已知f(x)=cosx+sin^2x,求f(派\6)求最大值最小值.

已知f(x)=cosx+sin^2x,求f(派\6)求最大值最小值
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f(x)=cosx+sin^2x
=cosx+1-cos^2x
=-cos^2x+cosx+1
f（π/6） =-cos^2（π/6）+cos（π/6）+1=-3/4+根号3/2+1=根号3/2+1/4
设t=cosx ,t∈[-1,1]
g(t)=-t^2+t+1
=-(t-1/2)^2+1/4+1
=-(t-1/2)^2+5/4
t=1/2时,取得最大值5/4
t=-1时,取得最小值-1
所以f（x）最大值5/4；最小值-1

1,根号3
2,求导数，f(x)'=-sinx+2cos2x
令f(x)'=0,得-sinx+2cos2x=0 sinx=2cos2x=2(1-2sinx^2)

f(π/6)=cos(π/6)+sin²(π/6)=√3/2+(1/2)²=(2√3+1)/4
f(x)=cosx+sin^2x
=cosx+1-cos²x
=-(cos²x-cosx+1/4)+5/4
=-(cosx-1/2)²+5/4
1°当cosx=1/2时，即x=±π/3...

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f(π/6)=cos(π/6)+sin²(π/6)=√3/2+(1/2)²=(2√3+1)/4
f(x)=cosx+sin^2x
=cosx+1-cos²x
=-(cos²x-cosx+1/4)+5/4
=-(cosx-1/2)²+5/4
1°当cosx=1/2时，即x=±π/3+2kπ，k为整数
f(x)取得最大值5/4
2°当cosx=-1时，即x=π+2kπ，k为整数
f(x)取得最小值f(π+2kπ)=-(-1-1/2)²+5/4=-1

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四分之一加上二分之根号三，f（x）=cosx+1-cos^2x=-(cosx-1/2)^2+5/4,最大值四分之五，最小值-1