考點(diǎn):函數(shù)零點(diǎn)的判定定理
專題:函數(shù)的性質(zhì)及應(yīng)用
分析:原題等價(jià)于函數(shù)g(x)=
,(x≠0)的值域,求導(dǎo)函數(shù)可得函數(shù)的單調(diào)性,可得值域,可得答案.
解答:
解:當(dāng)x=0時(shí),可得f(0)=-1,故x=0不是函數(shù)的零點(diǎn);
當(dāng)x≠0時(shí),由函數(shù)f(x)=kx-e
x有零點(diǎn)可得kx=e
x有解,
即k=
,故k的取值范圍為函數(shù)g(x)=
,(x≠0)的值域,
∵y′=
=
,
令y′<0可得x<1,故函數(shù)g(x)在(-∞,0)上單調(diào)遞減,(0,1)上單調(diào)遞減,(1,+∞)上單調(diào)遞增,
故當(dāng)x<0時(shí),函數(shù)值g(x)<0,
當(dāng)x>0時(shí),g(1)為函數(shù)的最小值,且g(1)=e,故g(x)≥e,
綜上可得g(x)的取值范圍為g(x)<0或g(x)≥e,
故k的取值范圍為:k<0或k≥e.
故答案為:k≥e或k<0.
點(diǎn)評(píng):本題考查函數(shù)零點(diǎn)的判斷,轉(zhuǎn)化為函數(shù)的值域是解決問題的關(guān)鍵,屬中檔題.