已知函数f(x)=√(1+sinx)+√(1-sinx)
1)用定义判断函数的奇偶性2)用定义判断函数的周期性,如果是周期函数,指出函数的最小正周期3)求出函数的值域及单调区间...
1)用定义判断函数的奇偶性
2)用定义判断函数的周期性,如果是周期函数,指出函数的最小正周期
3)求出函数的值域及单调区间 展开
2)用定义判断函数的周期性,如果是周期函数,指出函数的最小正周期
3)求出函数的值域及单调区间 展开
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1
f(-x)=(1+sinx)^0.5+(1-sinx)^0.5=f(x);
偶函数 .
2
f(x)=(1-sinx)^0.5+(1+sinx)^0.5
=(sin^2(x/2)+cos^2(x/2)-2sin(x/2)cos(x/2))^0.5+(sin^2(x/2)+cos^2(x/2)+2sin(x/2)cos(x/2))^0.5
=|sin(x/2)-cos(x/2)|+|sin(x/2)+cos(x/2)|
则f(x+pi)=|cos(x/2)+sin(x/2)|+|cos(x/2)-sin(x/2)|=f(x)
最小正周期为pi .
3
当2kpi-pi/4<=x/2<2kpi+pi/4
即4kpi-pi/2<=x<4kpi+pi/2时
sin(x/2)+cos(x/2)>0
sin(x/2)-cos(x/2)<0
f(x)=2cos(x/2)
当4kpi+pi/2<=x<4kpi+3pi/2时
sin(x/2)+cos(x/2)>0
sin(x/2)-cos(x/2)>0
f(x)=2sin(x/2)
当4kpi+3pi/2<=x<4kpi+5pi/2时
sin(x/2)+cos(x/2)<0
sin(x/2)-cos(x/2)>0
f(x)=-2cos(x/2)
当4kpi+5pi/2<=x<5kpi+7pi/2时
f(x)=-2sin(x/2)
根据f的以上表达不难知其单调性
f(-x)=(1+sinx)^0.5+(1-sinx)^0.5=f(x);
偶函数 .
2
f(x)=(1-sinx)^0.5+(1+sinx)^0.5
=(sin^2(x/2)+cos^2(x/2)-2sin(x/2)cos(x/2))^0.5+(sin^2(x/2)+cos^2(x/2)+2sin(x/2)cos(x/2))^0.5
=|sin(x/2)-cos(x/2)|+|sin(x/2)+cos(x/2)|
则f(x+pi)=|cos(x/2)+sin(x/2)|+|cos(x/2)-sin(x/2)|=f(x)
最小正周期为pi .
3
当2kpi-pi/4<=x/2<2kpi+pi/4
即4kpi-pi/2<=x<4kpi+pi/2时
sin(x/2)+cos(x/2)>0
sin(x/2)-cos(x/2)<0
f(x)=2cos(x/2)
当4kpi+pi/2<=x<4kpi+3pi/2时
sin(x/2)+cos(x/2)>0
sin(x/2)-cos(x/2)>0
f(x)=2sin(x/2)
当4kpi+3pi/2<=x<4kpi+5pi/2时
sin(x/2)+cos(x/2)<0
sin(x/2)-cos(x/2)>0
f(x)=-2cos(x/2)
当4kpi+5pi/2<=x<5kpi+7pi/2时
f(x)=-2sin(x/2)
根据f的以上表达不难知其单调性
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