Question

求函数y=(2x²-2x+3)/(x²-x+1)的值域

Answer 1

y=(2x²-2x+3)/(x²-x+1)
=( 2x²-2x+2+1)/(x²-x+1)
=[2((x²-x+1) +1]/(x²-x+1)
=2+1/((x²-x+1))
=2+1/(x²-x+1/4+3/4)
=2+1/[(x+1/2)平方 +3/4]

因为 (x+1/2)平方>=0
(x+1/2)平方 +3/4>=3/4
所以
0<1/[(x+1/2)平方 +3/4]<=4/3
所以 2 <2+1/[(x+1/2)平方 +3/4]<=2+4/3=10/3
即 2<y<=10/3

Answer 2

y=(2x²-2x+3)/(x²-x+1)
=[2(x²-x+1)+1]/(x²-x+1)
=2+1/[(x-1/2)²+3/4)
分母>=3/4
当x=1/2时y=10/3
函数值域(2,10/3]

Answer 3

y=(2x²-2x+3)/(x²-x+1)=[2(x²-x+1)+1]/(x²-x+1)=2+[1/(x²-x+1)].∵对任意x∈R,x²-x+1=[x-(1/2)]²+(3/4)≥3/4.∴0＜1/(x²-x+1)≤4/3.∴2＜2+[1/(x²-x+1)]≤10/3.即值域是(2,10/3].