∵函数f(x)=(k2-3k+2)x+b在R上是减函数,
∴k2-3k+2<0,即(k-1)(k-2)<0,
解不等式可得1<k<2
∴k的取值范围为:(1,2)
故答案为:(1,2)
若函数f(x)=(k2-3k+2)x+b在R上是减函数,则k的取值范围为_.
若函数f(x)=(k2-3k+2)x+b在R上是减函数,则k的取值范围为______.
其他人气:871 ℃时间:2020-02-01 09:27:00
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