Question

求值域y=-2cos^2x+2sinx+3/2,x∈[-π/4,π/4]

Answer 1

y=-2cos^2x+2sinx+3/2
=-2(1-sin^2x)+2sinx+3/2
=-2sin^2x+2sin^2x-1/2
=-2(sinx-1/2)^2

-π/4<=x<=π/4
则-√2/2<=sinx<=√2/2
-(√2+1)/2<=sinx-1/2<=(√2-1)/2
0<=(sinx-1/2)^2<=[-(√2+1)/2]^2=(3+2√2)/4
-(3+2√2)/2<=-2(sinx-1/2)^2<=0
值域[-(3+2√2)/2,0]

Answer 2

x∈[-π/4,π/4]
那么sinx∈[-根号2/2,根号2/2]
y=-2[1-2(sinx)^2]+2sinx+3/2
=4(sinx)^2-2+2sinx+3/2
=4(sinx)^2+2sinx-1/2
我们假设t=sinx∈[-根号2/2,根号2/2]
y=4t^2+2t-1/2
=4(t^2+t/2)-1/2
=4(t^2+t/2+1/16-1/16)-1/2
=4(t+1/4)^2-3/4
对称轴是t=-1/4,开口向上
所以当t=-1/4时，y最小=-3/4
当t=根号2/2时，y最大=根号3+3/2
不知是否明白了\(^o^)/~