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a.maths 只請教最末部份
發問:
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
最佳解答:
(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
a.maths 只請教最末部份
發問:
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
最佳解答:
(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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