Question

已知函数f(x)在（0,+∞）上有意义,且单调递增,若f(2)=1,f(xy)=f(x)+f(y)求f(1）f(4)的值

Answer 1

f(xy)=f(x)+f(y)

令y=1得f(x) = f(x) + f(1)

∴f(1) = 0

令x=y=2得f(4) = f(2)+f(2) = 2

f(x) + f(x+3) ≤ 2

f[x(x+3)] ≤ f(4)

f(x)在（0,+∞）上有意义,且单调递增

∴x>0,x+3>0,x(x+3) ≤4

解得 0 < x ≤ 1