f.4 amathz...
2008-10-21 1:48 am
Show that f(x)=1/(2x^2+6x+8) is always positive, and then find the maximum value of f(x).
回答 (1)
2008-10-21 1:58 am
✔ 最佳答案
f(x)=2x^2+6x+8=2(x^2+3x+(3/2)^2-(3/2)^2)+8
=2(x+(3/2))^2 -(9/2) +8
=2(x+(3/2))^2 +(7/2)
the minimum value of 2(x+(3/2))^2 can only be 0
so the minimum value of 2x^2+6x+8 is 7/2
so f(x) can't be negative.
and the maximum value of 1/(2x^2+6x+8) is 2/7
參考: me (F4)
收錄日期: 2021-04-13 16:11:10
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20081020000051KK01206