The symmetry of a function

One, the function of the symmetry of the given way:

1, given in the form of images;

Interpreting the image, we can find the axis of symmetry from the image.

2, in the form of parity gives [odd and even is a special case of symmetry];

For example, the odd function,\ (f (-X) =-f (x) \) or \ (f (x) +f (x) =0\longrightarrow\) The symmetry Center is \ ((0,0) \)

such as even function,\ (f (x) =f (x) \) or \ (f (-X)-F (x) =0\longrightarrow\) The axis of symmetry is \ (x=0\)

3, the development of the odd and even the form given;

For example \ (f (2+x) +f (-X) =2\), the symmetry Center is \ (() \);

For example \ (f (x) =f (4-x) \), the axis of symmetry is \ (x=2\)

Why?

4, given in the form of periodic + odd-even;

For example, the known function \ (f (x) \) is an odd function and satisfies \ (f (x+4) =-f (x) \),

Then by \ (\begin{align*} f (x+4) &=-f (x) \ \ f (-X) &=-f (x) \end{align*}\) \ (\big\}\longrightarrow f (x+4) = F (-X) \longrightarrow\) The axis of symmetry is \ (x=2\)

Second, the application of functional symmetry example

\ (\fbox{case 1}\) (2016 college Entrance Examination Science Mathematics National Volume 2 the 12th question) (Shared symmetry Center)
Known function \ (f (x) (x\in R) \) satisfies \ (f (x) =2-f (x) \), if function \ (y=\cfrac{x+1}{x}\) and function \ (y=f (x) \) The intersection of the image is \ ((x_1,y_1), (x_2,y_2), \cdots, (x_m,y_m) \), then \ (\sum\limits_{i=1}^m{(x_i+y_i)}\) The value is ""

A,\ (0\) \ (\hspace{2cm}\) B,\ (m\) \ (\hspace{2cm}\) C,\ (2m\) \ ( \hspace{2cm}\) D, \ (4m\)

Analysis: By the topic know \ (f (x) +f (×) =2\), that is, function \ (f (x) \) image about point \ ((0,1) \) symmetry,

and the function \ (y=\cfrac{x+1}{x}=1+\cfrac{1}{x}\) image also about point \ ((0,1) \) symmetry, that is, two function images have the same symmetry center,

Then the number of intersections must have an even number of points, it is known to the horizontal axis (\sum\limits_{i=1}^m{x_i}=0\),

For the ordinate, the number of paired points is \ (\cfrac{m}{2}\) , each of them satisfies \ (\cfrac{y_1+y_m}{2}=1\),

That is \ (y_1+y_m=2\), so \ (\sum\limits_{i=1}^m{y_i}=2\cdot \cfrac{m}{2}=m\),

therefore \ (\sum\limits_{i=1}^m{(x_i+y_i)}=\sum\limits_{i=1}^m{x_i}+\sum\limits_{i=1}^m{y_i}=m\), so select B.

\ (\fbox{case 2}\) (2016 college Entrance Examination Arts Mathematics National Vol. 2 the 12th question) (Common axis of symmetry)
Known function \ (f (x) (x\in R) \) satisfies \ (f (x) =f (2-x) \), if function \ (y=|x^2-2x-3|\) and function \ (y=f (x) \) The intersection of the image is \ ((x_1,y_1), (x_2,y_2), \cdots, (x_m,y_m) \), and the value of \ (\sum\limits_{i=1}^m{x_i}\) is ""

A,\ (0\) \ (\hspace{2cm}\) B,\ (m\) \ (\hspace{2cm}\) C,\ (2m\) \ ( \hspace{2cm}\) D, \ (4m\)

Analysis: function \ (f (x) (x\in R) \) satisfies \ (f (x) =f (2-x) \), then the symmetric axis of the function is a straight line \ (x=1\),

and functions \ (y=|x^2-2x-3|=| ( X-1) The symmetric axis of ^2-4|\) is also a straight line \ (x=1\), making the image of the function as shown in the picture on the right,

Then the number of intersections \ (m\) must be an even number, 22 pairs of the number of \ (\cfrac{m}{2}\), such as AB pairing,

Then there \ (\cfrac{x_1+x_m}{2}=1\),\ (x_1+x_m=2\), so \ (\sum\limits_{i=1}^m{x_i}=\cfrac{m}{2}\cdot 2 = m\), so select B.

"2017 National Volume 1 Liberal arts 9th question of the college entrance Examination"

Known functions \ (f (x) =lnx+ln (2-x) \), then

A,\ (f (x) \) in \ ((0,2) \) monotonically increment \ (\hspace{5cm}\) B,\ (f (x) \) in \ (((0,2) \) monotonically decreasing \ (\hspace{2cm}\)

C,\ (y=f (x) \) images about the line \ (x=1\) symmetric \ (\hspace{2cm}\) D,\ (y=f (x) \) about the point \ ((1,0) \ ) Symmetrical

Analysis: Because the function \ (f (x) \) is a compound function, define the domain to make \ (x>0,2-x>0\), that is, define the domain is \ ((0,2) \),

also \ (f (x) =ln[x (2-x)]=ln[-(x-1) ^2+1]\), it is known by the monotone law of the complex function,

On the \ ((0,1) \) on the monocytogenes, in \ (() \) on the single minus, so exclude A, B;

If the function \ (y=f (x) \) is symmetric about point \ ((1,0) \), then the function \ (f (x) \) must satisfy the relationship:\ (f (x) +f (2-x) =0\);

If the function \ (y=f (x) \) is symmetric about a line \ (x=1\) , then the function \ (f (x) \) must satisfy the relationship:\ (f (x) =f (2-x) \);

We then use the above conclusions to verify that, because \ (f ( x) =lnx+ln (2-x) \),\ (f (2-x) =ln (2-x) +ln ((2-x)) =ln (2-x) +lnx\), which satisfies \ (f ( x) =f (2-x) \),

So the function \ (y=f (x) \) of the image about the line \ (x=1\) symmetry, select C;

Again to verify D, Discovery \ (f (x) +f (2-x) =2[lnx+ln (2-x)]\neq 0\), the D option is not met.

So choose c.

The symmetry of a function