人教A版高中數(shù)學(xué)(必修第一冊)同步講義第19講 3.3冪函數(shù)（含解析）

第05講3.3冪函數(shù)課程標(biāo)準(zhǔn)學(xué)習(xí)目標(biāo)①了解冪函數(shù)的概念，會(huì)求冪函數(shù)的解析式；②掌握常見冪函數(shù)的圖像；③利用冪函數(shù)的單調(diào)性比較指數(shù)式大小。④利用冪函數(shù)的性質(zhì)解不等式及待定參數(shù)的求解通過本節(jié)課的學(xué)習(xí)，要求掌握冪函數(shù)的概念，能根據(jù)冪函數(shù)的要求求出冪函數(shù)的解析式，并能根據(jù)冪函數(shù)的性質(zhì)求待定參數(shù).知識(shí)點(diǎn)01一：冪函數(shù)的概念1、定義：一般地，函數(shù)SKIPIF1<0叫做冪函數(shù)，其中SKIPIF1<0是自變量，SKIPIF1<0是常數(shù).2、冪函數(shù)的特征①SKIPIF1<0中SKIPIF1<0前的系數(shù)為“1”②SKIPIF1<0中SKIPIF1<0的底數(shù)是單個(gè)的自變量“SKIPIF1<0”③SKIPIF1<0中SKIPIF1<0是常數(shù)【即學(xué)即練1】（2023·全國·高一專題練習(xí)）現(xiàn)有下列函數(shù)：①SKIPIF1<0；②SKIPIF1<0；③SKIPIF1<0；④SKIPIF1<0；⑤SKIPIF1<0；⑥SKIPIF1<0；⑦SKIPIF1<0，其中冪函數(shù)的個(gè)數(shù)為（

）A．1 B．2 C．3 D．4【答案】B【詳解】冪函數(shù)滿足SKIPIF1<0形式，故SKIPIF1<0，SKIPIF1<0滿足條件，共2個(gè)故選：B知識(shí)點(diǎn)02：冪函數(shù)的圖象與性質(zhì)1、五個(gè)冪函數(shù)的圖象(記憶五個(gè)冪函數(shù)的圖象）當(dāng)SKIPIF1<0時(shí)，我們得到五個(gè)冪函數(shù)：SKIPIF1<0；SKIPIF1<0；SKIPIF1<0；SKIPIF1<0；SKIPIF1<02、五個(gè)冪函數(shù)的性質(zhì)SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0定義域SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0值域SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0奇偶性奇函數(shù)偶函數(shù)奇函數(shù)非奇非偶奇函數(shù)單調(diào)性在SKIPIF1<0上單調(diào)遞增在SKIPIF1<0上單調(diào)遞減在SKIPIF1<0單調(diào)遞增在SKIPIF1<0上單調(diào)遞增在SKIPIF1<0單調(diào)遞增在SKIPIF1<0上單調(diào)遞減在SKIPIF1<0上單調(diào)遞減定點(diǎn)SKIPIF1<03、拓展：①SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0單調(diào)遞增；②SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0單調(diào)遞減.【即學(xué)即練2】（江西省贛州市2022-2023學(xué)年高一上學(xué)期11月期中考試數(shù)學(xué)試題）冪函數(shù)SKIPIF1<0在SKIPIF1<0上為減函數(shù)，則SKIPIF1<0的值為______.【答案】SKIPIF1<0【詳解】由函數(shù)SKIPIF1<0是冪函數(shù)，則SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，在SKIPIF1<0上為減函數(shù)，滿足題意；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，在SKIPIF1<0上為增函數(shù)，不合題意.故答案為：SKIPIF1<0.題型01判斷函數(shù)是否為冪函數(shù)【典例1】（2023·高一課時(shí)練習(xí)）在函數(shù)①SKIPIF1<0，②SKIPIF1<0，③SKIPIF1<0，④SKIPIF1<0，SKIPIF1<0，⑥SKIPIF1<0中，是冪函數(shù)的是（

）A．①②④⑤ B．③④⑥ C．①②⑥ D．①②④⑤⑥【答案】C【詳解】冪函數(shù)是形如SKIPIF1<0（SKIPIF1<0，SKIPIF1<0為常數(shù)）的函數(shù)，①是SKIPIF1<0的情形，②是SKIPIF1<0的情形，⑥是SKIPIF1<0的情形，所以①②⑥都是冪函數(shù)；③是指數(shù)函數(shù)，不是冪函數(shù)；⑤中SKIPIF1<0的系數(shù)是2，所以不是冪函數(shù)；④是常函數(shù)，不是冪函數(shù).故選：C.【典例2】（2023秋·云南德宏·高一統(tǒng)考期末）下列函數(shù)既是冪函數(shù)又是奇函數(shù)的是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】對于A，由冪函數(shù)的定義知SKIPIF1<0是冪函數(shù)，由題意可知SKIPIF1<0的定義域?yàn)镾KIPIF1<0,SKIPIF1<0，所以SKIPIF1<0是奇函數(shù)，符合題意；故A正確；對于B，由冪函數(shù)的定義知SKIPIF1<0是冪函數(shù)，由題意可知SKIPIF1<0的定義域?yàn)镾KIPIF1<0,SKIPIF1<0，所以SKIPIF1<0是偶函數(shù)，不符合題意；故B錯(cuò)誤；對于C，由冪函數(shù)的定義知SKIPIF1<0不是冪函數(shù)，不符合題意；故C錯(cuò)誤；對于D，由冪函數(shù)的定義知SKIPIF1<0不是冪函數(shù)，不符合題意；故D錯(cuò)誤；故選：A.【變式1】（2023·高一課時(shí)練習(xí)）給出下列函數(shù)：①SKIPIF1<0；②SKIPIF1<0；③SKIPIF1<0；④SKIPIF1<0；⑤SKIPIF1<0；⑥SKIPIF1<0，其中是冪函數(shù)的有（

