Let y = f (x) be an even function defined on R, and its image is symmetric with respect to x = 2. When x ∈ [- 2,2] is known, the function f (x) = - x ^ 2 + 1, then when x ∈ [- 6, - 2], f (x)=

Let y = f (x) be an even function defined on R, and its image is symmetric with respect to x = 2. When x ∈ [- 2,2] is known, the function f (x) = - x ^ 2 + 1, then when x ∈ [- 6, - 2], f (x)=

If f (x) = f (- x), then f (x) = f (4-x), so f (4-x) = f (- x), that is, f (x + 4) = f (x)
So for X ∈ [- 6, - 2], f (x) = f (x-4) = (x-4) ^ 2 + 1 = x ^ 2-8x + 17

If the function y = f (x) is even, then one of the symmetry axes of the function y = f (x + 1) is______ .

∵ the function y = f (x) is an even function
The function is symmetric about the y-axis, i.e., x = 0
The image of y = f (x) can be translated one unit to the left to obtain the image of function y = f (x + 1),
The graph of the function y = f (x + 1) is symmetric with respect to x = - 1
The answer is: X

If the function y = f (x-1) is even, then the graph of function y = f (x) is about () A. The line x = - 1 symmetry B. Line x = 1 symmetry C. Straight line x = 1 2 symmetry D. Straight line x = − 1 2 symmetry

If the function y = f (x-1) is even, then its image is symmetric about the y-axis,
The image of function y = f (x) can be obtained by shifting the image of function y = f (x-1) to the left by one unit
Therefore, the image of function y = f (x) is symmetric with respect to the line x = - 1,
Therefore, a

Given that y = f (x + 1) is an even function, then the symmetry axis of the image of the function y = f (x) is () A. x=2 B. x=-2 C. x=1 D. x=-1

Because y = f (x + 1) is an even function,
So the image of y = f (x + 1) is symmetric about the y-axis,
If y = f (x + 1) is shifted to the right by one unit, the image of y = f (x) can be obtained,
Therefore, x = 1,
Therefore, C

If the function y = f (x + 1) is even, then the image of function y = f (x) is symmetric with respect to ()

The axis of symmetry of F (x + 1) is x = 0
Move him one unit to the right, which is f [(x-1) + 1] = f (x)
So the axis of symmetry is x = 1

It is known that the function y = f (x + 2) is an even function whose domain is r, and when x > = 2, f (x) = - 1 + 3 ^ x, then if x

When X2, we use f (x) = - 1 + 3 ^ x when x > = 2
f(4-x)=-1+3^(4-x)=-1+81(1/3)^x
When x

It is known that the function y = f (x + 2) is an even function whose domain is r, and when x ≥ 2, f (x) = 3 ^ X-1, then if x = 2, then if

Y = f (x + 2) is the even function of defined domain on R,
Then f (x + 2) = f (- x + 2),
So f (x) = f ((X-2) + 2) = f (- (X-2) + 2) = f (4-x)
When x > = 2, f (x) = 3 ^ X-1
f(4-x)=3^(4-x)-1
When x

It is known that the function y = f (x) is an even function on R and X ≥ 0, f (x) = (1) 2）x-1． (1) Find the analytic expression of F (x); (2) Draw the graph of this function

(1) When x < 0, - x > 0
∴f（-x）=（1
2）-x-1，
∵ the function y = f (x) is even
∴f（x）=f（-x）=（1
2）-x-1=2x-1
So f (x)=
(1
2)x−1，x≥0
2x−1，x＜0
(2) The image of the function is shown in the figure

It is known that the function y = f (x) is an increasing function on (0,2) and that the function f (x + 2) is an even function? A:f(1)

Function f (x + 2) is even function, so f (2 + x) = f (2-x)
So:
f(5/2)=f(2+1/2)=f(2-1/2)=f(3/2)
f(7/2)=f(2+3/2)=2(2-3/2)=f(1/2)
Y = f (x) is an increasing function on (0,2), so f (1 / 2)

Let f (x) be an even function. If y = 2 ^ f (x) is an increasing function when x > 0, then if x = 2 ^ f (x) is an increasing function, then if y = 2 ^ f (x) is an increasing function, then if y = 2 ^ f (x) is an increasing function, then if y = 2 ^ f (x

Set X10
So f (- x1) > F (- x2)
2^f(-x1)>2^f(-x2)
g(x1)-g(x2)>0
So in X