已知ex求ex2
作者：户外装备党
KAILAS从2016年推出FUGA越野跑系列，至今已有八年，根据使用者的反馈持续改进。FUGA EX3作为EX缓震系列最新型号，在两次迭代后，真正成为KAILAS旗下最均衡能打的越野跑鞋，在性能上正面硬刚目前主流的越野跑鞋。
从左到右：FUGA EX1代（红）、EX2代（橙）、EX3代（白）
相对于FUGA DU大坡王为特定路况的激进设计，FUGA PRO系列的竞速轻量，EX系列则强调长距离舒适和缓震，口碑在越野跑圈一直不错。EX3则在缓震、抓地、包裹、舒适、耐用几个方面做得更均衡，更适合大体重选手，长距离赛程，也不为强化缓震和抓地过分牺牲灵活的操控脚感，能应对更复杂的路况。
KAILAS和几款主流越野跑鞋抓地的横向对比
EX3继承了前两代的明显特点，前后掌仍然有8mm落差，双扣快调鞋带，外侧鞋带快解扣，鞋套连接点等。
1、重量
三代都保持在285克左右。
同为42码的EX3是284.3克，EX2是288.9克，EX1为284.4克。
2、大底
EX3相较前两代明显加大了Vibram Megagrip大底的触地面积，增强抓地摩擦力。仍然采用全大底，而非有些越野跑鞋偷轻的半底，全大底的耐磨、耐用性、抓地止滑效果更好。
前脚掌宽EX3为118.32mm，比EX2增加近8mm，比EX1增加近7mm。
后底EX3则加宽更多，为96.03mm。比前两代增加了整整13mm。
大底纹路也有较明显调整，EX3前脚掌由四段式改为五段式，脚后部则排列了更密集的止滑齿纹。大底刻有专利号PAT NO.202330091563.6。
3、中底
EX3的中底采用了目前流行的突出外包造型，脚跟更明显，提升缓震和稳定性。下坡时，显现缓震优势。
相较于一二代，EX3的大底在脚掌和脚跟外包到中底，提供更好的保护支撑。
EX3突出的脚跟
4、包裹性
EX3的鞋跟有卡位填充突起，更包裹跟腱，提供保护也更“跟脚”，减少晃动摩擦。这也是较前两代明显提升脚感的一点。
EX3鞋舌的连接固定纱网回到一代设计，更靠上。鞋舌材料则和二代相同。综合了前两代的长处。
鞋舌上鞋带收纳口更合理，加大容量，加高位置，脚瘦的使用者也能把多余鞋带收纳进去，上坡时这位置也不会再卡脚脖子。
EX3鞋舌最大的变化，终于抛弃了鞋舌上的纱网，这应该是宽脚的福音吧。在长距离比赛中，脚面肿胀，确实需要松松鞋带或者穿厚点的袜子时，这层纱网就特别有束缚感。
取消鞋舌上纱网的连带好处，鞋口更大，更方便穿脱。
新款调整了前脚面鞋带调扣位置，做了预留弯折切口，移动了鞋带孔和打枣加固位，更符合人体工程学。应是吸取了真实用户的建议而改进。
EX3的鞋头直观上更上翘，提速更省力，大底前伸面积更大。
5、鞋面
EX3鞋头保护包胶面积更大，强化保护性。在大拇指处做了加厚。
侧面则保留了二代的设计，经编工程网面，厚度明显强于一代。
鞋套绊扣也同二代，做工优于一代。
KAILAS听取了大量跑者实际穿着反馈意见后，改进推出全新一代FUGA EX3，性能均衡，完全可以与一众国际品牌掰掰手腕一较高下。
KAILAS 的FUGA系列这几年已经陪许多越野跑者征战在各个赛场，今年KAILAS运动员Franco在意大利TORX巨人之旅330公里组摘得桂冠，并打破了纪录。
阅读更多越野跑鞋精彩内容，可前往什么值得买查看
","gnid":"9722cccf58f8a7061","img_data":[{"flag":2,"img":[{"desc":"","height":"810","title":"","url":"https://p0.ssl.img.360kuai.com/t01c918b7d498085c63.jpg","width":"1080"},{"desc":"","height":"2430","title":"","url":"https://p0.ssl.img.360kuai.com/t01b45cc61900582777.jpg","width":"1080"},{"desc":"","height":"162","title":"","url":"https://p0.ssl.img.360kuai.com/t017be2df0118eb1c56.jpg","width":"1080"},{"desc":"","height":"2430","title":"","url":"https://p0.ssl.img.360kuai.com/t01d43f43fd4da3f1b3.jpg","width":"1080"},{"desc":"","height":"162","title":"","url":"https://p0.ssl.img.360kuai.com/t017be2df0118eb1c56.jpg","width":"1080"},{"desc":"","height":"2430","title":"","url":"https://p0.ssl.img.360kuai.com/t017d8939a6d4ce8e1a.jpg","width":"1080"},{"desc":"","height":"2430","title":"","url":"https://p0.ssl.img.360kuai.com/t01715c78ed36a08d9b.jpg","width":"1080"},{"desc":"","height":"1620","title":"","url":"https://p0.ssl.img.360kuai.com/t01626608c1b2d3f7e4.jpg","width":"1080"},{"desc":"","height":"162","title":"","url":"https://p0.ssl.img.360kuai.com/t017be2df0118eb1c56.jpg","width":"1080"},{"desc":"","height":"1439","title":"","url":"https://p0.ssl.img.360kuai.com/t01665505d2c7bff898.jpg","width":"1080"},{"desc":"","height":"1439","title":"","url":"https://p0.ssl.img.360kuai.com/t0118b7cd3413781a55.jpg","width":"1080