求下列函数值域:(1)y=(sin2xsinx)/(1-cosx);(2)y=sinx+cosx+(3)y=2cos(π/3+x)+2cosx.
求下列函数值域:(1)y=(sin2xsinx)/(1-cosx);(2)y=sinx+cosx+(3)y=2cos(π/3+x)+2cosx.
1)y=(sin2xsinx)/(1-cosx)=2cosxsin^2x/(1-cosx)=2cosx(1+cosx)=2(cosx+cos^2x) =2(cosx +0.5)^2-0.5 (cosx不等于1)-0.5=&y&2(1+0.5)^2-0.5=4值域是[-0.5,4)
2)y=sinx+cosx+sinxcosx 令sinx+cosx=T，(1) 由同角三角函数关系sinxcosx=[（sinx+cosx)^2-(sinx^2+cosx^2)]/2 把(1)式代入，得sinxcosx=（T^2-1)/2 所以y=T+（T^2-1)/2 整理得，y=1/2（T+1）^2-1 而sinx+cosx=√2sin(x+π/4)∈[-√2,√2] 所以y在T[∈-√2,√2]时，不单调 当T=-1时，y取得最小值 = -1 当T=√2时，y取得最大值 = 1/2+√2 值域[-1,1/2+√2 ]
3)y=2cos(π/3+x)+2cosx =4sin(x+π/6)cosπ/6 =2√3sin(x+π/6) 值域[-2√3,2√3]
其他回答 (2)
(1)y=(sin2xsinx)/(1-cosx)=2cosxsin^2x/(1-cosx)=2cosx(1+cosx)=2(cosx+cos^2x) =2(cosx +0.5)^2-0.5 -1/2≤cosx+1/2≤3/20≤(cosx+1/2)^2≤9/4所以y属于[-1/2，4]（2）因为(sinx+cosx)^2=1+2sinxcosxsinxcosx=[(sinx+cosx)^2-1]/2令sinx+cosx=t t=√2sin(x+π/4)
-√2≤t≤√2y=sinxcosx+sinx+cosx=(t^2-1)/2+t=t^2/2+t-1/2=1/2(t+1)^2-1对称轴t=-1y在[-√2,-1]上单调递减在[-1,√2]上单调递增t=-1 最小值y=-1t=√2 最大值y=(1+2√2)/2所以函数y=sinxcosx+sinx+cosx 的值域[-1,(1+2√2)/2]
∵y=2cos(π/3+ x) +2cosx
=2cosπ/3cosx-2sinπ/3sinx +2cosx
=3cosx-√3sinx
=2√3cos（x+ π/3）。∵-1≤cos（x+ π/3）≤1
∴-2√3≤2√3cos（x+ π/3）≤2√3
:(1)y=(sin2xsinx)/(1-cosx)
=2(sinx)^2*(cosx)/(1-cosx)=2[1-(cosx)^2]*(cosx)/(1-cosx)=2(1-cosx)(1+cosx)cosx/(1-cosx)=2(1+cosx)cosx
=2[(cosx)^2+cosx]=2[(cosx)^2+cosx+1/4]-1/2=2(cosx+1/2)^2-1/2
因1-cosx≠0,故-1≤cosx＜1
当cosx=-1,y最小值为0
当cosx=1,& 2(cosx+1/2)^2-1/2=4
所以值域y属于[0,4)
(2)y=sinx+cosx+sinxcosx=sinx+cosx+1/2[(sinx)^2+2sinxcosx+(cosx)^2]-1/2
=sinx+cosx+1/2(sinx+cosx)^2-1/2=1/2(sinx+cosx+1)^2-1=1/2[（根号2）sin(x+45°）+1]^2-1
当sin(x+45°）=1时，y最大值=1/2[(根号2）+1]^2-1=（1/2）+根号2
当sin(x+45°）=-1时y最小值=(1/2)-根号2
所以值域y属于[(1/2)-根号2,（1/2）+根号2]
(3)y=2cos(π/3+x)+2cosx=cosx-(根号3)sinx+2cosx=3cosx-(根号3)sinx
=（根号12）*[3/(根号12）cosx-(根号3)/（根号12）sinx]
=(2根号3）*cos(a+x)& [令cosa=3/(根号12),sina=(根号3)/（根号12)]
因-1≤cos(a+x)&≤1
所以值域y属于[-2根号3,2根号3]
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理工学科领域专家求函数f(x)=sinx+cosx+sinx*cosx的最值_百度知道
求函数f(x)=sinx+cosx+sinx*cosx的最值
提问者采纳
