新高考數(shù)學(xué)一輪復(fù)習(xí)講與練第05講 復(fù)合函數(shù)與冪函數(shù)（講）（解析版）

第2講復(fù)合函數(shù)與冪函數(shù)本講為重要知識點，題型主要圍繞函數(shù)的思想以及函數(shù)的性質(zhì)考察，主要通過對函數(shù)定義的理解解決抽象函數(shù)相關(guān)的問題，包括定義域和值域的一系列題型思想，對學(xué)生的邏輯思維能力包括對函數(shù)的理解需要透徹。此外增加一個基本初等函數(shù)中的冪函數(shù)，也是高中需要學(xué)習(xí)的函數(shù)之一?？键c一復(fù)合函數(shù)1.復(fù)合函數(shù)一般地，對于兩個函數(shù)y＝f(u)和u＝g(x)，如果通過變量u，y可以表示成x的函數(shù)，那么稱這個函數(shù)為函數(shù)y＝f(u)和u＝g(x)的復(fù)合函數(shù)，記作y＝f(g(x))，其中y＝f(u)叫做復(fù)合函數(shù)y＝f(g(x))的外層函數(shù)，u＝g(x)叫做y＝f(g(x))的內(nèi)層函數(shù).2.函數(shù)的復(fù)合單調(diào)性的變化:f(x)+g(x)=h(x)f(x)-g(x)=h(x)增+增=增增-增增+減增-減=增減+減=減減-減減+增減-增=減注意：加同不變，減異隨前。3.復(fù)合函數(shù)的單調(diào)性變化SKIPIF1<0SKIPIF1<0內(nèi)函數(shù)SKIPIF1<0外函數(shù)SKIPIF1<0增增增增減減減增減減減增注意：同增異減，即內(nèi)外函數(shù)單調(diào)性相同時，單調(diào)性遞增；反之，遞減。考點二冪函數(shù)1.冪函數(shù)的定義一般地，形如y＝xα(α∈R)的函數(shù)稱為冪函數(shù)，其中x是自變量，α為常數(shù)．2.5個常見冪函數(shù)的圖象與性質(zhì)函數(shù)y＝xy＝x2y＝x3SKIPIF1<0y＝x－1定義域RRR{x|x≥0}{x|x≠0}值域R{y|y≥0}R{y|y≥0}{y|y≠0}奇偶性奇函數(shù)偶函數(shù)奇函數(shù)非奇非偶函數(shù)奇函數(shù)單調(diào)性在R上單調(diào)遞增在(－∞，0)上單調(diào)遞減，在(0，＋∞)上單調(diào)遞增在R上單調(diào)遞增在(0，＋∞)上單調(diào)遞增在(－∞，0)和(0，＋∞)上單調(diào)遞減圖象過定點(0,0)，(1,1)(1,1)高頻考點一已知f(x)定義域，求f(g(x))的定義域例1、已知函數(shù)SKIPIF1<0的定義域為SKIPIF1<0，則函數(shù)SKIPIF1<0的定義域為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】∵SKIPIF1<0的定義域為SKIPIF1<0，∴SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，則函數(shù)SKIPIF1<0的定義域為SKIPIF1<0故選：A.【變式訓(xùn)練】1、若函數(shù)SKIPIF1<0的定義域為SKIPIF1<0，則函數(shù)SKIPIF1<0的定義域為______．【答案】SKIPIF1<0【解析】SKIPIF1<0的定義域為SKIPIF1<0SKIPIF1<0SKIPIF1<0即SKIPIF1<0的定義域為SKIPIF1<0故答案為：SKIPIF1<0【基本規(guī)律】1、抽象函數(shù)求解解析式：2、求解定義域就是求解x的取值范圍；3、f()相同(對應(yīng)法則相同條件)下“（）”內(nèi)取值范圍相同。注意：抽象函數(shù)求解定義域問題時要考慮定義域范圍內(nèi)的單調(diào)性問題。高頻考點二已知f(g(x))定義域，求f(x)的定義域例2、已知函數(shù)y=f(x+1)的定義域是[-2，3]，則y=f(x)的定義域是（

）A．[0，5] B．[-1，4] C．[-3，2] D．[-2，3]【答案】B【解析】∵函數(shù)y=f(x+1)的定義域是[-2，3]，∴-2≤x≤3，∴-1≤x+1≤4，∴函數(shù)y=f(x)的定義域是[-1，4].故選：B【變式訓(xùn)練】1、已知SKIPIF1<0的定義域為SKIPIF1<0，則SKIPIF1<0的定義域為

