The monotone decreasing interval of - x 2 + 2x + 3 under the function y = radical is

The monotone decreasing interval of - x 2 + 2x + 3 under the function y = radical is

Y = under radical - x 2 + 2x + 3
-x²+2x+3=-(x-1)²+4
x≥1
-x²+2x+3≥0
x²-2x-3≤0
(x-3)(x+1)≤0
-1≤x≤3
So, 1 ≤ x ≤ 3
Therefore, the monotone decreasing interval is x ∈ [1,3]

Find the continuous interval y = radical x? - 3x + 2

By y = √ (x? - 3x + 2)
There are x? - 3x + 2 ≥ 0,
（x-1)（x-2)≥0,
The solution is x ≥ 1, X ≥ 2, X ≥ 2,
Or X-1 ≤ 0, X-2 ≤ 0,
X ≤ 1, X ≤ 2, X ≤ 1,
The continuous interval of function y (- ∞, 1] ∪ [2, + ∞)

The monotone interval of the function y = - 3x / (2x + 1) is the monotone recurrence of (x ^ 2-3x + 2) under the radical sign of function y = 1 The monotone interval of the function y = - 3x / (2x + 1) is The monotone decreasing interval of (x ^ 2-3x + 2) under the function y = 1 / radical is Detailed process... A tutorial is best Thank you

1. The monotone interval of the function y = - 3x / (2x + 1) is
Definition domain: X ≠ - 1 / 2
The monotone interval of Y '= [- 3 (2x + 1) + 6x] / (2x + 1) 2 = - 3 / (2x + 1) 2 < 0 holds for any X in the defined domain
(- ∞, - 1 / 2) ∪ (- 1 / 2, + ∞) are monotonically decreasing in both intervals. X = - 1 / 2 is its vertical asymptote
2. The monotone decreasing interval of the function y = 1 / √ (x? - 3x + 2) is
y=1/√(x²-3x+2)=1/√(x-1)(x-2)
Definition domain: (x-1) (X-2) > 0, that is, x < 1 or x > 2 is its definition domain
Let y '= [- (2x-3) / 2 √ (x? - 3x + 2)] / (x? - 3x + 2) = - (2x-3) / [2 (x? - 3x + 2) ^ (3 / 2)] = 0, the stagnation point x = 3 / 2 (outside the definition domain)
When x ≤ 3 / 2, y '≥ 0; when x ≥ 3 / 2, y' ≤ 0; therefore, the function is monotonically reduced in the interval (2, + ∞)

How to find the monotone Zeng interval of the square of the root of a function x minus 3x + 2

Let the radical be t
The original formula can be written as √ T, and the curve can be drawn. When t ≥ 0, the equation increases monotonically
Then x ^ 2-3x + 2 = (x-3 / 2) ^ 2-1 / 4 ≥ 0
(x-3/2)^2≥1/4
Then x-3 / 2 ≥ 1 / 2 or x-3 / 2 ≤ 1 / 2
That is, X ≥ 2 or X ≤ - 1

Find the monotone interval of function f (x) = radical (2x squared - 3x + 1)

F (x) = under the radical ((x-1) (2x-1)), define the field (x-1) (2x-1) "0
That is, X is less than or equal to 1 / 2 and greater than or equal to 1, monotone interval, decreasing interval (negative infinity, 1 / 2), increasing interval [1, positive interval]

The definition domain of 1-x under the root sign of function y = 2x square-3x-2 is

Greater than or equal to 0 under root sign
1-x≥0
x≤1
Denominator is not equal to 0
2x²-3x-2≠0
(2x+1)(x-2)≠0
x≠-1/2,x≠2
So the domain (- ∞, - 1 / 2) ∪ (- 1 / 2,1]

The monotone increasing interval of - x ^ 2 + 3x-2 is

F (x) = the monotone increasing interval of - x ^ 2 + 3x-2 is
Axis of symmetry = 3 / 2
Satisfy - x ^ 2 + 3x-2 ≥ 0
x^2-3x+2

The monotone interval procedure of the function f (x) = 5sinxcosx-5 times the root sign 3cos ^ 2x

(x) = 5sinxcosx-5 times root 3cos ^ 2x = (5 / 2) sin2x - (5 / 2) root 3 (cos2x + 1) = 5 sin (2x + Pai / 3) - (5 / 2) root sign 3 monotonically increasing interval Pie / 2 + 2K pie < = 2x + Pie / 3 < = Pie / 2 + 2K pie, K is an integer, solve x from above, solve by yourself

Let f (x) = sin (x / 2) cos (x / 2) + radical 3cos ^ (x / 2). If the three sides of a triangle ABC satisfy a, B, C satisfy B ^ 2 = AC and the angle of edge B is X Find the value range of angle X and the value range of function f (x) at this time

f(x)=sin(x/2)cos(x/2)+√3(cos(x/2))^2
=(1/2)sinx+√3(1+cosx)/2
=√3/2+sin(x+π/3)
According to the cosine theorem:
cosx=(a^2+c^2-b^2)/(2ac)
=(a^2+c^2-ac)/(2ac)
≧(2ac-ac)/(2ac)
That is, cosx ≥ 1 / 2
X is the inner angle of the triangle, so the range of X is:
0＜x≦π/3
The monotone increasing range of F (x) is 0 < x ≤ π / 6
When x = 0, f (x) = √ 3
When x = π / 6, f (x) = 1 + √ 3 / 2
So the range of F (x) is as follows:
√3＜f(x)≦1+√3/2

The symmetric axis of the function f (x) = 2Sin (Wx + π / 4) (W > 0) is exactly the same as that of the function g (x) = cos (2x + φ) (│φ≤ π / 2), Then what is the value of φ,

The image symmetry axis of W = 2; f (x) = 2Sin (2x + π / 4) is: 2x + π / 4 = k π + π / 2; that is, 2x = k π + π / 4; and the image symmetry axis of G (x) = cos (2x + φ) is: 2x + φ = k; that is, 2x = k π - φ symmetry axes are exactly the same