Question

15．已知函数f（x）=$\left\{\begin{array}{l}{x+1，x≤0}\\{lo{g}_{2}x，x＞0}\end{array}\right.$ 则函数y=f[f（x）+1]的零点个数是4．

Answer 1

分析 求出f（x）的零点，设为x₀，令f（x）+1=x₀即可解出y=f[f（x）+1]的零点．

解答 解：令f（x）=0得$\left\{\begin{array}{l}{x+1=0}\\{x≤0}\end{array}\right.$或$\left\{\begin{array}{l}{lo{g}_{2}x=0}\\{x＞0}\end{array}\right.$，解的x=-1或x=1．
即f（x）有两个零点x=1，x=-1．
令f[f（x）+1]=0，则f（x）+1=1或f（x）+1=-1．
即f（x）=0或f（x）=-2．
当f（x）=0时，x=±1，
当f（x）=-2时，$\left\{\begin{array}{l}{x+1=-2}\\{x≤0}\end{array}\right.$或$\left\{\begin{array}{l}{lo{g}_{2}x=-2}\\{x＞0}\end{array}\right.$，解的x=-3或x=$\frac{1}{4}$．
综上，f[f（x）+1]共有4个零点．
故答案为：4．

点评 本题考查了函数的零点计算，属于中档题．