Let f (x) be an even function and an increasing function on (0, + ∞). Judge whether f (x) is an increasing function or a decreasing function on (- ∞, 0), and prove your judgment

Let f (x) be an even function and an increasing function on (0, + ∞). Judge whether f (x) is an increasing function or a decreasing function on (- ∞, 0), and prove your judgment

F (x) is a decreasing function on (- ∞, 0) (1 point)
Proof: if X1 < x2 < 0, then - X1 > - x2 > 0 (3 points)
∵ f (x) is an increasing function on (0, + ∞)
﹤ f (- x1) ﹥ f (- x2) (7 points)
And f (x) is even function
∴f（-x1）=f（x1），f（-x2）=f（x2）
∴f（x1）＞f（x2）
ν f (x) is a decreasing function on (- ∞, 0) (12 points)

It is known that f (x) is an even function and f (x) is a decreasing function on (0, + ∞). It is proved that f (x) is an increasing function on (- ∞, 0)

It is proved that any x1, X2 ∈ (- ∞, 0) and x1 ＜ x2 are taken,
Then - x1, - x2 ∈ (0, + ∞) and - x1 ＞ - x2,
∵ f (x) is a decreasing function on (0, + ∞),
∴f（-x1）＜f（-x2）
And ∵ f (x) is an even function,
f（x1）=f（-x1），f（x2）=f（-x2）
∴f（x1）＜f（x2）
That is, f (x) is an increasing function on (- ∞, 0)

The function y = f (x) is an increasing function on (0, 2), and the function y = f (x + 2) is an even function. Try to compare the sizes of F (1), f (2.5), f (3.5) () A. f （3.5）＞f （1）＞f （2.5） B. f （3.5）＞f （2.5）＞f （1） C. f （2.5）＞f （1）＞f （3.5） D. f （1）＞f （2.5）＞f （3.5）

∵ the function y = f (x + 2) is an even function, and f (x + 2) = f (- x + 2), ᙽ the function is symmetric about x = 2.

Let y = f (x) be an even function, and its image on [0, 1] is shown in the figure, then its analytic formula on [- 1, 0] is______ .

 the function y = f (x) is even, ᙽ the image of the function is symmetric about the y-axis,

The power function f (x) = x − 1 is known 2p2+p+3 2 (P ∈ z) is an increasing function on (0, + ∞) and even in its definition domain. Find the value of P and write the corresponding function (x)

∵ power function f (x) = x − 1
2p2+p+3
2 is an increasing function on (0, + ∞),
So - 1
2p2+p+3
2＞0，
The result is - 1 < p < 3
By P ∈ Z,
So p = 0, 1, 2
When p = 0, f (x) = X3
2. It is not even function, which does not conform to the meaning of the question;
When p = 1, f (x) = X2 is an even function;
When p = 2, f (x) = X3
2, not even function, not in line with the meaning of the question
So p = 1, the corresponding function is f (x) = x2

The function f (x) is an even function on R, and when x > 0, the analytic formula of the function is f (x) = 2 / X - 1 (1) It is proved by definition that f (x) is a minus function on (0. + 00) (2) When x < 0, find the analytic expression of the function

(1) The expression of F (x) in the domain of (0. + 00) can be derived, and the expression of derivative function is - (2 / x ^ 2) is a negative value in the definition domain, so f (x) is a minus function (if you have not learned the derivative, you can also solve it according to the definition of monotonicity of function)
2. Let x > 0, so - X

Given the proposition p: x > 0 and proposition q: x? > 0, then p is () a of Q, sufficient and unnecessary condition Given the proposition p: x > 0 and proposition q: x? > 0, then p is () a of Q, sufficient but unnecessary condition B, necessary and insufficient condition C, sufficient and necessary condition D, neither sufficient nor necessary condition

Then p is () a of Q, sufficient and unnecessary

Given the proposition p|x + 1 | 2, Q: 5x-6 > x? Then what is the condition for non-p to be non-Q

Proposition p: | x + 1 > 2
X + 1 > 2 or x + 11 or XX
x²-5x+6

Given a, B ∈ (0, + ∞), if the proposition p: a 2 + B 2 < 1 and the proposition q: ab + 1 ≤ a + B, then p is a non-Q condition

a,b∈(0,+∞) p:a²+b²

It is known that proposition p: 1 belongs to {x 2 (1). If "PVQ" is true, find the range of real number a (2) If "P Λ Q" is a true proposition, find the value range of real number a

(1) PVQ is true, which means that as long as one is satisfied, the value of a is [- 2,2]
(2) P Λ q is true, which means that the two conditions must be satisfied at the same time, and the value of a is [- 1,1]