）A．1個(gè) B．2個(gè)C．3個(gè) D．4個(gè)【答案】B【詳解】由冪函數(shù)的定義：形如SKIPIF1<0（SKIPIF1<0為常數(shù)）的函數(shù)為冪函數(shù)，則可知①SKIPIF1<0和④SKIPIF1<0是冪函數(shù).故選；B.題型02求冪函數(shù)的值【典例1】（2023秋·江西萍鄉(xiāng)·高一統(tǒng)考期末）已知冪函數(shù)SKIPIF1<0的圖像過點(diǎn)SKIPIF1<0，則SKIPIF1<0的值為（

）A．2 B．3 C．4 D．5【答案】A【詳解】根據(jù)題意，設(shè)冪函數(shù)為SKIPIF1<0，則可得SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0故選:A【典例2】（2023春·上海楊浦·高三復(fù)旦附中校考階段練習(xí)）已知冪函數(shù)SKIPIF1<0的圖像過點(diǎn)SKIPIF1<0，則SKIPIF1<0的值為___________．【答案】SKIPIF1<0【詳解】設(shè)冪函數(shù)為SKIPIF1<0，由題意，SKIPIF1<0，解得SKIPIF1<0，所以冪函數(shù)解析式為SKIPIF1<0，所以SKIPIF1<0.故答案為：SKIPIF1<0【變式1】（2023秋·寧夏吳忠·高一統(tǒng)考期中）若SKIPIF1<0是冪函數(shù)，且SKIPIF1<0，則SKIPIF1<0__________【答案】9【詳解】解：因?yàn)镾KIPIF1<0是冪函數(shù)，記SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，故SKIPIF1<0，所以SKIPIF1<0.故答案為：9題型03求冪函數(shù)的解析式【典例1】（2023·浙江·統(tǒng)考模擬預(yù)測）已知SKIPIF1<0是冪函數(shù)，且滿足：①SKIPIF1<0；②SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，請寫出符合上述條件的一個(gè)函數(shù)SKIPIF1<0___________.【答案】SKIPIF1<0（答案不唯一）(形如SKIPIF1<0，SKIPIF1<0為正奇數(shù)，SKIPIF1<0為正偶數(shù)，均可)【詳解】因?yàn)镾KIPIF1<0是冪函數(shù)，且SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，故可設(shè)SKIPIF1<0，（SKIPIF1<0，SKIPIF1<0互質(zhì)），又SKIPIF1<0，所以SKIPIF1<0為奇數(shù)，SKIPIF1<0為偶數(shù)，故SKIPIF1<0為符合條件的一個(gè)函數(shù)，故答案為：SKIPIF1<0(形如SKIPIF1<0，SKIPIF1<0為正奇數(shù)，SKIPIF1<0為正偶數(shù)，均可).【典例2】（2023·高一課時(shí)練習(xí)）冪函數(shù)SKIPIF1<0是偶函數(shù)，且在SKIPIF1<0上為增函數(shù)，則函數(shù)解析式為_________．【答案】SKIPIF1<0或SKIPIF1<0【詳解】SKIPIF1<0是冪函數(shù)，也是偶函數(shù)，且在SKIPIF1<0上為增函數(shù)，SKIPIF1<0且SKIPIF1<0為偶數(shù)，解得SKIPIF1<0或SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0.故答案為：SKIPIF1<0或SKIPIF1<0【變式1】（2023秋·遼寧·高一大連二十四中校聯(lián)考期末）已知冪函數(shù)SKIPIF1<0在第一象限單調(diào)遞減，則SKIPIF1<0______.【答案】SKIPIF1<0【詳解】由題知，冪函數(shù)SKIPIF1<0在第一象限單調(diào)遞減，所以SKIPIF1<0，解得SKIPIF1<0（舍去），或SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，故答案為：SKIPIF1<0題型04根據(jù)函數(shù)是冪函數(shù)求參數(shù)【典例1】（2023·遼寧錦州·渤海大學(xué)附屬高級(jí)中學(xué)?？寄M預(yù)測）若冪函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，則SKIPIF1<0（

