解:(1)
,y=x
3是對(duì)等函數(shù);
(2)研究對(duì)數(shù)函數(shù)y=log
ax,其定義域?yàn)椋?,+∞),所以log
a|x|=log
ax,又|log
ax|≥0,所以當(dāng)且僅當(dāng)log
ax≥0時(shí)f(|x|)=|f(x)|成立.所以對(duì)數(shù)函數(shù)y=log
ax在其定義域(0,+∞)內(nèi)不是對(duì)等函數(shù).
當(dāng)0<a<1時(shí),若x∈(0,1],則log
ax≥0,此時(shí)y=log
ax是對(duì)等函數(shù);
當(dāng)a>1時(shí),若x∈[1,+∞),則log
ax≥0,此時(shí)y=log
ax是對(duì)等函數(shù);
總之,當(dāng)0<a<1時(shí),在(0,1]及其任意非空子集內(nèi)y=log
ax是對(duì)等函數(shù);當(dāng)a>1時(shí),在[1,+∞)及其任意非空子集內(nèi)y=log
ax是對(duì)等函數(shù).
(3)對(duì)任意x∈D,討論f(x)與f(-x)的關(guān)系.
1)若D不關(guān)于原點(diǎn)對(duì)稱(chēng),如
雖是對(duì)等函數(shù),但不是奇函數(shù)或偶函數(shù);
2)若D={0},則f(0)=|f(0)|≥0.當(dāng)f(0)=0時(shí),f(x)既是奇函數(shù)又是偶函數(shù);當(dāng)f(0)>0時(shí),f(x)是偶函數(shù).
3)以下均在D關(guān)于原點(diǎn)對(duì)稱(chēng)的假設(shè)下討論.
當(dāng)x>0時(shí),f(|x|)=f(x)=|f(x)|≥0;
當(dāng)x<0時(shí),f(|x|)=f(-x)=|f(x)|,若|f(x)|=f(x),則有f(-x)=f(x);此時(shí),當(dāng)x>0時(shí),-x<0,令-x=t,則x=-t,且t<0,由前面討論知,f(-t)=f(t),從而f(x)=f(-x);
綜上討論,當(dāng)x<0時(shí),若f(x)≥0,則f(x)是偶函數(shù).
若當(dāng)x<0時(shí),f(x)≤0,則f(|x|)=f(-x)=|f(x)|=-f(x);此時(shí),當(dāng)x>0時(shí),-x<0,令-x=t,則x=-t,且t<0,由前面討論知,f(-t)=-f(t),從而f(x)=-f(-x);
若f(0)=0,則對(duì)任意x∈D,都有f(-x)=-f(x).
綜上討論,若當(dāng)x<0時(shí),f(x)≤0,且f(0)=0,則f(x)是奇函數(shù).若f(0)≠0,則f(x)不是奇函數(shù)也不是偶函數(shù).
分析:(1)根據(jù)對(duì)等函數(shù)的定義,我們判斷
,y=x
3是對(duì)等函數(shù);
(2)要想一個(gè)函數(shù)不是“對(duì)等函數(shù)”關(guān)鍵是根據(jù)題中條件對(duì)任意x∈D,f(|x|)=|f(x)|,或舉出反例;
(3)對(duì)任意x∈D,對(duì)集合D分類(lèi)討論f(x)與f(-x)的關(guān)系,最后給出結(jié)論.
點(diǎn)評(píng):本小題主要考查進(jìn)行簡(jiǎn)單的合情推理、對(duì)數(shù)函數(shù)的性質(zhì)等基礎(chǔ)知識(shí),考查運(yùn)算求解能力,考查化歸與轉(zhuǎn)化思想.要想判斷f(x)為“對(duì)等函數(shù)”,要經(jīng)過(guò)嚴(yán)密的論證說(shuō)明f(x)滿足“對(duì)等函數(shù)”的概念,但要判斷f(x)不為“對(duì)等函數(shù)”,僅須要舉出一個(gè)反例即可.