《貴州省遵義市私立貴龍中學(xué)高三數(shù)學(xué)總復(fù)習(xí) 函數(shù)的奇偶性和周期性課件 新人教A版》由會(huì)員分享，可在線閱讀，更多相關(guān)《貴州省遵義市私立貴龍中學(xué)高三數(shù)學(xué)總復(fù)習(xí) 函數(shù)的奇偶性和周期性課件 新人教A版（35頁(yè)珍藏版）》請(qǐng)?jiān)谘b配圖網(wǎng)上搜索。

1、 20112011考綱下載考綱下載1 1了解奇函數(shù)、偶函數(shù)的定義，并能運(yùn)用奇偶性的定義判了解奇函數(shù)、偶函數(shù)的定義，并能運(yùn)用奇偶性的定義判斷一些簡(jiǎn)單函數(shù)的奇偶性斷一些簡(jiǎn)單函數(shù)的奇偶性2 2掌握奇函數(shù)與偶函數(shù)的圖象對(duì)稱關(guān)系，并熟練地利用對(duì)掌握奇函數(shù)與偶函數(shù)的圖象對(duì)稱關(guān)系，并熟練地利用對(duì)稱性解決函數(shù)的綜合問(wèn)題稱性解決函數(shù)的綜合問(wèn)題 新課標(biāo)新課標(biāo)考試大綱考試大綱把函數(shù)的奇偶性又提到與函數(shù)的單把函數(shù)的奇偶性又提到與函數(shù)的單調(diào)性同等地位，因此，函數(shù)的奇偶性在新高考中占有重要調(diào)性同等地位，因此，函數(shù)的奇偶性在新高考中占有重要的地位，成為新的熱點(diǎn)，在命題時(shí)主要是與函數(shù)的概念、的地位，成為新的熱點(diǎn)，在命題時(shí)主要

2、是與函數(shù)的概念、圖象、性質(zhì)綜合在一起考查而近幾年的高考中加大了對(duì)圖象、性質(zhì)綜合在一起考查而近幾年的高考中加大了對(duì)非三角函數(shù)的周期性和抽象函數(shù)的奇偶性，周期性的考查非三角函數(shù)的周期性和抽象函數(shù)的奇偶性，周期性的考查力度力度. 請(qǐng)注意請(qǐng)注意! ! 1 1奇函數(shù)、偶函數(shù)、奇偶性奇函數(shù)、偶函數(shù)、奇偶性 對(duì)于函數(shù)對(duì)于函數(shù)f f( (x x) )，其定義域關(guān)于原點(diǎn)對(duì)稱：，其定義域關(guān)于原點(diǎn)對(duì)稱： 如果對(duì)于函數(shù)定義域內(nèi)任意一個(gè)如果對(duì)于函數(shù)定義域內(nèi)任意一個(gè)x x，都有，都有f f( (x x) )f f( (x x) )，那么函數(shù)，那么函數(shù)f f( (x x) )就是奇函數(shù)；就是奇函數(shù)； 如果對(duì)于函數(shù)定義域內(nèi)任

3、意一個(gè)如果對(duì)于函數(shù)定義域內(nèi)任意一個(gè)x x，都有，都有f f( (x x) )f f( (x x) )，那么函數(shù)，那么函數(shù)f f( (x x) )就是偶函數(shù)；就是偶函數(shù)； 如果一個(gè)函數(shù)是奇函數(shù)如果一個(gè)函數(shù)是奇函數(shù)( (或偶函數(shù)或偶函數(shù)) )，則稱這個(gè)函數(shù)在其定義域內(nèi)具，則稱這個(gè)函數(shù)在其定義域內(nèi)具有奇偶性有奇偶性課前自助餐課前自助餐課本導(dǎo)讀課本導(dǎo)讀 2 2證明函數(shù)奇偶性的方法步驟證明函數(shù)奇偶性的方法步驟 確定函數(shù)定義域關(guān)于確定函數(shù)定義域關(guān)于原點(diǎn)原點(diǎn)對(duì)稱；對(duì)稱； 判定判定f f( (x x) )f f( (x x)()(或或f f( (x x) )f f( (x x)，從而證得函數(shù)是奇，從而證得函數(shù)