）A．SKIPIF1<0 B．3 C．SKIPIF1<0或3 D．1或SKIPIF1<0【答案】A【詳解】因?yàn)楹瘮?shù)SKIPIF1<0為冪函數(shù)，且在區(qū)間SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0且SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0或SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，滿足題意；當(dāng)SKIPIF1<0時(shí)，足SKIPIF1<0，不符合題意.綜上SKIPIF1<0.故選：A.【典例2】（2023·寧夏銀川·銀川一中?？级＃┮阎瘮?shù)SKIPIF1<0是冪函數(shù)，且為偶函數(shù)，則實(shí)數(shù)SKIPIF1<0______．【答案】2【詳解】由函數(shù)SKIPIF1<0是冪函數(shù)，則SKIPIF1<0，得SKIPIF1<0或SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0，其定義域?yàn)镾KIPIF1<0，SKIPIF1<0，則SKIPIF1<0是偶函數(shù)，滿足條件；當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0是奇函數(shù)，不合題意.故答案為：2.【變式1】（2023秋·陜西咸陽·高一統(tǒng)考期末）已知冪函數(shù)SKIPIF1<0滿足SKIPIF1<0，則SKIPIF1<0______.【答案】SKIPIF1<0【詳解】因?yàn)楹瘮?shù)SKIPIF1<0為冪函數(shù)，則SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，又因?yàn)镾KIPIF1<0，所以SKIPIF1<0，故答案為：SKIPIF1<0.【變式2】（2023春·上海楊浦·高一上海市控江中學(xué)?？奸_學(xué)考試）已知冪函數(shù)SKIPIF1<0的圖像不經(jīng)過原點(diǎn)，則實(shí)數(shù)SKIPIF1<0__________.【答案】SKIPIF1<0【詳解】由已知函數(shù)SKIPIF1<0為冪函數(shù)，得SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，定義域?yàn)镾KIPIF1<0，函數(shù)圖像不經(jīng)過原點(diǎn)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，定義域?yàn)镾KIPIF1<0，且SKIPIF1<0，函數(shù)圖像經(jīng)過原點(diǎn)，綜上所述：SKIPIF1<0，故答案為：SKIPIF1<0.題型05求冪函數(shù)的定義域【典例1】（2023秋·浙江·高一校聯(lián)考期末）已知冪函數(shù)SKIPIF1<0，則此函數(shù)的定義域?yàn)開_______．【答案】SKIPIF1<0.【詳解】由冪函數(shù)SKIPIF1<0，可得SKIPIF1<0，解得SKIPIF1<0，即SKIPIF1<0，則滿足SKIPIF1<0，即冪函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0.故答案為：SKIPIF1<0.【典例2】（2023·高一課時(shí)練習(xí)）函數(shù)SKIPIF1<0的定義域?yàn)開______．【答案】SKIPIF1<0【詳解】SKIPIF1<0，所以，SKIPIF1<0.因此，函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0.故答案為：SKIPIF1<0.【變式1】（2023·高一課時(shí)練習(xí)）若冪函數(shù)SKIPIF1<0的圖象經(jīng)過點(diǎn)SKIPIF1<0，求SKIPIF1<0的定義域.【答案】SKIPIF1<0【詳解】因?yàn)镾KIPIF1<0為冪函數(shù)，所以設(shè)SKIPIF1<0．又SKIPIF1<0的圖象經(jīng)過點(diǎn)SKIPIF1<0，可得SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0．故f(x)的定義域?yàn)镾KIPIF1<0．題型06求冪函數(shù)的值域【典例1】（2023·全國·高三專題練習(xí)）函數(shù)SKIPIF1<0的值域?yàn)開_______.【答案】SKIPIF1<0【詳解】SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0的值域?yàn)镾KIPIF1<0.故答案為：SKIPIF1<0【典例2】（2023·全國·高三專題練習(xí)）若冪函數(shù)SKIPIF1<0的圖象過點(diǎn)SKIPIF1<0，則SKIPIF1<0的值域?yàn)開___________．【答案】SKIPIF1<0【詳解】設(shè)SKIPIF1<0，因?yàn)閮绾瘮?shù)SKIPIF1<0的圖象過點(diǎn)SKIPIF1<0，所以SKIPIF1<0所以SKIPIF1<0，所以SKIPIF1<0故答案為：SKIPIF1<0【典例3】（2023·高一課時(shí)練習(xí)）已知函數(shù)SKIPIF1<0，若函數(shù)SKIPIF1<0的值域?yàn)镾KIPIF1<0，則實(shí)數(shù)SKIPIF1<0的取值范圍為__________．【答案】SKIPIF1<0【詳解】由函數(shù)SKIPIF1<0單調(diào)遞增，①當(dāng)SKIPIF1<0時(shí)，若SKIPIF1<0，有SKIPIF1<0，而SKIPIF1<0，此時(shí)函數(shù)SKIPIF1<0的值域不是SKIPIF1<0；②當(dāng)SKIPIF1<0時(shí)，若SKIPIF1<0，有SKIPIF1<0，而SKIPIF1<0，若函數(shù)SKIPIF1<0的值域?yàn)镾KIPIF1<0，必有SKIPIF1<0，可得SKIPIF1<0．則實(shí)數(shù)SKIPIF1<0的取值范圍為SKIPIF1<0．故答案為：SKIPIF1<0【變式1】（2023·全國·高三專題練習(xí)）已知冪函數(shù)SKIPIF1<0的圖像過點(diǎn)SKIPIF1<0，則SKIPIF1<0的值域是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】SKIPIF1<0冪函數(shù)SKIPIF1<0的圖像過點(diǎn)SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0的值域是SKIPIF1<0.故選：D.題型07冪函數(shù)的圖象的判斷及應(yīng)用【典例1】（2023·全國·高三對口高考）已知冪函數(shù)SKIPIF1<0（SKIPIF1<0且SKIPIF1<0與SKIPIF1<0互質(zhì)）的圖像如圖所示，則（

