函数f(x)=4x3+ax2+bx+5的图象在x=1处的切线方程为y=-12x, 发布时间:2017-05-14 23:17 │ 来源:www.tikuol.com 题型:解答题 问题: 函数f(x)=4x3+ax2+bx+5的图象在x=1处的切线方程为y=-12x,(1)求函数f(x)的解析式;(2)求函数f(x)的单调递减区间.
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