已知函数y=cos^2x+sinx,求在区间[-π/4,π/4]上的最小值.需要详细过程

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已知函数y=cos^2x+sinx,求在区间[-π/4,π/4]上的最小值.需要详细过程

已知函数y=cos^2x+sinx,求在区间[-π/4,π/4]上的最小值.需要详细过程
已知函数y=cos^2x+sinx,求在区间[-π/4,π/4]上的最小值.
需要详细过程

已知函数y=cos^2x+sinx,求在区间[-π/4,π/4]上的最小值.需要详细过程
y=(cosx)^2+sinx
=1-(sin^2x)+sinx
=-(sinx)^+sinx-1/4+5/4
=-(sinx-1/2)^2+5/4
-π/4≤x≤π/4
sin(-π/4)≤sinx≤sin(π/4)
-√2/2≤sinx≤√2/2
当sinx=-√2/2时,ymin=1-1/2-√2/2=√2-1/2

y=(cosx)^2+sinx
=1-(sinx)^2+sinx
=-(sinx)^+sinx-1/4+5/4
=-(sinx-1/2)^2+5/4
-π/4<=x<=π/4
-1/2<=sinx<=1/2
-1<=sinx-1/2<=0
0<=(sinx-1/2)^2<1
-1<=-(sinx-1/2)^2<=0
1/4<=-(sinx-1/2)^2+5/4<=5/4
y在x∈[-π/4,π/4]区间内最小值为1/4