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A．SKIPIF1<0、SKIPIF1<0均為奇數(shù)且SKIPIF1<0 B．SKIPIF1<0為奇數(shù)，SKIPIF1<0為偶數(shù)且SKIPIF1<0C．SKIPIF1<0為奇數(shù)，SKIPIF1<0為偶數(shù)且SKIPIF1<0 D．SKIPIF1<0為偶數(shù)，SKIPIF1<0為奇數(shù)且SKIPIF1<0【答案】D【詳解】由圖像知函數(shù)為偶函數(shù)，所以p為偶數(shù)，且由圖像的形狀判定SKIPIF1<0，又因?yàn)閜與q互質(zhì)，所以q為奇數(shù)，故選：D.【典例2】（2023·全國·高三對口高考）給定一組函數(shù)解析式：①SKIPIF1<0；②SKIPIF1<0；③SKIPIF1<0；④SKIPIF1<0；⑤SKIPIF1<0；⑥SKIPIF1<0；⑦SKIPIF1<0．如圖所示一組函數(shù)圖象．圖象對應(yīng)的解析式號(hào)碼順序正確的是（

）

A．⑥③④②⑦①⑤ B．⑥④②③⑦①⑤C．⑥④③②⑦①⑤ D．⑥④③②⑦⑤①【答案】C【詳解】圖象（1）關(guān)于原點(diǎn)對稱，為奇函數(shù)，且不過原點(diǎn)、第一象限遞減，故SKIPIF1<0滿足；圖象（2）關(guān)于SKIPIF1<0軸對稱，為偶函數(shù)，且不過原點(diǎn)、第一象限遞減，故SKIPIF1<0滿足；圖象（3）非奇非偶函數(shù)，且不過原點(diǎn)、第一象限遞減，故SKIPIF1<0滿足；圖象（4）關(guān)于SKIPIF1<0軸對稱，為偶函數(shù)，且過原點(diǎn)、第一象限遞增，故SKIPIF1<0滿足；圖象（5）關(guān)于原點(diǎn)對稱，為奇函數(shù)，且過原點(diǎn)、第一象限遞增，故SKIPIF1<0滿足；圖象（6）非奇非偶函數(shù)，且過原點(diǎn)、第一象限遞增，而增長率隨SKIPIF1<0增大遞減，故SKIPIF1<0滿足；圖象（7）非奇非偶函數(shù)，且過原點(diǎn)、第一象限遞增，而增長率隨SKIPIF1<0增大遞增，故SKIPIF1<0滿足；故圖象對應(yīng)解析式順序?yàn)棰蔻堍邰冖撷佗?故選：C【變式1】（2023·新疆阿勒泰·統(tǒng)考三模）已知函數(shù)則函數(shù)SKIPIF1<0，則函數(shù)SKIPIF1<0的圖象大致是（

）A． B．C． D．【答案】B【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0圖像與SKIPIF1<0的圖像關(guān)于SKIPIF1<0軸對稱，由SKIPIF1<0解析式，作出SKIPIF1<0的圖像如圖從而可得SKIPIF1<0圖像為B選項(xiàng).故選：B.【變式2】（2023·全國·高三專題練習(xí)）函數(shù)SKIPIF1<0的圖像大致為（

