望数学帝指教设函数f(x)是定义域为R上的增函数 且f(x)不等于0,对于任意的x1 x2属于R,都有f(x1+x2)=f(x1)×f(x2)求证f(x)>0 f(x1-x2)=f(x1)/f(x2)
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![望数学帝指教设函数f(x)是定义域为R上的增函数 且f(x)不等于0,对于任意的x1 x2属于R,都有f(x1+x2)=f(x1)×f(x2)求证f(x)>0 f(x1-x2)=f(x1)/f(x2)](/uploads/image/z/13938779-11-9.jpg?t=%E6%9C%9B%E6%95%B0%E5%AD%A6%E5%B8%9D%E6%8C%87%E6%95%99%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%E6%98%AF%E5%AE%9A%E4%B9%89%E5%9F%9F%E4%B8%BAR%E4%B8%8A%E7%9A%84%E5%A2%9E%E5%87%BD%E6%95%B0+%E4%B8%94f%28x%29%E4%B8%8D%E7%AD%89%E4%BA%8E0%2C%E5%AF%B9%E4%BA%8E%E4%BB%BB%E6%84%8F%E7%9A%84x1+x2%E5%B1%9E%E4%BA%8ER%2C%E9%83%BD%E6%9C%89f%EF%BC%88x1%2Bx2%EF%BC%89%3Df%EF%BC%88x1%EF%BC%89%C3%97f%EF%BC%88x2%EF%BC%89%E6%B1%82%E8%AF%81f%EF%BC%88x%EF%BC%89%EF%BC%9E0+f%EF%BC%88x1-x2%EF%BC%89%3Df%EF%BC%88x1%EF%BC%89%2Ff%EF%BC%88x2%EF%BC%89)
望数学帝指教设函数f(x)是定义域为R上的增函数 且f(x)不等于0,对于任意的x1 x2属于R,都有f(x1+x2)=f(x1)×f(x2)求证f(x)>0 f(x1-x2)=f(x1)/f(x2)
望数学帝指教
设函数f(x)是定义域为R上的增函数 且f(x)不等于0,对于任意的x1 x2属于R,都有f(x1+x2)=f(x1)×f(x2)
求证f(x)>0 f(x1-x2)=f(x1)/f(x2)
望数学帝指教设函数f(x)是定义域为R上的增函数 且f(x)不等于0,对于任意的x1 x2属于R,都有f(x1+x2)=f(x1)×f(x2)求证f(x)>0 f(x1-x2)=f(x1)/f(x2)
令x1=x2=x/2
f(x)=f^2(x/2)≥0,又因为f(x)不等于0,所以f(x)>0
f(x1-x2)*f(x2)=f(x1-x2+x2)=f(x1)
所以f(x1-x2)=f(x1)/f(x2)
在f(x1+x2)=f(x1)×f(x2)中令x1=x2=0,得:
f(0)=[f(0)]^2 ([f(0)]^2表示f(0)的平方)
∵?f(x)不等于0
∴f(0)=1,又f(x)是增函数
∴当x>0时,f(x)>f(0)>0
在f(x1+x2)=f(x1)×f(x2)中令x1=x,x2=-x(设x<0),得:
f(x)×f(-x)=1
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在f(x1+x2)=f(x1)×f(x2)中令x1=x2=0,得:
f(0)=[f(0)]^2 ([f(0)]^2表示f(0)的平方)
∵?f(x)不等于0
∴f(0)=1,又f(x)是增函数
∴当x>0时,f(x)>f(0)>0
在f(x1+x2)=f(x1)×f(x2)中令x1=x,x2=-x(设x<0),得:
f(x)×f(-x)=1
∴f(x)=1/[f(-x)]
由-x>0得f(-x)>0
∴f(x)=1/[f(-x)]>0
即当 x<0时,f(x)>0也成立
综上所述,f(x)>0 对一切x属于R都成立
2.在f(x1+x2)=f(x1)×f(x2)中令x1=x1,x2=-x2,得:
f(x1-x2)=f(x1)×f(-x2)
由上一问f(x)×f(-x)=1可知:f(-x2)=1/[f(x2]
∴ f(x1-x2)=f(x1)/f(x2)
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