In the intersection of the image of the function y = - 2Sin (4x + two-thirds) and the x-axis, what is the closest point to the origin?

In the intersection of the image of the function y = - 2Sin (4x + two-thirds) and the x-axis, what is the closest point to the origin?

-2sin(4x+2π/3)=0
sin(4x+2π/3)=0
4x+2π/3=kπ
x=kπ/4-π/6
k=0,x=-π/6
k=1,x=π/12
So when k = 1, it's the closest
So it's (π / 12,0)

The point nearest to the origin is ()
A. （π12，0）B. （-π12，0）C. （-π6，0）D. （π6，0）

From 4x + 2 π 3 = k π (K ∈ z), we can get x = k π 4 - π 6. When k = 0, we can get point a (- π 6, 0). When k = 1, we can get point (π 12, 0). It is obvious that the nearest point between the image of π 12 ＜ π 6 and the x-axis intersection of the function y = - 52sin (4x + 2 π 3) is (π 12, 0)

In the intersection of the image of the function y = - 2Sin (4x + two thirds pie) and the X axis, what is the coordinate of the point nearest to the origin?

From y = Sint image, we know that the period of the intersection of function and X is pie
Then the image period of the function y = - 2Sin (4x + two-thirds pie) is Pie / 4
Then when x = Pie / 12, the intersection of the image and X axis is closest to the origin

Among the intersections of the image of the function y = - 5 / 2Sin (4x + 2 / 3 Π) and the X axis, the closest point to the origin is () a (- Π / 6,0) B (- Π / 12)

Find the intersection point of X axis, that is y = 0, that is sin (4x + 2pai \ \ 3) = 0, that is 4x + 2pai \ \ 3 = 2kpai, k = 0, so the point (- Pai \ \ 6,0) is closest to the origin

Let f (x) = ex / 1 + AX2, where a is a positive real number
Let f (x) = ex / 1 + AX2, where a is a positive real number
(1) When a = 4 / 3, find the extremum of F (x)
(2) If f (x) is a monotone function on R, the value range of a is obtained

(1) If f '(x) = e ^ x {1 + (4 / 3) x ^ 2 - (8 / 3) x} / {1 + (4 / 3) x ^ 2} ^ 2, then x = 0.5 or 1.50 and x = 0.5 or 1.5, then the extreme point is x = 0.5 or 1.5 (2) f' (x) = e ^ x (AX ^ 2-2ax + 1) / (1 + ax ^ 2) ^ 2, because it is a monotone function, so long as ax ^ 2-2ax + 1 is constant or greater than 0

Let a be a real number, f (x) = a - (2 / 2 x power + 1)
Try to determine the value of a so that f (x) is an odd function

F (x) = - f (- x) f (x) = a - (x power of 2 / 2 + 1)
=-F (- x) = (2 / 2 to the power of - x + 1) - A
=(x of 2 + 1 / X of 2 + 1) - A
We can get a = 1

The interval of solutions of all real numbers with F (x) = 5 to the x power plus x minus 3

In the method of combination of number and shape, Let f (x) = 5 to the x power plus x minus 3 = 0, and get 5 to the x power = 3-x
In the same coordinate system, we can draw the x power of 5 and the image of 3-x

If f (x) = (x power of a · 2 + A-2) / (x power of 2 + 1) is an odd function, what is the value of real number a

If f (x) = (x power of a · 2 + A-2) / (x power of 2 + 1), then f (- x) = (x power of a · 2 + A-2) / (x power of 2 + 1) = [(A-2) · x power of 2 + A] / (x power of 2 + 1) f (x) is an odd function, so f (- x) = - f (x) is [(A-2) · x power of 2 + a] / (x power of 2 + 1) = - (x power of a · 2 + A-2) / (x power of 2 + 1), then a = 2-A, If it is a choice or fill in the blank, we only need to use a simple verification method, that is, if f (x) is an odd function and is meaningful when x = 0, then the function image must pass through the origin, then a = 1 is obtained from F (0) = 0

If f (x plus 1 / 2 of x) = the square of X divided by (the fourth power of x plus 1), then f (x)=
I forgot how to solve it
I hope to write out the specific steps and ideas of solving the problem, as well as the error prone links that should be paid attention to,

f(x+1/x)=x^2/(x^4+1)=1/(x^2+1/x^2)=1/[(x+1/x)^2-2]
So f (x) = 1 / (x ^ 2-2) is the domain of the function
The domain of this function should be the value range of X + 1 / x, obviously x > = 2 or X

Let f (x) = the second power of X divided by the sum of the squares of L + X

1. F (1-x) = 4 ^ 1-x / (4 ^ 1-x + 2) = 4 / (4 + 2 * 4 ^ x) = 2 / (4 ^ x + 2), so f (x) + F (1-x) = 1
2.f(1/101)+f(2/101)+f(3/101)+------+f(99/101)+f(100/101)=50[f(1/101)+f(1/100)]=50