已知函数f(x)=(e∧x)/|x|，关于x的方程f²(x)-2af(x)+a=1(a∈R)有四个_百度知道
已知函数f(x)=(e∧x)/|x|，关于x的方程f²(x)-2af(x)+a=1(a∈R)有四个
已知函数f(x)=(e∧x)/|x|，关于x的方程f²(x)-2af(x)+a=1(a∈R)有四个相异的实数根，则a的取值范围为？
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选 D1)x∈(-∞,0)时,f(x)=-e^x/xf'(x)=(1-x)(e^x/x²)&0f(x)在其上单增,值域是(0,+∞)2)x∈(0,+∞)时,f(x)=e^x/xf'(x)=(x-1)(e^x/x²)x∈(0,1)时,f'(x)&0,f(x)在其上单减,值域是(e,+∞)x∈(1,+∞)时,f'(x)&0,f(x)在其上单增,值域是(e,+∞)f'(1)=0,f(x)在x=1处取极小值f(1)=e设g(t)=t²-2at+a-1根据f(x)的图象(可先作出大致图象)得:f²(x)-2af(x)+a-1=0有4个相异实根的充要条件是:方程g(t)=0在(0,e)和(e,+∞)上各有一实根。得a可取的充要条件是:g(e)=e²-2ae+a-1=-(2e-1)+(e²-1)&0且g(0)=a-1&0即 a&(e²-1)/(2e-1) 且a&1又(e²-1)/(2e-1)-1=e(e-2)/(2e-1)&0即(e²-1)/(2e-1)&1所以 a的取值范围是 ((e²-1)/(2e-1),+∞)希望能帮到你！
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我们会通过消息、邮箱等方式尽快将举报结果通知您。e^x的麦克劳林展开式为什么是对的?就是那个e^x=1+x+x^2/2!+.的那个
分类：数学
因为迈克劳林展开式是泰勒公式在x=0处展开得到的,e^x的所有阶导数都为e^x,所以f‘（0),f''(0)等等都为1,因此e^x的迈克劳林展开式就是e^x=1+x+x^2/2!+……
158+262+138 378+44+114+242+222 276+228+353+219 (375+1034)+(966+125) ()+1017 99+999+ 7755－()
－1065 899+344 －317－357 －214 497－299
12×25 75×24 138×25×4 (13×125)×(3×8) (12+24+80)×50 704×25 25×32×125 32×(25+125) 88×125 102×76 58×98 178×101－178 84×36+64×84 75×99+2×75 83×102－83×2 98×199 123×18－123×3+85×123 50×(34×4)×3 25×（24+16） 100+1-100+1 48X99+10+8 5+95X28178×99+178
f(x)=x?+ax+1函数对称轴x=-a/2分类讨论：(1)当-a/2＞1,即a＜-2时,f(x)递减,∴f(x)max=f(-1)=1-a+1=2-af(x)min=f(1)=1+a+1=2+a(2)当-a/2＜-1,即a＞2时,f(x)递增,∴f(x)max=f(1)=1+a+1=2+af(x)min=f(-1)=1-a+1=2-a(3)当-a/2∈[-1,0]时,a∈[0,2],f(x)min=f(-a/2)=a?/4 -a?/2 +1=1 -a?/4f(x)max=f(1)=1+a+1=2+a(4)当-a/2∈[0,1]时,a∈[-2,0]f(x)min=f(-a/2)=1 -a?/4f(x)max=f(-1)=1-a+1=2-a完毕
（1）当其图象经过原点时：4k-2=0,k= 1/2；（2）当y随着x的增大而减小时：k＜0．故答案为：k= 1/2；k＜0．
求MATLAB大神,帮我解释一下这段代码的具体意思,用到什么知识,做PPT用%%%%%Reading of a RGB imagei=imread('touxiang.jpg');I=rgb2gray(i);BW=im2bw(I);% 二值化figure,imshow(BW)[n1n2]=size(BW);r=floor(n1/10);% 分成10块,行c=floor(n2/10);% 分成10块,列 %取最近的较小整数x1=1;x2=r;% 对应行初始化s=r*c;%块面积fori=1:10 %以图像尺寸百分之一的窗口大小滑动y1=1;y2=c; % 对应列初始化for j=1:10if (y2=9*c) | (x1==1 |x2==r*10)%这一判断条件就是把图像最外围全部归零loc=find(BW(x1:x2,y1:y2)==0);%find查找矩阵中满足条件的位置,[r,c,v]=find(X),r,c为不为零元素的行列,v为其数值,X也可由逻辑表达式替代%这里loc应该是BW窗口中的为黑像素的个数,就是后面的o值,在下面的if中做了一个条件判断[o p]=size(loc);pr=o*100/s; % 黑色像素所占的比例数if prmx &(BB2(1,k)/BB2(1,k+1))
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函数f（x）=（1-a/x）*e^x （x&0）有极值的充要条件是什么？
f（x）=（1-a/x）*e^x 导函数有根，且无重根 ,要算吗？ 对f（x）=（1-a/x）*e^x 求导得，我直接通分了。f‘（x）=（（X²-aX+1)/X²）*e^x导函数有根，且无重根 不用管X²，e^x，因为大于0.只要使X²-aX+1有根，且无重根就行了那么只需△=b²-4ac=a²-4&0得a&2或a&-2
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我们会通过消息、邮箱等方式尽快将举报结果通知您。limx趋近于正无穷[x/1+e^-x-ln(1+e^x)_百度知道
limx趋近于正无穷[x/1+e^-x-ln(1+e^x)
limx趋近于正无穷[x/1+e^-x-ln(1+e^x)
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因为是无穷大比无穷大,所以可以分子分母同时开导,limx→∞ ln(1+e^x)/√（1+x^2）=[e^x/(1+e^x)]/[√(1+x²)/x]当x趋近于无穷,[e^x/(1+e^x)]=1当x趋近于无穷,[√(1+x²)/x]是等阶,所以也是等于1所以原式=1
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