(1)f(x+2)=-1/f(x+1)=f(x)
f(x+3)=-1/f(x+2)=-1/f(x)
f(x+4)=-1/f(x+3)=f(x)
.
f(x+2k)=f(x)
f(x+2k+1)=-1/f(x+2k)=-1/f(x)
f(-x+2k+1)=-1/f(-x)=1/f(x)
x∈(0,1/2),-x∈(-1/2,0)
-x+2k+1∈(2k+1/2,2k+1)
現(xiàn)令x=-x+2k+1 即x∈(2k+1/2,2k+1)
當(dāng)x∈(2k+1/2,2k+1)時,f(x)=1/f(x)=8^(-x)
(3) 假設(shè)不等式log8f(x)>x^2-(k+3)x-k+2有解
即 x>x^2-(k+3)x-k+2
設(shè)g(x)=x^2-(k+4)x-k+2
即g(2k+1)=(2k+1)^2-(k+4)(2k+1)-k+2