Given that y in the first-order function y = KX + 2K decreases with the increase of X, then the inequality KX + 2K

Given that y in the first-order function y = KX + 2K decreases with the increase of X, then the inequality KX + 2K

From the fact that y in the function y = KX + 2K decreases with the increase of X, we can know K

Given the function f (x) = ln ^ 2 (1 + x) - [x ^ 2 / (1 + x)], find the extremum of function f (x)
g(x)=2(1+x)ln(1+x)-x^2-2x
It is proved that when x is less than 0, G (x) is less than 0

The derivative of F (x) is [2 (1 + x) ㏑ (1 + x) - 2x-x & # 178;] / (1 + x) # 178;];,
Let the molecule be h (x) and its derivative be 2 ㏑ (1 + x) - 2x
㏑ (1 + x) ≤ x is constant, so h (x) decreases monotonically,
H (0) = 0, so f (x) increases at (- 1,0) and decreases at (0, + ∞)
The maximum of F (x) is f (0) = 0

Let f (x) = / X-1 / + / x-a / (1) when a = 4, find the solution set (2) of the inequality f (x) > = 5. If f (x) > = 4 is r constant for X, find the value range of A
Solving additive points

（1）f(x)=/x-1/+/x-4/
When x > 4, f = 2x - 5 ≥ 5, so x ≥ 5
When 1 ≤ x ≤ 4, f = 3 ≤ 5, the constant holds
When x < 1, f = 5 - 2x ≥ 5, so x ≤ 0
In conclusion, the solution set is {x | x ≤ 0 or 1 ≤ x ≤ 4 or X ≥ 5}
(2) If f (x) > = 4 is r-constant for X
It can be seen from the number axis that when the value of X is between 1 and a, f is the minimum,
When a > 1, a - 1 ≥ 4, so a ≥ 5
When a < 1, 1 - a ≥ 4, so a ≤ - 3
A = 1 does not make f ≥ 4 constant
So a ≤ - 3 or a ≥ 5

Let f (x) = | x + 1 | + | X-5 |, X ∈ R. (1) find the solution set of the inequality f (x) ＜ x + 10; (2) if the inequality f (x) ≥ a - (X-2) 2 about X is constant on R, find the value range of real number a

(1) Remove the absolute value, f (x) = − 2x + 4, & nbsp; X ＜− 16, & nbsp; − 1 ≤ x ＜ 52X − 4, & nbsp; X ≥ 5; when x ＜ - 1, from - 2x + 4 ≤ x + 10, the solution is x ≥ - 2, ｜ - 2 ≤ x ＜ - 1; when - 1 ≤ x ＜ 5, from 6 ≤ x + 10, the solution is x ≥ - 4, ｜ - 1 ≤ x ＜ 5; when x ≥ 5, from 2X-4 ≤ x + 10