Find the range of F (x) = - 2cos ^ 2x-2sinx + 3~~~~~

Find the range of F (x) = - 2cos ^ 2x-2sinx + 3~~~~~

A:
f(x)=-2(cosx)^2-2sinx+3
=-2[1-(sinx)^2]-2sinx+3
=2(sinx)^2-2sinx-2+3
=2(sinx-1/2)^2+1/2
-1

Find the value range log2 as the bottom (- the square of X + 4x + 12)

∵-x²+4x+12=-(x-2)²+16≤16
∴log2(-x²+4x+12)≤log2(16)=4
The value range is (- ∞, 4]
Hope to adopt

Find the range of y = log2 (x ^ 2-4x-5)
Please give the specific process and explain why

The value field of log2 (x) is R and the definition field is r+
In other words, if the value of (x ^ 2-4x-5) can include all positive numbers, then the range of Y is r
F (x) = x ^ 2-4x-5 is a parabola with the opening upward. The minimum value is taken when x = 2, and the minimum value is - 9 < 0, so its value can take all positive numbers

Monotone decreasing interval of FX = log2 (x squared-4x-12)

(－∞,6）

How to find the range of y = Log1 / 2 (2x / x + 1) + 2

2x/x+1=(2x+2-2)/(x+1)=2-2/(x+1)≠2
therefore
y≠log1/2 2 +2
=-1+2
=1
So the range is (- ∞, 1) U (1, + ∞)

The range of function f (x) = Log & nbsp; 13 (2 + 2x-x2) is______ ．

Let t = 2 + 2x-x2 = - (x-1) 2 + 3 ≤ 3, ∵ function y = log13t monotonically decrease on (0, + ∞) ∵ log13 (2 + 2x-x2) ≥ log133 = - 1. So the range is [- 1, + ∞). So the answer is: [- 1, + ∞)

The range of function f (x) = - x2 + 2x + 3, X ∈ [- 2,2]

f(x)=-x²+2x+3
=-(x-1)²+4
Axis of symmetry x = 1 ∈ [- 2,2]
f(-2)=-5,f(1)=4,f(2)=3
The maximum value is f (1) = 4, and the minimum value is f (- 2) = - 5
So the range is [- 5,4]
I hope I can help you! Happy study_ ∩)O~】

Finding the range of the function g (x) = x2 + 2x + 3 / x + 1 on [1,2]

g(x)=(x^2+2x+3)/(x+1)=[(x+1)^2+2]/(x+1)=(x+1)+2/(x+1)
Because x ∈ [1,2]
Then x + 1 > 0
So g (x) = (x + 1) + 2 / (x + 1) ≥ 2 √ [(x + 1) * 2 / (x + 1)] = 2 √ 2
If and only if x + 1 = 2 / (x + 1), i.e. x = √ 2-1
Because x ∈ [1,2], it is obvious that the minimum value cannot be obtained
According to the characteristics of the double hook function, on the right side of the fixed point is the increasing function
That is to say, it is an increasing function on [1,2]
So the minimum value of G (x) is g (1) = 3, and the maximum value is g (2) = 11 / 3
If you don't understand, please hi me, I wish you a happy study!

The range of function FX = cos (x-3pai / 4) + sin2x is known

FX = cos (x-3 π / 4) + sin2x = - √ 2 / 2cosx - √ 2 / 2sinx + 2sinxcosx, let t = SiNx + cosx t belong to [- √ 2, √ 2] get 2sinxcosx = T ^ - 1, get y = - √ 2 / 2T + T ^ 2-1y, the symmetry axis is t = √ 2 / 4, so when t = √ 2 / 4, f (x) has the minimum value of - 9 / 8, when t = - √ 2, it has the maximum value of 2F (x)

Given the function FX = sin (2x + π / 6), where x ∈ [- π / 6, α], what is the range of FX when α = π / 3
Given the function FX = sin (2x + π / 6), where x ∈ [- π / 6, α],
《1》 What is the range of FX when α = π / 3
《2》 If the value range of FX is [- 1 / 2,1], what is the value range of real number α

1) A = π / 3, 2x + π / 6 ∈ [- π / 6,5 π / 6], thus f (x) ∈ [- 1 / 2,1];