It is known that f (x) is an increasing function defined on R. Let f (x) = f (x) - f (A-X) prove that the image of y = f (x) is centrosymmetric about the point (A / 2,0)

It is known that f (x) is an increasing function defined on R. Let f (x) = f (x) - f (A-X) prove that the image of y = f (x) is centrosymmetric about the point (A / 2,0)

Proof. Let m (x, y) be any point on y = f (x), then the symmetric point of m point with respect to point (A / 2,0) is
M '(A-X, - y), then
y=F(x)=f(x)-f(a-x)
F(a-x)=f(a-x)-f[a-(a-x)]=f(a-x)-f(x)=-y
So the point m 'is also on y = f (x)
That is, the image of y = f (x) is centrosymmetric about the point (A / 2,0)

Given that f (x) is an increasing function defined on R, Let f (x) = f (x) - f (A-X), it is proved that the image of F (x) is a centrosymmetric graph with respect to the point (A / 2,0)

F(a-x)=-F(x)

The image of the function f (x) defined on R is centrosymmetric about the point (- 3 / 4,0), For any real number x, f (x) = - f (x + 3 / 2), and f (- 1) = 1, f (0) = - 2, then f (1) + F (2) + F (3) +. + F (2008)= In this question, what function equation can be obtained by centrosymmetry?

F (x) = - f (x + 3 / 2) = f (x + 3) so period is 3
The image is centrosymmetric about the point (- 3 / 4,0), that is, f (x) = - f (- x-3 / 2)
sof(x)=f(-x)
f(1)=1,f(2)=f(-1)=1,f(3)=f(0)=-2
According to periodicity, f (1) + F (2) + F (3) +. + F (2008) = f (1) = 1

It is known that the image of the function f (x) defined on R is symmetrical about the center of the point (- 3 / 4,0), and f (x) = - 1 / F (x + 3 / 2), f (- 1) = 1, f (0) = - 2, What is the value of F (1) + F (2) + F (3) +... + F (2009)

From F (x) = - 1 / F (x + 3 / 2), f (x + 3 / 2) = - 1 / F (x + 3), so f (x) = - 1 / F (x + 3 / 2) = f (x + 3) f (x) is a periodic function with a period of 3, and from F (x) is symmetrical about the center of the point (- 3 / 4,0), f (x) = - f (- 3 / 2-x) = 1 / F (- x), so f (1) = 1 / F (- 1) = 1, f (0) = 2, so the sum of the three values of the function in a period

The function y = f (x) defined on R is a subtractive function, and the image of function y = f (x-1) is centrosymmetric about (1,0). If s and t satisfy the inequality f (s2-2s) ≤ - f (2t-t2), when 1 ≤ s ≤ 4, t The value range of S is () A. [−1 2，1) B. [−1 4，1) C. [−1 2，1] D. [−1 4，1]

Analysis: the image of F (x-1) is equivalent to the image of F (x) shifted one unit to the right
The image of F (x-1) is symmetrical about the center of (1,0)
Know that the image of F (x) is symmetrical about the center of (0,0),
That is, the function f (x) is an odd function
F (s2-2s) ≤ f (t2-2t),
Thus, t2-2t ≤ s2-2s, simplified to (T-S) (T + S-2) ≤ 0,
And 1 ≤ s ≤ 4,
Therefore, 2-s ≤ t ≤ s, so 2
s−1≤t
S ≤ 1, while 2
s−1∈[−1
2，1]，
So t
s∈[−1
2，1]．
Therefore, C

Let the definition domain of function y = f (x) be r, and for any x ∈ R, there is f (1 + x) = - f (1-x). It is proved that the image of function f (x) is symmetrical about point (1,0) Let the domain of the function y = f (x) be r. for any x ∈ R, there is f (1 + x) = - f (1-x) Verification: the image of function f (x) is symmetrical about point (1,0)

Let 1 + x = a, x = A-1
So f (a) = - f (1-A + 1) = - f (2-A)
f(x)=-f(2-x)
f(2-x)=-f(x)
So take two points on the function, and the abscissa is a and 2-A respectively
Then the ordinates are f (a) and f (2-A) respectively, and f (2-A) = - f (a)
So the coordinates of the two points are [a, f (a)], [2-A, - f (a)]
(a+2-a)/2=1,[f(a)-f(a)]/2=0
So the midpoint of these two points is (1,0)
So they are symmetrical about (1,0)