求函数f(x)=sinxcosx/(1+sinx+cosx)的最大值和最小值
答案:2 悬赏:10
解决时间 2021-03-01 09:35
- 提问者网友:恋你成殇
- 2021-02-28 16:50
求函数f(x)=sinxcosx/(1+sinx+cosx)的最大值和最小值
最佳答案
- 二级知识专家网友:白日梦制造商
- 2021-02-28 18:09
f(x)=sinxcosx/(1+sinx+cosx)=2sin(x/2)cos(x/2)(cos^(x/2)-sin^(x/2))/(2sin(x/2)cos(x/2)+2cos^(x/2))
=2sin(x/2)cos(x/2)(cos^(x/2)-sin^(x/2))/2cos(x/2)(sin(x/2)+cos(x/2))
=sin(x/2)(cos(x/2)-sin(x/2))
=sin(x/2)cos(x/2)-sin^(x/2)
=(sinx+cosx-1)/2
=(√2sin(x+45度)-1)/2
最大值为√2+1/2,最小值为√2-1/2
=2sin(x/2)cos(x/2)(cos^(x/2)-sin^(x/2))/2cos(x/2)(sin(x/2)+cos(x/2))
=sin(x/2)(cos(x/2)-sin(x/2))
=sin(x/2)cos(x/2)-sin^(x/2)
=(sinx+cosx-1)/2
=(√2sin(x+45度)-1)/2
最大值为√2+1/2,最小值为√2-1/2
全部回答
- 1楼网友:偏爱自由
- 2021-02-28 18:44
解:f(x)=sinxcosx[1-(sinx+cosx)]/(1+sinx+cosx)[1-(sinx+cosx)] =(sinxcosx-sin^2xcosx-sinxcos^2x)/1-(sinx+cosx)^2 =(sinxcosx-sin^2xcosx-sinxcos^2x)/1-(1+2sinxcosx) =(sinxcosx-sin^2xcosx-sinxcos^2x)/-2sinxcosx =-1/2+1/2sinx+1/2cosx =√2/2sin(x+π/4)-1/2 ∵-1≤sin(x+π/4)≤1
∴f(x)的最大值为(√2-1)/2,最小值为-(√2+1)/2
(ps:我不希望提问的得不到答案,所以挑靠后的零回答;
采纳时回答速度选很快,回答态度选很认真,谢谢。)
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