函数f(x)=(ax+b)/(x*2+1)的最大值是4,最小值是-1,求实数a,b的值

令f(x)=y,则y=(ax+b)/(x2+1)
yx2+y=ax+b
∴yx2-ax+y-b=0
∵f(x)的定义域是R
∴yx2-ax+y-b=0恒有实数根,即
△=a2-4y(y-b)≥0
-4y2+4by+a2≥0
∵y∈[-1,4]由韦达定理,得
-4b/(-4)=4-1,a2/(-4)=-4
∴a=正负4,b=3