Let f (x) = asin (Wx + phi) (a ≠ 0, w > 0, Phi > 0, │ phi │ 0, Phi > 0, │ phi │

Let f (x) = asin (Wx + phi) (a ≠ 0, w > 0, Phi > 0, │ phi │ 0, Phi > 0, │ phi │

When a > 0, f (x) is a decreasing function on [5 π / 12,2 π / 3],
And when a

The function f (x) = asin (Wx + φ) (a > 0, w > 0, | φ) is known|

Given the function f (x) = asin (Wx + φ) (a > 0, w > 0, | φ | 0, w > 0, | φ | t = 4 π = = > W = 2 π / T = 1 / 2  f (x) = 2Sin (1 / 2x + φ) ∫ f (x) image is symmetric with respect to the straight line x = π / 3, and the first axis of symmetry on the right side of Y axis is: Wx + φ = π / 2 = = > x = (π - 2 φ) / (2W) such that (π - 2 φ) / (2W) = π / 3 = = > (π

The known function f (x) = asin (ω x + ψ) (a > 0, ψ)

Then f = 2 is the maximum of the function;
If the distance between two adjacent symmetrical axes is 4, then t / 2 = 4, so t = 8;
That is, w = pi / 4;
And the function passes through the point (1,2);
Then f (1) = 2;
The solution ψ = pi / 4;
So f (x) = 2Sin (PI / 4 * x + pi / 4)
So f (1) + F (2) +. F (2011) = - 1

The image of given function f (x) = asin (ω x + φ) x ∈ R (a > 0, ω > 0, 0 < φ < π / 2) is shown in the figure
When x = 5 π / 12, y = 0 π / 6 and 5 π / 12 are separated by 1 / 4 cycles, and the lowest point of the image is - 2
(1) Find f (x)
2. When x ∈ [0, π / 2], the equation f (x) = 2a-3 has two unequal real roots x1, x2. Find the value range of real number a, and find the value of X1 + x2

When 1 x = π / 6, the maximum value of the function is 2, & nbsp;, & nbsp; & nbsp;, that is, a = 2 π / 6 and 5 π / 12 are separated by 1 / 4 period, & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp;, when 1 x = π / 6

Given the function f (x) = asin (ω x + φ) (a > 0, ω > 0, | φ | < π / 2), the image passes through (0,1) at the two adjacent maximum points
On (x0,2), (x + 3 / 2, - 2) (x0 ＞ 0), f (x) has maximum and minimum values respectively
(1) Find the analytic expression of F (x);
(2) If the maximum and minimum values of G (x) = AF (x) + B are 6 and 2 respectively, the values of a and B are obtained

f(0)=Asin(φ)=1
maxf=2----A=2,φ=π/6
ωx'+φ=π/2+2kπ-----1
w(x'+3/2)+φ=-π/2+2kπ------2
Simultaneous 1 and 2 results in w = 2 / 3 π?
f(x)=2sin( 2/3πx+π/6)
2A + B = 6 - 2A + B = 2, a = 1, B = 4
-2A + B = 6, 2A + B = 2, a = 1, B = 4
So a = 1, B = 4
I almost forgot my knowledge, right

Let f (x) = asin (Wx + Q), (a = / 0, w > 0, - Pai / 2

Let f (x) = asin (Wx + Q), (a = / 0, w > 0, - Pai / 2

Let f (x) = asin (Wx + φ) (a > 0, w > 0, φ > 0, │

The minimum positive period is π, w = 2, the image is symmetric about the straight line x = 2 π / 3, 4 π / 3 + φ = 3 π / 2, φ = π / 6
f(x)=Asin(2x+π/6)
-π/2

Let f (x) = asin (ω x + φ), (a ≠ 0, ω＞ 0, - π 2 ＜φ＜π 2) be symmetric with respect to the straight line x = 2 π 3, and its period is π, then ()
A. The maximum value of C. f (x) is ad. a symmetry center of F (x) is point (5 π 12,0)

The image is symmetric with respect to the straight line x = 2 π 3, sin (φ + 2 π 3 × 2) = ± 1, i.e. 2 π 3 × 2 + φ = π 2 + K π, K ∈ Z, and ∵ - π 2 ∵ π 2 ∵ φ = π 6 ∵ f (x) = asin (2x + π 6)

If f (1-m) > F (m) holds, the value range of M is obtained

∵ f (x) is an even function
And ∵ f (1-m) > F (m) holds
∴ f(-|1-m|)>f(-|m|)
Moreover, f (x) increases monotonically over the interval [- 2,0],
∴ -|1-m|>-|m|≥-2
That is | 1-m|

How to judge function monotonicity and parity

Parity and monotonicity of functions and their distinguishing methods
Monotonicity of general functions
1. Definition method: set in the definition domain x1