Let f (x) = (2x + 1) ln (2x + 1); (1) find the minimum of F (x); (2) if x ≥ 0, f (x) ≥ 2aX holds, find the value range of real number a

Let f (x) = (2x + 1) ln (2x + 1); (1) find the minimum of F (x); (2) if x ≥ 0, f (x) ≥ 2aX holds, find the value range of real number a

(1) ∵ f ′ (x) = 2ln (2x + 1) + 2, f ′ (x) = 0, ∵ x = 12 (1E − 1) when x ∈ (12 (1E − 1), & nbsp; +If x ∈ (− 12, (12 (1E − 1)), the minimum value of F ′ (x) ＜ 0 is f (12 (1E − 1) + = - 1E (2) Let g (x) = (2x + 1) ln (2x + 1) - 2axg ′ (x) = 2 [ln (2x + 1) + 1-A] = 0, x = 12ea − 1 − 1, f (x) ≥ 2aX holds when a ≤ 1, A-1 ≤ 0, 12ea − 1 − 1 ≤ 0g ′ (x) ≥ 0, G (x) increases only on [0, + ∞), and G (x) ≥ g (0) = 0. For X ≥ 0, f (x) ≥ 2aX holds. For a > 1, A-1 > 0, 12 (EA − 1) When x ∈ [0,12 (EA − 1 − 1)), G ′ (x) ＜ 0 is always true, and G (0) = 0, when x ∈ [0,12 (EA − 1 − 1)), G (x) ≤ g (0) = 0 is true, that is, when a > 1, not all x ≥ 0 have f (x) ≥ 2aX, so a ≤ 1