"},{"desc":"","height":"810","title":"","url":"https://p0.ssl.img.360kuai.com/t01a854b0d9edd6e468.jpg","width":"1080"},{"desc":"","height":"162","title":"","url":"https://p0.ssl.img.360kuai.com/t017be2df0118eb1c56.jpg","width":"1080"},{"desc":"","height":"270","title":"","url":"https://p0.ssl.img.360kuai.com/t01334c749872b46f9b.jpg","width":"1080"},{"desc":"","height":"810","title":"","url":"https://p0.ssl.img.360kuai.com/t0135e89623bad6fa65.jpg","width":"1080"},{"desc":"","height":"810","title":"","url":"https://p0.ssl.img.360kuai.com/t015fdda917b8c16dc9.jpg","width":"1080"},{"desc":"","height":"479","title":"","url":"https://p0.ssl.img.360kuai.com/t01a1da74a18acc1367.jpg","width":"1080"},{"desc":"","height":"810","title":"","url":"https://p0.ssl.img.360kuai.com/t016a200deb2f317832.jpg","width":"1080"},{"desc":"","height":"479","title":"","url":"https://p0.ssl.img.360kuai.com/t0151b5d75239ddc76c.jpg","width":"1080"},{"desc":"","height":"810","title":"","url":"https://p0.ssl.img.360kuai.com/t019c59745b6c4ae10f.jpg","width":"1080"},{"desc":"","height":"2430","title":"","url":"https://p0.ssl.img.360kuai.com/t010bd75af4e6caaed7.jpg","width":"1080"},{"desc":"","height":"810","title":"","url":"https://p0.ssl.img.360kuai.com/t014f1b4d879501ae34.jpg","width":"1080"},{"desc":"","height":"810","title":"","url":"https://p0.ssl.img.360kuai.com/t019811f9de1444c4a3.jpg","width":"1080"},{"desc":"","height":"810","title":"","url":"https://p0.ssl.img.360kuai.com/t01e3a635d7f1edc7b1.jpg","width":"1080"},{"desc":"","height":"162","title":"","url":"https://p0.ssl.img.360kuai.com/t017be2df0118eb1c56.jpg","width":"1080"},{"desc":"","height":"810","title":"","url":"https://p0.ssl.img.360kuai.com/t0126977b404030a13c.jpg","width":"1080"},{"desc":"","height":"2430","title":"","url":"https://p0.ssl.img.360kuai.com/t019f08ec3d8e3848aa.jpg","width":"1080"},{"desc":"","height":"270","title":"","url":"https://p0.ssl.img.360kuai.com/t013ba63c3cd2bf687c.jpg","width":"1080"},{"desc":"","height":"162","title":"","url":"https://p0.ssl.img.360kuai.com/t017be2df0118eb1c56.jpg","width":"1080"}]}],"original":0,"pat":"zzc,art_src_1,fts0,sts0","powerby":"hbase","pub_time":1695979263000,"pure":"","rawurl":"http://zm.news.so.com/b52d2f967b2bc1945ad58d8ed48c1e9e","redirect":0,"rptid":"eabb5e1287ff2af3","rss_ext":[],"s":"t","src":"什么值得买","tag":[{"clk":"kfashion_1:越野","k":"越野","u":""}],"title":"户外装备党 篇三十七：持续进化，KAILAS FUGA EX 最能打的越野跑鞋
诸晶曲1657设函数f(x)=ex - ln(x+1).(1)求函数f(x)的最小值;(2)已知0≤x1<x2,求证:ex2 - x1>lne(x2+1)x -
滑施向15238238358 ______ 解答:(1)解:函数f(x)=ex-ln(x+1)的导数 f′(x)=ex-1 x+1 = xex+ex?1 x+1 (x>-1),令h(x)=xex+ex-1,h′(x)=(x+2)ex>0,则h(x)在(-1,+∞)递增,由于h(0)=0,当-10时,f′(x)>0,则f(x)min=f(0)=e0-ln1=1;(2)证明:由(1)得当x>0时,ex>ln(x+1)+1,由于...
