|
f'(x)=3ax2+2bx-a2,a>0f'(x)=0有两不等实根x1,x2,x1≠x2?△>0?b2+3a3>0,成立∵|x1|+|x2|=2√2x1=[-b-√(b2+3a3)]/3a<0,x2=[-b+√(b2+3a3)]/3a可以得到-b-√(b2+3a3)<0?x2<0∴|x1|+|x2|=x1-x2=2√2因为韦达定理:x1x2=-a/3,x1+x2=-2b/3a进而x1+x2=2√[b2/a2+3a]/3=2√2进而b2=18a2-3a3对t=18a2-3a3t'=36a-9a2?t'=0?a=0舍去,a=4讨论得到:a=4取得最大即t=96即b2=96?bmax=4√6 |
