题目内容
若函数f(x)=ax3+bx2+cx+d满足f(0)=f(x1)=f(x2)="0" (0<x1<x2),且在[x2,+∞上单调递增,则b的取值范围是_________.
(-∞,0)
∵f(0)=f(x1)=f(x2)=0,
∴f(0)=d=0. f(x)=ax(x-x1)(x-x2)=ax3-a(x1+x2)x2+ax1x2x,
∴b=-a(x1+x2),又f(x)在[x2,+∞单调递增,故a>0.
又知0<x1<x,得x1+x2>0,
∴b=-a(x1+x2)<0.
∴f(0)=d=0. f(x)=ax(x-x1)(x-x2)=ax3-a(x1+x2)x2+ax1x2x,
∴b=-a(x1+x2),又f(x)在[x2,+∞单调递增,故a>0.
又知0<x1<x,得x1+x2>0,
∴b=-a(x1+x2)<0.
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