The monotone increasing interval of the function y = log2 (x2-2x) is______ .

The monotone increasing interval of the function y = log2 (x2-2x) is______ .

Let t (x) = x2-2x, then from t (x) > 0, the definition domain of the function is {x | x < 0, or x > 2}, and y = log2t,
This problem is to find the increasing interval of function T (x) in the definition domain
By using the properties of quadratic function, the increasing interval of function T (x) in the definition domain is (2, + ∞),
So the answer is: (2, + ∞)

The function y = log2sin (2x + π) 6) The monotone decreasing interval of is () A. [kπ−π 12，kπ+5π 12)（k∈Z） B. (kπ+π 6，kπ+2π 3)（k∈Z） C. [kπ−π 3，kπ+π 6]（k∈Z） D. [kπ+π 6，kπ+5π 12)（k∈Z）

According to the title, sin (2x + π)
6)＞0，
The monotone decreasing interval of function satisfies
sin(2x+π
6)＞0
2kπ+π
2≤2x+π
6≤2kπ+3π
2，k∈Z ，
therefore
2kπ＜2x+π
6＜2kπ+π，k∈Z
2kπ+π
2≤2x+π
6≤2kπ+3π
2，k∈Z ，
The solution is 2K π + π
2≤2x+π
6＜2kπ+π ，k∈Z，
That is, X ∈ [K π + π
6，kπ+5π
12)，k∈Z．
Therefore, D

Monotone increasing interval of function y = log2 | X-1 |____________________

(1,＋∞)

Is the monotone increasing interval of the function y = | log2 (x-3) |?

The domain of F (x) = log (2, x-3) is (3, + ∞). Obviously, f (x) is monotonically increasing in the domain
When x ≥ 4, log (2, x-3) > 0, | | log (2, x-3) | = log (2, x-3), thus y = | log2 (x-3) | increases monotonically
But 3

Is the monotone increasing interval of the function y = log2 ^ (6-x-x ^ 2)?

Let t = 6-x-x ^ 2 = - (x + 1 / 2) ^ 2 + 25 / 4
t> 0 gives - 3

What is the value range of the function y = sin ^ 4x + cos ^ 4x + 1?

y=sin^4x+cos^4x+1
=[(sinx)^2+(cosx)^2]^2-2(sinxcosx)^2+1
=1-2(sinxcosx)^2+1
=2-[(sin2x)^2]/2
=2-[1-(cos4x)]/4
=(7+cos4x)/4
Max = 2
Minimum = 3 / 2
Range [3 / 2,2]

Find the minimum positive period, maximum value and minimum value of the function f (x) = (SiNx + cos ^ 4x + sin ^ 2xcos ^ 2x) / 2-2sinxcosx-1 / 2sinxcosx + 1 / 4cos2x

The title is a little strange! Can you type it wrong?
Hint: sin ^ 2xcos ^ 2x = sin ^ 2 (2x) / 4
2sinxcosx=sin2x
1/2sinxcosx=sin2x/4
1/4cos2x=1/4-sin^2x/2

(1-(sin^4x-sin^2xcos^2x+cos^4x)/sin^2x +3sin^2x

sin^4x-sin^2xcos^2x+cos^4x
=sin^4x+2sin^2xcos^2x+cos^4x-3sin^2xcos^2x
=(sin^2x+cos^2x)^2-3sin^2xcos^2x
=1-3sin^2xcos^2x
So molecule = 3sin ^ 2xcos ^ 2x
So the original formula = 3sin ^ 2xcos ^ 2x / sin ^ 2x + 3sin ^ 2x
=3cos^2x+3sin^2x
=3(sin^2x+cos^2x)
=3

Find the function y = sin (π The monotone increasing interval of 6 − 2x) is______ .

y=sin（π
6-2x）=-sin（2x-π
6）；
∵ function y = sin (2x - π)
6) The monotone interval of sin = π is monotone
6-2x);
∴2kπ+π
2≤2x-π
6≤2kπ+3π
2⇒kπ+π
3≤x≤kπ+5π
6，k∈Z．
The function y = sin (π)
The monotone increasing interval of 6 − 2x) is: K π + π
3，kπ+5π
6]，k∈Z．
So the answer is: [K π + π
3，kπ+5π
6]，k∈Z．

The increasing interval of the function y = LG [sin (π / 3-2x)] is

In order to make the function y = LG [sin (π / 3-2x)] monotonically increase, it is necessary to make the function y = LG [sin (π / 3-2x)] monotonically increase
Sin (π / 3-2x) > 0 and monotone increasing function in a certain interval
Sin (π / 3-2x) = - sin (2x - π / 3) > 0, sin (2x - π / 3) < 0 and is a monotone decreasing function in a certain interval
In this case, 2K π < 2x - π / 3 ≤ 2K π + π / 2
kπ+π/6＜x≤kπ+5π/12 k∈z