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 (1) If | B | > 1, it is proved that for any C there is m > 2 (2) If M > = k holds for any B, C. 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 (1) If | B | > 1, it is proved that for any C there is m > 2 (2) If M > = k holds for any B, C. try to find the maximum value of K

If f (x) = X3 + bx2 + CX + D is a decreasing function in the interval [- 1,2], then B + C ()
A. There is a maximum of 152b. There is a maximum of - 152C. There is a minimum of 152D. There is a minimum of - 152

If f (x) is a decreasing function on [- 1, 2], we know that f ′ (x) = 3x2 + 2bx + C ≤ 0, X ∈ [- 1, 2], then f ′ (− 1) = 3 − 2B + C ≤ 0f ′ (2) = 12 + 4B + C ≤ 0 {15 + 2B + 2C ≤ 0} B + C ≤ - 152

F (x) = x / (1 + X & sup2;), which is an odd function on (- 1,1). It is proved that f (x) is an increasing function and an odd function on (- 1,1)

-1

The function f (x) defined on R satisfies log2 (1-x) when x ≤ 0 and f (x-1) - f (X-2) when x > 0, then f (2014)=
2 is the base

f(-1)=log2(1-(-1))=1
f(0)=log2(1-0)=0
f(1)=f(0)-f(-1)=-1
f(2)=f(1)-f(0)=-1
f(3)=f(2)-f(1)=0
f(4)=f(3)-f(2)=1
f(5)=f(4)-f(3)=1
f(6)=f(5)-f(4)=0
f(7)=f(6)-f(5)=-1
f(8)=f(7)-f(6)=-1
f(9)=f(8)-f(7)=0
f(10)=f(9)-f(8)=1
f(11)=f(10)-f(9)=1
f(12)=f(11)-f(10)=0
When x > 0, the minimum period of F (x) is 6
2014÷6=335.4
So f (2010) = - 1
It is also proved that:
f(x)=f(x-1)-f(x-2)
f(x-1)=f(x-2)-f(x-3)
f(x)=f(x-1)-f(x-2)=f(x-2)-f(x-3)-f(x-2)=-f(x-3)
f(x+3)=-f(x)
f(x+6)=f[(x+3)+3]=-f(x+3)=-[-f(x)]=f(x)
So f (x) period is 6

Given the function f (x) = alg2x-blog3x + 2, if f (12014) = 4, then the value of F (2 & nbsp; 014) is______ ．

∵ function f (x) = alg2x-blog3x + 2, ∵ f (12014) = alg212014-blog312014 + 2 = - alg22014 + blog32014 + 2 = 4, ∵ f (2014) = alg22014-blog32014 + 2 = - 2 + 2 = 0

Given the function f (x) = - x + log2 (1-x / 1 + x) (1), find the value of F (1 / 2012) + F (- 1 / 2012)
Given function f (x) = - x + log2 (1-x / 1 + x)
(1) Find the value of F (1 / 2012) + F (- 1 / 2012)

f(x)=-x+log2[（1-x)/(1+x﹚]
f(-x)=x+log2[(1+x)/(1-x)]
=x-log2[(1-x)/(1+x)
=-f(x)
So f (x) is an odd function
So f (1 / 2012) + F (- 1 / 2012) = 0

Given the function f (x) = x ^ 2 + (a + 1) x + B and f (3) = 3, and f (x) ≥ x is constant, find the value of a, B

f(3)=3
Then 9 + 3 (a + 1) + B = 3
3a+b=-9
b=-9-3a
F (x) ≥ x
That is, x ^ 2 + (a + 1) x + B ≥ x holds
X ^ 2 + ax + B ≥ 0
Δ=a^2-4b

Given the function f (x) = x2 + (a + 1) x + B, and f (3) = 3, and f (x) > = x is constant, find the value of a and B
Detailed process

F (x) > = x holds
That is x2 + ax + b > = 0
∴△=a²-4b

Let f (x) = A-2 / (2 ^ x + 1) determine the value of a so that f (x) satisfies the condition f (- x) = - f (x) holds
Early technetium the answer is good

Let f (x) satisfy the condition f (- x) = - f (x)
It is an odd function
Then f (0) = 0
So a = 1

Given the function f (x) = AX2 + 2a-2, if there is FX < 0 for any real number x, the value range of a is obtained

Given the function f (x) = AX2 + 2a-2, if there is FX < 0 for any real number x, the value range of a is obtained
When a = 0; f (x) = - 2; coincidence;
When a < 0, 2a-2 < 0, 2a-2 < 1, 2a-2 < 0, 2a-2 < 0, 2a-2 < 0, 2a-2 < 0, 2a-2 < 0, 2a-2 < 0, 2a-2 < 0, 2a-2 < 0, 2a-2 < 0, 2a-2 < 0;
The range of a is a ≤ 0;
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