1.
f(xy)=f(x)+f(y)取y = 1 ,x = 1
则有 f(1*1)=f(1)+f(1) 既2f(1)= f(1)
则f(1)= 0
2.
因为 x属于(0,正无穷),所以有 1 = 1/x * x
则 f(1)= f(1/x) + f(x) = 0
所以 f(1/x) = - f(x)
3.设a>1,x>1,f(a*x) = f(a)+f(x)
则有 f(a*x)-f(x) = f(a)
a*x > x,f(a) > 0
所以f(x)在(0,正无穷)上是增函数
1.
f(xy)=f(x)+f(y)取y = 1 ,x = 1
则有 f(1*1)=f(1)+f(1) 既2f(1)= f(1)
则f(1)= 0
2.
因为 x属于(0,正无穷),所以有 1 = 1/x * x
则 f(1)= f(1/x) + f(x) = 0
所以 f(1/x) = - f(x)
3.设a>1,x>1,f(a*x) = f(a)+f(x)
则有 f(a*x)-f(x) = f(a)
a*x > x,f(a) > 0
所以f(x)在(0,正无穷)上是增函数