It is known that the function f (x) is an even function defined on the interval [- 2,2]. When x belongs to [0,2], f (x) is a decreasing function if the inequality f (1-m) is less than f (m) Find the value range of M

It is known that the function f (x) is an even function defined on the interval [- 2,2]. When x belongs to [0,2], f (x) is a decreasing function if the inequality f (1-m) is less than f (m) Find the value range of M

f(1-m)f(m)
Because it is an even function, that is, the solution f (1-m) f (- M)
Because 1-m-m is constant, it is in the range of decreasing interval, so
01-m2 and - 1m1
0-m2 and - 2m0
So: - 1m0

The even function y = f (x) is an increasing function in the interval [- 4, - 1]. The following inequality holds
A.f(-2)

If y = f (x) is an even function, we obtain that:
F (3) = f (- 3) in a, f (PAI) = f (- PAI) in B, f (1) = f (- 1) in C, f (radical 3) = f (- radical 3) in D
In addition, if the interval [- 4, - 1] is an increasing function, then:
f(-pai)

The even function y = f (x) is an increasing function in the interval [- 4 - 1]. The following inequality holds ()
A. F (- 2) < f (3) B.F (- π) < f (π) C.F (1) < f (- 3) d f (- radical 2) < f (radical 3)
D is a negative root, 2 is greater than 3, and the correct answer is d

In [- 4, - 1] is an increasing function, in [1,4] is a decreasing function
Please refer to the definition of even function
And f (x) = f (- x)
Please refer to the definition of even function again
The next refuelling, Ms did not see an option is right

Given the function f (x) = 23sinxcosx + cos2x (1), find the value of F (π 6); (2) let x ∈ [0, π 4], find the range of F (x)

(1) ∵ f (x) = 23sinxcosx + cos2x = 3sin2x + cos2x = 2Sin (2x + π 6), so f (π 6) = 2Sin (2 π 6 + π 6) = 2Sin π 2 = 2. (2) because 0 ≤ x ≤ π 4, so π 6 ≤ 2x + π 6 ≤ 2 π 3, so 1 ≤ 2Sin (2x + π 6) ≤ 2, that is, the range of function f (x) is [1,2]

When x = what, the value of 3x-x-5 / 2 and X + 2 / 3 are opposite to each other

3X-X-5/2=-(X+2/3)
2X-5/2=-X-2/3
3X=5/2-2/3
3X=5/6
X=5/18
A: when x = 5 / 18, the value of 3x-x-5 / 2 and X + 2 / 3 are opposite

Given that the opposite number of 3x-3 is - 15, find X

According to the meaning of the question, we get 3x-3 + (- 15) = 0, and the solution is x = 6

Given that the value of the algebraic formula 12x - (X-2) / 3 + 1 is zero, find the value of the algebraic formula (3x-1) / 4 + (2x + 1) / 3
The answer is 15!

12x - (X-2) / 3 + 1 = 0, then x = - 1 / 7
Taking x = - 1 / 7 into the algebraic formula (3x-1) / 4 + (2x + 1) / 3, the result is - 120 / 7

Given that the value of the algebraic formula 12x-x-2 / 3 + 1 is 0, find the value of another algebraic formula 3x-1 / 4 + 2x + 1 / 3

From the known algebraic formula 12x-x-2 / 3 + 1 = 0, we get x = - 1 / 33
So 3x-1 / 4 + 2x + 1 / 3 = 5 x + 1 / 12 = - 5 / 33 + 1 / 12 = - 3 / 44

The product of (x2 + PX + 8) (x2-3x + Q) does not contain the terms of x2 and X5, so the value of P Q can be obtained
2 and 5 after X are powers

（x2+px+8）(x2-3x+q)
=x^4 +(p-3)x^3 +(8-3p+q)x^2 +(pq-24)x+8q
Because it has no x2 and X5 terms, 8-3p + q = 0
q-3p=8

How to change (3x-1) (x + 1) = 0 to x = 1 / 3, x = - 1, can you tell me how to calculate

X+1=0;3x-1=0
x=-1;x=1/3