(2011安徽省赛)
$f(x)=ax^3+bx+c(a,b,cin R),$当$0le x le 1$时,$0le f(x)le 1$,求$b$的可能的最大值.
提示:取三个点$f(0),f(1),f(dfrac{sqrt{3}}{3})$,反解系数得
$2sqrt{3}b=9f(dfrac{sqrt{3}}{3})-sqrt{3}f(1)-(9-sqrt{3})cle9 $得$bledfrac{3sqrt{3}}{2}$
注:关键的$dfrac{sqrt{3}}{3}$由图像可得.
方法2：$f(0)=c,f(1)=a+b+c,f(t)=at^3+bt+c,(0<t<1)$,
得$(t-t^3)b=f(t)-t^3f(1)-(1-t^3)f(0)le f(t)le1$恒成立.
故$ble left(dfrac{1}{t-t^3}
ight)_{min}$故$ble dfrac{3sqrt{3}}{2}$当$c=0,t=dfrac{sqrt{3}}{3}$时取到.