數學:求最大值..............
回答 (2)
2010-04-18 5:01 am
✔ 最佳答案
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2010-04-17 21:01:28 補充：
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2010-04-18 5:09 am
1/(x^2 + bx + a^2) is a maximum when x^2 + bx + a^2 is a minimum which is equal to - [ b^2 - 4(1)(a^2)]/4 = (4a^ - b^2)/4.
so maximum value of 1/(x^2 + bx + a^2) = 4/(4a^2 - b^2).
so maximum value of 1/(x^2 + bx + a^2) = 4/(4a^2 - b^2).
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原文連結 [永久失效]:
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