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若|x|≤π/4,那么函数F(x)=cos^2x+sinx的最大值为
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若|x|≤π/4,那么函数F(x)=cos^2x+sinx的最大值为
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答案和解析
答：
|x|<=π/4,-π/4<=x<=π/4
f(x)=cos²x+sinx
=1-sin²x+sinx
=-(sinx-1/2)²+1/4
sin(-π/4)<=sinx<=sin(π/4)
-√2/2<=sinx<=√2/2
所以：当sinx-1/2=0即sinx=1/2时f(x)取得最大值1/4
|x|<=π/4,-π/4<=x<=π/4
f(x)=cos²x+sinx
=1-sin²x+sinx
=-(sinx-1/2)²+1/4
sin(-π/4)<=sinx<=sin(π/4)
-√2/2<=sinx<=√2/2
所以：当sinx-1/2=0即sinx=1/2时f(x)取得最大值1/4
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