Question

这个函数的奇偶性怎么证明.>0

Answer 1

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

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

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

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

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

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

f(x)=f(-x)

所以是偶函数