小女不才,请各位大虾帮忙!求函数f(x)=(sin^4x+cos^4x+sin^2xcos^2x)/(2-sin2x) 的最大值和最小值~ o(∩_∩)o. 感谢啦...
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![小女不才,请各位大虾帮忙!求函数f(x)=(sin^4x+cos^4x+sin^2xcos^2x)/(2-sin2x) 的最大值和最小值~ o(∩_∩)o. 感谢啦...](/uploads/image/z/14092819-43-9.jpg?t=%E5%B0%8F%E5%A5%B3%E4%B8%8D%E6%89%8D%2C%E8%AF%B7%E5%90%84%E4%BD%8D%E5%A4%A7%E8%99%BE%E5%B8%AE%E5%BF%99%21%E6%B1%82%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%3D%28sin%5E4x%2Bcos%5E4x%2Bsin%5E2xcos%5E2x%29%2F%282-sin2x%29+%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC%E5%92%8C%E6%9C%80%E5%B0%8F%E5%80%BC%7E++++++++++++++++++++o%28%E2%88%A9_%E2%88%A9%29o.+%E6%84%9F%E8%B0%A2%E5%95%A6...)
小女不才,请各位大虾帮忙!求函数f(x)=(sin^4x+cos^4x+sin^2xcos^2x)/(2-sin2x) 的最大值和最小值~ o(∩_∩)o. 感谢啦...
小女不才,请各位大虾帮忙!
求函数f(x)=(sin^4x+cos^4x+sin^2xcos^2x)/(2-sin2x) 的最大值和最小值~ o(∩_∩)o. 感谢啦...
小女不才,请各位大虾帮忙!求函数f(x)=(sin^4x+cos^4x+sin^2xcos^2x)/(2-sin2x) 的最大值和最小值~ o(∩_∩)o. 感谢啦...
f(x)=(sin^4x+cos^4x+sin^2xcos^2x)/(2-sin2x)
=((sin^2x+cos^2x)^2-sin^2xcos^2x)/(2-sin2x)
=(1-sin^2xcos^2x)/(2-sin2x)
=(1-sinx cosx)(1+sinxcosx)/(2-sin2x)
=(1-0.5xin2x)(1+0.5sin2x)/(2-sin2x)
=(1+0.5sin2x)/2=0.5+0.25sin2x
sin2x的最大值为1,最小值为-1
所以
y最大=0.75
y最小=0.25
解 :
f(x)
=[(sin^2 x+cos^2 x)^2-sin^2 x*cos^2 x]/(2-sin2x)
=(1-1/4 sin^2 2x)/(2-sin 2x)
设 t=sin 2x ,t∈[-1,1]
则f(x)可化为
g(t)=1/4*(t^2-4)/(t-2)=1/4*(t+2)
t=-1时,g(t)取min=...
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解 :
f(x)
=[(sin^2 x+cos^2 x)^2-sin^2 x*cos^2 x]/(2-sin2x)
=(1-1/4 sin^2 2x)/(2-sin 2x)
设 t=sin 2x ,t∈[-1,1]
则f(x)可化为
g(t)=1/4*(t^2-4)/(t-2)=1/4*(t+2)
t=-1时,g(t)取min=1/4;
t=1时,g(t)取max=3/4;
在原函数中
sin2x=-1,即x=kπ-1/4*π 时,f(x)取min=1/4
sin2x=1, 即x=kπ+1/4*π 时,f(x)取max=3/4
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