How to prove that f x is symmetrical with f x with respect to the y axis

To prove the symmetry of two images, we must first show that the symmetry of each point on the y=f(x) image with respect to the y-axis is on the y=f(-x) image.

Therefore, it is necessary to set (for y=f(x) to any point, then y0=f(x0), (the symmetry point about the y-axis is (, satisfying y0=f(-(x0)), and explain (on the image of y=f(-x). So the symmetry point of each point on the y=f(x) image with respect to the y-axis is on the y=f(-x) image. In the same way, it can be proved that the symmetry of each point on the y=f(-x) image with respect to the y-axis is also on the y=f(x) image, so the two-function image is symmetrical with respect to the y-axis.

i.e. prove f(x) = f(-x).

I'm sorry, but what the guys above said is to prove the symmetry of the origin.

Abstract functions can only be proved by f(x)=f(-x), which can be proved by periodicity, reductivity, etc.

There are many more ways to prove specific functions. Most of the proofs about y-axis symmetry, that is, even functions, will not be directly proved, the most common and practical method is the periodic proof, about the y-axis symmetry function has a period of 0, you can extend it, as long as you use the proof method of the periodic function, it is no problem to prove the function on the y-axis.

f(x)=f(-x), the function image is symmetrical with respect to the y-axis for the following reasons:

With respect to y-axis symmetry, the ordinates are equal, the abscissa is the opposite of each other, f(x), f(-x) are the ordinates of the points on the function image, and f(x)=f(-x) are equal.

x and -x are the abscissa of the points on the function image, and x and -x are opposite to each other, so their images are symmetrical with respect to the y axis!

In order to prove the symmetry of two images, we must first show that the symmetry of y=f(x) is on y=f(x) image, and the symmetry point of y-axis is on y=f(-x) image.

If you want to set (for y=f(x) at any point, then y0=f(x0), (the symmetry point about the y-axis is (, satisfying y0=f(-(x0)), and the description (on the image of y=f(-x).) So the symmetry point of each point on the y=f(x) image with respect to the y-axis is on the y=f(-x) image. In the same way, it can be proved that the symmetry of each point on the y=f(-x) image with respect to the y-axis is also on the y=f(x) image, so the two-function image with respect to the y-axis is called by the Zen Hu Shi.

Since f(-x) and f(x) are symmetrical with respect to the y-axis, i.e., f(x) is symmetrical with respect to the y-axis, and because f(x) and g(x) have symmetry points with respect to the y-axis, then the image of f(x) symmetrical with respect to the y-axis has an intersection point with g(x), that is, f(-x) has an intersection point with g(x).

It might be helpful to review the definition of even function f(-x) = f(x). A point symmetrical with respect to the y-axis means a mirror point with respect to the y-axis. That is, if y1 = f(-x) and y2 = g(x), symmetry with respect to y-axis means y1 = y2, i.e. f(-x) = g(x).

Note that the x here is just a point greater than zero.

The definition of the even function f(-x) = f(x) may be helpful. A point symmetrical with respect to the y-axis means a mirror point with respect to the y-axis. That is, if y1 = f(-x) and y2 = g(x), symmetry with respect to y-axis means y1 = y2, i.e. f(-x) = g(x).

Note that the x here is just a point greater than zero.

The image of f(-x) and f(x) is symmetrical with respect to the y-axis.

In the title, it is known that f(x) and g(x) have a point about y-axis symmetry, that is, f(-x) and g(x) have a common point, that is, there is an intersection point. Hope.

This should make high school. The math teacher thinks a lot.

If the function f(x) is symmetric with respect to x=2 and t=2 (with a period of 2), we can derive the property of f(x) with respect to the y-axis symmetry by following the following steps.

1.The function is symmetrical with respect to x=2:

For arbitrary x, if f(x) exists, then f(x+4) = f(x). This is because the period is 2, so f(x+2) and f(x) have the same value as brightness.

Specifically, when x=2, we have f(2+2)=f(4)=f(2), which means that the function has the same value as x=2 at x=4.

2.Derive f(x) with respect to y-axis symmetry:

Since the function f(x) is symmetrical with respect to x=2, i.e., f(2+h)=f(2-h) is permeated by any h.

If we define a new variable y=x-2 (i.e., move the x-axis in the original coordinate system to x=2), then the original function f(x) becomes g(y)=f(y+2).

We can further express the symmetry of g(y) as: g(h) = g(-h) holds for any h.

Now let's look at the symmetry of g(y) with respect to the y-axis:

When y=h, g(h) = g(h+(-2)) g(h-(-2)) g(-h).

This means that the function g(y) is symmetrical with respect to the y-axis, that is, the function f(x) is symmetric with respect to the y-axis.

To sum up, based on the properties of the function f(x) symmetrical with respect to x=2 and t=2, we can conclude that f(x) is symmetric with respect to the y-axis.

f(x) is symmetrical with respect to x=2 and t=2, how do you get that f(x) is off and symmetrical on the y axis?

f(x) for the symmetrical chaotic celery of x=2, then f(2+x)=f(2-x)t=2, f(x+2)=f(x).

f(x)=f(2-x)，f(-x)=f(2+x)f(x)=f(-x)

With respect to x 2 symmetry, then f(2 x) f(2 x) period is 2, then f(x) is 2) f(x) so f( e.g. return to the slag section x) f[2 (x 2)]f[2 (x 2)].

f(x＋2)＝f(x)

So about y-axis symmetry.

This condition does not mean that there is a symmetry center of the function, but only that the half-period of the function is 1;

f(x+1)=

f(x)f(x+2)=

f(x+1)=

f(x)]=f(x)

t=2 If you add the function even, there will be a symmetry center;

f(x+1)=

f(-x) replaces x with x-1 2

f(x+1/2)=

f(-x+1/2)

The center of symmetry is: (1 2,0).

Because f(a+x)=f(b-x) is true for any x.

Therefore, the x of the above equation is unified with [x+(b-a) 2] instead of collapse.

Get. f[(b+a) 2+x]=f[(b+a) 2-x], then f(x) is x=(b+a) 2