Let x be a real number.Find the range of values of
(x^2-15)/(x^2+1)
A. maths (F.4)
2006-11-02 3:51 am
回答 (1)
2006-11-02 4:20 am
✔ 最佳答案
(x2-15)/(x2+1) = (x2+1 - 16)/(x2+1)
= 1 – 16/(x2+1)
當 x = 0
x2 + 1 的極小值為 1
所以
16 / (x2 + 1) 的極大值為 16
所以
(x2-15)/(x2+1)
= 1 – 16/(x2+1)
的極小值為 – 15
而 x 趨近於無限大小
16 / (x2+1) 趨近於 0
所以
(x2-15)/(x2+1) 的極大值為 1
因此
1 ≧ (x2-15)/(x2+1) ≧ -16
收錄日期: 2021-04-23 16:18:22
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