As shown in the figure, it is known that y = f (x) is an even function defined on R. when x ≥ 0, f (x) = x2-2x (1) Find the values of F (1) and f (- 2); (2) Find the analytic formula of F (x) and draw a diagram; (3) The root of equation f (x) = k is discussed

As shown in the figure, it is known that y = f (x) is an even function defined on R. when x ≥ 0, f (x) = x2-2x (1) Find the values of F (1) and f (- 2); (2) Find the analytic formula of F (x) and draw a diagram; (3) The root of equation f (x) = k is discussed

(1) ∵ when x ≥ 0, f (x) = x2-2x
∴f（1）=12-2=-1
f（2）=22-2×2=0
And ∵ y = f (x) is an even function defined on R
∴f（-2）=f（2）=0    … .. (3 points)
(2) When x ≤ 0, - x ≥ 0
So f (- x) = (- x) 2-2 (- x) = x2 + 2x
And ∵ y = f (x) is an even function defined on R
∴f（x）=x2+2x（x≤0）
∴f（x）=
x2−2x，x≥0
x2+2x，x＜0  … .. (7 points)
The image is shown in the following figure:
(3) From the graph of function f (x) in (2), we can get the following results
When k ＜ - 1, the equation has no real root
When k = - 1, or k > 0, there are two roots;
When k = 0, there are three roots;
When - 1 ＜ K ＜ 0, there are four roots .. (14 points)

It is known that y = f (x) is an even function defined on R. when x is greater than or equal to 0, f (x) = bungalow-2x of X, then when x is less than 0, the analytic formula of the function is

F (x) and f (- x) are symmetric about the y-axis, f (x) = x-2x, the axis of symmetry is x = 1, and the axis of symmetry of F (- x) is x = - 1, and then substituted by the common point (0,0), f (- x) = x-square + 2x

(1 / 2) the known function y = f (x) is an even function defined on (negative one, one). When x is greater than or equal to zero, f (x) = x squared minus 2x minus 1 (1 / 2) the known function y = f (x) is an even function defined on (negative one, one). When x is greater than or equal to zero, f (x) = x squared minus 2x minus 1. Find the expression of F (x) when x is less than zero

When x < 0, - x > 0
f(-x)=(-x)^2-2(-x)=xx+2x
Because it's even,
So f (x) = f (- x) = the square of X + 2x

It is known that the function y = f (x) defined on R is even function and X ≥ 0, f (x) = ln (x2-2x + 2), (1) When x < 0, find the analytic formula of F (x); (2) Write the monotone increasing interval of F (x)

(1) When x < 0, - x > 0 ∵ x ≥ 0, f (x) = ln (x2-2x + 2) ∵ f (- x) = ln (x2 + 2x + 2) (2 points) ∵ y = f (x) is an even function,

Let f (x) be an even function defined on R and satisfy f (x + 2) = - 1 / F (x). When 2 ≤ x ≤ 3, then f (x) = x, then f (105.5) is equal to A.-2.5 B.2.5 C.5.5 D-5.5

F (x + 2 + 2) = - 1 / F (x + 2), that is: F (x + 4) = - 1 / F (x + 2)
So f (x + 4) = f (x), function with period 4
Even function, f (- x) = f (x)
When x = 2.5, f (- 2.5) = 2.5
-2.5+4*27=105.5
So f (- 2.5) = f (105.5) = 2.5
B

It is known that f (x) is an even function of defined domain on R and satisfies f (x + 2) = - 1 / F (x). If 2 ≤ x ≤ 3, f (x) = x, then f (105.5) is equal to

f(x+2)=-1/f(x)
f(x+4)=-1/(f(x+2))=-1/(-1/f(x))=f(x)
F (x) is a function with a period of 4
f(105.5)=f(105.5-26*4)=f(1.5)
[f (x) even function]
=f(-1.5)=f(-1.5+4)=f(2.5)
[when 2 ≤ x ≤ 3, f (x) = x]
=2.5

It is known that f (x) is an even function defined on R and f (x + 2) = - [1 / F (x)] when x is greater than or equal to 2 and less than or equal to 3, then f (105.5) =? Step

F (x + 4) = - [1 / F (x + 2)] = f (x), so it is a periodic function
f(105.5)=f(108-2.5)=f(-2.5)
F (x) is an even function defined on R. when x is greater than or equal to 2 and less than or equal to 3, f (x) = X
f(-2.5)=f(2.5)=2.5

It is known that f (x) is an even function of defined domain on R and satisfies f (x + 2) = - 1 / F (x). If 2 ≤ x ≤ 3, f (x) = x, then f (105.5) is equal to

f(x+2+2)=-1/f(x+2)=-1/[-1/f(x)]=f(x)
T=4
f(105.5)=f(104+1.5)=f(1.5)=f(-1.5)=f(4-1.5)=f(2.5)
f(2.5)=2.5

F (x) is an even function defined on R and satisfies f (x + 2) = - 1 / F (x). When 2 ≤ x ≤ 3, f (x) = x + 1, then f (5.5) is equal to ()

First of all, you can know that this function is a periodic function by F (x + 2) = - 1 / F (x), that is to say, there should be the following relationship: F (x) = f (x + 4). The proof is as follows: F (x + 2) = - 1 / F (x), f (x) = f [(X-2) + 2] = - 1 / F (X-2)

Let f (x) be a function defined on the real number set R. if the function y = f (x + 1) is an even function, and if x ≥ 1, there is f (x) = 1-2x, then f (3) 2)，f(2 3)，f(1 3) The size relationship of is______ .

If the function y = f (x + 1) is even, then f (- x + 1) = f (x + 1), so the function is symmetric about x = 1,
When x ≥ 1, there is f (x) = 1-2x, which is a monotone decreasing function,
When x ≤ 1, the function f (x) increases monotonically
Because f (3
2)＝f(1+1
2)＝f(1−1
2)＝f(1
2) , and 1
3＜1
2＜2
3，
So f (1)
3)＜f(1
2)＜f(2
3) F (1)
3)＜f(3
2)＜f(2
3)．
So the answer is: F (1)
3)＜f(3
2)＜f(2
3)．