The range of the function y = 5-6cosx-sin & sup2; X is Writing process

The range of the function y = 5-6cosx-sin & sup2; X is Writing process

y=5-6cosx-sin²x
=5-6cosx-(1-cos²x)
=cos²x-6cosx+4
=(cosx-3)²-5
∵-1≤cosx≤1
When cosx = 1, y has a minimum value of - 1
When cosx = - 1, the maximum value of Y is 11,
The value range is [- 1,11]

The range of the function f (x) = sin2x-3cosx (x ∈ [0, π]) is______ ．

∵ f (x) = sin2x-3cosx = 1-cos2x-3cosx = - (cosx + 32) 2 + 74, ∵ x ∈ [0, π], ∵ 1 ≤ cosx ≤ 1, ∵ when cosx = 1, f (x) gets the minimum value, that is, f (x) min = - 3; when cosx = - 32, f (x) gets the maximum value, f (x) max = 74; ∵ the range of function f (x) = sin2x-3cosx (x ∈ [0, π]) is [- 3, 74]. So the answer is: [- 3, 74]

Let the graph of function y = f (x) = (1-3x) / (X-2) and function y = g (x) be symmetric with respect to the straight line y = x, then what is g (x) equal to?

If f (x) = (1-3x) / (X-2), and G (x) are symmetric with respect to y = x, then they are inverse functions of each other
Y * (X-2) = yx-2y = 1-3x, then x = (1 + 2Y) / (y + 3)
The position to be changed is: y = (1 + 2x) / (x + 3)
So g (x) = (1 + 2x) / (x + 3)

Let the image of the functions y = ax + 2 and y = 3x-b be symmetric with respect to the line y = x, then what are a and B equal to respectively,

Because the function y = ax + 2 passes through the point (0,2) by the meaning y = 3x-b passes through the point (2,0), so 0 = 6-b B = 6, so y = 3x-6 because the function y = 3x-6 passes through the point (1, - 3), so the function y = ax + 2 passes through the point (- 3,1), so 1 = - 3A + 2A = 1 / 3

The known function f (x) = sin (ω x + φ x) (ω > 0, | φ)|
Wrong number. There's no X behind it

Do you have the wrong number Phi? Shouldn't there be an X after it
Because the distance between a center of symmetry of a function and the nearest axis of symmetry is π / 4
So the period T = π
ω=2
Image passing point (- π / 6,0) of F (x) = sin (2x + φ) f (x)
Substituting sin (- π / 3 + φ) = 0
-π / 3 + φ = KxK is a positive integer
|φ|

The function f (x) = sin (ω x + φ) (ω > 0, | φ) is known|

(1) T / 4 = Π / 4, t = Π = 2 Π / ω, ω = 2 F (x) = sin (2x + Φ)
0=sin(2*(-Π/6)+Φ)=sin(Φ-Π/3)
Φ - Π / 3 = k Π, Φ = k Π + Π / 3, k = 0, Φ = Π / 3
So f(x)=sin(2x+Π/3)
2kΠ-Π/2

If the image of function sin (Wx + V) is only shifted 3 / 4Pi to the right or 2 / 3pi to the left, the image obtained is symmetrical about the origin, then how much is w
(the value range of V is - pi / 2 to pi / 2)

Shift 3 / 4 π to the right and get sin (Wx + v-3 / 4 * π * W)
Shift 2 / 3 π to the left and get sin (Wx + V + 2 / 3 * π * W)
When the image is symmetrical about the origin, it is deduced
V-3 / 4 * π * w = k π and V + 2 / 3 * π * w = k π
V = 3 / 4 * π * W + K π and V = k π - 2 / 3 * π * W
-2/3-4/3*k

The known function f (x) = sin (ω x + φ) (ω > 0, - π / 2)

f(x)=sin(2x+φ)(|φ|

A:
F (x) = sin (2x + φ) passing point (0, - √ 3 / 2)
f(0)=sinφ=-√3/2
Because: | φ|

Given that the image of the function f (x) = 2Sin (ω x + φ) (ω > 0) is symmetric with respect to the straight line x = π 3 and f (π 12) = 0, then the minimum value of ω is ()
A. 2B. 4C. 6D. 8

∵ π 3 - π 12 = π 4 = T4, ∵ t = π, ∵ ω = 2