Question

设fx是偶函数，gx是奇函数，且fx+gx=1/(x-1),求fx,gx的解析式

步骤 谢谢

Answer 1

f(x)是偶函数,g(x)是奇函数
f(x)+g(x)=1/(x-1),(1)
所以有;
f(x)=f(-x)
g(x)=-g(-x)
令(1)中x=-x,
则有:
f(-x)+g(-x)=-1/(x+1)
f(x)-g(x)=-1/(x+1),(2)
由(1)(2)把f(x),g(x)当作两个未知数,解出
(1)+(2),
2f(x)=1/(x-1)-1/(x+1)=2/(x^2-1)
f(x)=1/(x^2-1)
所以
g(x)=x/(x^2-1)

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f(x)是偶函数，g(x)是奇函数
则f(-x)=f(x) g(-x)=-g(x)

f(x)+g(x)=1/(x+1)①
取x为-x
f(-x)+g(-x)=1/(1-x)
f(x)-g(x)=1/(1-x)②

①+②得
2f(x)=1/(x+1)+1/(1-x)=2/(1-x^2)
f(x)=1/(1-x^2)
①-②得
2g(x)=1/(x+1)-1/(1-x)=-2x/(1-x^2)
g(x)=-x/(1-x^2)