Given that f (x 1) is an even function, then the symmetry axis of the image of function y = f (2x) is____ .

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Given that the function y = f (2x + 1) is an even function, then the straight line of the symmetry axis of the image of the function y = f (2x) must be () A. x=-1 Two B. x=0 C. x=1 Two D. x=1

If (x + 1) of (y + 1) is shifted to the right (F = 1), then (x + 1) = 1
The image of y = (2x) can be obtained by two units,
Because y = (2x + 1) is an even function, its image is symmetric about the y-axis, while the image of y = f (2x) is about the straight line x = 1
2 symmetry
Therefore, C

Given that y = f (2x + 1) is an even function, then it must be a function. The line of symmetry axis of y = f (2x) image is x = 1 / 2. Why?

It is known that y = f (2x + 1) is an even function
The symmetry axis is: x = 0
y=f(2x)=f[2(x-1/2)+1]
The image is y = f (2x + 1) shifted to the right by 1 / 2 units
The axis of symmetry also shifts 1 / 2 units to the right
therefore
The line of symmetry axis is x = 1 / 2

If the function y = f (2x-1) is even, then the symmetry axis of function y = f (2x) is () A. x=0 B. x=-1 C. x=1 Two D. x=-1 Two

∵ the function y = f (2x-1) is even, and the graph of the function is symmetric about the y-axis
∵ the function y = f (2x) is shifted to the left by the image of function y = f (2x-1)
2 units
The symmetry axis of the function y = f (2x) is a straight line x = - 1
Two
Therefore, D

If f (2x + 1) is known to be an even function, then the symmetry axis of F (2x) image is

If the function y = f (2x-1) + 2 even function, then the symmetry axis of the image of the function y = f (x + 2) is?

Y = f (2x-1) + 2 even function
That is, the axis of symmetry of F (x) is - 1 / 2
The symmetry axis of the image of the function y = f (x + 2) is x = - 1 / 2-2 = - 5 / 2

Given that the function y = f (2x + 1) is an even function, then the straight line of the symmetry axis of the image of the function y = f (2x) must be () A. x=-1 Two B. x=0 C. x=1 Two D. x=1

If the function f (2x + 1) is even, then the image of F (2x) is symmetric with respect to the straight line x = The solution is f (- 2x + 1) = f (2x + 1), let 2x = t, f (T + 1) = f (- t + 1), then t = 1, x = 1 / 2, let 2x = t, f (T + 1) = f (- t + 1), then t = 1, x = 1 / 2!

f（t+1）=f（-t+1）
That is, f (1 + T) = f (1-T)
So the axis of symmetry is t = 1
That is, 2x = 1
x=1/2

Given the proposition p: | 4-x | ≤ 6, Q: x2-2x + 1-a2 ≥ 0 (a > 0), if P is not a sufficient and unnecessary condition of Q, the value range of a is obtained

, P: | 4-x | 6, X | 10, or X | 2,
A = {x | x > 10, or X ＜ - 2}
q: X2-2x + 1-a2 ≥ 0, X ≥ 1 + A, or X ≤ 1-A,
Let B = {x | x ≥ 1 + A, or X ≤ 1-A}
And ⊂ P ⊂ Q, ⊂ a ⊂ B, i.e
1-a≥-2
1+a≤10
a＞0 ，∴0＜a≤3．

According to the definition of even function, the inference process of "function f (x) = X2 is even function" is () A. Inductive reasoning B. Analogical reasoning C. Deductive reasoning D. Not the above answer

According to the definition of even function, the reasoning process of "function f (x) = X2 is even function" can be deduced as follows: for function y = f (x), if there is f (- x) = f (x) for any X in the definition domain, then function f (x) is even function; minor premise: function f (x) = x2 satisfies that for any X in defined domain R, there is f (...)

Given that f (x 1) is an even function, then the symmetry axis of the image of function y = f (2x) is____ .