Given that the function f (x) = kx3-4x2-8 is a monotone function in the interval [2,8], find the value range of real number K

∵f（x）=kx3-4x2-8
∴f'（x）=3kx2-8x
∵ f (x) is monotonic on [2, 8]
‡ on [2, 8], f '(x) ≥ 0 or F' (x) ≤ 0
If f '(x) ≥ 0, 3kx2-8x ≥ 0 is established,
Then K ≥ 8
3x
∴k≥4
three
If f '(x) ≤ 0, 3kx2-8x ≤ 0 holds
Then K ≤ 8
3x
∴k≤1
three
To sum up, the value range of K is (− ∞, 1)
3]∪[4
3，+∞)．

If the function f (x) = |x| / (X-2) - (KX ^ 3) has three different zeros, the value range of real number k is

First, f (0) = 0. This is a zero point. When x is not equal to 0. The formula F (x) = 0 can be reduced to k = 1 / [(X-2) (x) | x|]. That is, the function image on the right and y = k have another two intersections. Take the derivative, and then draw the approximate image. I won't help you calculate it

If the definition field of function f (x) = LG (SiNx + a) is R and there is zero, the value range of real number a is () A. [1，2] B. （1，2] C. [2，3） D. [2，3]

The domain of F (x) is r, that is, SiNx + a > 0 is always true,
∴a＞1，
∵ function f (x) = LG (SiNx + a) has zero,
That is, LG (SiNx + a) = 0 has a solution,
There is a solution for SiNx + a = 1, and the solution is 0 ≤ a ≤ 2
∴1＜a≤2．
Therefore, B

Given that f (x) = ln (x + 1) - A / (x + 1) has exactly two different zeros in its definition field, what is the range of real number a? You should be able to use lobida's law

f(x)=ln(x+1)-a/(x+1)
The definition domain of F (x) is (- 1, + ∞)
f'(x)=1/(x+1)+a/(x+1) ²
=(x+a+1)/(x+1) ²
If a ≥ 0, x + A + 1 > 0 is constant, f '(x) > 0, f (x) is an increasing function
F (x) will not have two different zeros on (- 1, + ∞)
When a

The domain of function f (x) is R. for any real number x, there is f (x + 1) = - f (- x). And f (x) has three zeros. Find the sum of all zeros

Let's do it like this. Since there are three zeros and the definition field is r, I can assume that this function is a cubic polynomial, Let f (x) = ax ^ 3 + BX ^ 2 + CX + D;
By substituting f (x + 1) = - f (- x) into the polynomial, we can get:
a(x+1)^3+b(x+1)^2+c(x+1)+d=-(-ax^3+bx^2-cx+d）
That is, ax ^ 3 + (3a + b) x ^ 2 + (3a + 2B + C) x + A + B + C + D = ax ^ 3-bx ^ 2 + cx-d
So we can get two equations
3a+2b=0 （1）
a+b+c+2d=0 （2）
Moreover, because f (x + 1) = - f (- x), it is easy to know that f (1 / 2) = 0;
Substitute in and get 1 / 8A + 1 / 4B + 1 / 2C + D = 0 (3)
Since (2) and (3) in the three equations are equivalent, the three unknowns cannot be eliminated and the four unknowns cannot be solved, but the function f (x) must satisfy (1) and (2), so in the special case, d = 0, the solution is b = - 3 / 2A and C = 1 / 2A
By substituting the original polynomial, f (x) = ax ^ 3-3 / 2aX ^ 2 + 1 / 2aX = ax (x-1) (x-1 / 2)
Moreover, it can be verified that the polynomial satisfies all the conditions of the problem, so the zero is 0,1,1 / 2, and the sum is 3 / 2
I don't know if the period calculation is wrong, but the idea is like this, and only a special method is adopted

Let the function f (x) = ln (x + 1) + AE ^ (- x) - A, a belongs to R (1) , when a = 1, it is proved that f (x) is an increasing function (2) at (0, positive infinity). If x belongs to [0, positive infinity), f (x) is greater than or equal to 0, find the value range of A

(1) If a = 1, f (x) = ln (x + 1) - e ^ (- x) - 1, x > 0, let X1 be less than x2 and bring in known monotonicity
This is the definition
You can also look directly at the monotonicity of the function
Ln (x + 1) is an increasing function, e ^ (- x) is a decreasing function, so - e ^ (- x) is an increasing function, an increasing function, an adding function or an increasing function
∴f(x)↑.
（2）ln(x+1)+ae^(-x)-a>=0(x>=0),
When x = 0, the above formula holds;
x> When 0, 1-e ^ (- x) > 0,
a0,
∴h'(x)↑,h'(x)>h'(0)=0,
∴h(x)↑,h(x)>h(0)=0,
∴g'(x)>0,g(x)↑,
∴g(x)>g(0)=0,
To sum up, a

Given that the function f (x) = kx3-4x2-8 is a monotone function in the interval [2,8], find the value range of real number K