Given that w is a positive real number and the function f (x) = 2sinwx is an increasing function on [- Pai / 3, Pai / 3], then the value range of W is Given that w is a positive real number and the function f (x) = 2sinwx is an increasing function on [- Pai / 3, Pai / 3], then the value range of W is 3 / 2], can't understand the analysis in the book, Y = 2sinwx is an increasing function on [- Pai / 2, Pai / 2]. To make FX = sinwx an increasing function on [- Pai / 3, Pai / 3], we only need t / 2 > = Pai / 3 - (- Pai / 3) = 2 Pai / 3, that is, 2 Pai / W > = 4 Pai / 3, so the answer is as above

Given that w is a positive real number and the function f (x) = 2sinwx is an increasing function on [- Pai / 3, Pai / 3], then the value range of W is Given that w is a positive real number and the function f (x) = 2sinwx is an increasing function on [- Pai / 3, Pai / 3], then the value range of W is 3 / 2], can't understand the analysis in the book, Y = 2sinwx is an increasing function on [- Pai / 2, Pai / 2]. To make FX = sinwx an increasing function on [- Pai / 3, Pai / 3], we only need t / 2 > = Pai / 3 - (- Pai / 3) = 2 Pai / 3, that is, 2 Pai / W > = 4 Pai / 3, so the answer is as above

Known - π / 3

If the function f (x) = 3sin (Wx + φ) has f (π / 3 + x) = f (- π) for any real number x, it is urgent to find f (one sixth)

Maybe the title is wrong?
If so, we can make x = - (π / 3),
Let x = 0,
Let x = 2 π / 3,
And so on
So we get W and φ,
So f (the sixth) is easy to get

If f (x) = 3sin (ω x + φ) has f (π / 3 + x) = f (- x) for any x, then f (π / 6)=

The solution is f (π / 3 + x) = f (- x)
The axis of symmetry of the known function is x = (π / 3) / 2 = π / 6
Let x = π / 6 be the axis of symmetry of the function f (x) = 3sin (ω x + φ)
f(π/6)=±3.