）A．B．C．D．【答案】B【詳解】由SKIPIF1<0，排除A，D．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0，排除C．故選：B．題型08冪函數(shù)過定點(diǎn)問題【典例1】（2023·高一課時(shí)練習(xí)）函數(shù)SKIPIF1<0恒過定點(diǎn)______．【答案】SKIPIF1<0【詳解】當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0函數(shù)恒過定點(diǎn)SKIPIF1<0.故答案為：SKIPIF1<0.【典例2】（2023·高一課時(shí)練習(xí)）冪函數(shù)SKIPIF1<0的圖象過點(diǎn)SKIPIF1<0，則函數(shù)SKIPIF1<0的圖象經(jīng)過定點(diǎn)__________．【答案】SKIPIF1<0【詳解】因?yàn)閮绾瘮?shù)SKIPIF1<0過點(diǎn)SKIPIF1<0，可解得SKIPIF1<0，所以SKIPIF1<0，故SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故SKIPIF1<0恒過定點(diǎn)SKIPIF1<0.故答案為SKIPIF1<0【變式1】（2023·高一課時(shí)練習(xí)）冪函數(shù)SKIPIF1<0的圖像一定經(jīng)過第______象限【答案】一、三【詳解】因?yàn)镾KIPIF1<0為自然數(shù)，所以SKIPIF1<0為偶數(shù)，所以SKIPIF1<0為奇數(shù)，所以SKIPIF1<0是奇函數(shù)，且函數(shù)的圖像經(jīng)過SKIPIF1<0和點(diǎn)SKIPIF1<0并且在SKIPIF1<0單調(diào)遞增，所以冪函數(shù)SKIPIF1<0的圖像一定經(jīng)過第一、三象限．故答案為：一、三題型09冪函數(shù)的單調(diào)性【典例1】（多選）（2023秋·重慶長壽·高一統(tǒng)考期末）下列函數(shù)既是冪函數(shù)，又在SKIPIF1<0上單調(diào)遞減的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】CD【詳解】對于A，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減但不是冪函數(shù)，故選項(xiàng)A錯(cuò)誤；對于B，函數(shù)SKIPIF1<0是冪函數(shù)，在SKIPIF1<0上單調(diào)遞增，故選項(xiàng)B錯(cuò)誤；對于C，函數(shù)SKIPIF1<0是冪函數(shù)且在SKIPIF1<0上單調(diào)遞減，故選項(xiàng)C正確；對于D，函數(shù)SKIPIF1<0是冪函數(shù)且在SKIPIF1<0上單調(diào)遞減，故選項(xiàng)D正確，故選：CD.【典例2】（2023·高一課時(shí)練習(xí)）已知函數(shù)：①SKIPIF1<0，②SKIPIF1<0，③SKIPIF1<0，④SKIPIF1<0，既是偶函數(shù)，又在SKIPIF1<0上為增函數(shù)的是_________．【答案】①④【詳解】對于①SKIPIF1<0，設(shè)SKIPIF1<0，定義域?yàn)镾KIPIF1<0，滿足SKIPIF1<0，故SKIPIF1<0為偶函數(shù)，又SKIPIF1<0，在SKIPIF1<0上為增函數(shù)，符合題意；對于②，SKIPIF1<0定義域?yàn)镽，且為偶函數(shù)，在SKIPIF1<0上為增函數(shù)，故在SKIPIF1<0上為減函數(shù)，不符題意；對于③SKIPIF1<0，定義域?yàn)镽，設(shè)SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0為奇函數(shù)，不符題意；對于④SKIPIF1<0，定義域?yàn)镾KIPIF1<0，設(shè)SKIPIF1<0，滿足SKIPIF1<0，故SKIPIF1<0為偶函數(shù)，在SKIPIF1<0上為減函數(shù)，故在SKIPIF1<0上為增函數(shù)，符合題意，故答案為：①④【變式1】（2023·四川成都·石室中學(xué)?？寄M預(yù)測）冪函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減，則下列說法正確的是（

）A．SKIPIF1<0 B．SKIPIF1<0是減函數(shù)C．SKIPIF1<0是奇函數(shù) D．SKIPIF1<0是偶函數(shù)【答案】C【詳解】函數(shù)SKIPIF1<0為冪函數(shù)，則SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，不滿足條件，排除A；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減，滿足題意.函數(shù)SKIPIF1<0在SKIPIF1<0和SKIPIF1<0上單調(diào)遞減，但不是減函數(shù)，排除B；因?yàn)楹瘮?shù)定義域關(guān)于原點(diǎn)對稱，且SKIPIF1<0，所以函數(shù)SKIPIF1<0是奇函數(shù)，不是偶函數(shù)，故C正確，D錯(cuò)誤.故選：C.【變式2】（2023·全國·高三專題練習(xí)）已知SKIPIF1<0，若冪函數(shù)SKIPIF1<0奇函數(shù)，且在SKIPIF1<0上為嚴(yán)格減函數(shù)，則SKIPIF1<0__________.【答案】-1【詳解】解：因?yàn)閮绾瘮?shù)SKIPIF1<0在SKIPIF1<0上為嚴(yán)格減函數(shù)，所以SKIPIF1<0，所以SKIPIF1<0，又因?yàn)閮绾瘮?shù)SKIPIF1<0奇函數(shù)，且SKIPIF1<0，所以SKIPIF1<0，故答案為：-1題型10冪函數(shù)的奇偶性【典例1】（2023·高一課時(shí)練習(xí)）已知冪函數(shù)SKIPIF1<0的圖像關(guān)于SKIPIF1<0軸對稱，則SKIPIF1<0等于（

