作业帮y=sinx+cosx+sinxcosx+1的最值
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/04 05:35:16
y=sinx+cosx+sinxcosx+1的最值
y=sinx+cosx+sinxcosx+1的最值
y=sinx+cosx+sinxcosx+1的最值
y = sinx+cosx+sinxcosx+1
y' = cosx - sinx - (sinx)^2 + (cosx)^2
= (cosx+1/2)^2- (sinx+1/2)^2
= (cosx+sinx+1)(cosx-sinx)
y' =0
=> sinx = cosx or cosx+sinx+1=0
=> tanx = 1 or sin(π/4 +x ) = -√2/2
=> x = π/4 or x = π
max y , at x= π/4
max y = sinπ/4 + cosπ/4 + sinπ/4cosπ/4 + 1
= √2/2+√2/2+(√2/2)√2/2+1
= 3/2+ √2
miny, at x= π
miny = sinπ + cosπ+ sinπcosπ+1
= 0-1+(-1).0+1
= 0
设 t=sinx+cosx=√2sin（x+45º） ∴-√2≤t≤√2
所以 sinxcosx=（t²-1）/2
y=sinx+cosx+sinxcosx+1=t+（t²-1）/2+1=（t+1）²/2
∵-√2≤t≤√2
所以y的最大值为（3+2√2）/2
最小值为0
y=sinx+cosx+sinxcosx+1
=(sinx+1)+cosx(sinx+1)
=(1+sinx)(1+cosx)
由于-1≤sinx≤1，-1≤cosx≤1
因此1+sinx和1+cosx恒非负，y恒非负，当sinx=-1或cosx=-1时，有ymin=0
y=sinx+cosx+sinxcosx+1
=√2sin(x+π/4...
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y=sinx+cosx+sinxcosx+1
=(sinx+1)+cosx(sinx+1)
=(1+sinx)(1+cosx)
由于-1≤sinx≤1，-1≤cosx≤1
因此1+sinx和1+cosx恒非负，y恒非负，当sinx=-1或cosx=-1时，有ymin=0
y=sinx+cosx+sinxcosx+1
=√2sin(x+π/4)+sin(2x)/2+1
当x=2kπ+π/4时，sin(x+π/4)和sin(2x)同时取到最大值1，此时，有ymax=√2+1/2+1=√2+(3/2)
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y=sinx+cosx+sinxcosx+0.5+0.5
=sinx+cosx+0.5*(sinx+cosx).^2+0.5
=0.5*a^2+a+0.5
=0.5*(a+1)^2
其中 a=sinx+cosx
-根号2<=a<=根号(2)
max(y)=0.5×（根号2 +1）²=1.5 +根号2