已知函数f(x)=x(x-a)(x-b),点A(s,f(s)),B(t,f(t)).
(1)若a=0,b=3,函数f(x)在(t,t+3)上既能取到极大值,又能取到极小值,求t的取值范围;
(2)当a=0时,
![manfen5.com 满分网](http://img.manfen5.com/res/GZSX/web/STSource/20131024183252755616652/SYS201310241832527556166020_ST/0.png)
对任意的
![manfen5.com 满分网](http://img.manfen5.com/res/GZSX/web/STSource/20131024183252755616652/SYS201310241832527556166020_ST/1.png)
恒成立,求b的取值范围;
(3)若0<a<b,函数f(x)在x=s和x=t处取得极值,且
![manfen5.com 满分网](http://img.manfen5.com/res/GZSX/web/STSource/20131024183252755616652/SYS201310241832527556166020_ST/2.png)
,O是坐标原点,证明:直线OA与直线OB不可能垂直.
考点分析:
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