）A．1 B．2 C．1或2 D．3【答案】A【詳解】由于函數(shù)是冪函數(shù)，所以SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，是偶函數(shù)，圖像關(guān)于SKIPIF1<0軸對稱，符合題意.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，是奇函數(shù)，圖像不關(guān)于SKIPIF1<0軸對稱，不符合題意.所以SKIPIF1<0的值為SKIPIF1<0.故選：A【典例2】（2023秋·山東棗莊·高一棗莊八中?？计谀┮阎獌绾瘮?shù)SKIPIF1<0的圖像關(guān)于SKIPIF1<0軸對稱.(1)求SKIPIF1<0的值;(2)若函數(shù)SKIPIF1<0，求SKIPIF1<0的單調(diào)遞增區(qū)間.【答案】(1)SKIPIF1<0(2)SKIPIF1<0【詳解】（1）解:由題意知SKIPIF1<0，解得SKIPIF1<0，或SKIPIF1<0．又因?yàn)镾KIPIF1<0的圖像關(guān)于y軸對稱，所以SKIPIF1<0為偶函數(shù)，從而SKIPIF1<0．所以，SKIPIF1<0.（2）解：由（1）知，SKIPIF1<0SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，對稱軸為SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，對稱軸為SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增．所以，SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0．【變式1】（2023春·河北保定·高二?？茧A段練習(xí)）設(shè)SKIPIF1<0，若冪函數(shù)SKIPIF1<0定義域?yàn)镽，且其圖像關(guān)于SKIPIF1<0軸成軸對稱，則SKIPIF1<0的值可以為（

）A．1 B．4 C．7 D．10【答案】C【詳解】解：由題意知SKIPIF1<0，因?yàn)槠鋱D像關(guān)于y軸成軸對稱，則SKIPIF1<0.故選：C．【變式2】（2023春·江蘇南京·高一江蘇省高淳高級(jí)中學(xué)校聯(lián)考階段練習(xí)）請寫出一個(gè)滿足條件①和②的冪函數(shù)SKIPIF1<0，條件：①SKIPIF1<0是偶函數(shù)；②SKIPIF1<0為SKIPIF1<0上的減函數(shù)．則SKIPIF1<0________．【答案】SKIPIF1<0(答案不唯一)【詳解】設(shè)SKIPIF1<0，根據(jù)冪函數(shù)為偶函數(shù)，則SKIPIF1<0為偶數(shù)，又SKIPIF1<0為SKIPIF1<0上單調(diào)遞減，故SKIPIF1<0，故可取SKIPIF1<0，故答案為：SKIPIF1<0（答案不唯一）題型11根據(jù)冪函數(shù)的單調(diào)性求參數(shù)【典例1】（2023秋·浙江杭州·高一杭州市長河高級(jí)中學(xué)?？计谀┮阎獌绾瘮?shù)SKIPIF1<0在SKIPIF1<0上是減函數(shù)，則SKIPIF1<0的值為（

