已知函数fx=x+1,x≤0.=log2x,x＞0则函数f(fx)+1的零点个数是

Question

已知函数fx=x+1,x≤0.=log2x,x＞0则函数f(fx)+1的零点个数是

Answer 1

当x＞1时,f(x)=log(2)x＞0
f(f(x))+1=log(2)(log(2)x)+1=0
x=√2
当0＜x≤1时,f(x)=log(2)x≤0
f(f(x))+1=log(2)x+1+1=0
x=1/4
当-1＜x≤0时,f(x)=x+1＞0
f(f(x))+1=log(2)(x+1)+1=0
x=-1/2
当x≤-1时,f(x)=x+1≤0
f(f(x))+1=(x+1)+1+1=0
x=-3
综上所述,f(f(x))+1共有4个零点.