Given the function y = sin (2x) - 8 (SiNx + cosx) + 19 (0 < = x < = π), find the maximum and minimum value of function y

Given the function y = sin (2x) - 8 (SiNx + cosx) + 19 (0 < = x < = π), find the maximum and minimum value of function y

Let t = SiNx + cosx = (√ 2) SAIN (x + π / 4) (0 < = x < = π)
π/4<=x+π/4<=5π/4,
t∈[-1,√2],
sin2x=t^2-1,
∴y=t^2-8t+18=(t-4)^2+2,↓,
When t = √ 2, the minimum value of Y is 20-8 √ 2,
When t = - 1, the maximum value of Y is 27

The function y = sin ^ 2-2asinx + 1 + A ^ 2 obtains the maximum value when x = 2K π + π / 2 (k belongs to Z), and gets the minimum value when SiNx = a, and calculates the value range of real number a

y=(sinx-a)²+1
The opening is upward and the axis of symmetry SiNx = a
SiNx = a has a minimum
-1

If the square of function y = sinx-2asinx + 1 = a is taken as the maximum value when SiNx = - 1. When SiNx = a, the minimum value is taken, then the value range of a is（

y=(sinx-a)²+1
SiNx = a has a minimum
That is, SiNx = a can be obtained
So - 1 < = a < = 1
Sina = - 1 is the largest
Because the farther SiNx is away from the axis of symmetry SiNx = a, the larger the function value
So - 1 is farther from a than 1
So a - (- 1) > 1-A
A>0
So 0
Homework help users 2017-09-17
report

The function f (x) = - Sin ^ 2x + SiNx + A, for any x belongs to R, 1 = solution`

F(X)=-SIN^2X+SINX+A
=-(sinx-1/2)^2+A+1/4
Because: - 1

Given the function y = - Sin ^ 2x + SiNx + A, if 1 ≤ y ≤ 4 holds for all x ∈ R. find the value range of real number a

f(x)=-(sinx-1/2)^2+1/4+a
-1

If the function y = sin? X-2asinx + 2 + a? Obtains the maximum value at SiNx = - 1 and the minimum value at SiNx = a, then the value range of a is

A:
Y = sin? X-2asinx + 2 + a? Has the maximum value at SiNx = - 1 and the minimum value at SiNx = a
=(sinx-a)²+a²+2
The maximum value is obtained when SiNx = - 1
The minimum value is obtained when SiNx = a
The symmetry axis SiNx = a is closer to SiNx = 1
So: 0

1. The difference between the maximum and the minimum of the function y = sin ^ 2 x-sinx + 4 is () A.2 B.7/4 C.9/4 D.11/4

Let SiNx = t, - 1 ≤ t ≤ 1y = t? - t + 4 = (t? - t + 1 / 4) + 15 / 4 = (t-1 / 2) 2 + 15 / 4, the image is a parabola with the opening up and the symmetry axis t = 1 / 2. Because - 1 ≤ t ≤ 1, when t = 1 / 2, y (min) = 15 / 4, when t = - 1, y (max) = 6y (max) - Y (min) = 6-15 / 4 = 9 / 4

Find the maximum and minimum values, and obtain the maximum and minimum values of X by the function (1) y = (sinx-3 / 2) ^ 2-2 (2) y = - Sin ^ 2x + √ 3sinx + 5 / 4

(1) When (sinx-3 / 2) ^ 2 has the maximum, y max = (25 / 4) - 2 = 17 / 4x = - (25 / 4) - 2 = 17 / 4x = - (25 / 4) - 2 = 17 / 4x = - Π / 2 + 2K Π, K ∈ Z (Π is the rate of the circle) SiNx = 1, when the SiNx = 1, the minimum value of Y is the minimum value = - 7 / 4x = / 2 + 2 + 2K Π, K ∈ Z (2) y = (SiNx + 3 / 3 / 2) ^ 2 + 1 / 2sinx = - 3 / 2, when the Y minimum value = 1 / 2x = 4 / 3 + 3 + 2K Π or 5 Π / 3 + 3 + 3 + K Π, K

The function y = (sinx-a) ^ 2 + 1 has a minimum value when SiNx = A and a maximum value when SiNx = 1? The answer is [- 1,0]

Let SiNx = t, - 1

The difference between the maximum value minus the minimum value of the function y = sin ^ 2 * x-sinx + 4 is

y=(sinx-1/2)^2+15/4
-1