4、是奇( (偶偶) )函函數(shù)數(shù) 3 3奇偶函數(shù)的性質(zhì)奇偶函數(shù)的性質(zhì) 奇函數(shù)圖象關(guān)于奇函數(shù)圖象關(guān)于原點(diǎn)原點(diǎn)對(duì)稱，對(duì)稱， 偶函數(shù)圖象關(guān)于偶函數(shù)圖象關(guān)于y y軸軸對(duì)稱；對(duì)稱； 若奇函數(shù)若奇函數(shù)f f( (x x) )在在x x0 0處有意義，則處有意義，則f f(0)(0)0 0； 奇函數(shù)在關(guān)于原點(diǎn)對(duì)稱的兩個(gè)區(qū)間上分別單調(diào)，則其單調(diào)性奇函數(shù)在關(guān)于原點(diǎn)對(duì)稱的兩個(gè)區(qū)間上分別單調(diào)，則其單調(diào)性一致一致； 偶函數(shù)在關(guān)于原點(diǎn)對(duì)稱的兩個(gè)區(qū)間上分別單調(diào)，則其單調(diào)性偶函數(shù)在關(guān)于原點(diǎn)對(duì)稱的兩個(gè)區(qū)間上分別單調(diào)，則其單調(diào)性相反相反 若函數(shù)若函數(shù)f f( (x x) )為偶函數(shù)，則為偶函數(shù)，則f f( (x x) )f f(|

5、(|x x|)|)，反之也成立，反之也成立 4 4周期函數(shù)周期函數(shù) 若若f f( (x x) )對(duì)于定義域中任意對(duì)于定義域中任意x x均有均有f f( (x xT T) )f f( (x x) )( (T T為不等于為不等于0 0的常數(shù)的常數(shù)) )，則，則f f( (x x) )為周期函數(shù)為周期函數(shù) 1 1對(duì)任意實(shí)數(shù)對(duì)任意實(shí)數(shù)x x，下列函數(shù)中的奇函數(shù)是，下列函數(shù)中的奇函數(shù)是( () ) A Ay y2 2x x3 3B By y3 3x x2 2 C Cy yln5ln5x x D Dy y| |x x|cos|cosx x 答案答案C C 2 2若函數(shù)若函數(shù)y yf f( (x x)()(

6、x xR)R)是奇函數(shù)，則下列坐標(biāo)表示的點(diǎn)一定在函數(shù)是奇函數(shù)，則下列坐標(biāo)表示的點(diǎn)一定在函數(shù)y yf f( (x x) )圖象上的是圖象上的是 ( () ) A A( (a a，f f( (a a) B) B( (a a，f f( (a a) C C( (a a，f f( (a a) D) D( (a a，f f( (a a) 答案答案B B 4 4(2010(2010廣東卷廣東卷) )若函數(shù)若函數(shù)f f( (x x) )3 3x x3 3x x與與g g( (x x) )3 3x x3 3x x的定義域均為的定義域均為R R，則則( () ) A Af f( (x x) )與與g g( (x

7、x) )均為偶函數(shù)均為偶函數(shù) B Bf f( (x x) )為偶函數(shù)，為偶函數(shù)，g g( (x x) )為奇函數(shù)為奇函數(shù) C Cf f( (x x) )與與g g( (x x) )均為奇函數(shù)均為奇函數(shù) D Df f( (x x) )為奇函數(shù)，為奇函數(shù)，g g( (x x) )為偶函數(shù)為偶函數(shù) 答案答案B B 解析由解析由f f( (x x) )3 3x x3 3x xf f( (x x) )可知可知f f( (x x) )為偶函數(shù)，由為偶函數(shù)，由g g( (x x) )3 3x x3 3x x(3(3x x3 3x x) )g g( (x x) )可知可知g g( (x x) )為奇函數(shù)為奇函

