已知函数,f（x）=log2 1-mx/x-1 的图像关于原点对称

已知函数,f（x）=log2 1-mx/x-1 的图像关于原点对称
已知函数,f（x）=log2 1-mx/x-1 的图像关于原点对称

已知函数,f（x）=log2 1-mx/x-1 的图像关于原点对称
（1）函数f(x) = log 2 [(1 – mx)/(x – 1)]的图像关于原点对称,说明函数f(x)是奇函数,定义域关于原点对称,求f(x)的定义域：(1 – mx)/(x – 1) > 0 => (1 – mx)(x – 1) > 0,所以当且仅当 m = -1 时,对应f(x) = log 2 [(1 + x)/(x – 1)],定义域是(-∞,-1)∪(1,+∞),f(x)的图像关于原点对称,符合题意； （2）f(x)在(1,+∞)上单调递减.证明：f(x) = log 2 [(1 + x)/(x – 1)],x∈(1,+∞),任取x 1 ,x 2 ∈(1,+∞),而且x 1 < x 2 ,计算f(x 2 )– f(x 1 ) = log 2 [(1 + x 2 )/(x 2 –1)] – log 2 [(1 + x 1 )/(x 1 –1)] = log 2 {[(1 + x 2 )/(x 2 –1)] / [(1 + x 1 )/(x 1 –1)]} = log 2 {[(x 2 + 1)(x 1 –1)] / [(x 2 –1)(x 1 + 1)]} = log 2 [(x 1 x 2 –x 2 + x 1 –1) / (x 1 x 2 + x 2 –x 1 –1)]①,因为x 1 < x 2 ,x 2 –x 1 > 0 > x 1 –x 2 ,所以(x 1 x 2 + x 2 –x 1 –1) > (x 1 x 2 –x 2 + x 1 –1) > 0,所以(x 1 x 2 –x 2 + x 1 –1) / (x 1 x 2 + x 2 –x 1 –1) ∈(0,1),进而log 2 [(x 1 x 2 –x 2 + x 1 –1) / (x 1 x 2 + x 2 –x 1 –1)] ∈(-∞,0),所以f(x 2 )– f(x 1 ) = log 2 [(x 1 x 2 –x 2 + x 1 –1) / (x 1 x 2 + x 2 –x 1 –1)] < 0,即f(x 2 ) < f(x 1 ),可知f(x)在(1,+∞)上单调递减.