Question

函数f(x)=cos²x +sinx在区间[-π/4，π/4]上的最小值是

Answer 1

f(x)=(cosx)^2+sinx
=1-(sinx)^2+sinx
=-(sinx-1/2)^2+5/4

x∈[-π/4,π/4]
所以sinx∈[-√2/2,√2/2]
故sinx-1/2∈[-√2/2-1/2,√2/2-1/2]
那么(1-√2)/2≤f(x)≤5/4

所以最小值是(1-√2)/2

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Answer 2

f=1-snx*sinx+sinx=3/4-(sinx-1/2)*sin(x-1/2)
[-π/4，π/4] -√2/2<=sinx<=√2/2
f(-π/4)=1/2-√2/2 f(π/4)=1/2+√2/2 f(π/6)=3/4
最小值是f(-π/4)=1/2-√2/2

Answer 3

答案看图片