(1)f(2)=f(1+1)=f(1)*f(1)=k^2f(4)=f(2+2)=f(2)*f(2)=k^4f(8)=f(4+4)=f(4)*f(4)=k^8f(10)=f(2+8)=f(2)*f(8)=k^10(2)设x1>=0,x1+p>0(p>0)f(-p)=f(x1-x1-p)=f(x1)*f(-x1-p)-p<0,f(-p)>1;-x1-p<0,f(-x 展开
(1)f(2)=f(1+1)=f(1)*f(1)=k^2f(4)=f(2+2)=f(2)*f(2)=k^4f(8)=f(4+4)=f(4)*f(4)=k^8f(10)=f(2+8)=f(2)*f(8)=k^10(2)设x1>=0,x1+p>0(p>0)f(-p)=f(x1-x1-p)=f(x1)*f(-x1-p)-p<0,f(-p)>1;-x1-p<0,f(-x1-p)>1f(x1)>0即当x>=0,有f(x)>0,又x<0,f(x)>1,所以x∈R,f(x)>0设x2>x3,则x3-x2<0,f(x3-x2)>1f(x3)=f(x2+x3-x2)=f(x2)*f(x3-x2)>f(x2)f(x)在R上是减函数f(x)=f(x+0)=f(x)*f(0)f(0)=1f(x+5)>1/f(x)f(x+5)f(x)>1,f(2x+5)>f(0)2x+5<0,x<-5/2 收起
