求函数f(x)=x/1+x²的单调区间于极值

f(x)=x/(1+x²)f '(x)=[(1+x²)-x·2x]/(1+x²)²=(1-x²)/(1+x²)²令f '(x)=0得1-x²=0解得x1=-1,x2=1当x＜-1时,f '(x)＜0,为减函数当-1＜x＜1时,f '(x)＞0,为增函数当x＞1时,f '...