（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】因為SKIPIF1<0的定義域為SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0的定義域為SKIPIF1<0．故選：C高頻考點三已知f(g(x))定義域，求f(h(x))的定義域例3、已知函數(shù)SKIPIF1<0的定義域為SKIPIF1<0，則SKIPIF1<0的定義域為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】依題意函數(shù)SKIPIF1<0的定義域為SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，所以SKIPIF1<0的定義域為SKIPIF1<0.故選：B【變式訓(xùn)練】1、已知函數(shù)SKIPIF1<0定義域為SKIPIF1<0，則函數(shù)SKIPIF1<0的定義域為_______.【答案】SKIPIF1<0【解析】因SKIPIF1<0的定義域為SKIPIF1<0，則當(dāng)SKIPIF1<0時，SKIPIF1<0，即SKIPIF1<0的定義域為SKIPIF1<0，于是SKIPIF1<0中有SKIPIF1<0，解得SKIPIF1<0，所以函數(shù)SKIPIF1<0的定義域為SKIPIF1<0.故答案為：SKIPIF1<0高頻考點四復(fù)合函數(shù)定義域例4、已知函數(shù)SKIPIF1<0的定義域為SKIPIF1<0，則函數(shù)SKIPIF1<0的定義域是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】解：由題意得：SKIPIF1<0，解得SKIPIF1<0，由SKIPIF1<0解得SKIPIF1<0，故函數(shù)的定義域是SKIPIF1<0

.故選：D【變式訓(xùn)練】1、已知函數(shù)SKIPIF1<0的定義域為SKIPIF1<0，若SKIPIF1<0，則函數(shù)SKIPIF1<0的定義域為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】由題意得：SKIPIF1<0，即SKIPIF1<0，又SKIPIF1<0，∴SKIPIF1<0．故選：B高頻考點五抽象函數(shù)的值域例5：SKIPIF1<0是SKIPIF1<0上的奇函數(shù)，SKIPIF1<0是SKIPIF1<0上的偶函數(shù)，若函數(shù)SKIPIF1<0的值域為SKIPIF1<0，則SKIPIF1<0的值域為_____________．【答案】SKIPIF1<0【解析】解：由SKIPIF1<0是SKIPIF1<0上的奇函數(shù)，SKIPIF1<0是SKIPIF1<0上的偶函數(shù)得到SKIPIF1<0，SKIPIF1<0因為函數(shù)SKIPIF1<0的值域為SKIPIF1<0即SKIPIF1<0所以SKIPIF1<0又SKIPIF1<0，SKIPIF1<0得SKIPIF1<0所以SKIPIF1<0的值域為：SKIPIF1<0.故答案為：SKIPIF1<0.【變式訓(xùn)練】1.已知定義在R上的函數(shù)SKIPIF1<0滿足對任意實數(shù)x，都有SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0________．【答案】2021【解析】由題意，函數(shù)SKIPIF1<0滿足對任意實數(shù)x，都有SKIPIF1<0，且SKIPIF1<0，當(dāng)SKIPIF1<0且SKIPIF1<0時，可得SKIPIF1<0，則SKIPIF1<0SKIPIF1<0，所以SKIPIF1<0．故答案為：SKIPIF1<0.高頻考點六復(fù)合函數(shù)的值域例6、已知定義在R上的函數(shù)SKIPIF1<0滿足SKIPIF1<0，若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上的值域為SKIPIF1<0，則SKIPIF1<0在區(qū)間SKIPIF1<0上的值域為__________.【答案】SKIPIF1<0【解析】因為SKIPIF1<0，故對任意的整數(shù)SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，而SKIPIF1<0且SKIPIF1<0，故SKIPIF1<0，故SKIPIF1<0在區(qū)間SKIPIF1<0上的值域為：SKIPIF1<0，即為SKIPIF1<0.故答案為：SKIPIF1<0.【變式訓(xùn)練】1.若函數(shù)SKIPIF1<0的值域是SKIPIF1<0，則函數(shù)SKIPIF1<0的值域為__．【答案】SKIPIF1<0【解析】因為函數(shù)SKIPIF1<0的值域是SKIPIF1<0，所以函數(shù)SKIPIF1<0的值域為SKIPIF1<0，則SKIPIF1<0的值域為SKIPIF1<0，所以函數(shù)SKIPIF1<0的值域為SKIPIF1<0．故答案為：SKIPIF1<0．高頻考點七冪函數(shù)例7、現(xiàn)有下列函數(shù)：①SKIPIF1<0；②SKIPIF1<0；③SKIPIF1<0；④SKIPIF1<0；⑤SKIPI