First, explore the symmetry of the function itself.
Theorem 1. The necessary and sufficient condition for the image of function y = f (x) to be symmetric about point A (a, b) is that
f (x) + f (2a-x) = 2b
It is proved that (necessary) point P( x, y) is any point on the image of y = f (x), and the symmetrical point P' (2a-x, 2b-y) of point P(x, y) about point A (a, b) is also on the image of y = f (x).
That is, y+f (2a-x) = 2b, so f (x)+f (2a-x) = 2b, which proves the necessity.
(Sufficiency) If the set point P(x0, y0) is any point in the image where y = f (x), y0 = f (x0).
∫ f (x)+f (2a-x) = 2b ∴ f (x0)+f (2a-x0) = 2b, that is 2b-y0 = f (2a-x0).
Therefore, the point P' (2a-x0, 2b-y0) is also on the image of y = f (x), and the points p and p' are symmetric about the point A (a, b), and the sufficiency is obtained.
Inference: The necessary and sufficient condition for the image of the function y = f (x) to be symmetrical about the origin o is f (x)+f (-x) = 0.
Theorem 2. The necessary and sufficient condition for the image of the function y = f (x) to be symmetrical about the straight line x = a is
F (a+x) = f (a-x) means f (x) = f (2a-x-x) (for readers to prove).
Inference: The necessary and sufficient condition for the image of function y = f (x) to be symmetric about y axis is f (x) = f (-x).
Theorem 3. (1) if the image of function y = f (x) is centrosymmetric (a≠b) about A (a, c) and B (b, c), then y = f (x) is a periodic function, and 2a-b is one of its periods.
(2) If the image of the function y = f (x) is axisymmetrical (a≠b) about two straight lines x = a and x = b, then y = f (x) is a periodic function and 2a-b is one of its periods.
(3) If the image of function y = f (x) is both symmetric about the center of point A (a, c) and symmetric about the straight line x =b (a≠b), then y = f (x) is a periodic function and 4a-b is one of its periods.
① The proof of ② is reserved for readers, and ③ is as follows:
∫ Function y = f (x) The image is not only symmetric about the center of point A (a, c),
∴ f (x)+f (2a-x) = 2c, and replace x with 2b-x to obtain:
f(2 b-x)+f[2a-(2 b-x)]= 2c……………………(*)
The function y = f (x) and the image straight line x =b are symmetrical.
∴ f (2b-x) = f (x) Substitution (*):
F (x) = 2c-f [2 (a-b)+x] ...................................... (* *), and x is represented by 2 (a-b)-x.
F [2 (a-b)+x] = 2c-f [4 (a-b)+x] substitute (* *):
F (x) = f [4 (a-b)+x], so y = f (x) is a periodic function, and 4a-b is one of its periods.
Second, explore the symmetry of different functions.
Theorem 4. Images with functions y = f (x) and y = 2b-f (2a-x) are centrosymmetric about point A (a, b).
Theorem 5. ① the images with functions y = f (x) and y = f (2a-x) are symmetrical about the straight line x = a. ..
② Images with functions y = f (x) and a-x = f (a-y) are symmetrical about the straight line x+y = a..
③ Images with functions y = f (x) and x-a = f (y+a) are symmetrical about the straight line x-y = a..
The proof of ① ② in Theorem 4 and Theorem 5 is left to readers, and ③ in Theorem 5 has been proved now.
If the point P(x0, y0) is any point in the image where y = f (x), y0 = f (x0). If the point P( x, y) is P'(x 1, y 1), then x 1 = a+y0, y 1 = x0-a, ∴ x-y = a+y66. Y0 = X 1-A substitute y0 = f (x0), X 1-A = F (A+y 1) ∴ point P'(x 1, y 1) in the function.
Similarly, it can be proved that any point on the image of function X-A = f (y+a) is also on the image of function y = f (x). Therefore, ③ in Theorem 5 holds.
Inference: the image of function y = f (x) and the image of x = f (y) are symmetrical about the straight line x = y.
Third, the symmetry list of trigonometric function images
Note: ① k∈Z in the above table.
② In the first volume of 2 1 Century Senior High School Mathematics edited by Cen Can and Wang, and the New Teaching Plan of Senior High School Mathematics published by Guangxi Normal University Press edited by Chen, the symmetrical central coordinates of Y = Tan X should be (kπ/2,0), while the symmetrical central coordinates of y = tan x should be (kπ,0).
Fourth, examples of functional symmetry application
Example 1: The non-constant function defined on R satisfies that f (10+x) is an even function, and f (5-x) = f (5+x), then f (x) must be () (the second question of xx Hope Cup Grade Two).
(a) is an even function and a periodic function; (b) is an even function, but not a periodic function.
(c) is a odd function sum periodic function. (d) It is a odd function, but not a periodic function.
Solution: ∫f( 10+x) is an even function, ∴ f (10+x) = f (10-x).
∴f (x) has two symmetry axes x = 5, x = 10, so f (x) is a periodic function with 10 as the period, and∴ x = 0 means that the y axis is also the symmetry axis of f (x), so f (x) is still an even function.
So choose (a)
Example 2: Let the functions y = f (x) and y = g(x) have inverse functions with R, and the images of the functions F (x- 1) and G- 1 (x-2) are symmetrical about the straight line y = x, if G (5) = 10.
(A) 1999； In 2000; (C)200 1； In 2002.
Solution: The images of ∵ y = f (x- 1) and y = g- 1 (x-2) functions are symmetrical about the straight line y = x,
The inverse function of y = g- 1 (x-2) is y = f (x- 1) and the inverse function of y = g- 1 (x-2) is y = 2+g (x), ∴ f.
So f(4) = 200 1, and (c) should be chosen.
Example 3. Let f(x) be an even function defined on r, and f (1+x) = f (1-x). -1 ≤ x ≤ 0,
F (X) =-X, then F (8.6) = _ _ _ _ (XX Hope Cup, the first test in Grade Two)
Solution: ∵f(x) is the even function defined on r ∴x = 0 is the symmetry axis of y = f(x);
And ∵ f (1+x) = f (1-x) ∴ x =1is also the symmetry axis of y = f (x). So y = f(x) is a periodic function with a period of 2, ∴ f (8.6) = f (8+0.6) = f (0.6) = f (-0.6) = 0.3.
Example 4. The equation of an axis of symmetry of the image with function y = sin (2x+) is () (92 college entrance examination) (a) x =-(b) x =-(c) x = (d) x =
Solution: The equation of all symmetry axes of the image with function y = sin (2x+) is 2x+= k+
∴ x =-, obviously the symmetric equation when k = 1 is x =-, so choose (a).
Example 5. Let f(x) be a odd function defined on R, and f (x+2) =-f (x). When 0≤x≤ 1,
F (x) = x, then f (7.5) = ()
(A)0.5(B)-0.5(C) 1.5(D)- 1.5
Solution: ∵y = f (x) is the odd function defined on R, and ∴ point (0,0) is its symmetry center;
And ∵ f (x+2) =-f (x) = f (-x), that is, f (1+x) = f (1-x), ∴ straight line x = 1 is y = f (x.
∴ f (7.5) = f (8-0.5) = f (-0.5) =-f (0.5) =-0.5, so choose (b).