已知f(x)是偶函数,g(x)是奇函数,且在公共定义域{x丨x∈R,x≠±1}上有f(x)+g(x)=1/x-1,
问题描述:
已知f(x)是偶函数,g(x)是奇函数,且在公共定义域{x丨x∈R,x≠±1}上有f(x)+g(x)=1/x-1,
求f(x)的解析式.
问题发的有出入,正确的是.....f(x)+g(x)=1/(x-1)....抱歉...再次求正确答案...
答
x≠±1时有
f(x)+g(x) = 1/(x-1)
f(-x)+g(-x) = f(x)-g(x) = -1/(x+1)
两式相加有 2f(x) = 1/(x-1) - 1/(x+1) = 2/( (x-1)(x+1) )
得到 f(x) = 1/( (x-1)(x+1) )