求y=(3cosx+1)/(cosx+2)的最值
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/30 14:57:02
![求y=(3cosx+1)/(cosx+2)的最值](/uploads/image/z/14913777-57-7.jpg?t=%E6%B1%82y%3D%283cosx%2B1%29%2F%28cosx%2B2%29%E7%9A%84%E6%9C%80%E5%80%BC)
x){V89BPS_0|>ٜ
{l +/!,{ajL#Md4\L^XL%@hS$X~,GK`r@1Dc&PT(ą(7 h4X@A}Q*+utD cma<ٻz=mm~dm!Ĩ^ O;ڞھ
`|6pCĎ
求y=(3cosx+1)/(cosx+2)的最值
求y=(3cosx+1)/(cosx+2)的最值
求y=(3cosx+1)/(cosx+2)的最值
y=(3cosx+1)/(cosx+2)
y=[(3cosx+6)-5]/(cosx+2)
y=3-[5/(cosx+2)]
因为:1≤cosx+2≤3
则:5/3≤5/(cosx+2)≤5
得:-2≤3-[5/(cosx+2)]≤4/3
即:y∈[-2,4/3]
使用换元,t=cosx,-1<=t<=1
再换成分式函数求极值