0 or x≥ 1/(2+a)"ie x>1/a

If the function g (x) = x3-ax2 + 1 is a monotone decreasing function in the interval [1,2], then the value range of real number a is______ ．

Let g (x) = x3-ax2 + 1, then G ′ (x) = 3x2-2ax, because & nbsp; G (x) = x3-ax2 + 1 is a monotone decreasing function in the interval [1,2], so g ′ (x) = 3x2-2ax ≤ 0 is constant in X ∈ [1,2]. That is to say, 2aX ≥ 3x2, a ≥ 32x is constant in X ∈ [1,2]