Given the function f (x) = LNX ax. (1) when a > 0, judge the monotonicity of F (x) in the domain of definition; (2) if the minimum value of F (x) in [1, e] is 32, find the value of A

(1) The definition field of function is (0, + ∞), and f ′ (x) = x + AX2 ∵ a > 0, f ′ (x) > 0, f (x) increases monotonically in the definition field; (2) from (1), we know that f ′ (x) = x + AX2. If a ≥ - 1, then x + a ≥ 0, that is, f ′ (x) ≥ 0 is constant on [1, e], then f (x) on [1, e] is

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