已知函数f(x)=x3+ax2+bx+c在x=-23与x=1时都取得极值(1)求a,b的值和...

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热心网友 时间：20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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