确定函数y＝x/（x²＋1）的单调区间,并求它的最大值、最小值.

y=x/(x²+1)y'=[1(x²+1)-x*2x]/(x²+1)²=(1-x²)/(x²+1)²=-(x-1)(x+1)/(x²+1)²由y'>0解得-1