f(x)=(-1/2)x²+bln(x+2)
f'(x)=-x+b/(x+2)
函數(shù)f(x)在(-1,∞)上是減函數(shù)
那么當(dāng)x>-1時,f'(x)≤0恒成立
即-x+b/(x+2)≤0
b/(x+2)≤x
b≤x^2+2x恒成立
∵x^2+2x=(x+1)^2-1
x>-1
∴(x+1)^2-1>-1
∴b≤-1