A series of corresponding values of the known function f (x) = asin (Wx + φ) + B (a > 0; W > 0) are as follows: For detailed explanation,

A series of corresponding values of the known function f (x) = asin (Wx + φ) + B (a > 0; W > 0) are as follows: For detailed explanation,

(1) From the chart, the maximum value is 3, and the minimum value is - 1 ᙨ B = (3-1) / 2 = 2 = 1A = [3 - (- 1)] / 2 = 2 cycle, t = 2 π, when x = - π / 6, the minimum value is obtained ᙨ f (x) = 2Sin (x-π / 3) + 1 (2) f (KX) = 2Sin (kx-π / 3) + 1 (1) f (KX) = 2Sin (kx-π / 3) + 1 + 1 cycle is t = 2 π / k = 2 π / 3 K = 3 9 F (KX) = 2Sin (3x - π / 3) + 1 x ∈ 1 x ∈ 1 ∈ ∈ 1 ∈ 1 ∈ [0, π / 3

A series of corresponding values of the known function f (x) = asin (Wx + φ) (a > 0, w > 0) + B (a > 0, w > 0) are as follows: X -π/6 π/3 5π/6 4π/3 11π/6 7π/3 17π/6 Y -1 1 3 1 -1 1 3 Ask God to answer, score added

The period of this function is t = (11 π / 6) - (- π / 6) = 2 π
W=1
If the maximum value of the function is 3 and the minimum value is - 1, then:
－A＋B=－1、A＋B=3
The results are as follows
A=2、B=1
When x = 5 π / 6, the function gets the maximum value, then:
φ=－π/3
The results are as follows
f(x)=2sin(x－π/3)＋1

A series of corresponding values of the function f (x) = asin (Wx + θ), (a > 0, w > 0, θ < π / 2) are as follows: x:… -π/8 0 π/8 3π/8 π/2 5π/8 7π/8… y：… 0 1 √2 0 -1 -√2 0… 1. The analytic formula of F (x) is obtained according to the data in the table; 2. It is pointed out that the image of function f (x) is obtained from the image of function y = SiNx (x ∈ R); 3. Let g (x) = f (x + π / 8) - A, if G (x) has two zeros when x ∈ [- π / 6, π / 3], find the value range of A Technical numbness! I see other people fight like this.

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Let f (x) = asin (Wx + b), (a > 0, w > 0), and (1) what value of B is f (x) an odd function (2) When B is taken, the function is even

(1) If f (x) is an odd function, then f (x) = - f (- x) is introduced to have
Asin (Wx + b) = - asin (b-wx) = asin (WX-B) [because SiNx itself is an odd function]
So if we make asin (Wx + b) = asin (WX-B)
Then we get (Wx + b) - (WX-B) = 2K π
So if B = k π, then f (x) is odd
(2) If f (x) is an even function, then f (x) = f (- x) is introduced to have
Asin(wx+b)=Asin(b-wx)
Then we get (Wx + b) + (b-wx) = 2K π + π
So if B = k π + π / 2, then f (x) is even
If you don't understand, you can ask questions

How to find the three coefficients in y = asin (Wx + b) How to get the three coefficients of the analytic formula of this function according to the image of trigonometric function?

Find a according to the maximum value, find B according to x = 0, that is, find B at the intersection of Y axis, and find w according to the period

If the image of the function f (x) = sin (Wx + π / 3) is shifted to the right by π / 3 units and is symmetric with the image of the original function about the x-axis, then the minimum positive value of W is A1/2 B.1 C.2 D.3

The image with F (x) = sin (Wx + π / 3) shifts π / 3 units to the right
The image of the function and the image of the original function are symmetric about the x-axis
Then the translation is (2k + 1) * t / 2, K ∈ n [odd times of half period]
That is (2k + 1) * (π / W) = π / 3
∴w=3(2k+1) (k∈N)
When k = 0, the minimum positive value of W is 3
Choose D

If the image of the function f (x) is shifted to the right by 1 unit length, the resulting image is symmetrical to the curve y = ex with respect to the y-axis, then f (x) =? The x power of the curve y = e

By changing x into - X and y-axis symmetry, the resulting image is y = e ^ (- x) and shifted to the left by 1 unit to obtain f (x), and left plus f (x) = e ^ (- (x + 1)) = e ^ (- x-1)

The translation of the image of function [y = asin (ω x + φ)] By extending the abscissa of each point on the image of function y = sin (6x + π 4) to three times of the original value, and then shifting π 8 units to the right, a symmetry center of the function is obtained () A．(π/2,0) B．(π/4,0) C．(π/9,0) D．(π/16,0)

By extending the abscissa of each point on the image of function y = sin (6x + π / 4) to three times of the original, the image of y = sin (2x + π / 4) is obtained;
Then shift π / 8 units to the right, and the image of function y = sin [2 (x - π / 8) + π / 4] = sin2x is obtained. One of its symmetry centers is a

The function s = asin (ω T + φ) (a > 0, ω > 0) represents a vibration quantity. If the amplitude is 2, the frequency is 3 / 2 π, and the initial phase is π / 12, then this function is It's about the process

Answer: S = 2Sin (3T + π / 120)
A is the amplitude, a = 2
F is the frequency (converted into radians) ω = 2 π f = 3 / 2 π * 2 π
φ is the primary phase, and φ = π / 12
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If the image of the function y = asin (Wx + R) (R is FAI) (W > 0) and the absolute value Fai ≤ π / 2 is shown in the figure, then an expression of the function is A.y=-4sin(π／8+π／4） B. Y = 4sin (π divided by 8 - π divided by 4) C. y = - 4sin (π divided by 8 - π divided by 4) d. 4sin (π divided by 8 + π divided by 4) the image is similar to y = SiNx (- 2,0) (6,0) (2, - 4) (12,4) w = π divided by 8. T = 16. A = ± 4. How to calculate a if it is equal to 4fai can't be illustrated

According to the image, a and W can be obtained
Then we put a point in and we get R
Here a = - 4, w = π / 8
(6,0)
Then 0 = sin (3 π / 4 + R)
3π/4+r=kπ
r=kπ-3π/4
|r|<=π/2
So k = 1
r=π/4