Question

已知函数fx=(1/(2^x-1)+1/2)x判断奇偶性?和证明fx＞0

Answer 1

已知函数f（x）=(1/(2^x-1)+1/2)*x^3.
（1）判断f(x)的奇偶性
f(-x)=(1/(2^(-x)-1)+1/2)*(-x)^3
=-(2^x/(1-2^x)+1/2)*x^3
=-(-2^x/(2^x-1)+1-1/2)*x^3
=-((-2^x+2^x-1)/(2^x-1)-1/2)*x^3
=-(-1/(2^x-1)-1/2)*x^3
=(1/(2^x-1)+1/2)*x^3.
所以f(x)为偶函数.
（2）证明f（x）>0
满足f(x)成立，需2^x-1≠0,即x≠0.
x>0时，x^3>0
又因为2^x-1>0,所以1/(2^x-1)>0
1/(2^x-1)+1/2>0
则f(x)=(1/(2^x-1)+1/2)*x^3>0
因为f(x)为偶函数，
x<0时,有f(-x)=f(x)>0
所以f(x)>0