Let the image of function f (x) = AX2 + BX + A-3 be symmetrical about the Y axis, and its definition domain is [A-4, a] (a, B belong to R) to find the value domain of F (x)

Function f (x) = ax ²+ The image of BX + a - 3 is symmetrical about the Y axis, indicating that the function is an even function
f(-x)=f(x),
ax ²- bx+a-3= ax ²+ BX + A-3, then B = 0
And because the definition field of even function must be symmetrical about the origin, A-4 + a = 0,
a=2.
∴f(x)=2 x ²- 1,x∈[-2,2]
Its value range is [- 1,7]

The image crossing (- 1, - 6) of function f (x) = X3 + MX2 + nx-2 is known, and the image of function g (x) = f '(x) + 6x is about the y-axis ； (2) If a > 0, find the extreme value of function y = f (x) in the interval (A-1, a + 1)

Find the value of M, N and the monotone interval of function y = f (x)
Is that the question?
I tried to write it down
∵ f (x) over point (- 1, - 6)
∴f(-1)=-6
That is: M-N = - 3
∵g(x)=3x^2+2mx+n+6x
And ∵ g (x) is symmetric about the y-axis
∴g(-x)=g(x)
That is: m=-3
∴n=0
f(x)=x^3-6x-2
f'(x)=3x^2-6
Let f '(x) = 0, i.e. x = ±√ 2
The monotonic increasing interval of F (x) is (- ∞, - √ 2), (√ 2, + ∞)
Monotonic decreasing interval is (- √ 2, √ 2)

Let f (x) be an odd function defined on R, and the image of y = f (x) is about line x = 1 3 symmetric, then f (- 2 3)=（　　） A. 0 B. 1 C. -1 D. 2

∵ f (x) is an odd function defined on R, then f (0) = 0
And the image of ∵ y = f (x) is about line x = 1
3 symmetry,
∴f（2
3）=f（0）=0．
∴f(-2
3)=-f(2
3)=0，
So choose a

Let f (x) be an odd function defined on R, and the image of y = f (x) is x = 1 with respect to the straight line 2 symmetric, then f (1) + F (2) + F (3) + F (4) + F (5) = ___

F (x) is an odd function defined on R, and the image of y = f (x) is x = 1 with respect to the straight line
2 symmetrical,
∴f（-x）=-f（x），f(1
2+x)＝f(1
2−x)⇒f(x)＝f(1−x)，
∴f（-x）=f（1+x）=-f（x）f（2+x）=-f（1+x）=f（x），
∴f（0）=f（1）=f（3）=f（5）=0，f（0）=f（2）=f（4）=0，
So f (1) + F (2) + F (3) + F (4) + F (5) = 0
So the answer is: 0

Let f (x) be an odd function defined on R, and the image of y = f (x) is symmetrical about the straight line x = 0.5, then f (1) + F (2) + F (3) + F I want the process

(x) Is an odd function defined on R, with F (0) = 0, f (- x) = - f (x)
If the image of y = f (x) has a straight line x = 0.5, then f (x + 0.5) = f (x-0.5)
f(x+1)=f(x+0.5+0.5)=f(x+0.5-0.5)=f(x),
So f (x) is a periodic function of T = 1
f(0)=f(1)=f(2)=f(3)=……＝f(n)
Original formula = 0
Xie Bi

F (x) all functions have f (x + 6) = f (x) + 2F (3) on the definition field R. if the image of function f (x + 1) is symmetrical about the straight line x = - 1, and f (- 2) = 2012, then f (2012) =?

The image of function f (x + 1) is symmetrical about the straight line x = - 1,
Let t = x + 1
The image of function f (T) is symmetric about t = 0
So the function f (T) is an even function
So the function f (x) is an even function
So f (3) = f (- 3)
F (x + 6) = f (x) + 2F (3) make x = - 3
Get f (3) = f (- 3) + 2F (3) = 3f (3)
Get f (3) = 0
So f (x + 6) = f (x)
f（2012）=f（335*6+2）=f（2）=f（-2）=2012

Let the image of function f (x) = AX2 + BX + A-3 be symmetrical about the Y axis, and its definition domain is [A-4, a] (a, B belong to R) to find the value domain of F (x)