Given the function f (x) = 2Sin (2x + π / 6), if f (x0) = 2, find the set of values composed of all x0 that satisfy the condition

Given the function f (x) = 2Sin (2x + π / 6), if f (x0) = 2, find the set of values composed of all x0 that satisfy the condition

f(x0)=2sin(2x0+π/6)=2
sin(2x0+π/6)=1
2x0+π/6=π/2+2kπ,k∈Z
∴x0=π/6+kπ,k∈Z
{x0|x0=π/6+kπ,k∈Z}

Known set a = {x | x + 1 / 2x-1 ≤ 2}, B = {a | known function f (x) = A / X - 1 + LNX
&#If f (x0) ≤ 0}, then a ∩ B=

It can be calculated that a = {1 / 2}

Given the function f (x) = LNX, G (x) = ax, Let f (x) = f (x) + G (x). (I) when a = 1, find the monotone interval of function f (x); (II) if the tangent slope k ≤ 12 of any point P (x0, Y0) on the image of function y = f (x) (0 < x ≤ 3) is constant, find the minimum value of real number a

(I) given a = 1, f (x) = f (x) + G (x) = LNX + LX, the domain of function is (0, + ∞), then f ′ (x) = LX − 1x2 = x − 1x2, from F ′ (x) = LX − 1x2 = x − 1x2 > 0, f (x) monotonically increases in the interval (1, + ∞), and f ′ (x) = LX − 1x2 = x − 1x2 < 0, f (x) monotonically decreases in the interval (0, 1); (II) from the meaning of the question, k = f ′ (x0) = x0 − ax20 ≤ 12 Let t = x0 − 12x20 = − 12 (X20 − 2x0) = − 12 (x0 − 1) 2 + 12 ≤ 12, then a ≥ 12, that is, the minimum value of real number a is 12

Given the points a (1, ya), B (- radical 2, Yb), C (- 2, YC), on the image of function y = 2 (x + 1) 2 - (1 / 2), then the size relationship of ya, Yb, YC is

y=2(x+1)^2-1/2
y=2(x^2+2x+1)-1/2
=2x^2+4x+1/2
According to the vertex formula, the curve is a parabola with an opening upward, and the minimum value is
When x = - B / 2a, that is, when x = - 1, y is the smallest, and the closer to x = - 1, the smaller y is
therefore
yb

Given that the vertex of parabola y = (x + a)? + 2A? + 3a-5 is on the coordinate axis, find the value of letter A, and point out the vertex coordinates

Do you know where the vertex coordinates of the image y = x are? It should be at the origin! The vertex coordinates of the image y = (x + a) should be translating the image y = x to the left or right by | a | units, so the image y = (x + a) should also be on the X axis, and the image y = (x + a) + 2A + 3a-5 is also on the X axis, so 2A + 3a-5 = 0

It is known that the vertex of parabola y = (x + a) ² + 2A & #178; + 3a-5 is on the coordinate axis. Find the value of letter A and point out the vertex coordinates

The vertex coordinates of y = (x + a) &# 178; + 2A & # 178; + 3a-5 are (- A, 2A & # 178; + 3a-5). There are two cases where the known vertex is on the coordinate axis: on the X axis, 2A & # 178; + 3a-5 = 0, a = - 5 / 3 or a = 1. When a = - 5 / 3, the vertex coordinates (- 5 / 3,0) when a = 1, the vertex coordinates (1,0) on the Y axis, a = 0, the vertex sits

Let a > 0, when x belongs to [- 1,1], the function f (x) = - x ^ 2-ax + B has a minimum value of - 1 and a maximum value of 1
Make the maximum and minimum values of the function be the corresponding values of X

No point
Let's talk about it briefly
f(x)=-x^2-ax+b
=-（x^2+ax-b）
The minimum value of x ^ 2 + ax-b is obtained at - A / 2
The maximum value of F (x) is obtained at - A / 2. By substituting x = - A / 2, y = 1 into the equation, an equation about a and B is obtained
a>0,-a/2

Given the function f (x) = ax & # 178; + BX + C (a > 0, B ∈ R, C ∈ R), if the minimum value of F (x) is f (- 1) = 0, and f (0) = 1, the axis of symmetry is x = - 1
(1) Find the analytic expression of F (x);
(2) Under the condition of (1), find the minimum value of F (x) in the interval [T, t + 2] (t ∈ R)

1) If the minimum value is f (- 1) = 0, then f (x) = a (x + 1) ^ 2
f(0)=a(0+1)^2=a=1
So f (x) = (x + 1) ^ 2 = x ^ 2 + 2x + 1
2) If the axis of symmetry x = - 1 is in the interval [T, t + 2], that is - 3=

Find the axis of symmetry of the function g (x) = f (x) - x, f (x) = ax & # 178; + BX + C

Compound function, the f (x) into the solution
G(x)=ax^2+(b-1)x+c
So, the axis of symmetry is (1-B) / 2A

The axis of symmetry of the parabola y = ax, square + BX + C is a straight line x = 2, and it passes through (0, - 1), (3, - 4). At this time, the analytic formula of the function is given
2. The vertex coordinates of parabola y = ax square + BX + C (- 2,3) and pass through (- 1,5), the analytic expression of this function is obtained
Urgent, urgent, thank you very much!

1. Let the parabola be y = a (X-2) ^ 2 + C
Substituting (0, - 1), (3, - 4) yields:
-1=a*(0-2)^2+c
-4=a*(3-2)^2+c
The solution is: a = 1, C = - 5
y=(x-2)^2-5=x^2-4x-1
2. Let the parabola be y = a (x + 2) ^ 2 + 3
Substituting (- 1,5) gives:
5=a*(-1+2)^2+3
The solution is a = 2
y=2(x+2)^2+3=2x^2+8x+11