Question

求函数y=(x^2+x+2)/(2x^2-x+1)的最大值和最小值

Answer 1

y=(x^2+x+2)/(2x^2-x+1)

y(2x^2-x+1)=x^2+x+2
(2y-1)x^2-(y+1)x+y-2=0
当2y-1不=0时,方程有解,则判别式>=0
即有(y+1)^2-4(2y-1)*(y-2)>=0
y^2+2y+1-4(2y^2-4y-y+2)>=0
y^2+2y+1-8y^2+20y-8>=0
7y^2-22y+7<=0
y^2-22/7y+1<=0
(y-11/7)^2<=121/49-1=72/49
-6/7根号2+11/7<=Y<=6/7根号2+11/7
当2y-1=0,即y=1/2时X有解,则成立
所以,最大值是(6根号2+11)/7,最小值是(-6根号2+11)/7

Answer 2

y=(x^2+x+2)/(2x^2-x+1)
y(2x^2-x+1)=x^2+x+2
(2y-1)x^2-(y+1)x+y-2=0
当2y-1不=0时,方程有解,则判别式>=0
即有(y+1)^2-4(2y-1)*(y-2)>=0
y^2+2y+1-4(2y^2-4y-y+2)>=0
y^2+2y+1-8y^2+20y-8>=0
7y^2-22y+7<=0
y^2-22/7y+1<=0
(y-11/7)^2<=121/49-1=72/49
-6/7根号2+11/7<=Y<=6/7根号2+11/7
当2y-1=0,即y=1/2时X有解,则成立
所以,最大值是(6根号2+11)/7,最小值是(-6根号2+11)/7