诸晶曲1657已知函数f(x)=ex - ln(x+1)(1)求f(x)最小值;(2)已知:0≤x1
滑施向15238238358 ______[答案] (1)f′(x)=ex− 1 x+1(x>−1),x>0时f′(x)>0,-1
诸晶曲1657关于随机变量的一些题麻烦各位大虾了.1、已知X~B〔8,1/4〕,(1)EX,DX;(2)若Y=6X,2、已知离散型随机变量的X的分布列为X 1 2 3 4P 1/6 1/3 1/6 a(1)求EX(2)... -
滑施向15238238358 ______[答案] 1. (1). X B[8 ,1/4] , E(X) = 8 * 1/4 = 2 ,D(X) = 8 * 1/4 * (1 - 1/4) = 3/2 , (2). E(Y) = E(6X) = 6 E(X) = 12 . 2. (1). 1/6 + 1/3 + 1/6 + a = 1 ==> a = 1/3 . E(X) = 1 * 1/6 + 2 * 1/3 + 3 * 1/6 + 4 * 1/3 = 8/3 . (2). E(N) = E(2X+3) = 2 E(X) + 3 = 25/3 . 3. E(Y) = 0 + 1 * 0.2 +...
诸晶曲1657数学帝!已知EX=3,E(X+2) -
滑施向15238238358 ______ E(X+2)=EX+E2=3+2=5
诸晶曲1657关于概率论问题 已知EX= - 1,DX=3,则E[3(X*X - 2)]等于多少?? -
滑施向15238238358 ______ DX=EX^2-(EX)^2 EX^2=DX+(EX)^2=10 E[3(X*X-2)]=3E(x^2)=30
诸晶曲1657已知函数y=ex(1)求这个函数在x=e处的切线方程;(2)过原点作曲线y=ex的切线,求切线的方程. -
滑施向15238238358 ______[答案] (1)函数y=ex,f(e)=ee,则切点坐标为(e,ee), 求导y′=ex,则f′(e)=ee,即切线斜率为ee, 则切线方程为y-ee=ee(x-e), 化简得y=eex-ee+1+ee; (2)y=ex,y′=ex, 设切点的坐标为(x0,ex0), 则切线的斜率为f′(x0)=ex0, 故切线方程为y-ex0=ex0(x-x0)...
诸晶曲1657设函数f(x)=ex.(I)求证:f(x)≥ex;(II)记曲线y=f(x)在点P(t,f(t))(其中t<0)处的切线为l,若l与x轴、y轴所围成的三角形面积为S,求S的最大值. -
滑施向15238238358 ______[答案] (I)证明:设g(x)=ex-ex,∴g′(x)=ex-e,由g′(x)=ex-e=0,得x=1,∴在区间(-∞,1)上,g′(x)<0,函数g(x)在区间(-∞,1)上单调递减,在区间(1,+∞)上,g′(x)>0,函数g(x)在区间(1,+∞...
诸晶曲1657设函数f(x)=ex - ln(x+1).(Ⅰ)求函数f(x)的最小值;(Ⅱ)已知0≤x1<x2.求证:ex2?x1>lne(x2+1 -
滑施向15238238358 ______ (Ⅰ)f′(x)=ex?1 x+1 ;∴-11 e 1 x+1 >1,∴f′(x)x>0时,ex>1,01 x+1 0,∴x=0时,f(x)取到最小值1. (Ⅱ)由题意知:x2-x1>0;∴f(x2-x1)>f(0),即e(x2?x1)?ln(x2?x1+1)>1,即e(x2?x1)>lne(x2?x1+1);∴要使:e(x2?x1)>ln e(x2+1) x1+1 ,我们来证lne(x2?x1+...
诸晶曲1657设随机变量X服从参数为1的泊松分布,则P{X=EX2}=______. -
滑施向15238238358 ______[答案] 由于随机变量X服从参数为1的泊松分布, 所以:E(X)=D(X)=1 又因为:DX=EX2-(EX)2, 所以:EX2=2, X 服从参数为1的泊松分布, 所以:P{X=2}= 1 2e−1, 故答案为: 1 2e−1.
诸晶曲1657已知EX=4,则E(2X - 3)= -
滑施向15238238358 ______[答案] 已知EX=4,则E(2X-3)= 0 - 离问题结束还有 14 天 23 小时 解释一下为什么好吗 SEEK_KEY - 魔法学徒 一级 回答: E(2X-3)=2EX-3=5 因为E(ax+b)=aEx+b