✔ 最佳答案
y = (t^2 + 2t + 2)/(3t^2 + 2t + 1)
y(3t^2 + 2t + 1) - (t^2 + 2t + 2) = 0
t^2(3y - 1) + t(2y - 2) + (y - 2) = 0
For t to be real, delta > = 0
that is
(2y - 2)^2 - 4(3y - 1)(y - 2) > = 0
y^2 + 1 - 2y - (3y^2 + 2 - 7y) > = 0
- 2y^2 + 5y - 1 > = 0
2y^2 - 5y + 1 < = 0
y = [5 +/- sqrt(25 - 8)]/4 = [5 +/- sqrt 17]/4 = 2.28 or 0.2192
That is (y - 2.28)(y - 0.2192) < = 0
so y < = 2.28 or y > = 0.2192
That is (t^2 + 2t + 2)/(3t^2 + 2t + 1) < = 2.28 or > = 0.2192
2012-05-07 05:00:58 補充：
Using differentiation is also possible but could be a bit clumsy.
2012-05-07 15:41:55 補充：
(1) If t can be unreal, what is the meaning of maximum and minimum of the expression? What is the max. and min. of a complex expression such as x^2 + (2x + 1)i?
(2) Since your question is max. AND min., so answer should be y < = 2.28 AND y > = 0.2192. Anyway, one is max. and the other is min.
