Given x^2-6x+11=(x+a)^2+b, where x is real.
Find the range of possible values of 1/(x^2-6x+11).
AMath
2007-10-20 11:00 pm
回答 (2)
2007-10-20 11:46 pm
✔ 最佳答案
x2 - 6x + 11= x2 - 6x + 9 + 2
= (x - 3)2 + 2
So a = -3 and b = 2
Let y = 1 / (x2 - 6x + 11). Then,
yx2 - 6yx + 11y - 1 = 0
Since x is real, Δ ≧ 0
36y2 - 4y(11y - 1) ≧ 0
9y2 - 11y2 + y ≧ 0
-2y2 + y ≧ 0
y(-2y + 1) ≧ 0
So 0 ≦ y ≦ 2
i.e. 0 ≦ 1 / (x2 - 6x + 11) ≦ 2
2007-10-23 7:02 pm
x^2-6x+11 = x^2 - 6x + 9 + 2 = (x - 3)^2 + 2
Therefore a = -3 and b = 2, and its mimimum is 2 when x = 3
For 1/(x^2-6x+11) = 1/[(x-3)^2+2], the maximum is 1/2 when x = 3 and is positive since (x-3)^2+2 >=2 > 0.
In conclusion, 0 < 1/(x^2-6x+11) <= 1/2.
Therefore a = -3 and b = 2, and its mimimum is 2 when x = 3
For 1/(x^2-6x+11) = 1/[(x-3)^2+2], the maximum is 1/2 when x = 3 and is positive since (x-3)^2+2 >=2 > 0.
In conclusion, 0 < 1/(x^2-6x+11) <= 1/2.
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