It is known that f(x) is a odd function symmetric about (1, 0).
F(-x)=-f(x) Deduction: f(-x+ 1)=-f(x+ 1) Here is a property about (1, 0) symmetry.
About (1, 0) symmetry, half of the sum of abscissas is equal to 1. Here is not only about (1, 0) symmetry, but also about the origin (0, 0) symmetry, the characteristics of odd function, so f(-x)=-f(x).
Then X=-x is brought into the original formula to get a new formula: f(X+ 1)=-f( 1-X).
At this time, we notice that after adding 1 to the independent variable, the latter part -f( 1-X) becomes -f(-X)=f(X).
But this is the result of the independent variable plus 1, so the independent variable of the previous formula should also be added with 1, that is, f(X+2)=f(X).
So the period is calculated.
As for why 1 was added or because the abscissa of the symmetry point was 1, suppose that 0.5 was added at that time.
So can x+0.5 and 0.5-x guarantee the symmetry of these two points about 1?
Suppose x=0.2, one is 0.7 and the other is 0.3. They are not symmetric about 1.
Obviously not, if these two points are on the same side of 1, then there is no universal law, which leads to errors.