Question

已知函数f(x)=ax/(x^2+1)(a≠0）,求f(x)的单调区间.

Answer 1

f'(x)=a(x^2+1)-2xax/(x^2+1)^2=a(1-x^2)/(x^2+1)^2

令上式等于0 解得x=+1或-1

若a>0

x>1 或x0所以f(x)在 【-1,1】 单调递增

当a1 或x