设f(x)=ax2+(b-8)x-a-ab,不等式f(x)>0的解集是(-3,2 发布时间:2017-05-06 20:48 │ 来源:www.tikuol.com 题型:解答题 问题: 设f(x)=ax2+(b-8)x-a-ab,不等式f(x)>0的解集是(-3,2).(1)求f(x);(2)当函数f(x)的定义域是[0,1]时,求函数f(x)的值域.
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