已知函数f(x)=ax^3+bx^2+cx在x=0和x=1时取得极值,又f'(1/2)=3/2.(1)求f(x)的解析式;(2)求其单调区间
人气:310 ℃ 时间:2019-08-19 15:49:21
解答
求出f'(x)=3ax^2+2bx+c
极值在0和1处,故f‘(0)=f'(1)=0
解得:a=-2 b=3 c=0
(2)f(x) 在(-无穷,0) (1,+无穷下降);(0,1)上升.
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