发布网友 发布时间:2024-10-24 11:17
共1个回答
热心网友 时间:18小时前
(1)∵f(x)=x3+ax2+bx+c,∴f′(x)=3x2+2ax+b,
∵函数f(x)在x=-23与x=1时都取得极值,
∴f′(-23)=0,f′(1)=0,
即f′(-23)=129-4a3+b=0,f′(1)=3+2a+b=0
得a=-12,b=-2,
经检验,a=,b=-2符合题意.
则f(x)=x3-12x2-2x+c,
∴f′(x)=3x2-x-2=(3x+2)(x-1),
列表
x(-∞,-23)-23(-23,1)1(1,+∞)f′(x)+0-0+f(x)↑极大值↓极小值↑,则函数的单调增区间为(-∞,-23)和(1,+∞).
(2)由(1)可得函数f(x)的极大值为f(-23)=2227+c,极小值f(1)=-32+c,
要使y=b=-2与函数y=f(x)的图象有3个交点,
则2227+c>-2-32+c<-2,即c>-7627c<-12,即-已赞过已踩过你对这个回答的评价是?评论收起 ._1uevpeq{zoom:1;background-color:#fff;border:0;margin-bottom:10px;padding:30px 0 20px 42px;position:relative}._1uevpeq.ec-1841{padding:20px 0}._1uevpeq.ec-2246{padding:20px 0 10px}.ec-1841 .y7we4hu{font-size:16px;margin-bottom:-5px}.y7we4hu{color:#7a8f9a;height:25px;line-height:25px;overflow:hidden;position:relative}.y7we4hu h2{margin:0;padding:0}.y7we4hu:after{clear:both;content:" ";display:block;height:0;visibility:hidden}a.tycfu7u{color:#666;float:right;font-size:12px;margin-left:8px;text-decoration:none}.hhhv6ex{color:#666;font-size:13px;line-height:normal;line-height:20px;margin-top:10px}.vnsdjzp{margin-top:15px;position:relative}.vnsdjzp h3{font-weight:400;padding:0}.vnsdjzp a{text-decoration:none}.vnsdjzp em{color:#d81419;font-style:normal}.ec-2246 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