如图为函数f(x)=x(0<x<1)的图象,其在点M(t,f(t))处的切线为l,l与y轴...

发布网友 发布时间:2024-10-19 06:06

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热心网友 时间:2024-10-19 06:08

对函数求导可得,f′(x)=12x,由题意可得M(t,t),
切线的斜率k=f′(t)=12t
过点M的切线方程为y-t=12t(x-t)
则可得P(0,t2),N(0,1),Q(2t-t,1),
s△PNQ=12PN?NQ=12(2t-t)(1-t2)=t-t+tt4,
令g(t)=t-t+tt4(0<t<1)
g′(t)=38t+12t-1=3t?8t+48t=(3t?2)(t?2)8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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._5dn6uyv{margin-right:8px;max-width:35px;min-width:35px;vertical-align:top}body .qrpnxfs{background:url(//nv00.cdn.bcebos.com/nv01/static/ecom/img/pc/head-img-535c333798.png) no-repeat 50%;background-size:100% 100%;height:100%;width:100%}body .nmkjhaf{height:35px;min-width:0}body .v1ftvns{color:#000;font-size:16px;font-weight:inherit;line-height:1;margin-bottom:8px;overflow:hidden;text-overflow:ellipsis;white-space:nowrap}body ._7pt3pe5{color:#999;font-size:12px;line-height:1}body .c69sfzc .m11ehk9 img{width:100%}body .c69sfzc ._7enmqsz{margin-right:15px}body ._5dn6uyv{border-radius:50%;max-width:40px;min-width:40px;overflow:hidden}body .v1ftvns{color:#333;font-size:14px;font-weight:700;line-height:20px;margin-bottom:0}body ._7pt3pe5{color:#9eacb6;line-height:17px;margin-top:3px} 百度教育广告2024-10-16【答案】有关函数知识点搜题学习,秒出答案,试卷解析,强化学习。【解析】有关函数知识点,各版本教材的试题试卷,尽在百度教育。www.baidu.com !function(t){"use strict";function e(t,e){if(t){var n=new Date().getTime(),i=new 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热心网友 时间:2024-10-19 06:08

对函数求导可得,f′(x)=12x,由题意可得M(t,t),
切线的斜率k=f′(t)=12t
过点M的切线方程为y-t=12t(x-t)
则可得P(0,t2),N(0,1),Q(2t-t,1),
s△PNQ=12PN?NQ=12(2t-t)(1-t2)=t-t+tt4,
令g(t)=t-t+tt4(0<t<1)
g′(t)=38t+12t-1=3t?8t+48t=(3t?2)(t?2)8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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热心网友 时间:2024-10-19 06:10

对函数求导可得,f′(x)=12x,由题意可得M(t,t),
切线的斜率k=f′(t)=12t
过点M的切线方程为y-t=12t(x-t)
则可得P(0,t2),N(0,1),Q(2t-t,1),
s△PNQ=12PN?NQ=12(2t-t)(1-t2)=t-t+tt4,
令g(t)=t-t+tt4(0<t<1)
g′(t)=38t+12t-1=3t?8t+48t=(3t?2)(t?2)8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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