If f (x) satisfies f (x + 2) = f (x), f (2 + x) = f (2-x), and X belongs to [2,3], f (x) = (X-2) ^ 2, find the expression of F (x) in the interval [4,6]

If f (x) satisfies f (x + 2) = f (x), f (2 + x) = f (2-x), and X belongs to [2,3], f (x) = (X-2) ^ 2, find the expression of F (x) in the interval [4,6]

First, we define f (x) = f (x + T) (t is a real number) as a periodic function. T is the period of this function. That is to say, if any value x in the domain is taken, then the function f (x + T) = f (x) corresponding to the value x + T ∈ D
Because f (2 + x) = f (2-x), f (x) is symmetric with respect to the straight line x = 2. So the analytic expression of the function in [1,3] image is still f (x) = (X-2) ^ 2. And 1 + 2 = 3. So 1 and 3 are exactly a period. So f (1) = f (3). And 1 increases gradually. As long as the increasing value of 3 is the same as that of 1, the shape of the function will not change
So 1 + 3 = 4.3 + 3 = 6. In the interval [4,6]. The shape of F (x) does not change, but it shifts 3 units to the right, and the analytical formula becomes f (x) = (x-2-3) ^ 2 = (X-5) ^ 2
I wonder if this will solve your problem