已知f(3x+1)=9x^2-6x+5,则函数y(x)=[f(x)-4]/[f(x)+4x]的值域为_____.

f(3x+1)=9x^2-6x+5
=9x^2+6x+1-12x+4
=(3x+1)^2-12x-4+8
=(3x+1)^2-4(3x+1)+8
f(x)=x^2-4x+8
y=(x^2-4x+4)/(x^2+8)
=(x-2)^2/(x^2+8)
f'=(2(x-2)(x^2+8)-2x(x-2)^2)/(x^2+8)^2
=2(x-2)(x^2+8-x^2+2x)/(x^2+8)^2
=4(x-2)(x+4)/(x^2+8)^2=0
x=2 取最小值0
x=-4取最大值1.5
所以值域是[0,1.5]