Question

已知函数f(x)=x/(x+1)+1/(x-1) 判断函数f(x)的奇偶性 比较f（a²+a+3）与f(-2)

Answer 1

f(x)=x/(x+1) +1/(x-1)

=[x(x-1) +x+1]/(x²-1)

=(x²+1)/(x²-1)

=1+2/(x²-1),

（1）易得f(-x)=f(x)

从而 f(x)是偶函数.

(2)易得f(x)在(1,+∞)上是减函数,

而 a²+a+3=(a+1/2)²+11/4 >2

从而 f(a²+a+3)