Symmetry to find the function analytical, how to use symmetry to solve?

This question is simple, mark it, no one will come to get points again, and I won't join in the fun if there are people.

It can only be proved that f(x) is a periodic function that passes through three points (2a-b, c), (b,c), (3b-2a,c).

His cycle is (4b-4a).Many functions can meet this condition, such as polylines, straight lines, after sinx deformation, and so on. As long as you construct a [a,b] monotonic function, you can extend these four non-one periods to the function on the entire set of real numbers r.

Above, I assume that AB can adjust the positions of A and B, and the period becomes (4A-4B).Construct a [b,a] monotonic function and then extend it to r.

What I'm trying to say is that the analytic formula of the function is uncertain, and what is certain is that there are some special points in its period.

f(x)=f(x+4b-4a) and f[b+k(2b-2a)]=c, k z.

1.Function y=f《x》With respect to x=a symmetry, let any point on the symmetry function be (x,y), then the point (2a-x,y) is on f(x), and y=f(2a-x) is substituted for y=f(2a-x).

2.The function y=f"x", the point "b,c" is symmetrical, and if any point on the symmetry function is (x,y), then the point (2b-x,2c-y) is on f(x), and 2c-y=f(2a-x) is substituted

That is, y=2c-f(2a-x) is sought.

<> "Using symmetry, select a semi-rigid frame (the middle point of the beam is a sliding bearing), and use the displacement method to solve only one unknown quantity, which can be missed to solve it simply (note that the line dust stiffness of the beam is doubled), you can make a bending moment diagram of the semi-rigid frame, and symmetrically draw another half-return skin calendar.

It's useless to look for more, the key is that you have to master the principle.

1.Symmetry f(x+a)=f(b x) Remember that this equation is a general form of symmetry. As long as x has a positive and a negative. There is symmetry. As for the axis of symmetry, you can find x=a+b 2 by eating the formula

For example, f(x+3)=f(5 x) x=3+5 2=4 and so on. This formula is common to those who do not know the equation but know the relationship between the two equations. You can apply it, but I won't give you an example here.

For known equations that require the axis of symmetry, first of all, you have to keep in mind some common symmetry equations for the axis of symmetry. For example, a primitive quadratic equation f(x) = ax2 + bx + c axis of symmetry x b 2a

The axis of symmetry of the original function and the inverse function is y x

And for some functions, it is difficult to say that their axes of symmetry are not only x 90 but also 2n if they are not restricted, such as trigonometric functions, and its axes of symmetry are not only x 90 but also 2n! 90 degrees and so on because his definition is r

f(x) x and his axis of symmetry is x 0, and it should also be noted that some axes of symmetry required after translation by simple functions can be reversed to the original and so on, and then the number of translations can be added

If f(x 3) x 3 makes t x 3, then f(t) t shows that the original equation is shifted by 3 units to the right by the elementary function, and similarly, the axis of symmetry is also shifted to the right by 3 units x 3 (remember that translation is in the form of left addition and right subtraction, as x 3 in this problem illustrates the shift by direction).

2. As for periodicity, let's first start with the general form f(x) f(x t).

Note that the x in this formula is the same sign, not like the symmetry equation, which is positive and negative, and this difference is also the key to determining whether symmetry or periodicity

Also keep in mind some common periodic functions such as trigonometric functions, what sine functions, cosine functions, tangent functions, etc., of course, their minimum periods are 2 , 2 , of course.

Their period is more than that, as long as it is a positive multiple of their minimum period, it can be the period of the problem, e.g. f(x) sinx t 2 (t 2 w).

But if it is f(x) sinx, its period is t, because after adding the absolute value, the graph below the y-axis is all turned to the top, and it is not difficult to see from the graph that the minimum symmetry week t

y1=(sinx)^2=(1-cos2x)/2

y2=(cosx)^2=(1+cos2x)/2

The above 2 equations t (t 2 w).

And for the addition and subtraction composite equation of the equation of the 2 periodic functions, if their periods are the same, then its period is still the same period e.g. y=sin2x+cos2x because they have a common period t, so its period is t

For periods that are not identical, then its period is the least common multiple of their respective periods, such as.

y=sin3 x+cos2 x t1 2 3 t2 1 then t 2 3

This should already be given to the center of symmetry, for example, (a, b), and the main idea of the derivation is that the value of y corresponding to the distance c on both sides of the symmetry point should be equal, that is, f(a+c)=f(a-c); Or in the case of any x a, there is f(x) = f(2a-x).

This should have been given a symmetry center, for example, (a, b), and the main idea of the derivation is that the value of y corresponding to the distance c on both sides of the symmetry point should be equal, that is, f(a+c)=f(a-c); Or in the case of any x a, there is Qin Liang Chun f(x)=f(2a-x).

y=3-2x is the straight-line resistance rule.

Then f(x) is also a straight line.

Take two points on y=3-2x, then they are about.

1,3) of the symmetry point of the car is eliminated on the closed f(x).

Take any two points: a(0,3) and b(1,1).

Let the symmetry point of a be c(a,b).

then (1,3) is the AC midpoint.

So (a+0) 2=1, a=2

3+b)/2=3,b=3

c(2,3)

In the same way, the point of symmetry of b is (1,5).

Pass. 2, 3) and (1, 5).

y-3)/(5-3)=(x-2)/(1-2)f(x)=y=-2x+7

1.The hail number of the Han Dan key y=f《x》With respect to the symmetry of x=a, let any point on the symmetry function be (x,y).

Then the point (2a-x, y) is on f(x), and y=f(2a-x) is what Liang regrets wants.

2.The function mode sail number y=f"x", the point "b,c", is symmetrical, and any point on the symmetry function is (x,y).

Then the points (2b-x, 2c-y) are on f(x), and 2c-y=f(2a-x) is substituted

That is, y=2c-f(2a-x) is sought.

Provide a more general method than the cautious fight on the second floor of Kuan Qinxiang.

Let y=f(x)(x),y).

y=3-2x

y=f(x) with.

y=3-2x symmetry about (first calendar 1,3).

So (a+x) 2=1

b+y)/2=3

Therefore, a=2-x

b=6-y brings in b=3-2a

Get: y=7-2x

Suppose that (x,y) is symmetrical with respect to (x1,y1) and is (m,n), then (x-m) 2=x1,(y-n) 2=y1

Same when the definition field is r.

In f(x)+f(2a-x)=2b, replace x with (a+x), i.e., get:

f(a+x)+f(2a-(a+x))=2b, i.e.: f(a+x)+f(a-x)=2b

Both mean that y=f(x) (x r) is symmetrical in the image with respect to points (a,b).

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

Replace x with (t-a).

f(t) + f(2a-t) = 2b can be obtained

Then let's change t to x and we're good to go.

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

Because sinkx*sinmx is an even function, so.

Primitive = 2 (0, )sinkx*sinmxdx(1)If k=m, then primitive = 2 (0, )sinkx) 2dx= (0, )1-cos2kx)dx

x-(1/2k)*sin2kx]|(0, ) = (2) if k = -m

then the original formula =-2 (0, )sinkx) 2dx=- (0, )1-cos2kx)dx=-[x-(1 2k)*sin2kx]|(0, )=- (3)If k≠ m

then the original formula = (0, )cos(kx-mx)-cos(kx+mx)]dx