1、f'(x)=a/x+2bx+11和2是极值点所以1和2是f'(x)=0的解代入解得b=-1/6,a=-2/3f'(x)=-2/(3x)-x/3+1=-(x^2-3x+2)/(3x)定义域x>0所以02时,-(x^2-3x+2)0,f'(x)>0,增函数所以单调增区间(1,2)单调减区间(0,1),(2,+∞)2、f'(x)=3x^2+6mx+n,在x=-1时有极值0则x=-1,f'(-1)=0且f(-1)=0f'(-1)=3-6m+n=0f(-1)=-1+3m-n+m^2=0m=1,m=2则... |