Question

已知f(x)=x^3×(1÷(2^)x-1)+1/2)判断函数的奇偶性 证明f(x)>0

Answer 1

f(x)+f(-x)=x^3*[1/(2^x - 1)+ 1/2] + (-x)^3*[1/(2^(-x) - 1)+ 1/2]
=x^3*[1/(2^x - 1)- 1/(2^(-x) - 1)] = x^3*[1/(2^x - 1)- 2^x/(1 - 2^x)]
= x^3*[1/(2^x - 1)+ 2^x/(2^x -1)] = x^3*[ (2^x + 1)/(2^x -1)]
= x^3*[ 1 + 2/(2^x -1)]=2*{x^3*[ 1/2 + 1/(2^x - 1)]} = 2f(x)

所以f(x)是偶函数。
f(x) = x^3*[ (2^x + 1)/(2^x -1)]
当x>0时，(2^x-1)>0, x^3>0, (2^x + 1)>0 ，所以此时f(x)>0
f(x)是偶函数，则x<0时，f(x)也大于0
f(0) = 0;
综上，f(x)>=0（如果刨除零点，就满足f(x)>0）