2022版新教材高考數(shù)學(xué)一輪復(fù)習(xí)第2章函數(shù)的概率與基本初等函數(shù)Ⅰ第2節(jié)函數(shù)的單調(diào)性與最值學(xué)案含解析新人教A版

1、第二節(jié)函數(shù)的單調(diào)性與最值一、教材概念結(jié)論性質(zhì)重現(xiàn)1單調(diào)遞增、單調(diào)遞減一般的，設(shè)函數(shù)f (x)的定義域為i，區(qū)間di：(1)如果x1，x2d，當(dāng)x1x2時，都有f (x1)f (x2)，那么就稱函數(shù)f (x)在區(qū)間d上單調(diào)遞增(2)如果x1，x2d，當(dāng)x1f (x2)，那么就稱函數(shù)f (x)在區(qū)間d上單調(diào)遞減2增函數(shù)、減函數(shù)(1)當(dāng)函數(shù)f (x)在定義域上單調(diào)遞增時，我們就稱它是增函數(shù)；(2)當(dāng)函數(shù)f (x)在定義域上單調(diào)遞減時，我們就稱它是減函數(shù)1單調(diào)遞增(減)函數(shù)定義中的x1，x2的三個特征一是任意性；二是有大小，即x1x2)；三是同屬于一個單調(diào)區(qū)間，三者缺一不可2增、減函數(shù)定義的等價形式對

2、于x1，x2d，都有(x1x2)f (x1)f (x2)0(0(0，解得x4或x0”的是()af (x)2xbf (x)|x1|cf (x)xdf (x)ln(x1)ad解析：由(x1x2)f (x1)f (x2)0可知，f (x)在(0，)上是增函數(shù)，a，d選項中，f (x)為增函數(shù)；b中，f (x)|x1|在(0，)上不單調(diào)；對于f (x)x，因為y與yx在(0，)上單調(diào)遞減，因此f (x)在(0，)上是減函數(shù)故選ad4試討論函數(shù)f (x)(a0)在(1,1)上的單調(diào)性解：(方法一：定義法)設(shè)1x1x21，f (x)aa，則f (x1)f (x2)aa.因為1x1x20，x110，x210

3、時，f (x1)f (x2)0，即f (x1)f (x2)，函數(shù)f (x)在(1,1)上單調(diào)遞減；當(dāng)a0時，f (x1)f (x2)0，即f (x1)0時，f (x)0，函數(shù)f (x)在(1,1)上單調(diào)遞減；當(dāng)a0，函數(shù)f (x)在(1，1)上單調(diào)遞增判斷函數(shù)的單調(diào)性和求單調(diào)區(qū)間的方法定義法一般步驟為設(shè)元作差變形判斷符號得出結(jié)論圖象法若f (x)是以圖象形式給出的，或者f (x)的圖象易作出，則可由圖象的上升或下降判斷函數(shù)的單調(diào)性導(dǎo)數(shù)法先求導(dǎo)數(shù)，再利用導(dǎo)數(shù)值的正負(fù)確定函數(shù)的單調(diào)區(qū)間性質(zhì)法對于由基本初等函數(shù)的和、差構(gòu)成的函數(shù)，根據(jù)各基本初等函數(shù)的增減性及“增增增，增減增，減減減，減增減”進(jìn)行判斷

4、復(fù)合法對于復(fù)合函數(shù)，先將函數(shù)f (g(x)分解成f (t)和tg(x)，然后討論(判斷)這兩個函數(shù)的單調(diào)性，再根據(jù)復(fù)合函數(shù)“同增異減”的規(guī)則進(jìn)行判斷考點2函數(shù)的最值(值域)綜合性(1)函數(shù)y|x1|x2|的最小值為_3解析：(方法一)函數(shù)y作出函數(shù)的圖象如圖所示根據(jù)圖象可知，函數(shù)y|x1|x2|的最小值為3.(方法二)利用絕對值不等式的性質(zhì)y|x1|x2|x1|2x|x12x|3.故函數(shù)的最小值為3.(2)函數(shù)f (x)log2(x2)在區(qū)間1,1上的最大值為_3解析：由于y在1,1上單調(diào)遞減，ylog2(x2)在1,1上單調(diào)遞增，所以f (x)在1,1上單調(diào)遞減，故f (x)在1,1上的最大

