如果函数f(x)在区间D上有定义,且对任意x1,x2∈D,x1≠x2,都有f(x...
如果函数f(x)在区间D上有定义,且对任意x1,x2∈D,x1≠x2,都有f(x1+x22)<f(x1)+f(x2)2,则称函数f(x)在区间D上的“凹函数”.(Ⅰ)已知...
如果函数f(x)在区间D上有定义,且对任意x1,x2∈D,x1≠x2,都有f(x1+x22)<f(x1)+f(x2)2,则称函数f(x)在区间D上的“凹函数”. (Ⅰ)已知f(x)=ln(1+ex)-x(x∈R),判断f(x)是否是“凹函数”,若是,请给出证明;若不是,请说明理由; (Ⅱ)已知f(x)=ln(1+ex)-x是定义域在R上的减函数,且A、B、C是其图象上三个不同的点,求证:△ABC是钝角三角形.
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解:(Ⅰ)函数f(x)是凹函数,证明如下:设x1,x2∈R,且x1<x2,
则f(x1)+f(x2)-2f(x1+x22)
=ln(1+ex1)+ln(1+ex2)-x1-x2-2[ln(1+ex1+x22)-x1+x22]
=ln(1+ex1)(1+ex2)-ln(1+ex1+x22)2
=ln(1+ex1+ex2+ex1+x2)-ln(1+2ex1+x22+ex1+x2)
∵ex1>0,ex2>0,且x1≠x2
∴ex1+ex2>2√ex1ex2=2ex1+x22
∴1+ex1+ex2+ex1+x2>1+2ex1+x22+ex1+x2
∴ln(1+ex1+ex2+ex1+x2)>ln(2+2ex1+x22+ex1+x2)
∴ln(1+ex1+ex2+ex1+x2)-ln(1+2ex1+x22+ex1+x2)>0
∴f(x1)+f(x2)>2f(x1+x22)∴f(x)是凹函数(7分)
(Ⅱ)证明:(Ⅱ)设A(x1,y1),B(x2,y2),C((x3,y3),
且x1<x2<x3,∵f(x)是x∈R上的单调减函数∴f(x1)>f(x2)>f(x3)
∴BA•BC=(x1-x2)(x3-x2)+(f(x1)-f(x2))(f(x3)-f(x2))
∵x1-x2<0,x3-x2>0,f(x1)-f(x2)>0,f(x3)-f(x2)<0
∴BA•BC<0,∴cosB<0,∠B为钝角
故△ABC为钝角三角形.(13分)
则f(x1)+f(x2)-2f(x1+x22)
=ln(1+ex1)+ln(1+ex2)-x1-x2-2[ln(1+ex1+x22)-x1+x22]
=ln(1+ex1)(1+ex2)-ln(1+ex1+x22)2
=ln(1+ex1+ex2+ex1+x2)-ln(1+2ex1+x22+ex1+x2)
∵ex1>0,ex2>0,且x1≠x2
∴ex1+ex2>2√ex1ex2=2ex1+x22
∴1+ex1+ex2+ex1+x2>1+2ex1+x22+ex1+x2
∴ln(1+ex1+ex2+ex1+x2)>ln(2+2ex1+x22+ex1+x2)
∴ln(1+ex1+ex2+ex1+x2)-ln(1+2ex1+x22+ex1+x2)>0
∴f(x1)+f(x2)>2f(x1+x22)∴f(x)是凹函数(7分)
(Ⅱ)证明:(Ⅱ)设A(x1,y1),B(x2,y2),C((x3,y3),
且x1<x2<x3,∵f(x)是x∈R上的单调减函数∴f(x1)>f(x2)>f(x3)
∴BA•BC=(x1-x2)(x3-x2)+(f(x1)-f(x2))(f(x3)-f(x2))
∵x1-x2<0,x3-x2>0,f(x1)-f(x2)>0,f(x3)-f(x2)<0
∴BA•BC<0,∴cosB<0,∠B为钝角
故△ABC为钝角三角形.(13分)
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