已知函数fx=(1/(2^x-1)+1/2)x^3,判断奇偶性,证明fx>0

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已知函数fx=(1/(2^x-1)+1/2)x^3,判断奇偶性,证明fx>0
已知函数fx=(1/(2^x-1)+1/2)x^3,判断奇偶性,证明fx>0

已知函数fx=(1/(2^x-1)+1/2)x^3,判断奇偶性,证明fx>0
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,
x0
所以f(x)>0