Question

定义在R上的函数f(x)满足f(x)=log2(1-x) ,x≤0

定义在R上的函数f(x)满足f(x)=log2(1-x) ,x≤0，f(x-1)-f(x-2),x>0 ,则f（2014）的值为

Answer 1

f(x)=f(x-1)-f(x-2) ; x>0

put x =2014
f(2014) = f(2013) -f(2012)
= f(2012)-f(2011) - f(2012)
= -f(2011)
=...
=...
= (-1)^(671) . f(1)
= -f(1)
= -( f(0) - f(-1) )
= f(-1)
= log<2>2
= 1

Answer 2

f2014=f2013=。。。=f0=log2(1-0)=0
注意定义域，前面递推的时候只能推到0，因为x>0. ,X-1>-1即只有大于-1的数才能参与递推