Question

求函数y=(2x^2-2x+3)/(x^2-x+1)的值域

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Answer 1

解：y=(2x^2-2x+3)/(x^2-x+1)
=2+1/(x^2-x+1)
=2+1/[(x-0.25)^2+0.75]
因为(x+0.25)^2+0.75≥0.75
所以0＜1/[(x-0.25)^2+0.75]≤4/3
所以
2＜2+1/[(x-0.25)^2+0.75]≤10/3
即2＜y≤10/3
所以函数的值域为（2，10/3]

Answer 2

y=(2x²-2x+3)/(x²-x+1)
=[2(x²-x+1)+1]/(x²-x+1)
x²-x+1≠0,(Δ＜0),可约去一个x²-x+1
y=2+1/(x²-x+1)
求导
y'=-(2x-1)/(x²-x+1)²
-(2x-1)/(x²-x+1)²=0得x=1/2
当2x-1＜0即x＜1/2时,y'=-(2x-1)/(x²-x+1)²＞0,y单调递增
当2x-1＞0即x＞1/2时,y'=-(2x-1)/(x²-x+1)²＜0,y单调递减
y在x=1/2处有最大值
y<max>=f(1/2)=2+1/[(1/2)²-1/2+1]=10/3
当x→∞时,求极限lim<x→∞>(2x²-2x+3)/(x²-x+1)
lim<x→∞>(2x²-2x+3)/(x²-x+1)
=lim<x→∞>(2-2/x+3/x²)/(1-1/x+1/x²)
=2
y的值域是(2,10/3]

Answer 3

y=(2x^2-2x+2+1)/(x^2-x+1)
=2+1/(x^2-x+1)
=2+1/((x-1/2)^2+3/4)
当x=1/2时，(x-1/2)^2+3/4取最小值3/4，但没有最大值，所以1/((x-1/2)^2+3/4)的值域为（0，4/3】，
原函数值域为（2，10/3】。