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判断f(x)=[1/(2^x-1)+1/2]x奇偶性,

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判断f(x)=[1/(2^x-1)+1/2]x奇偶性,
f(x) = [1/(2^x-1)+1/2]x
f(-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 /(2^x -1) -1 +1/2]x
=[2^x /(2^x -1) - (2^x -1)/(2^x -1) +1/2]x
=[1/(2^x -1) +1/2]x
=f(x)
故f(x)是偶函数.
再问: f(-x) =[1/(2^-x -1)+1/2](-x) = [2^x / (1- 2^x) +1/2](-x) 能不能再详细些
再答: 1/(2^-x -1)= 2^x / (1- 2^x) 【分子分母同乘以 2^x 】
再答: 1/(2^-x -1)= 2^x / (1- 2^x) 【分子分母同乘以 2^x 】
再问: = [2^x /(2^x -1) -1 +1/2]x =[2^x /(2^x -1) - (2^x -1)/(2^x -1) +1/2]x =[1/(2^x -1) +1/2]x 这一步是否有问题
再答: 你好!只是通分,没有问题: 2^x /(2^x -1) - 1 = 2^x /(2^x -1) - (2^x -1)/(2^x -1) = [2^x - (2^x -1)] / (2^x -1) = 1/(2^x -1)