5、值為f (1)3.(3)函數(shù)f (x)的最大值為_4解析：當(dāng)x0時，f (x)x24x(x2)24，而2(，0，此時f (x)在x2處取得最大值，且f (2)4；當(dāng)x0時，f (x)sin x，此時f (x)在區(qū)間(0，)上的最大值為1.綜上所述，函數(shù)f (x)的最大值為4.求函數(shù)最值的五種常用方法及其思路(1)單調(diào)性法：先確定函數(shù)的單調(diào)性，再由單調(diào)性求最值(2)圖象法：先作出函數(shù)的圖象，再觀察其最高點、最低點，得出最值(3)換元法：對比較復(fù)雜的函數(shù)可通過換元轉(zhuǎn)化為熟悉的函數(shù)，再用相應(yīng)的方法求最值(4)分離常數(shù)法：求形如y(ac0)的函數(shù)的值域或最值常用分離常數(shù)法求解(5)基本不等式法：先對解

6、析式變形，使之具備“一正二定三相等”的條件后用基本不等式求出最值1函數(shù)y的值域為_1,1)解析：由y，可得x2.由x20，知0，解得1yf (3)f (2)bf ()f (2)f (3)cf ()f (3)f (2)df ()f (2)f (3)f (2)，即f ()f (3)f (2)比較函數(shù)值大小的解題思路比較函數(shù)值的大小時，若自變量的值不在同一個單調(diào)區(qū)間內(nèi)，要利用函數(shù)的性質(zhì)，轉(zhuǎn)化到同一個單調(diào)區(qū)間內(nèi)進(jìn)行比較對于選擇題、填空題能數(shù)形結(jié)合的盡量用圖象法求解考向2解不等式已知函數(shù)f (x)ln x2x，若f (x24)2，則實數(shù)x的取值范圍是_(， 2)(2, )解析：因為函數(shù)f (x)ln x

7、2x在(0，)上單調(diào)遞增，且f (1)ln 122.由f (x24)2得f (x24)f (1)所以0x241，解得x2或2xf (h(x)的形式，再根據(jù)函數(shù)的單調(diào)性去掉“f ”，得到一般的不等式g(x)h(x)(或g(x)0成立，則實數(shù)a的取值范圍為_4,8)解析：由題意，函數(shù)f (x)在(，1和(1，)上分別單調(diào)遞增，且f (x)在(，1上的最高點不高于其在(1，)上的最低點，即解得4a8.利用單調(diào)性求參數(shù)的范圍(或值)的方法(1)視參數(shù)為已知數(shù)，依據(jù)函數(shù)的圖象或單調(diào)性的定義，確定函數(shù)的單調(diào)區(qū)間，與已知單調(diào)區(qū)間比較求參數(shù)；(2)需注意，若分段函數(shù)在r上是單調(diào)的，則該函數(shù)在每一段上具有相同的

8、單調(diào)性，還要注意分界點處的函數(shù)值大小1定義在r上的偶函數(shù)f (x)滿足f (x)f (x2)，且在1,0上單調(diào)遞減設(shè)af ()，bf (2)，cf (3)，則a，b，c的大小關(guān)系是()abca babc cbac dacbc解析：因為偶函數(shù)f (x)滿足f (x2)f (x)，所以函數(shù)f (x)的周期為2，則af ()f (2)，bf (2)f (0)，cf (3)f (1)因為120，且函數(shù)f (x)在1,0上單調(diào)遞減，所以ba0，x1x2，且f (a2a)f (2a2)，則實數(shù)a的取值范圍為_0,1)解析：因為函數(shù)f (x)滿足(x1x2)f (x1)f (x2)0，x1x2，所以函數(shù)f (x)在2,2上單調(diào)遞增，所