Question

已知f(x)定义域为R,且对任意m,n属于R都有f(m+n)=f(m)+f(n)-1,且f(-1/2)=0.当x>-1/

Answer 1

f(0)=f(0)+f(0)-1=2f(0)-1,所以f(0)=1

f(0)=f(1/2-1/2)=f(1/2)+f(-1/2)-1=f(1/2)-1=1,所以f(1/2)=2

设R上有x1,x2,且x10

f(x2)-f(x1)=f(x2-x1)-1

=f(x2-x1-1/2+1/2)-1

=f(x2-x1-1/2)+f(1/2)-2

=f(x2-x1-1/2)

x2-x1-1/2>0-1/2=-1/2,所以f(x2-x1-1/2)>0

f(x2)-f(x1)>0

所以f(x)单调递增

举例f(x)=2x+1

2(m+n)+1=(2m+1)+(2n+1)-1

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

x>-1/2时,f(x)=2x+1>2(-1/2)+1=0