Question

函数y=(x^2-2x+3)/(x^2-x+1)的值域是多少？求详细过程

Answer 1

设(x^2-2x+3)/(x^2-x+1)=k
x^2-2x+3=kx^2-kx+k
(1-k)x^2+(k-2)x+3-k=0
△=(k-2)²-4(1-k)(3-k)
=k^2-4k+4-4k^2+16k-12
=-3k^2+12k-8
=-3(k-2)²+4≥0
(k-2)²≤4/3
-2√3/3≤k-2≤2√3/3
2-2√3/3≤k≤2+2√3/3
值域为 【2-2√3/3,2+2√3/3】

Answer 2

去分母：x^2(y-1)-x(y-2)+y-3=0
y=1时，有：x=2
y<>1时，delta>=0
delta=(y-2)^2-4(y-1)(y-3)=y^2+4-4y-4y^2+16y-12=-3y^2+12y-8>=0, 解得：
2-2/√3=<y<=2+2/3
因此值域为：[2-2/√3, 2+2/√3]