f(0)=f(0+0)=f(0)*f(0)
f(0)=1
f(0)=f[x+(-x)]=f(x)*f(-x) (x>0)
f(-x)>1>0
f(x)>0
任取x1>x2 设x2=x1+t (t0
所以f(x2)>f(x1)
f(x)是减函数
2)当f(4)=1/16时,令a=2=b,则f(4)=f(2)*f(2),所以f(2)=1/4
若f(x-3)*f(5-X^2)0,则
f(0)=f(0+0)=f(0)*f(0)
f(0)=1
f(0)=f[x+(-x)]=f(x)*f(-x) (x>0)
f(-x)>1>0
f(x)>0
任取x1>x2 设x2=x1+t (t0
所以f(x2)>f(x1)
f(x)是减函数
2)当f(4)=1/16时,令a=2=b,则f(4)=f(2)*f(2),所以f(2)=1/4
若f(x-3)*f(5-X^2)0,则