(2011安徽省赛)设$f(x)=ax^3+bx+c(a,b,cin R)$,当$0le x le1$时,$0le f(x)le1$,求$b$的可能的最大值.
分析:$f(0)=c,f(1)=a+b+c,f(t)=at^3+bt+c(0<t<1)$
得$(t-t^3)b=f(t)-t^3f(1)-(1-t^3)f(0)le f(t)le1$ 恒成立.
故$ble (dfrac{1}{t-t^3})_{min}$故$ble dfrac{3sqrt{3}}{2}$当$c=0,t=dfrac{sqrt{3}}{3}$时取到.