When f (x) = (x + 1) ln (x + 1) + m (x ^ 2 + 2x) x > = 0, f (x)

When f (x) = (x + 1) ln (x + 1) + m (x ^ 2 + 2x) x > = 0, f (x)

F '(x) = ln (x + 1) + 1 + 2m (x + 1) = ln (x + 1) + 2mx + 2m + 1. If x ≥ 0, ln (x + 1) ≥ ln1 = 0 and f (x) ≤ 0 is constant, then f' (x) ≤ 0 is necessary. So 2mx + 2m + 1 ≤ 0 is m ≤ - 1 / 2 (x + 1) ≤ - 1 / 2. At this time, the function only needs f (0) = 0 ≤ 0, so m ≤ - 1 / 2. I hope it can help you O (∩_ ∩)O...

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