=cos²x+sinx
=1-sin²x+sinx
=-sin²x+sinx-1/4+1/4+1
=-(sinx-(1/2))²+(5/4)當(dāng)|x|≤π/4,
-√2/2≤sinx≤√2/2∴當(dāng)sinx=1/2時,
y有最大值,
y(max)=5/4,
此時,x=π/6當(dāng)sinx=-√2/2時(離對稱軸sinx=1/2最遠),
y有最小值y(min)
=1-(1/2)+(-√2/2)
=(1-√2)/2,
此時,x=-π/4