Question

4．已知函數(shù)f（x）=$\left\{\begin{array}{l}{1-|x|，x≤1}\\{（x-1）^{2}，x＞1}\end{array}\right.$，函數(shù)g（x）=$\frac{4}{5}$-f（1-x），則y=f（x）-g（x）零點(diǎn)的個(gè)數(shù)為4．

Answer 1

分析 求出函數(shù)f（1-x）的解析式，推出f（x）-g（x）的表達(dá)式，然后求解函數(shù)的零點(diǎn)．

解答 解：函數(shù)f（x）=$\left\{\begin{array}{l}{1-|x|，x≤1}\\{（x-1）^{2}，x＞1}\end{array}\right.$，$f（1-x）=\left\{\begin{array}{l}1-|1-x|，x≥0\\{x^2}，x＜0\end{array}\right.$，
則$f（x）+f（1-x）=\left\{\begin{array}{l}{x^2}+x+1，x＜0\\ 1，0≤x≤1\\{x^2}-3x+3，x＞1\end{array}\right.$，
令f（x）-g（x）=0，
可得$f（x）+f（1-x）=\frac{4}{5}$，
畫出y=f（1-x）+f（x）與y=$\frac{4}{5}$的圖象如圖所示：
由圖可得：y=f（1-x）+f（x）與y=$\frac{4}{5}$有4個(gè)交點(diǎn)
故y=f（x）-g（x）有4個(gè)零點(diǎn)．
故答案為：4．

點(diǎn)評(píng) 本題考查函數(shù)的解析式的應(yīng)用，函數(shù)的零點(diǎn)的求法，考查數(shù)形結(jié)合，轉(zhuǎn)化思想的應(yīng)用，考查計(jì)算能力．