Question

已知函数f(x)=x[1/(2^x-1)+1^2].1.判断f(x)的奇偶性,2.求证f(x)大于0

Answer 1

(1)偶函数

f(x)=x[1/(2^x-1)+1/2]=x*[(2^x+1)/2(2^x-1)]

f(-x)=(-x)[2^x/(1-2^x)+1/2]=x*[2^x/(2^x-1)-1/2]=x*[(2^x*2-2^x+1)/(2^x-1)]=x*[(2^x+1)/2(2^x-1)]

f(x)=f(-x),所以是偶函数

(2)首先x≠0

x>0时,

2^x>2^0=1

即2^x-1>0,所以f(x)>0

当x