If y + 2 is an increasing function, then x = 2 is an even function A、f（1）＜f（5/2）＜f（7/2） B、f（5/2）＜f（1）＜f（7/2） C、f（5/2）＜f（7/2）＜f（1） D、f（7/2）＜f（1）＜f（5/2） The process is because f (x + 2) is an even function, so the axis of symmetry of F (x) is x = 2. Why is the axis of symmetry x = 2?

The function y = f (x + 2) is even and the image is symmetric about the y-axis
Y = f (x) shifts 2 units to the left to get the image of y = f (x + 2)
Then y = f (x + 2) shifts 2 units to the right to get the image of y = f (x)
The image is symmetric with respect to the line x = 2
∵ y = f (x) is an increasing function on (0,2)
/ / y = f (x) is a decreasing function on (2,4)
Answer D, f (7 / 2) < f (1) < f (5 / 2)

If y = f (x) is an increasing function on (0,2), and the function f (x + 2) is an even function, then what is correct in the following conclusion? A. F (1) < f (2.5) < f (3.5) B, f (3.5) < f (1) < f (2.5) C, f (3.5) < f (2.5) < f (1) d, f (2.5) < f (1) < f (3.5) how to judge the size

Y = f (x) is an increasing function on (0,2), and the function f (x + 2) shifts f (x) to the left by 2 bits, so f (x + 2) increases in the definition domain (- 2,0), and decreases symmetrically on the definition domain (0,2)

Let y = f (x) be an even function, and its analytic formula on [0,1] is f (x) = X-1, then its analytic expression on [- 1,0] is

Let y = f (x) be an even function,
Its analytic expression on [0,1] is f (x) = X-1,
Then, when x ∈ [- 1,0],
-x∈[0,1]
f(x)=f(-x)=(-x)-1=-x-1

Given that the power function f (x) = x ^ (m ^ 2-2m-3), m belongs to Z, is an even function, and the interval (0, positive infinity) is a minus function. The analytic formula of Y is obtained and the monotonicity is discussed

First of all, by the subtraction function, M 2 - 2m-3 = (M-3) (M + 1) < 0
-And F is an even function thousand, which indicates that M 2 - 2m-3 = (M-3) (M + 1) is even
Then M is odd, so there can only be m = 1
So f (x) = x ^ (- 4)
Is a minus function at (0, + ∞)
It is an increasing function at (- ∞, 0)

The function y = f (x) is even function. If x0, the analytic formula of function f (x) is?

x> 0 then - X0
f(x)=1+x

Given the propositions P: X ≥ 1, Q: x? ≥ x, what condition is p for Q? A. Necessary and sufficient condition B. necessary and sufficient condition C. sufficient and necessary condition D. neither sufficient nor necessary condition

Proposition p: X ≥ 1
Proposition q: X ≥ 1 or X ≤ 0
P can push Q, but Q can't push P, so choose a

If the even function f (x) defined on R satisfies f (x + 1) = - f (x) and increases on [- 1,0], then... F (3)

Because it is an even function, f (x) = f (- x) is obtained by the condition f (x + 1) = - f (x) = - f (x + 1). From F (x) = - f (x + 1), f (x + 1) = - f (x + 2) can be derived from F (x) = - f (x + 2)

Properties of odd function and even function The formula what also said, also has the common question type

If f (x) is an odd function and G (x) is an even function, then there are f (- x) = - f (x), G (- x) = g (x). At the same time, the f (x) image is symmetric about the origin, and the G (x) image is symmetric about the y-axis

It is known that the proposition p: equation x2 + MX + 1 = 0 has two unequal negative real roots. Proposition q: equation 4x2 + 4 (m-2) x + 1 = 0 has no real roots. If P or q are true, P and Q are false, then the range of real number m is () A. （1，2]∪[3，+∞） B. （1，2）∪（3，+∞） C. （1，2] D. [3，+∞）

If P is true, then
m2−4＞0
- M < 0, M > 2;
If q is true, then △ = [4 (m-2)] 2-16 < 0, the solution is: 1 < m < 3;
∵ P or q is true, P and Q are false,
/ / P and Q are true and false,
When p is true and Q is false, m ≥ 3 is obtained; when p is false and Q is true, the solution is 1 < m ≤ 2
In conclusion, 1 ＜ m ≤ 2 or m ≥ 3;
Therefore, a

If the definition domain of an even function f (x) is x not equal to one, if x > 0, f (x) = log2 (x), then all x sums of F (x) = f (6 / (x + 5)) are satisfied When x > 0, f (x) = log2 (x), then what is the sum of all x satisfying f (x) = f (6 / (x + 5)) I figured it was - 10

If x is not equal to 1, even function f (x) = f (- x),
If f (x) = f (6 / (x + 5))
So we get (x = 6) x = 1
2. X = - 6 / (x + 5) gives x = - 2, x = - 3
All x and - 11 satisfied