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Given that the derivative of the function f (x) is f '(x) = 4 & sup3; x-4x, and the image passes through the point (0,3), when the function f (x) reaches the minimum, the value of X should be?
F '(x) = 4x & # 179; - 4x = 4x (X & # 178; - 1) = 4x (x + 1) (x-1) f (x) = x ^ 4-2x & # 178; + CF (0) = C = 3f (x) = x ^ 4-2x & # 178; + 3F' (x) = 0 x = 0 or x = 1 or x = - 1x0 f (x) when increasing the minimum value of 00 f (x), x = - 1 or x = 1
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