EM@Coordinates@Function@Symmetry and folding transformation of images

Here we transform the image indirectly by studying the points on the image. Let the equation of the image be $y = f (x)$ ,The domain of $f$ $($ $x$ $)$ $D_f$

Function $f (- x)$ can be viewed as the function $u = - x$ 和 $y =$ Function composed of $f$ $($ $u$ $)$
- $x\in{D_u}=\mathbb{R}$
Let the function The domain of $f$ $($ $x$ $)$ ${D_f}$ , for $g (x) = f (- x)$ , $-x\in{D_f}$ ,即 $x\in{-D_f}$ Or as $D_g=-D_f$ (representing $f,$ The domain of $g is symmetric about the origin)$
若 $a\in{D_f}$ , $x = At a$ , we can take the function $f (x)$ point $A (a, f (a))$ ;
$-a\in{D_g}$ ,There must be a point $on g$ $($ $x$ $)$ $B (- a, f (a))$ ;
Obviously $A, BAbout$ _ $y-$ axis symmetry, for allPoint corresponding to $x$ $(x, f (x))$ and $(x, f (- x))$ is symmetric about the y-axis
从而 $f (x), g (x)$ about $y-$ axis symmetry, that is, $f (x), f (- x)$ about $y-$ axis symmetry
For example:
- $f(x)=\sin(x)$ ,则 $f(-x)=\sin(-x)=-\sin{x}$ 和 $f(x)=\sin(x)$ with respect to $y-$ axis symmetry
- For $f(x)=\cos{x}$ , $f(-x)=\cos{(-x)}$ = $cos{x}$ , $f (- x), f (x)$ about $The y-$ axis is symmetrical, that is, the function $cos{x}$ itself about $y-$ axis symmetry

Similar to the above analysis, take a point analysis: if the function $f (x)$ , existence $A (a, f (a))$ , then the functionThere must be corresponding $on -$ $f$ $($ $x$ $)$ $B (a, - f (a))$
Obviously two points about $The x-$ axis is symmetrical, and $x$ is any point within the domain, so $- f (x)$ and $f (x)$ aboutxx $x-$ axis symmetry

Even function: if function $The domain of f (x)$ is symmetric about the origin and satisfies $f (- x) = f (x)$ , then the function $f (x)$ is an even function, obviously $f (x)$ about $y-$ axis symmetry
- 若 $f (- x) = f (x)$ , $f (- x), f (x)$ about $The y-$ axis symmetry becomes $f (x), f (x)$ about $y-$ axis symmetry ( $f (x)$ and $f (- x)$ coincides), that is, $f (x)$ about $y-$ axis symmetry
Odd function: if function $The domain of f (x)$ is symmetric about the origin and satisfies $f (- x) = - f (x)$ , then the function $f (x)$ is an odd function, obviously $f (x)$ is symmetric about the coordinate origin
- Can $The graph of f (x)$ that is symmetric about the origin is understood to be two parts: $f (x)$ about $Graph with y$ -axis symmetry and $f (x)$ aboutxx $If the x-$ axis symmetrical figures coincide, then $f (x)$ is an odd function that is symmetric about the origin

The graph of $f (- x)$ is the mirror image of the graph of $f (x)$ with respect to the vertical axis.
The graph of $- f (x)$ is the mirror image of the graph of $f (x)$ with respect to the horizontal axis.
A function is called even if $f (- x) = f (x)$ for all $x$ (For example, $\cos(x)$ ).
A function is called odd if $f (- x) = - f (x)$ for all $x$ (For example, $\sin(x)$ ).

If $f (x)$ 关于 $x = u$ Symmetry:
- The domain of $f$ $($ $x$ $)$ $x = u$ Symmetry
If $x_1,x_2$ 关于 $u$ name,则 $x_1+x_2=2u$ , vice versa
- 设 $A (a, f (a))$ ispoint on $f$ $($ $x$ $)$ $A$ is about the axis of symmetry $x = u$ 's reference point $B (2 and - a, f (a))$ must also be in $f (x)$ on
- Then $f (2 u - a)$ = $f (a)$
- due to $a$ is any point within the domain, so $f (2 u - x) = f (x)$
- That is, satisfy:
  - The domain is about $x = u$ Symmetry
  - $f (2 u - x)$ = $f (x)$
- The function is about $x = u$ symmetric function
For example $y(x)=(x-1)^2$ ; $y(2-x)=((2-x)-1)^2=(1-x)^2=(x-1)^2$ ,即 $y (x) = y (2 - x)$ , the axis of symmetry is $u=\frac{1}{2}\cdot2=1$
- In particular, even functions with respect to $x = 0$ symmetry, $f (x) = f (- x)$ ,axis of symmetry $x = u = 0$ ,因为 $x + (- x) = 2 and = 0; u = 0$

若 $f_1(x),f_2(x)$ $f_1(x)+f_2(x)=2v$ in the domain of definition $f_{1} (x) + f_{2} (x) = 2 v$ ,则 $f_1(x),f_2(x)$ 关于 $y = vSymmetry$ _