）A．SKIPIF1<0 B．1 C．3 D．1或SKIPIF1<0【答案】B【詳解】因?yàn)楹瘮?shù)SKIPIF1<0是冪函數(shù)，則SKIPIF1<0，所以SKIPIF1<0或SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上是增函數(shù)，不合題意.當(dāng)SKIPIF1<0時(shí)SKIPIF1<0在SKIPIF1<0上是減函數(shù)，成立.故選：B.【典例2】（2023春·四川廣安·高一?？茧A段練習(xí)）已知冪函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增．(1)求SKIPIF1<0的值及函數(shù)SKIPIF1<0的解析式；(2)若函數(shù)SKIPIF1<0在SKIPIF1<0上的最大值為3，求實(shí)數(shù)SKIPIF1<0的值．【答案】(1)SKIPIF1<0，SKIPIF1<0；(2)SKIPIF1<0.【詳解】（1）冪函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，故SKIPIF1<0，解得SKIPIF1<0，故SKIPIF1<0；（2）由（1）知：SKIPIF1<0，所以SKIPIF1<0，所以函數(shù)SKIPIF1<0的圖象為開口向下的拋物線，對稱軸為直線SKIPIF1<0；由于SKIPIF1<0在SKIPIF1<0上的最大值為3，①當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，故SKIPIF1<0，解得SKIPIF1<0；②當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，故SKIPIF1<0，解得SKIPIF1<0；③當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，故SKIPIF1<0，解得SKIPIF1<0（舍去）或SKIPIF1<0（舍去）．綜上所述，SKIPIF1<0．【變式1】（2023秋·河南許昌·高三校考期末）已知函數(shù)SKIPIF1<0是冪函數(shù)，且在SKIPIF1<0上是增函數(shù)，則實(shí)數(shù)SKIPIF1<0的值為______．【答案】1【詳解】因?yàn)楹瘮?shù)SKIPIF1<0是冪函數(shù)，則SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，又因?yàn)镾KIPIF1<0在SKIPIF1<0上是增函數(shù)，所以SKIPIF1<0，所以SKIPIF1<0.故答案為：1【變式2】（2023·高一課時(shí)練習(xí)）冪函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，則SKIPIF1<0的值為______．【答案】2【詳解】解：因?yàn)楹瘮?shù)SKIPIF1<0是冪函數(shù)，則有SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，不符合題意，當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，符合題意.所以SKIPIF1<0的值為SKIPIF1<0故答案為：SKIPIF1<0題型12根據(jù)冪函數(shù)的單調(diào)性解不等式【典例1】（2023·高三課時(shí)練習(xí)）已知冪函數(shù)SKIPIF1<0過點(diǎn)SKIPIF1<0，則滿足SKIPIF1<0的實(shí)數(shù)SKIPIF1<0的取值范圍是______．【答案】SKIPIF1<0【詳解】可得冪函數(shù)SKIPIF1<0，且函數(shù)在其定義域SKIPIF1<0上單調(diào)遞增．因?yàn)镾KIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，所以實(shí)數(shù)a的取值范圍是SKIPIF1<0．故答案為：SKIPIF1<0【典例2】（2023·高一課時(shí)練習(xí)）求不等式SKIPIF1<0的解.【答案】SKIPIF1<0【詳解】解：SKIPIF1<0等價(jià)于SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，故答案為：SKIPIF1<0．【變式1】（2023·江蘇淮安·江蘇省盱眙中學(xué)?？寄M預(yù)測）已知冪函數(shù)SKIPIF1<0，若SKIPIF1<0，則a的取值范圍是__________.【答案】SKIPIF1<0【詳解】由冪函數(shù)SKIPIF1<0，可得函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，且是遞減函數(shù)，因?yàn)镾KIPIF1<0，可得SKIPIF1<0，解得SKIPIF1<0，即實(shí)數(shù)SKIPIF1<0的取值范圍為SKIPIF1<0.故答案為：SKIPIF1<0.【變式2】（2023秋·湖南常德·高一漢壽縣第一中學(xué)?？计谀┤魞绾瘮?shù)SKIPIF1<0過點(diǎn)SKIPIF1<0，則滿足不等式SKIPIF1<0的實(shí)數(shù)SKIPIF1<0的取值范圍是__________．【答案】SKIPIF1<0【詳解】設(shè)冪函數(shù)SKIPIF1<0，其圖像過點(diǎn)SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0；∴SKIPIF1<0，函數(shù)定義域?yàn)镾KIPIF1<0，在SKIPIF1<0上單調(diào)遞增，不等式SKIPIF1<0等價(jià)于SKIPIF1<0，解得SKIPIF1<0；則實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0.故答案為：SKIPIF1<0題型13根據(jù)冪函數(shù)的單調(diào)性比較大小【典例1】（2023·高一課時(shí)練習(xí)）已知SKIPIF1<0，若SKIPIF1<0，則下列各式中正確的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，SKIPIF1<0，故SKIPIF1<0，故SKIPIF1<0.故選：B.【典例2】（2023·高一課時(shí)練習(xí)）比較下列各組數(shù)的大?。?1)SKIPIF1<0，SKIPIF1<0；(2)SKIPIF1<0，SKIPIF1<0；(3)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0．【答案】(1)SKIPIF1<0(2)SKIPIF1<0(3)SKIPIF1<0【詳解】（1）因?yàn)閮绾瘮?shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，且SKIPIF1<0，所以SKIPIF1<0．（2）因?yàn)閮绾瘮?shù)SKIPIF1<0在SKIPIF1<0上為增函數(shù)，且SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0．（3）SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，因?yàn)閮绾瘮?shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0．【變式1】（2023春·遼寧鞍山·高一校聯(lián)考階段練習(xí)）已知SKIPIF1<0，則SKIPIF1<0的大小關(guān)系為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】SKIPIF1<0，因?yàn)楹瘮?shù)SKIPIF1<0是實(shí)數(shù)集上的增函數(shù)，所以由SKIPIF1<0可得：SKIPIF1<0，即SKIPIF1<0，故選：CA夯實(shí)基礎(chǔ)B能力提升A夯實(shí)基礎(chǔ)一、單選題1．（2023·海南·統(tǒng)考模擬預(yù)測）已知SKIPIF1<0為冪函數(shù)，則（

