If the function f (x) = x ^ 3-3x ^ 2 + 3x + 1, the center of symmetry of the image is

f(x)=x^3-3x^2+3x+1
The derivative is f '(x) = 3x ^ 2-6x + 3 = 3 (x-1) ^ 2
We obtain that f (x) is increasing on R and K = 0 at x = 0
The center of symmetry of F (x) is (1,2)

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