It is proved that the periods of F (x + a) = - f (x) and f (x + a) = 1 / F (x) are 2A The approximate step is f (x + 2a) = f [(x + a) + a] = - f (x + a) = f (x), so f (x + 2a) = f (x), so the period is 2A F (x + 2a) = f [(x + a) + a] = 1 / F (x + a) = f (x), so f (x + 2a) = f (x), so the period is 2A These are the teacher's steps, but I can't understand them now,

It is proved that the periods of F (x + a) = - f (x) and f (x + a) = 1 / F (x) are 2A The approximate step is f (x + 2a) = f [(x + a) + a] = - f (x + a) = f (x), so f (x + 2a) = f (x), so the period is 2A F (x + 2a) = f [(x + a) + a] = 1 / F (x + a) = f (x), so f (x + 2a) = f (x), so the period is 2A These are the teacher's steps, but I can't understand them now,

I think the teacher's practice is very standard
Let x + a = t, then
F (x + 2a) = f (T + a) - this step is substitution
=-F (T) - this step uses the property of F (x + a) = - f (x)
=-F (x + a) - this step is to restore the part of the substitution
=-[- f (x)] - here we use the property of F (x + a) = - f (x) again
=f(x)
The essence is the same. The solution to another question is the same. If you still don't understand it, you can ask

Quadratic function f (x), for any x ∈ R
① Let f (2 + x) = f (2-x)
② There is f (x-4) = f (2-x)
Is the axis of symmetry 2 and - 1 respectively?
Is it possible to find the axis of symmetry by dividing all the expressions of periodic functions by 2?

You're right, but notice that the axis of symmetry is a straight line, not a number
If f (x) satisfies f (a + x) = f (b-X), then the axis of symmetry is x = (a + b) / 2
For any x ∈ R, f (x) is a quadratic function
① Let f (2 + x) = f (2-x)
② There is f (x-4) = f (2-x)
The axis of symmetry is x = 2 and x = - 1, respectively

If a function has two axes of symmetry, what is its period?

Its period is twice the absolute value of the abscissa subtraction of the two axes of symmetry

If a function has two axes of symmetry, then its period? Please prove

Let the period be a and B respectively, then f (x + a) = f (x), f (x + b) = f (x), that is, f (x + a) = f (x + b), then the period is | A-B |

1. What forms can explain the axis and center of symmetry of a function? 2. What forms can explain the period of a function?

For any x, if f (x) = f (2a-x), then f (x) is axisymmetric with respect to the line x = a; for any x, if f (x) = - f (2a-x), then f (x) is centrosymmetric with respect to the point (a, 0); for any x, if f (x) = f (x + T), then f (x) is a periodic function and t is its period

If f (x) is an even function and the image is symmetric with respect to x = 2, then the period of F (x) is_____________

f(x+2)=f(x-2)
f(x+2+2)=f(x+2-2)
That is, f (x + 4) = f (x)
So the period is four

If f (x) is an even function and the image is symmetric with respect to point (a, 0), it is proved that 4a is the period of F (x)
F (- x) = f (x), f (x) is symmetric about point (a, 0), f (2a-x) = - f (x) = f (x-2a), f (x + 2a) = - f (x); f (x + 4a) = - f (x + 2a) = f (x)
What's the meaning of "f (2a-x) = - f (x) = f (x-2a)" here? How to get it

If f (x) is symmetric with respect to (a, 0), then f (x) + F (- x + 2a) = 0. Generally, if f (x) is symmetric with respect to (a, b), then f (x) + F (- x + 2a) = 2B

How to draw the image of piecewise function in grade one of senior high school?
Draw the following function image:
F（x)=｛ 0,x≤0,
1,x＞0;

Draw a line parallel to the x-axis on y = 1. Draw a hollow point on y = 1 to the right

Piecewise function and mapping how to draw function image? (hope to give an example) thank you

How to make function image in senior one

Point drawing
Take a few key points first, and then draw