Find the value range of the following function: y = cosx / (2cosx + 1)

Find the value range of the following function: y = cosx / (2cosx + 1)

Y = cosx / (2cosx + 1) = (cosx + (1 / 2) - (1 / 2)) / (2cosx + 1) = (1 / 2) - (1 / (4cosx + 2)) - 1 / (4cosx + 2) has no maximum or minimum value, because 4cosx + 2 can approach 0, but it is certain that - 1 / (4cosx + 2) cannot be equal to 0, because neither denominator is 0, so the range of Y is (- ∞, 1 / 2) ∪

Find the value range of the function y = 3 + 2Sin ^ 2x-2sinx

Since the SiNx range is [- 1,1], assume t = SiNx,
The function becomes y = 2T ^ 2-2t + 3, where t is in the range of [- 1,1],
According to the value range method of quadratic function, its symmetry axis is 1 / 2, which is in the interval, so the minimum value is taken at 1 / 2,
Take the maximum value at - 1, so the value range is: [5 / 2,7]

Find the value range of the function f (x) = - 2Sin ^ x + 2sinx + 1

If the function is f (x) = - 2sinx + 2sinx + 1
Then let a = SiNx, - 1 ≤ a ≤ 1
f（x）=-2a²+2a+1=-2（a²-a）+1=-2（a-1/2）²+1+1/2=-2（a-1/2）²+3/2
This is a quadratic function with vertices (1 / 2,3 / 2)
If the coefficient of quadratic term is less than 0, then a = 1 / 2, that is, when SiNx = 1 / 2, the maximum value of F (x) is 3 / 2
When a = - 1, that is, SiNx = - 1, the minimum value of F (x) = - 3, because a is monotonically increasing on [- 1,1 / 2]
Range [- 3,3 / 2]

Find the value range of the function y = 1-2sin ^ 2x + cosx Good integral, big deal!

y=1-2(1-cos^2x)+cosx
=2cos^2x+cosx-1
Let t = cosx
Then y = 2T ^ 2 + T-1
And - 1 = t = - 1 / 4 belongs to [- 1,1]
Then y = 2T ^ 2 + T-1
The minimum value is - 9 / 8
The maximum value is 2
Then the range is [- 9 / 8,2]

The value range of function y = LG 2 (x 2 + 2) - LG (x 2 + 2) 2 + 3

Y
= lg²(x² + 2) - lg(x² ﹢ 2)² + 3
= lg²(x² + 2) - 2lg(x² ﹢ 2) + 3
= [lg(x² + 2) - 1]² + 2
Because x 2 + 2 ≥ 2
So LG (x? 2 + 2) ≥ LG2
So [LG (x 2 + 2) - 1] 2 ≥ 0
So [LG (x 2 + 2) - 1] 2 + 2 ≥ 2
So the range is [2, + ∞)

Given the function y = LG (2cosx + 1), find its value range

(-∞,lg3]

What is the range of y = LG (x2 + 1) and y = LG (x? - 1)

x²+1≥1
So y = LG (x? 2 + 1) ≥ LG1 = 0
So the range is [0, + ∞)
x²-1≥-1
So the range of y = LG (x? - 1) is r

Find the value range of the function y = 2cosx + 1 / 2cosx-1

[method 1]
Because y = (2cosx + 1) / (2cosx-1)
Therefore, y (2cosx-1) = 2cosx + 1
Therefore, cosx = (y + 1) / (2y-2)
Because - 1 ≤ cosx = (y + 1) / (2y-2) ≤ 1
Therefore: (1) (y + 1) / (2y-2) ≤ 1, y ≥ 3 or Y < 1
(2) (y + 1) / (2y-2) ≥ - 1, Y > 1 or Y ≤ 1 / 3
Therefore, the value range of the function is: {y | y ≥ 3 or Y ≤ 1 / 3}
[method 2]
y=(2cosx-1+2)/(2cosx-1)=1+2/(2cosx-1),
-1

If π / 4 ≤ π ≤ π / 3, what is the range of the function y = [2Sin (x + π / 6)] / cosx?

y=2(sinxcosπ/6+cosxsinπ/6)/cosx
=2sinxcosπ/6/cosx+2cosxsinπ/6/cosx
=√3tanx+1
π/4≤x≤π/3
tanπ/4≤tanx≤tanπ/3
√3≤√3tanx≤3
√3+1≤√3tanx+1≤4
Range [√ 3 + 1,4]

Function y = 2cosx + 1 The value range of 2cosx − 1 is______ .

Solution 1: the original function is transformed into y = 1 + 2
2cosx−1，
∵|cosx|≤1，
It can be directly obtained that y ≥ 3 or Y ≤ 1
3．
Then the value range of the function is (- ∞, 1)
3]∪[3，+∞）．
Solution 1: the original function is transformed into cosx = y + 1
2(y−1)，
∵|cosx|≤1，∴|y+1
2(y−1)|≤1，
ν y ≥ 3 or Y ≤ 1
3．
Then the value range of the function is (- ∞, 1)
3]∪[3，+∞）．
So the answer is: (- ∞, 1
3]∪[3，+∞）．