已知f（x）=（ax+b）/（x^2+1）的值域为[-1,4],求实数a、b的值

x属于R
y=(ax+b)/(x^2+1)
ax+b=yx^2+y
yx^2-ax+y-b=0
y=0,x=-b/a
有解，判别式是：
a^2-4y（y-b)>=0
4y^2-4by-a^2<=0
4(y+1)(y-4)<=0
4y^2-12y-16<=0
比较有：
4b=12,a^2=16
b=3,a=+/-4

yx^2-ax+y-b=0

判别式=a^2-4y(y-b)>=0
即4y^2-4by-a^2<=0

可知它的解集是[-1,4]

所以4b/4=-1+4=3,b=3
-a^2/4=-1*4=-4,a=4或-4

x^2+1>0
设y=f(x)
yx^2-ax+y-b=0
a^2-4y(y-b)>=0
4y^2-4by-a^2<=0
y^2-by-a^2/4<=0
解集是[-1,4]
y^2-3y-4<=0
b=3
a=4/-4