y`=2x/(x²+1)y`=[2x²+2-2x(2x)]/(x²+1)² =(2-2x²)/(x²+1)²=0x=±1 (-∝,-1) -1 (-1,1) 1 (1,+∝)y`` <0 0 >0 0 <0 y 凸 拐点 凹 拐点 凸f(1)=ln2 f(-1)=ln2凹区间 (-1,1)凸区间 (-∝,-1)U(1,+∝)