Question

已知函数f(x)的定义域为R,对任意实数m,n,满足f(1/2)=2,且f(m+n)=f(m)+f(n)-1,当x>-1

Answer 1

(1)由题意可知：f(1/2)=f(1/4+1/4)=f(1/4)+f(1/4)-1=2

得f(1/4)=1.5.

f(1/4)=f(1/2-1/4)=f(1/2)+f(-1/4)-1

f(-1/4)=f(1/4)-f(1/2)+1=0.5

f(-1/2)=f(-1/4-1/4)=f(-1/4)+f(-1/4)-1=0

(2)在定义域R中,设a>b,令a=b+c,此时必定有c>0.由题意可知

f(a)=f(b+c)=f(b)+f(c)-1

则f(a)-f(b)=f(c)-1

又f(c)=f(c-1/2+1/2)=f(c-1/2)+f(1/2)-1=f(c-1/2)+1

因c>0.故c-1/2>-1/2.即有f(c-1/2)>0

所以

f(a)-f(b)=f(c-1/2)+1-1=f(c-1/2)>0

即在x的定义域内,当a>b时,恒有f(a)>f(b)

所以f(x)为单调递增