已知函数f(x)=x^3+bx+cx+2在x=2/3处取得极值
确定函数f(x)的解析式 求函数f(x)的单调区间
人气:388 ℃ 时间:2019-08-20 16:42:26
解答
先求导,得f’(x)=3x^2+b+c
当取极值时,即f’(2/3)=0,解得b+c=-4/3
所以f(x)=x^3-(4/3)*x+2
单调增区间f’(x)大于0,解不等式即可.x大于2/3或x小于-2/3
减区间同理,导数小于0,解不等式.x小于2/3且x大于-2/3
(等号取不取均可,无过多要求!)
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