The monotone decreasing interval of the absolute value of the function y = a is

The monotone decreasing interval of the absolute value of the function y = a is

This function is a piecewise function y = {a (a > 0)
｛-a (a≤0）
The monotone interval of a function is (- ∞, 0]

If the function f (x) is a decreasing function on R, f (- 2) = 0, find the monotone interval of the absolute value of the function g (x) = f (x)

The function f (x) is a decreasing function on R, f (- 2) = 0
So x0
When x ≥ - 2, f (x) ≤ 0
g(x) = |f(x)| = f(x) ,x

It is known that the function f (x) is an odd function defined on the interval (- 1,1) and a decreasing function on the interval (0,1), f (1-A) + F (1-2a)

Let's assume that the function is f (x) = - x ^ 3, so f (1-A) + F (1-2a) A-1, so a

Let F X be an odd function defined on R, and a decreasing function on the interval (- ∞, 0), and the real number a satisfy the inequality f (3a2 + A-3) ＜ (3a2-2a)

Let f (3a & sup2; + A-3) < f (3a & sup2; - 2A)
A: the value range of real number a is a 3A & sup2; + A-3 < 3A & sup2; - 2A -- > a

The function f (x) defined on R is an odd function, and f (x) = f (2-x). If f (x) is a decreasing function in the interval (1,2), then f (x)
The function f (x) defined on R is an odd function, and f (x) = f (2-x). If f (x) is a decreasing function in the interval (1,2), it is proved that f (x) is an increasing function in the interval (- 2, - 1) and a decreasing function in the interval (3,4)
This kind of question seems very typical

1. Where (1,2) is a decreasing function,
So f (1) > F (2),
Because f (x) = f (2-x), f (x) = f (2-x),
So f (2) = f (0)
So f (1) > F (0); f (- 2) = f (4), f (- 1) = f (3), f (4) = f (- 2) = - f (2),
f(3)=f(-1)=-f(1),
Also because - f (1)

Let f (x) be a decreasing function defined on (0, + ∞), then the relation between F (2) and f (A2 + 2A + 2) is______ ．

A2 + 2A + 2 = (a + 1) 2 + 1 ≥ 1, let t = A2 + 2A + 2-2 = A2 + 2A = a (a + 2), so when - 2 < a < 0, A2 + 2A + 2 < 2; when a = 0 or a = - 2, A2 + 2A + 2 = 2; when a < - 2 or a > 0, A2 + 2A + 2 > 2; because f (x) is a decreasing function defined on (0, + ∞), when - 2 < a < 0, f (A2 +...)

Let f (x) be a decreasing function defined on (0, + ∞), then the relation between F (2) and f (A2 + 2A + 2) is______ ．

A2 + 2A + 2 = (a + 1) 2 + 1 ≥ 1, let t = A2 + 2A + 2-2 = A2 + 2A = a (a + 2), so when - 2 < a < 0, A2 + 2A + 2 < 2; when a = 0 or a = - 2, A2 + 2A + 2 = 2; when a < - 2 or a > 0, A2 + 2A + 2 > 2; because f (x) is a decreasing function defined on (0, + ∞), when - 2 < a < 0, f (A2 +...)

Let f (x) be a decreasing function defined on (0, + ∞), then the relation between F (2) and f (A2 + 2A + 2) is______ ．

A2 + 2A + 2 = (a + 1) 2 + 1 ≥ 1, let t = A2 + 2A + 2-2 = A2 + 2A = a (a + 2), so when - 2 < a < 0, A2 + 2A + 2 < 2; when a = 0 or a = - 2, A2 + 2A + 2 = 2; when a < - 2 or a > 0, A2 + 2A + 2 > 2; because f (x) is a decreasing function defined on (0, + ∞), when - 2 < a < 0, f (A2 +...)

Let f (x) be a decreasing function defined on (0, positive infinity), then the relation between F (2) and f (a ^ 2 + 2A + 3) is

Let f (x) be a decreasing function defined on (0, positive infinity), then the relation between F (2) and f (a ^ 2 + 2A + 3)
a^2+2a+3=（a-1）^2+2≧2
So a ^ 2 + 2A + 3 ≥ 2
Then f (2) ≥ f (a ^ 2 + 2A + 3)

Let f (x) be a decreasing function defined on (0, + OO), then the relationship between F (2) and f (A2 + 2A + 2) is

a^2 + 2a + 2 = (a-1)^2 + 1
0