求函数f x=x+2/x2+1,x属于[-2.2]的值域

f'(x)=1/(x^2+1)-(x+2)*2x/(x^2+1)^2=(1-x^2-4x)/(x^2+1)^2,
令f'(x)=0,解得：x=v5-2,x=-v5-2(舍去),
即f(v5-2)=(v5+2)/2为其极值点,
f(2)=4/5,f(-2)=0,
所以f(v5-2)为其极大值,
f(x)的值域为[0,(v5+2)/2].