Question

已知函数f(x)=sinπx,x<0, f(x-1)-1,x>0 则f(-11/6)+f(11/6)=

Answer 1

解：f(11/6)=f(11/6-1)-1=f(5/6)-1=f(5/6-1)-1-1=f(-1/6)-2=sin(-π/6)-2=-1/2-2=-5/2
f(-11/6)=sin(-11π/6)=sin(π/6)=1/2
所以f(-11/6)+f(11/6)=1/2-5/2=-2

Answer 2

f(11/6) = f(5/6)-1 = f(-1/6)-2

f(-11/6)+f(11/6) = sin(-11π/6）+ sin(-π/6) -2 = -2

Answer 3

x>0,f(x)=f(x-1)-1
f(11/6)=f(11/6-1)-1
=f(5/6)-1
=f(5/6-1)-1-1
==f(-1/6)-2
x<0,f(x)=sinπx
f(-11/6)+f(11/6)=f(-11/6)+f(-1/6)-2

=-sin11/6π-sin1/6π-2
=sin1/6π-sin1/6π-2
=-2