Why is the function symmetrical about 3 4,0, we get that there is f x 3 2 f x

The sufficient and necessary conditions for the symmetry of the function y = f (x) with respect to the point a (a ,b) are.

f (x) +f (2a－x) = 2b

2. The sufficient and necessary conditions for the image of the function y = f (x) with respect to the origin o symmetry are f (x) + f ( x ) = 0

3 The sufficient and necessary conditions for the symmetry of the function y = f (x) with respect to the straight line x = a are.

f (a +x) = f (a x) i.e. f (x) = f (2a x).

The sufficient and necessary condition for the image of the function y = f (x) with respect to y-axis symmetry is f (x) = f (x).

Theorem 3If the function y = f (x) the image is symmetrical with respect to both the points a (a,c) and the points b (b,c).

a≠b), then y = f (x) is a periodic function, and 2| a－b|is one of its cycles.

If the function y = f (x) the image is axisymmetric with respect to the line x = a and the line x = b at the same time.

a≠b), then y = f (x) is a periodic function, and 2| a－b|is one of its cycles.

If the function y = f (x) image is both centrally symmetric with respect to the point a (a ,c) and axisymmetric (a≠b) with respect to the line x = b, then y = f (x) is a periodic function and 4| a－b|is one of its cycles.

Theorem 4The function y = f (x) is symmetrical with respect to the point a (a ,b) in the image of y = 2b f (2a x).

Theorem 5The image of the function y = f (x) with y = f (2a x) is axisymmetric with respect to the straight line x = a.

The image of the function y = f (x) with a x = f (a y) is axisymmetric with respect to the line x + y = a.

The image of the function y = f (x) with x a = f (y + a) is axisymmetric with respect to the straight line x y = a.

Just tell one point, of course, it's central symmetry.

If the function is symmetrical with respect to (0,0), then this function is an odd function, so symmetry with respect to (3 4,0) is equivalent to moving the above function to the right, so it can be obtained.

Untie. From the question setting, it can be seen that for any real number x, there is always :

f(x+3)+f(-x+3)=0

Let x+3=k, then -x+3=-k+6

The above equation can be reduced to:

f(k)+f(-k+6)=0

In other words, there is always f(x)+f(-x+6)=0.

If p(a,b) is any point on the curve y=f(x), then b=f(a)

Combined with the above equation, we can see that f(a)+f(-a+6)=0 or b+f(-a+6)=0

f(-a+6)=-b

This shows that the point q(-a+6,-b) is also the point of the volcanic beat on the curve y=f(x), and if p(a,b) is any point on the curve y=f(x), then q(-a+6,-b) is also a point on the curve, and these two points p,q are symmetrical with respect to the point m(3,0), and the symmetry center of the curve y=f(x) is the point (3,0).

If the center of symmetry is the kernel of Hui(2,0), otherwise it is not a periodic function f(x-1) is an odd function, then f(-x-1)=-f(x-1) is symmetrical with respect to (2,0) since (x,y) and (4-x,y) are symmetrical with respect to (2,0), thus f(x)=f(4-x).

Replace x with x-1, get.

f(x-1)=f(5-x) ②

Compare f(-x-1)=f(5-x).

Replace Stupid Detective x with -x-1 to get the front gear dig.

f(x)=f(6+x)

Thus t=6

y=f(x+3) is an odd function.

So f(x+3)=—f(-x+3), so f(x+3)+f(-x+3)=0, so the center of symmetry is (3,0).

As for how to solve it, it is because there is a formula f(a+x)+f(a-x)=2b, the center of symmetry is (a,b); Proof of the following is Oak Royal:

Let the center of symmetry be (a,b) and the point on the function be (x,y), then the point of symmetry is (2a-x,2b-y).

So f(2a-x)=2b-y, y is f(x), so f(2a-x)+f(x)=2b, use a+x instead of x

then f(a-x)+f(a+x)=2b).

OK proof is complete, the landlord is a good person, give points, the computer Liang Youyan mills the world to play mathematical formulas.

It's really inconvenient.

Untie. Composed.

Questions. It can be seen that for any real number x, there is always :

f(x+3)+f(-x+3)=0

Let x+3=k, then -x+3=-k+6

The above equation can be reduced to:

f(k)+f(-k+6)=0

In other words, there is always f(x)+f(-x+6)=0.

If p(a,b) is any point on the curve y=f(x), then b=f(a)

Combined with the above equation, it can be seen that f(a)+f(-a+6)=0 or rather: Chi Chan b+f(-a+6)=0

Envy shirts. f(-a+6)=-b

This shows that the point q(-a+6,b) is also a point on the curve y=f(x), and if the brother cavity p(a,b) is any point on the curve y=f(x), then q(-a+6,b) is also a point on the curve, and the two points p,q are symmetrical with respect to the point m(3,0), and the center of symmetry of the curve y=f(x) is the point (3,0).

Because. f(x+3)

is an odd function on r, and therefore the finch is an arbitrary real number.

x base digging has it.

f(-x+3)=

f(x+3)

Sosun Feng was early, by.

f(3-x)+f(3+x)=0

Know, function. y=f(x)

The center of symmetry is.

Thank you. f(x+3 2)=-f(x) uses the malfunction x-3 4 to replace the key spine xf(x+3 4)=-f(x-3 4).

Again, f(x-3 4) is an odd function, manuscript percolation.

f(x-3 4) = f(-x-3 4), so the image with a function is symmetrical with respect to the point (-3 4, 0).

Thus, f(x+3 4) = f(-x-3 4) uses x-3 4 to replace x

f(x)=f(-x)

Therefore, f(x) is an even function.

is not true, because f(x-3 2) is an odd function, so the image is symmetrical with respect to the origin, and y=f(x) is translated to the right.

3 2 units give an image of f(x-3 2), so the center of symmetry of f(x) is (-3 2,0).

Proof: Since f(x-3 2) is an odd function, there is a: f(x-3 2) = f(-x+3 2).

And because f(x+3) +f(x) = 0, there is b: f(x + 3 2) +f(x-3 2) =0 (think of x+3 2 as the x in the formula f(x+3) +f(x) =0).

According to the A and B formulas, there are: f(x + 3 2) -f(-x + 3 2) = 0

So there is f(x + 3 2) = f(-x+3 2) and it is expressed as f((x-3 2) +3) =f(-(x-3 2)).

Let p = x - 3 2, there is f(p+3) = f(-p).

According to formula a, there is f(p+3) +f(p) =0

There is f(-p) +f(p) =0

Obviously, p = x - 3 2 can take any value on the x-axis, that is to say, for any coordinate p on the x-axis, f(p) +f(-p) =0 holds, so the image of the function f(x) is symmetrical with respect to the y-axis. Proven.

Since 1+1=2, the image of the function f(x) is symmetrical with respect to the y-axis.

Just prove that f(x) is an even function.

According to the question: f(x-3 2)=-f(-x-3 2) that is, f(x-3 2)+f(-x-3 2)=0 so that x=x+3 2 gives f(x)+f(-x-3)=0

And from f(x+3)+f(x)=0 we get f(x+3)=f(-x-3) so that x=x-3 i.e. f(x)=f(-x) so that f(x) is an even function, symmetric with respect to the y-axis.

The mathematical orange ridge line description of the central symmetry of the point x=2 in the function image is the field answer f(-x)=-f(x+4).

Let x=-2, then f(2)=-f(2), and the circle has f(2)=0