）．A．SKIPIF1<0在SKIPIF1<0上單調(diào)遞增 B．SKIPIF1<0在SKIPIF1<0上單調(diào)遞減C．SKIPIF1<0在SKIPIF1<0上單調(diào)遞增 D．SKIPIF1<0在SKIPIF1<0上單調(diào)遞減【答案】B【詳解】因?yàn)镾KIPIF1<0是冪函數(shù)，所以SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，所以SKIPIF1<0或SKIPIF1<0，對于SKIPIF1<0，函數(shù)在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減；對于SKIPIF1<0，函數(shù)在SKIPIF1<0上單調(diào)遞減，且為奇函數(shù)，故在SKIPIF1<0上單調(diào)遞減；故只有B選項(xiàng)“SKIPIF1<0在SKIPIF1<0上單調(diào)遞減”符合這兩個(gè)函數(shù)的性質(zhì)．故選：B2．（2023春·四川宜賓·高一?？茧A段練習(xí)）已知冪函數(shù)SKIPIF1<0的圖象過點(diǎn)SKIPIF1<0，則SKIPIF1<0等于（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】設(shè)SKIPIF1<0，因?yàn)閮绾瘮?shù)SKIPIF1<0的圖象過點(diǎn)SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0.故選：C.3．（2023春·湖北宜昌·高一校聯(lián)考期中）已知點(diǎn)SKIPIF1<0在冪函數(shù)SKIPIF1<0的圖象上，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】SKIPIF1<0函數(shù)SKIPIF1<0是冪函數(shù)，SKIPIF1<0，即SKIPIF1<0點(diǎn)SKIPIF1<0在冪函數(shù)SKIPIF1<0的圖象上，SKIPIF1<02，即SKIPIF1<0，故SKIPIF1<0.故選：D.4．（2023·全國·高三專題練習(xí)）函數(shù)SKIPIF1<0的圖象大致為（

）A． B．C． D．【答案】C【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0為偶函數(shù)，排除A,B選項(xiàng)；易知當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0為增函數(shù)，且增加幅度較為緩和，所以D不正確.故選：C.5．（2023·全國·高一專題練習(xí)）如圖，下列3個(gè)冪函數(shù)的圖象，則其圖象對應(yīng)的函數(shù)可能是（

）A．①SKIPIF1<0，②SKIPIF1<0，③SKIPIF1<0 B．①SKIPIF1<0，②SKIPIF1<0，③SKIPIF1<0C．①SKIPIF1<0，②SKIPIF1<0，③SKIPIF1<0 D．①SKIPIF1<0，②SKIPIF1<0，③SKIPIF1<0【答案】A【詳解】由函數(shù)SKIPIF1<0是反比例函數(shù)，其對應(yīng)圖象為①；函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，應(yīng)為圖②；因?yàn)镾KIPIF1<0的定義域?yàn)镾KIPIF1<0且為奇函數(shù)，故應(yīng)為圖③.故選：A.6．（2023秋·安徽·高一校聯(lián)考期末）若冪函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減，則SKIPIF1<0（

）A．3 B．1 C．SKIPIF1<0或3 D．1或SKIPIF1<0【答案】A【詳解】因?yàn)楹瘮?shù)SKIPIF1<0為冪函數(shù)，且在區(qū)間SKIPIF1<0上單調(diào)遞減，所以SKIPIF1<0且SKIPIF1<0，又SKIPIF1<0，可得SKIPIF1<0或SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，滿足SKIPIF1<0，舍去；當(dāng)SKIPIF1<0時(shí)，滿足SKIPIF1<0.綜上SKIPIF1<0.故選：A.7．（2023秋·廣東湛江·高一雷州市第一中學(xué)?？计谀┤鐖D所示，圖中的曲線是冪函數(shù)SKIPIF1<0在第一象限的圖象，已知SKIPIF1<0取SKIPIF1<0，SKIPIF1<0四個(gè)值，則相應(yīng)于SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的SKIPIF1<0依次為（

）A．SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0 B．SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0C．SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0 D．SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0【答案】B【詳解】解：根據(jù)冪函數(shù)SKIPIF1<0的性質(zhì)，在第一象限內(nèi)的圖象:當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0越大，SKIPIF1<0遞增速度越快，故SKIPIF1<0的SKIPIF1<0，SKIPIF1<0的SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0越大，曲線越陡峭，所以曲線SKIPIF1<0的SKIPIF1<0，曲線SKIPIF1<0的SKIPIF1<0.故選:B8．（2023·江蘇常州·江蘇省前黃高級(jí)中學(xué)校考模擬預(yù)測）如圖所示是函數(shù)SKIPIF1<0（SKIPIF1<0均為正整數(shù)且SKIPIF1<0互質(zhì)）的圖象，則（

）A．SKIPIF1<0是奇數(shù)且SKIPIF1<0B．SKIPIF1<0是偶數(shù)，SKIPIF1<0是奇數(shù)，且SKIPIF1<0C．SKIPIF1<0是偶數(shù)，SKIPIF1<0是奇數(shù)，且SKIPIF1<0D．SKIPIF1<0是奇數(shù)，且SKIPIF1<0【答案】B【詳解】由冪函數(shù)性質(zhì)可知：SKIPIF1<0與SKIPIF1<0恒過點(diǎn)SKIPIF1<0，即在第一象限的交點(diǎn)為SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0；又SKIPIF1<0圖象關(guān)于SKIPIF1<0軸對稱，SKIPIF1<0為偶函數(shù)，SKIPIF1<0，又SKIPIF1<0互質(zhì)，SKIPIF1<