Question

已知定义域为R的单调函数f(x)满足f(x+y)=f(x)+f(y),且f(1)=2

Answer 1

∵f(x+y)=f(x)+f(y),

∴f(1+0)=f(1)+f(0),f(0)=0

令x=-y,可得f(0)=f(-y)+f(y),即f(y)=-f(-y).

∴f(x)为奇函数

∵f(2)=f(1)+f(1)=4

又∵f(0)=0,f(1)=2,

∴f(x)在R上单调递增