Given x^2 - 6x + 11 = (x + a)^2 + b, where x is real.
(a) Find the values of a and b.
Hence write down the least values of x^2 - 6x + 11
(b) Using (a), or otherwise,
write down the range of possible values of 1/(x^2 - 6x + 11)
(a) a = -3 b = 2 the least value = 2
a.maths 只請教最末部份
2009-03-01 9:53 am
回答 (1)
2009-03-01 9:59 am
✔ 最佳答案
(a) x^2 - 6x + 11 =(x-3)^2+2
= > a=-3,b=2 The least value is 2
(b) Since x^2 - 6x + 11 > 0 for all x and least value is 2
0<1/(x^2 - 6x + 11)<1/2
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