Given that the function f (x) = 2Sin ω x monotonically decreases on [- π 4, π 4], then the value range of real number ω is______ ．

Given that the function f (x) = 2Sin ω x monotonically decreases on [- π 4, π 4], then the value range of real number ω is______ ．

From the monotonicity of sine function, we can get a simple decreasing interval of function f (x) = 2Sin ω x, which is [π 2 ω, − π 2 ω], we can get π 2 ω ≤ - π 4 π − 2 ω ≥ π 4-2 ≤ ω < 0, so the answer is: - 2 ≤ ω < 0

Given that w is a positive real number and the function f (x) = 2sinwx is an increasing function on [- π / 3, π / 4], then the value range of W is?
Why is the increasing interval of F (x) = 2sinx [- π / 2, π / 2]
Can't the monotone interval π / 2 + K π be added

In general, we use the minimum positive period to discuss it. The next one is [3 π / 2,5 π / 2]. Isn't it more troublesome? When you ask for all increasing intervals, you must add it. Note that it is 2K π

It is known that ω is a positive number and the function f (x) = 2Sin ω x is an increasing function in the interval [− π 3, π 4]. The value range of ω is obtained

In this paper, we obtain the - π 2 ω - π 2 ω + 2K π from - π 2 + 2K π (K ∈ z) from - π 2 + 2K π + 2K π (K ∈ z) from - π 2 + 2K π (K ∈ z) from - π 2 (2 + 2K π 2 + 2K π (K ∈ z) from - π 2 + 2K π + 2K π (K ∈ z) for the - π 2 π 2 π 2 π (2k π\883838⊆ [[ [\\\\\\\\\\\\\\\\\\π4, and ω ＞ 0, the solution is 0 ＜ω≤ 32, so the value range of ω is (0,32]

Given that the function f (x) = 2sinwx monotonically decreases on [- π / 4, π / 4], then the value range of real number W is:
That's what I did
Let w (- π / 4) and w (π / 4) belong to (π / 2 + 2K π, 3 / 2 π + 2K π) respectively
Since he is monotonic decreasing, he only needs to control the two ends of the interval within this range~
But it doesn't seem to work~

Obviously, w ≠ 0
If w > 0, Wx ∈ [- w π / 4, w π / 4], this interval contains 0,
It is impossible for f (x) = 2sinwx to decrease on [- π / 4, π / 4]
So W0
wx∈[wπ/4,-wπ/4],
f(x)=2sinwx=-2sin(-wx)
The function f (x) decreases monotonically on [- π / 4, π / 4],
That is, sin (- Wx) increases monotonically on [- π / 4, π / 4],
The interval [w π / 4, - w π / 4] contains 0,
The increasing interval containing 0 is [- π / 2, π / 2],
Ψ [w π / 4, - w π / 4] is contained in [- π / 2, π / 2],
So w π / 4 ≥ - π / 2, - w π / 4 ≤ π / 2
And because of W