如图所示，望采纳
f(x)=1/2*x²+2x+ln(-x)f'(x)=x+2+(1/x)=(x+1)²/x当-4≤x≤-1时，f'(x)≤0，∴f(x)在[-4,-1]上单调递减∴f(x)最大值=f(-4)=8-8+ln4=2ln2f(x)最小值=f(-1)=1/2-2+0=-3/2
