已知函数,0
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已知函数,0
已知函数,0
已知函数,0
根据已知条件,A、B、C三点坐标分别为(t,loga(t)),(t+2,log a(t+2)),
(t+4,log a(t+4)),
由图形,当妨令三点A,B,C在x轴上的垂足为E,F,N,则△ABC的面积为
SABC=S梯形ABFE+S梯形BCNF-S梯形ACNE
=-[log a(t)+log a(t+2)]-[log a(t+2)+log a(t+4))]+2[log a(t)+log a(t+4))]
=log a(t)+log a(t+4)-2log a(t+2)]=-loga((t+2)^2/(t^2+4t))
=l-oga(1+(t^2+4t))
即△ABC的面积为S=f(t)=-loga(1+(t^2+4t))(t≥1)
(2)f(t)=-loga(1+(t^2+4t))=log (1/a) (1+(t^2+4t)) (t≥1)是复合函数,其外层是一个递增的函数,t≥1时,内层是一个递减的函数,故复合函数是一个减函数,
(3)由(2)的结论知,函数在t=1时取到最大值,故三角形面积的最大值是
S=f(1)=-loga(1+4/(1+4))=-loga(9/5)=log a (5/9)
过A,B,C平行Y轴分别交X轴于D,E,F点,
则有梯形ACFD的面积=1/2×4[f(t)+f(t+4)],
S(ABED)=1/2×2(f(t)+f(t+2)),S(BCFE)=1/2×2(f(t+2)+f(t+4))
,所以S(ABC)
=S(ACFD)-S(ABED)-S(BCFD)
=f(t)+f(t+4)-2f(t+2)
=loga(1-...
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过A,B,C平行Y轴分别交X轴于D,E,F点,
则有梯形ACFD的面积=1/2×4[f(t)+f(t+4)],
S(ABED)=1/2×2(f(t)+f(t+2)),S(BCFE)=1/2×2(f(t+2)+f(t+4))
,所以S(ABC)
=S(ACFD)-S(ABED)-S(BCFD)
=f(t)+f(t+4)-2f(t+2)
=loga(1-4/(t+4)ˆ2)
因为logax是减函数
1-4/(t+4)ˆ2是增函数
loga(1-4/(t+4)ˆ2)是增函数为单调递减函数,
所以当t=0时,s最大,为loga(3/4)
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已知函数,0<a<1,y=loga x的图像上有A,B,C三点,它们的横坐标分别为t,t+2,t+4,设三角形ABC的面积为S,求S=f(t)的表达式和单调性,还有S的最大值 解析:∵函数y=log(a,x),0<a<1,其图像上A(t,log(a,t)),B(t+2,log(a,t+2)), C(t+4,log(a,t+4)) 则S(⊿ABC) =1/2{[t* log(a,t+2)-(t+2)*log(a,t)]+[(t+2)*log(a,t+4)-(t+4)*log(a,t+2)] +[(t+4)*log(a,t)-t*log(a,t+4)]} =1/2log(a,t^2*(t+4)^2/(t+2)^4) =log(a,t*(t+4)/(t+2)^2) ∵S(t) =log(a,t*(t+4)/(t+2)^2) 是复合函数,真数部分单调增;由于0<a<1,所以外层对数部分单调减,故复合函数是一个减函数,所以不存在最大值
A(t,log(a,t)),B(t+2,log(a,t+2)),C(t+4,log(a,t+4))
S(⊿ABC)= S(⊿ABD)+ S(梯形BDEC)- S(⊿AEC)
=1/2*2*[log(a,t)- log(a,t+2)]+1/2*2*[log(a,t)- log(a,t+2)+log(a,t)- log(a,t+4)]-1/2*4[log(a,t)-log(a,t+4)...
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A(t,log(a,t)),B(t+2,log(a,t+2)),C(t+4,log(a,t+4))
S(⊿ABC)= S(⊿ABD)+ S(梯形BDEC)- S(⊿AEC)
=1/2*2*[log(a,t)- log(a,t+2)]+1/2*2*[log(a,t)- log(a,t+2)+log(a,t)- log(a,t+4)]-1/2*4[log(a,t)-log(a,t+4)]
=1/2*2* log[a,t/(t+2)]+1/2*2*[log[a,t/(t+2)]+ log[a,t/(t+4)]]-1/2*4[log[a,t/(t+4)]
=2log[a,t/(t+2)]- log[a,t/(t+4)]
= log[a,t(t+4)/(t+2)^2)
=iog[a,1-4/(t+2)^2]
t>0时,(t+2)^2是t的增函数,是1-4/(t+2)^2的增函数
又0loga(1-4/(t+4)ˆ2)是增函数为t的单调递减函数,
t>0,S没有最大值!!
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