Question

15．已知函數(shù)f（x）=$\left\{\begin{array}{l}{x+1，x≤0}\\{lo{g}_{2}x，x＞0}\end{array}\right.$ 則函數(shù)y=f[f（x）+1]的零點個數(shù)是4．

Answer 1

分析 求出f（x）的零點，設(shè)為x₀，令f（x）+1=x₀即可解出y=f[f（x）+1]的零點．

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）有兩個零點x=1，x=-1．
令f[f（x）+1]=0，則f（x）+1=1或f（x）+1=-1．
即f（x）=0或f（x）=-2．
當f（x）=0時，x=±1，
當f（x）=-2時，$\left\{\begin{array}{l}{x+1=-2}\\{x≤0}\end{array}\right.$或$\left\{\begin{array}{l}{lo{g}_{2}x=-2}\\{x＞0}\end{array}\right.$，解的x=-3或x=$\frac{1}{4}$．
綜上，f[f（x）+1]共有4個零點．
故答案為：4．

點評 本題考查了函數(shù)的零點計算，屬于中檔題．