Find the function y = 2Sin (x + 6 / π) + 3, X belongs to the maximum and minimum value of [0, π]

Find the function y = 2Sin (x + 6 / π) + 3, X belongs to the maximum and minimum value of [0, π]

The image of sin (x + 6 times π) is obtained by moving the image of SiNx horizontally π / 6 to the left;
Then, the abscissa of all points in the image of sin (x + 6 / π) is unchanged, and the ordinate is expanded to 2 times of the original; then the obtained image is translated up 3 units to get the image of function y = 2Sin (x + 6 / π) + 3. From the graph, on [0, π], when x = π / 3, the maximum value is: 5;
When x = π, there is a minimum value of 2
This is already the answer. The maximum value is 5 and the minimum value is 2

Find the maximum and minimum values of the function y = 2Sin (2x + π / 3) (- π / 6 ≤ x ≤ π / 6)

First - π / 6 ≤ x ≤ π / 6
So 0 ≤ 2x + π / 3 ≤ 2 π / 3
We also know that y = 2Sin φ
When ∈ [0, π / 2] increases monotonically
So when φ = π / 2, that is 2x + π / 3 = π / 2
The function takes the maximum value y = 2
When ∈ [π / 2,2 π / 3] decreases monotonically
But when φ = 0, that is 2x + π / 3 = 0
The function takes the minimum value y = 0
Hope can help you, wish you study happily!

The minimum and maximum values of the function y = 2Sin (60 ° - x), X ∈ [π / 6,2 π / 3] are

x∈[π/6,2π/3]
therefore
-x∈[-2π/3,-π/6]
therefore
60°-x=π/3-x∈[-π/3,π/6]
So the minimum value of 2Sin (60 ° - x) is 2Sin (- π / 3) = - radical 3
The maximum value is 2Sin (π / 6) = 1

If 0 is less than or equal to X2, find the maximum and minimum values of the function y = 4 ^ X-1 / 2-3 * 2 ^ x + 5

Let t = 2 ^ x, 0

If x is greater than or equal to 0 and less than or equal to 2, find the maximum and minimum values of the function y = 4 ^ X-1 / 2 - 3 * 2 ^ x + 5

y=4^(x-1/2)-3*(2^x)+5
=[(2^x)²/2]-3*(2^x)+5
={(2^x)-6*(2^x)+10]/2
={[(2^x)-3]²+1}/2
∵0≤x≤2
∴1≤2^x≤4
When 2 ^ x = 3, Ymin = 1 / 2
When 2 ^ x = 1, ymax = 5 / 2

The function f (x) = 2Sin (1) is known 3x-π 6），x∈R． (1) Find the value of F (0); (2) Let α, β∈ [0, π 2]，f（3α+π 2）=10 13，f（3β+π 2）=6 5. Find the value of sin (α + β)

（1）f（0）=2sin（-π
6）=-1
（2）f（3α+π
2）=2sinα=10
13，f（3β+π
2）=2sinβ=6
5．
∴sinα=5
13，sinβ=3
Five
∵α，β∈[0，π
2]，
∴cosα=
1−25
169=12
13，cosβ=
1−9
25=4
Five
∴sin（α+β）=sinαcosβ+cosαsinβ=56
Sixty-five

Let the (2) f (2) of (2) f (2) be (2) f (2) f (2) f (2) be (2) f (2) f (2) f (2) f (2) f (2) be (2) f (2) and (2) be (2) f (2) and (2) be (2) f (2) and (2) be (2) f (2) and (2) be (2) f (2) and (2) f (2) respectively Given the function f (x) = 2Sin (1 / 3x - π / 6), X ∈ R (1), find the value of F (0); (2) Let α, β∈ [0, π / 2], f (3 α + π / 2) = 10 / 13, f (3 β + π / 2) = 6 / 5. Find the value of sin (α + β) 1.f(0)=2sin(-π/6)=2×(-1/2)=-1 2.f(3α+π/2)=2sin[1/3(3α+π/2)-π/6]=2sinα=10/13 sinα=5/13 α∈[0,π/2] cosα=12/13 f(3β+π/2)=2sin[1/3(3β+π/2)-π/6]=2sinβ=6/5 sinβ=3/5 β∈[0,π/2] cosβ=4/5 sin(α+β)=sinαcosβ+sinβcosα=5/13×4/ 5+12/13×3/5=56/65 2. F (3 α + π / 2) = 2Sin [1 / 3 (3 α + π / 2) - π / 6] = 2Sin α = 10 / 13 sin α = 5 / 13 α ∈ [0, π / 2] cos α = 12 / 13 f(3β+π/2)=2sin[1/3(3β+π/2)-π/6]=2sinβ=6/5 sinβ=3/5 β∈[0,π/2] cosβ=4/5 How is this part solved? What's the 1 / 3

f(x)=2sin(1/3x-π/6)
In F (3 α + π / 2), x = 3 α + π / 2
f(3α+π/2)=2sin[1/3(3α+π/2)-π/6]=2sin（α+π/6-π/6）=2sinα=10/13
sinα=5/13
α∈[0,π/2] cosα=12/13
f(3β+π/2)=2sin[1/3(3β+π/2)-π/6]=2sinβ=6/5
sinβ=3/5
β∈[0,π/2]
cosβ=4/5
sin(α+β)=sinαcosβ+sinβcosα=5/13×4/5+12/13×3/5=56/65

Let f (x) = 2Sin (x / 3-pai / 6), X belong to r-find the value of F (5 Pai / 4). Let a, B belong to [0, Pai / 2], f (3a + Pai / 2) = 10 / 13, f (3b + 2) = 6 / 5, and find the value of COS (a + b)

1、 F (5 Π / 4) = 2Sin (5 Π / 4 / 3 - Π / 6)
=2Sin (Π / 4) = radical 2
2、 F (3a + Π / 2) = 2Sin ((3a + Π / 2) / 3 - Π / 6) = 2Sin (a) = 10 / 13
sin(a)=5/13 cosa=12/13
f(3b+2Π)=2sin(（3b+2Π）/3-Π/6)=2sin(b+Π/2)=2cos(b)=6/5
cos(b)=3/5 sinb=4/5
cos(a+b)=cosa*cosb-sin(a)*sinb
=12/13*3/5+5/13*4/5
=56/65

The image of sine function y = SiNx is an axisymmetric graph, and its equation of symmetry axis is

The image of sine function y = SiNx is axisymmetric, and its equation of symmetry axis is x = k π + π / 2

Using the image of sine function, the definition domain of function y = LG (SiNx - (root 3 / 2)) is found

The problem of finding the definition domain is very common. Here is to find the true number of logarithmic function greater than 0
In other words, SiNx radical 3 / 2 > 0
Here is SiNx ＞√ 3 / 2
Then draw the image of y = SiNx, and then draw a horizontal line y = √ 3 / 2
After that, the upper part of the image is the domain you want to define as 2K π + π / 3