更新1:
最後面應該是"!"
更新2:
x的值大約是66.78
求sqrt(360/x)*sin(x)的最大值~~ 請附過程，謝謝~~?
回答 (1)
2017-11-01 1:22 pm
✔ 最佳答案
f(x)=sqrt(360/x)*sin(x)=(360)^(1/2)*x^(-1/2)* sin x
f'(x)=(360)^(1/2)*[(-1/2)*x^(-3/2)*sin x + x^(-1/2)*cos x]
=(360)^(1/2)*x^(-1/2)*[(-1/2)*x^(-1)*sin x + cos x]
f'(x)=0 時有最大值, (-1/2)*x^(-1)*sin x + cos x = 0
tan x - 2x = 0 ---> x ~ 66.78 度 (x要以弳度表示)
f(x)=sqrt(360/x)*sin(x)
f(66.78)~16.15116
收錄日期: 2021-05-03 03:30:43
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20171031120315AAyYUFG