1)取x=x'=1代入f(xx')=f(x)+f(x'),可得 f(1)=02)取x=x'=-1代入f(xx')=f(x)+f(x'),可得f(1)=2f(-1) ,f(-1)=03)取 x'=-1,代入f(xx')=f(x)+f(x'),可得f(-x)=f(x)+f(-1)=f(x)则为偶函数4)f(x)+f(1-1/x)≤0 利用f(xx')=f(x)+f(x'),可得f(x-1)≤0当x-1>0 由于f(x-1)≤f(1)=0并结合y=f(x)在(0,+∞)上是增函数, 可得0<=x-1<=1,1<=x<=2当x-1≤0,由于f(x-1)≤f(-1)=0,由于f(x)为偶函数,那么在(-∞,0)上为减函数,可得-1<=x-1<=0,0<=x<=1 综合上面分析并考虑到f(x)+f(1-1/x)≤0中x!=0可知 0<x<=2 |