12sinX-9sin^2X max and min value?
回答 (2)
2016-05-01 8:07 am
✔ 最佳答案
Ýi) y = 12*sin(x) - 9*{sin(x)}^2 + 4 - 4==> y = 4 - {3*sin(x) - 2}^2
ii) For all real x, {3*sin(x) - 2}^2 >/= 0
So Max value = 4
Since sin value lies in [-1, 1],
{3*sin(x) - 2}^2 lies in [0, 25] {it = 0 when sin(x) = 2/3}
So min value = 4 - 25 = -21
This Max of y = 4 & Min of y = -21.
2016-05-01 7:06 am
收錄日期: 2021-04-23 20:45:51
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160430225103AAMv7jw