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?

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?

If y = f (2x-1) + 2 even function, then y = f (2x-1) is even function, then it is easy to know
F (2x-1) = f (- 2x-1), that is, the function f (x) is symmetric with respect to x = - 1
Then y = f (x + 2) is symmetric with respect to x = - 3

Given that f (x + 1) is an even function, then the symmetry axis of the image with y = f (2x) is a straight line______ .

∵ f (x + 1) is an even function,
The graph of the function f (x + 1) is symmetric about the y-axis
The graph of function f (x) is symmetric with respect to the line x = 1
The graph of function f (2x) is about the line x = 1
2 symmetry
So the answer is: x = 1
Two

If the f (x + 1) function is even, then the image of y = f (2x) is symmetric about a straight line To process

F (x + 1) function is even function,
F (x + 1) = f (1-x), the symmetry axis of this function x = (1 + 1) / 2 = 1
F(2-X)=F(X),
The image of y = f (x) is symmetric with respect to x = (1 + 1) / 2. There is f (1 + MX) = f (1-mx), that is, f (1 + 1-mx) = f (MX),
Then the image of y = f (2x) is symmetric with respect to the straight line x = (1 + 1) / 2 = 1

If x is a symmetric function of Y-1, then it is equal to the graph That's f (x minus 1)

The symmetry axis of even function is y-axis, and the image of y = f (x-1) is obtained by moving the image of y = f (x) to the right by one unit, so the symmetry axis of the image of y = f (x) is x = - 1 or defined by the even function f (- x-1) = f (x-1), that is, f (- 1-x) = f (- 1 + x), so the symmetry axis is x = - 1

It is known that the function y = f (x) is an even function whose domain is r, and increases monotonically on [0, + infinity) and a is equal to f (sin2 π / 7) B is equal to f (cos5 π / 7) C is equal to f (tan5 π / 7). Compare the size of a B C

cos5π/7=cos(π-2π/7)=-cos2π/7
tan5π/7=tan(π-2π/7)=-tan2π/7
Even function, f (- x) = f (x)
So f (cos5 π / 7) = f (Cos2 π / 7)
f(tan5π/7)=f(tan2π/7)
The first quadrant sin is an increasing function and COS is a decreasing function
So sin 2 π / 7 > sin π / 4 = cos π / 4 > cos 2 π / 7
Because 0sin2 π / 7 > Cos2 π / 7
Increasing function
f(tan2π/7)>f(sin2π/7)>f(coss2π/7)
f(tan5π/7)>f(sin2π/7)>f(coss5π/7)

Given that the function f (x) = x ^ 2 + BX + 1 and y = f (x + 1) is an even function in the domain of definition, then the analytic expression of function f (x) is given

From the meaning of the title
y=f(x+1)=(x+1)^2+b(x+1)+1=x^2+(2+b)x+b+2
Because y is an even function, the axis of symmetry is y-axis, - (2 + b) / 2 = 0
B = - 2
So f (x) = x ^ 2-2x + 1

It is known that the function f (x) is an even function on the definition domain R. when x ≥ 0, f (x) = x (1 + x). Draw the image of function f (x) and find the function analytic formula When < 0, - x > 0 f（-x）=-x（1-x）=x（x-1） Because f (x) is an even function So f (x) = f (- x) = x (x-1) So the analytic expression of F (x) is f（x）=x（1+x）,x≥0 f（x）=x（x-1）,x＜0 I know the answer, but part 3. F (x) = f (- x) should refer to this even function, not that X

It is known that the function f (x) is an even function on the definition domain R. when x ≥ 0, f (x) = x (1 + x). Draw the image of the function f (x) and find the analytic formula of the function:

The function y = f (x) is an increasing function on (0, 2), and the function y = f (x + 2) is an even function. Try to compare the sizes of F (1), f (2.5), f (3.5) () A. f （3.5）＞f （1）＞f （2.5） B. f （3.5）＞f （2.5）＞f （1） C. f （2.5）＞f （1）＞f （3.5） D. f （1）＞f （2.5）＞f （3.5）

∵ the function y = f (x + 2) is an even function and f (x + 2) = f (- x + 2) is obtained,
The function is symmetric about x = 2
∵ y = f (x) is an increasing function on (0, 2),
/ / y = f (x) is a decreasing function on (2,4),
∵ f (1) = f (2-1) = f (2 + 1) = f (3), and 2.5 < 3 < 3.5,
∴f （2.5）＞f （3）＞f （3.5），
That is, f (2.5) > F (1) > F (3.5),
Therefore, C

The function y = f (x) is an increasing function on (0, 2), and the function y = f (x + 2) is an even function. Try to compare the sizes of F (1), f (2.5), f (3.5) () A. f （3.5）＞f （1）＞f （2.5） B. f （3.5）＞f （2.5）＞f （1） C. f （2.5）＞f （1）＞f （3.5） D. f （1）＞f （2.5）＞f （3.5）

∵ the function y = f (x + 2) is an even function, and f (x + 2) = f (- x + 2), ᙽ the function is symmetric about x = 2.

Let f (x) be an even function and an increasing function on (0, + ∞). Judge whether f (x) is an increasing function or a decreasing function on (- ∞, 0), and prove your judgment

F (x) is a decreasing function on (- ∞, 0) (1 point)
Proof: if X1 < x2 < 0, then - X1 > - x2 > 0 (3 points)
∵ f (x) is an increasing function on (0, + ∞)
﹤ f (- x1) ﹥ f (- x2) (7 points)
And f (x) is even function
∴f（-x1）=f（x1），f（-x2）=f（x2）
∴f（x1）＞f（x2）
ν f (x) is a decreasing function on (- ∞, 0) (12 points)