Move the factor outside the root sign into the root sign: a root sign minus 1 / A fast

Move the factor outside the root sign into the root sign: a root sign minus 1 / A fast

a√(-1/a)
Greater than or equal to 0 in root sign
So - 1 / a > = 0
So a < 0
So a √ (- 1 / a) < 0
So a √ (- 1 / a) = - √ [a ^ 2 (- 1 / a)]
=-√(-a)

It is known that f (x) = sin2x radical 3cos2x + 1 (1) Find the maximum and minimum value (2) If the inequality | f (x) - M belongs to [Pai / 4, Pai / 2] |, find the value range of M

First, the deformation is in one;
1.f(x)=sin2x-√3*cos2x+1
=2sin(2x-π/3)+1
So,
When 2x - π / 3 = π / 2 + 2K π, there is a maximum
When x = 5 π / 12 + 2K π, the maximum value is 3
When 2x - π / 3 = - π / 2 + 2K π, there is a minimum
When x = - π / 6 + 2K π, the minimum value - 1 is obtained
Because of π / 4

Function f (x) = LG (sin2x)+ The domain of 3cos2x-1) is______ .

To make a function meaningful, you need to
sin2x+
3cos2x−1＞0
That is sin (2x + π)
3)＞1
Two
So 2K π + π
6＜2x+π
3≤2kπ+5π
Six
The solution is {x | K π − π
12＜x＜kπ+π
4，k∈Z}
So the answer is {x| K π − π
12＜x＜kπ+π
4，k∈Z}

Find the value range of the function y = x / root (x square + 2x + 2)?

① When x = 0, y = 0
② When x ∈ (0, + ∞)
y=x/√(x²+2x+2)
=1/√（2/x²+2/x+1）
=1/√[2(1/x+1/2)²+1/2]
∵(1/x+1/2)²∈（1/4,+∞）
∴2(1/x+1/2)²+1/2∈（1,+∞）
∴√[2(1/x+1/2)²+1/2]∈（1,+∞）
y∈(0,1)
③ When x ∈ (- ∞, 0)
y=-x/√(x²+2x+2)
=-1/√（2/x²+2/x+1）
=-1/√[2(1/x+1/2)²+1/2]
∵(1/x+1/2)²∈（0,+∞）
∴2(1/x+1/2)²+1/2∈（1/2,+∞）
∴√[2(1/x+1/2)²+1/2]∈（√2/2,+∞）
∴y∈（-√2,0）
To sum up, y ∈ (- √ 2,1)

Using the method of substitution, we can find the following function value range y = x-radical (2x-1),

a=√(2x-1)
So a > = 0
2x-1=a²
x=(a²+1)/2
y=a²/2+1/2-a
=1/2(a-1)²
So a = 1, the minimum value is 0
So the range is [0, + ∞)

To find the value range of function y = x + radical 1 + 2x by substitution method

y=x+√(1+2x)
Let t = √ (1 + 2x) > = 0
Then x = (t? - 1) / 2
y=(t²-1)/2+t=(t²+2t-1)/2=1/2*(t+1)²-1
Because t > = 0, y increases monotonically with respect to t, and the minimum value is obtained when t = 0, which is y = - 1 / 2
X = - 1 / 2

The value range of (1-x ^ 2) under the function y = x-radical is [- radical 2,1]

Firstly, the definition domain is obtained, and then the value domain is obtained by using the trigonometric transformation method (note that the range of the angle in the trigonometric function should conform to the definition domain of the original function!). Let x = cos β, then √ (1-x ^ 2) = sin β [where β ∈ [2K π, 2K π + π]] (i.e., ensure that √ (1-x ^ 2) = sin β is not less than 0), then y = cos β - sin β =

The value range of 16-4 ^ x under y = radical is

∵ 4^x>0
∴ 0

What is the value range of the function f (x) = 2 + 1 under the radical sign and x ^ 2-2x + 3 under the radical sign? Notice there's another one\

First find the value range of x ^ 2-2x + 3
x^2-2x+3=(x-1)^2+2
1) The minimum value is: when x = 1, 2; at this time, the maximum value of x ^ 2-2x + 3 under the root sign of 1 is √ 2 / 2, and the maximum value of F (x) is 3 √ 2 / 2;
2) In this case, the minimum value of x ^ 2-2x + 3 is 0, and the minimum value of F (x) is √ 2
The value range of F (x) is (√ 2,3 √ 2 / 2)

It is known that the definition domain of the function f (x) = Radix 2Sin (2x + π / 4) is [0, π / 2], and its value range is

The domain of F (x) is [0, π / 2]
π/4≤2x+π/4≤5π/4
-√2/2≤sin(2x+π/4)≤1
The value range of function f (x) is [- 1, √ 2]