51问答网 > 判断函数f(x)=x⼀(2的x次方-1)+x⼀2的奇偶性

判断函数f(x)=x⼀(2的x次方-1)+x⼀2的奇偶性

2025-01-07 00:23:15
推荐回答（2个）
回答1：

f（-x）=-x/（2^-x-1)+(-x)/2
=-x[1/(2^-x-1)+1/2]
=-x[2^x/(1-2^x)+1/2]
=x[2^x(2^x-1)-1/2]
=x[(2^x-1+1)/(2^x-1)-1/2]
=x[1+1/(2^x-1)-1/2]
=x[(1/(2^x-1)+1/2]
=x/(2^x-1)+x/2=f(x)
所以f（x）为偶函数

回答2：

偶函数

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