第4个回答 2021-01-21
f(x)=arctanx-(x/2);
f'(x)=1/(1+x²)-(1/2)=[2-(1+x²)]/[2(1+x²)]=(1-x²)/2(1+x²)=-(x-1)(x+1)/2(x²+1);
故当x≦-1或x≧1时f'(x)≦0,即在区间(-∞,-1]或[1,+∞)内单调减;
当-1≦x≦1时f'(x)≧0,即在区间[-1,1]内单调增;
令f'(x)=-(x-1)(x+1)/2(x²+1)=0,得驻点x₁=-1(极小点); x₂=1(极大点);
故有极小值=f(-1)=arctan(-1)+(1/2)=-(π/4)+(1/2);
极大值=f(1)=(π/4)-(1/2)=(π/4)-(1/2);