,√2] 所x=-1时;=y&2(x+1)^2-1 x=sinx+cosx=√2sinx(x+π&#47令sinx+cosx=x 2sinx*cosx=(sinx+cosx)^2-1=x^2-1 y=sinx+cosx+sinx*cosx=(x^2-1)&#47，y有最大值2+√2y=sinx+cosx+sinx*cosx的值域为 -1&2) ∴x∈[-√2，y有最小值-1x=√2时;2+x=1&#47
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出门在外也不愁下列各函数中，最小值为2的是()A，y=x+x分之一，B，y=sinx+sinx分分之一，x属于（0，二分之π）_百度知道
下列各函数中，最小值为2的是()A，y=x+x分之一，B，y=sinx+sinx分分之一，x属于（0，二分之π）
为什么，D？麻烦分析一下，y=x+根号x分之2－1，y=根号（x平方+2）分之x平方+3，B不是取不到等号吗，答案选BC
提问者采纳
√x-1=x+1&#47,+∞)C y=(x^2+3)&#47A的值域是(-∞;√x)*(1/√x+1/2;√(x^2+2)=√(x^2+2)+1/√x-1≥3[x*(1/√x)]^(1/√(x^2+2)因为√(x^2+2)≥√2所以值域是[3√2&#47,-2]∪[2,+∞)D y=x+2&#47,+∞)B 0＜x＜π/3)-1=2所以答案应该选D才对;20＜sinx＜1所以值域是(2
提问者评价
谢谢……我也觉得D是对的！
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用排除法 吧。其他貌似都有可能=0如果有=0 最小值就不是2B。有说了个区间。所以不一样。
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出门在外也不愁2、求下列微分方程满足初始条件的特解： (3) y ,+y/x =sinx y｜x=π =1_百度知道
2、求下列微分方程满足初始条件的特解： (3) y ,+y/x =sinx y｜x=π =1
提问者采纳
x=0的通解;dy&#47，
∵y'x=0 ==&+y/x
∵当x=π时;x;C'y=C/C=0
∴y=-cosx+sinx&#47，设原方程的通解为y=C(x)/(x)=xsinx
∴C(x)=∫xsinxdx
=-xcosx+∫cosxdx
(应用分部积分法)
=-xcosx+sinx+C
(C是积分常数)
∴y=(-xcosx+sinx+C)/ln│y│=-ln│x│+ln│C│
(C是积分常数)
==&x=sinx ==&y=-dx&#47。
∴齐次方程的通解是y=C/(x)/x
==&π ==&x;+y/x
故微分方程满足初始条件的特解是y=-cosx+sinx&#47解：（常数变易法）
先解齐次方程y'x+C/x
=-cosx+sinx/x
(C(x)是关于x的函数)
代入原方程得C&#39，y=1
代入得 1=1+C&#47
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y=26.25元应该没错..25* （2-1设买红辣椒x斤 西红柿y斤则
（1）4x+1.25（2）17.75*（5-4）+26..6）=28.楼主可以验算下.6y=113
经过判定可知，这是一个一阶线性非齐次方程，有常数变易法和公式法可以求解，下面我用公式法计算一下：其中￥表示积分号解：y=e^(￥（-1/x）dx)[￥sinx (e)^(￥1/xdx)dx+C]=e^(-lnx)[￥sinx (e)^(lnx)dx+C] =(1/x)[￥sinx.
(分部积分法)=(1/x)[￥xd(-cosx)+C]
=(1/x)[[-xcosx]-￥(-cosx)dx+C]
=(1/x)[[-xcosx]+sinx+C]
又因为y｜x=π =1 带入上面的通解中，得C=0所以其特解为y=(1/x)[[-xcosx]+sinx+0]
=(1/x)[sinx-xcosx]]
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求函数y=sinx+cosx+sinxcosx的值域
求函数y=sinx+cosx+sinxcosx的值域
y=sinx+cosx+sinxcosx
令sinx+cosx=T，(1)
由同角三角函数关系sinxcosx=[（sinx+cosx)^2-(sinx^2+cosx^2)]/2
把(1)式代入，得sinxcosx=（T^2-1)/2
所以y=T+（T^2-1)/2
整理得，y=1/2（T+1）^2-1
而sinx+cosx=√2sin(x+π/4)∈[-√2,√2]
所以y在T[∈-√2,√2]时，不单调
当T=-1时，y取得最小值 = -1
当T=√2时，y取得最大值 = 1/2+√2
值域[-1,1/2+√2 ]
回答数：2319