Question

14．已知符號(hào)函數(shù)sgn（x）=$\left\{\begin{array}{l}{1，x＞0}\\{0，x=0}\\{-1，x＜0}\end{array}\right.$，f（x）=x²-2x，則函數(shù)F（x）=sgn[f（x）]-f（x）的零點(diǎn)個(gè)數(shù)為5．

Answer 1

分析 利用符號(hào)函數(shù)求出F（x）的解析式，然后求解函數(shù)的零點(diǎn)即可得到結(jié)果．

解答 解：符號(hào)函數(shù)sgn（x）=$\left\{\begin{array}{l}{1，x＞0}\\{0，x=0}\\{-1，x＜0}\end{array}\right.$，f（x）=x²-2x，
則函數(shù)F（x）=sgn[f（x）]-f（x）=$\left\{\begin{array}{l}{-{x}^{2}+2x+1，x∈（-∞，0）∪（2，+∞）}\\{-{x}^{2}+2x，x=0或x=2}\\{-{x}^{2}+2x-1，x∈（0，2）}\end{array}\right.$，
當(dāng)x∈（-∞，0）∪（2，+∞）時(shí)，-x²+2x+1=0，解得x=$\sqrt{2}+1$或x=1-$\sqrt{2}$滿(mǎn)足題意．
當(dāng)x=0或x=2時(shí)，-x²+2x=0，x=0或x=2是函數(shù)的零點(diǎn)．
當(dāng)x∈（0，2）時(shí)，-x²+2x-1=0，解得x=1滿(mǎn)足題意．
所以函數(shù)的零點(diǎn)個(gè)數(shù)是5．
故答案為：5．

點(diǎn)評(píng) 本題考查新函數(shù)的應(yīng)用，函數(shù)的零點(diǎn)個(gè)數(shù)的求法，考查分析問(wèn)題解決問(wèn)題的能力．