Question

已知函数为分段函数f x={-cosπx,x>0 f(x+1)+1,x<=0,则f[4/3]+f[-4/3]

要过程详细点呐，谢谢

Answer 1

解
f(x)={-cosπx x>0
{f(x+1)+1 x<=0
4/3>0，直接代入f(x)=-cosπx
f(4/3)=-cos4/3π=-cos(π+π/3)=cosπ/3=1/2

-4/3<0,代入f(x+1)+1=f(x)
∴f(-4/3)=f(-4/3+1)+1=f(-1/3)+1=f(2/3)+1+1=f(2/3)+2——代入f(x)=-cosπx
=-cos2/3π
=1/2
∴f(4/3)+f(-4/3)
=1/2+1/2
=1

追问

怎么一个等3一个等1

追答

我后面加错了,不好意思
f(-4/3)=f(2/3)+2
=-cos2/3π+2
=1/2+2
=5/2
∴f(4/3)+f(-4/3)
=1/2+5/2
=3

Answer 2

f[4/3]=-cos4/3π=-（-cosπ/3)=1/2
f[-4/3]=f(-4/3+1)+1=f(-1/3)+1=f(2/3)+2=-cos2/3π+2=5/2
f[4/3]+f[-4/3]=1/2+5/2=3