已知函数f(x)满足:对任意x,y∈R,都有f(x+y)=f(x)f(y)-f(x)-f(y)+2成立已知函数f(x)满足:对任意x,y∈R,都有f(x+y)=f(x)·f(y)-f(x)-f(y)+2成立 ,且x>0时,f(x)>2(1)求f(0)的值,并证明:当x<0时,1<f(x)<2(2)判断f(x)
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![已知函数f(x)满足:对任意x,y∈R,都有f(x+y)=f(x)f(y)-f(x)-f(y)+2成立已知函数f(x)满足:对任意x,y∈R,都有f(x+y)=f(x)·f(y)-f(x)-f(y)+2成立 ,且x>0时,f(x)>2(1)求f(0)的值,并证明:当x<0时,1<f(x)<2(2)判断f(x)](/uploads/image/z/8397331-43-1.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%E6%BB%A1%E8%B6%B3%3A%E5%AF%B9%E4%BB%BB%E6%84%8Fx%2Cy%E2%88%88R%2C%E9%83%BD%E6%9C%89f%28x%2By%29%3Df%28x%29f%28y%29-f%28x%29-f%28y%29%2B2%E6%88%90%E7%AB%8B%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%E6%BB%A1%E8%B6%B3%3A%E5%AF%B9%E4%BB%BB%E6%84%8Fx%2Cy%E2%88%88R%2C%E9%83%BD%E6%9C%89f%28x%2By%29%3Df%28x%29%C2%B7f%28y%29-f%28x%29-f%28y%29%2B2%E6%88%90%E7%AB%8B+%2C%E4%B8%94x%EF%BC%9E0%E6%97%B6%2Cf%28x%29%EF%BC%9E2%281%29%E6%B1%82f%280%29%E7%9A%84%E5%80%BC%2C%E5%B9%B6%E8%AF%81%E6%98%8E%EF%BC%9A%E5%BD%93x%EF%BC%9C0%E6%97%B6%2C1%EF%BC%9Cf%28x%29%EF%BC%9C2%282%29%E5%88%A4%E6%96%ADf%28x%29)
已知函数f(x)满足:对任意x,y∈R,都有f(x+y)=f(x)f(y)-f(x)-f(y)+2成立已知函数f(x)满足:对任意x,y∈R,都有f(x+y)=f(x)·f(y)-f(x)-f(y)+2成立 ,且x>0时,f(x)>2(1)求f(0)的值,并证明:当x<0时,1<f(x)<2(2)判断f(x)
已知函数f(x)满足:对任意x,y∈R,都有f(x+y)=f(x)f(y)-f(x)-f(y)+2成立
已知函数f(x)满足:对任意x,y∈R,都有f(x+y)=f(x)·f(y)-f(x)-f(y)+2成立 ,且x>0时,f(x)>2
(1)求f(0)的值,并证明:当x<0时,1<f(x)<2
(2)判断f(x)的单调性并加以证明
(3)若函数g(x)=|f(x)-k|在(负无穷,0)上递减,求实数k的取值范围
已知函数f(x)满足:对任意x,y∈R,都有f(x+y)=f(x)f(y)-f(x)-f(y)+2成立已知函数f(x)满足:对任意x,y∈R,都有f(x+y)=f(x)·f(y)-f(x)-f(y)+2成立 ,且x>0时,f(x)>2(1)求f(0)的值,并证明:当x<0时,1<f(x)<2(2)判断f(x)
(1)令x,y=0,设f(0)=t,则有方程t=t²-2t+2.解得t=1或t=2又∵x>0时f(x)>2,所以f(0)=2.呃然后那个证明先空着吧……
(2)设x1>x2.有f(x1+x2-x2)=f(x1-x2)f(x2)-f(x1)-f(x2-x2)+2.(这步好像有问题,不过想不出其他办法了,先这样吧……),因为f(0)=2.化简有2f(x1)/f(x2)=f(x1-x2).因为x1-x2>0.所以f(x1-x2)>2,即2f(x1)/f(x2)>2,要使其满足,则f(x1)/f(x2)>1,即x1,x2>0时f(x1)>f(x2),为单调增函数,x1,x2<0时为单调减函数