已知函数f loga(x)=(1/2x次方-1+...

已知定义在r上的函数f(x)等于2x次方+1分之a-2x次方是奇函数 求实数a的值 判断f(x)的单调性,并证明
f(x)=(a-2^x)/(2^x+1)
跪求帮解出来
f(x)是奇函数, 所以f(-x)=-f(x)
即 (a-2^(-x))/(2^(-x)x+1)=-(a-2^x)/(2^x+1)
a*2^x-1=2^x-a
(a-1)2^x=1-a
因上式对任何x都成立, 所以 a-1=0,
f(x)=(1-2^x)/(1+2^x)
f'(x)=((1-2^x)'(1+2^x)-(1-2^x)(1+2^x)')/(1+2^x)²
=(-2^xln2(1+2^x)-(1-2^x)(2^xln2))/(1+2^x)²
=2^xln2(2^x-1-1-2^x)/(1+2^x)²
=-2ln2*2^x/(1+2^x)²
因此, f'(x)<0
f(x) 单调递减
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/etc/nginx/nginx.conf.已知函数F(x)=A的2x次方+2乘以A的x次方再减一在闭区间-1到1上的最大值为14,求实数A的取值.
暗袭低调175
由题目,得到:F(x)=A^(2X)+2A^x-1=(A^x+1)^2-2-----------(1)设y(x)=A^x,-----------(2)则有:F(x)=(y(x)+1)^2-2,所以有:F(y)=(y+1)^2-2----------------(3)由于(2)式函数y(x)是单调函数,1.如果A>1,则为递增函数,y在[-1,1]上的值为:[A^(-1),A]而(3)式函数F(y)在[A^(-1),A]区间上也是单调上升的,所以,最大值为y=A时的值,即:F(A)=(A+1)^2-2,由题目已知为14,所以:(A+1)^2-2=14,即:A=32.如果A
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