作业帮已知函数 f(x)=sin2x+√2cos(x-π/4) 求f(x) 值域
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已知函数 f(x)=sin2x+√2cos(x-π/4) 求f(x) 值域
已知函数 f(x)=sin2x+√2cos(x-π/4) 求f(x) 值域
已知函数 f(x)=sin2x+√2cos(x-π/4) 求f(x) 值域
f(x)=sin(2x)+√2cos(x-π/4)
=sin(2x)+√2[cosxcos(π/4)+sinxsin(π/4)]
=sin(2x)+cosx+sinx
=sin(2x)+√2sin(x+π/4)
当x=2kπ+π/4时,sin(2x)和sin(x+π/4)同时取到最大值,此时f(x)max=1+√2
当x=2kπ-3π/4时,sin(2x)和sin(x+π/4)同时取到最小值,此时f(x)min=-1-√2
f(x)的值域为[-1-√2,1+√2]
√2cos(x-π/4)
=√2(cosxcosπ/4+sinxsinπ/4)
=√2((√2/2)cosx+(√2/2)sinx)
=cosx+sinx
f(x)=sin2x+√2cos(x-π/4)
=2sinxcosx+((sinx)^2+(cosx)^2)-((sinx)^2+(cosx)^2)+√2cos(x-π/4)
=(sinx+cos...
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√2cos(x-π/4)
=√2(cosxcosπ/4+sinxsinπ/4)
=√2((√2/2)cosx+(√2/2)sinx)
=cosx+sinx
f(x)=sin2x+√2cos(x-π/4)
=2sinxcosx+((sinx)^2+(cosx)^2)-((sinx)^2+(cosx)^2)+√2cos(x-π/4)
=(sinx+cosx)^2-1+√2cos(x-π/4)
=(√2cos(x-π/4))^2+√2cos(x-π/4)-1
为了使算式看起来简洁，令√2cos(x-π/4)=t
-√2=
=(t+1/2)^2-5/4
所以当t=-1/2时有最小值-5/4
当t=√2时有最大值√2+1
f(x)的值域为[-5/4,√2+1]
收起
f(x)=sin2x+√2cos(x-π/4)
当x=π/4时
f(x)取得最大为1+√2
当x=-3π/4时
f(x)取得最小为1-√2
所以值域为[1-√2,1+√2]