8、數(shù) 5 5(2010(2010安徽卷安徽卷) )若若f f( (x x) )是是R R上周期為上周期為5 5的奇函數(shù)，且滿足的奇函數(shù)，且滿足f f(1)(1)1 1，f f(2)(2)2 2，則，則f f(3)(3)f f(4)(4)( () ) 答案答案A A 解析由于函數(shù)解析由于函數(shù)f f( (x x) )的周期為的周期為5 5，所以，所以f f(3)(3)f f(4)(4)f f( (2)2)f f( (1)1)，又，又f f( (x x) )為為R R上的奇函數(shù)，上的奇函數(shù)，f f( (2)2)f f( (1)1)f f(2)(2)f f(1)(1)2 21 11.1.題型一題型一 判

9、斷函數(shù)的奇偶性判斷函數(shù)的奇偶性授人以漁授人以漁 探究探究1 1判斷函數(shù)的奇偶性，一般有以下幾種方法：判斷函數(shù)的奇偶性，一般有以下幾種方法： (1)(1)定義法：若函數(shù)的定義域不是關(guān)于原點(diǎn)對(duì)稱的區(qū)間，則立即可判斷該定義法：若函數(shù)的定義域不是關(guān)于原點(diǎn)對(duì)稱的區(qū)間，則立即可判斷該函數(shù)既不是奇函數(shù)也不是偶函數(shù)；若函數(shù)的定義域是關(guān)于原點(diǎn)對(duì)稱的區(qū)函數(shù)既不是奇函數(shù)也不是偶函數(shù)；若函數(shù)的定義域是關(guān)于原點(diǎn)對(duì)稱的區(qū)間，再判斷間，再判斷f f( (x x) )是否等于是否等于f f( (x x) ) (2)(2)圖象法：奇圖象法：奇( (偶偶) )函數(shù)的充要條件是它的圖象關(guān)于原點(diǎn)函數(shù)的充要條件是它的圖象關(guān)于原點(diǎn)( (

10、或或y y軸軸) )對(duì)稱對(duì)稱 (3)(3)性質(zhì)法：偶函數(shù)的和、差、積、商性質(zhì)法：偶函數(shù)的和、差、積、商( (分母不為零分母不為零) )仍為偶函數(shù)；奇函數(shù)仍為偶函數(shù)；奇函數(shù)的和、差仍為奇函數(shù)；奇的和、差仍為奇函數(shù)；奇( (偶偶) )數(shù)個(gè)奇函數(shù)的積、商數(shù)個(gè)奇函數(shù)的積、商( (分母不為零分母不為零) )為奇為奇( (偶偶) )函數(shù)；一個(gè)奇函數(shù)與一個(gè)偶函數(shù)的積為奇函數(shù)函數(shù)；一個(gè)奇函數(shù)與一個(gè)偶函數(shù)的積為奇函數(shù)( (注：利用上述結(jié)論時(shí)要注：利用上述結(jié)論時(shí)要注意各函數(shù)的定義域注意各函數(shù)的定義域) ) 思考題思考題1 1判斷下列函數(shù)的奇偶性判斷下列函數(shù)的奇偶性p(2)g(x)(2)g(x)的定義域?yàn)榈亩x域?yàn)?/p>

11、R Rp當(dāng)當(dāng)a a0 0時(shí)，時(shí)，g(x)g(x)x x2 2|x|x|pg(g(x)x)( (x)x)2 2| |x|x|x x2 2|x|x|g(x)g(x)p此時(shí)此時(shí)g(x)g(x)為偶函數(shù)為偶函數(shù)p當(dāng)當(dāng)a0a0時(shí)，時(shí)，g(a)g(a)a a2 2，g(g(a)a)a a2 22|a|2|a|p顯然顯然g(a)g(g(a)g(a)a)，g(a)g(a)g(g(a)a)p此時(shí)此時(shí)g(x)g(x)既不是奇函數(shù)，也不是偶函數(shù)既不是奇函數(shù)，也不是偶函數(shù)題型二題型二 奇偶性的應(yīng)用奇偶性的應(yīng)用 (3)(3)f f( (x x1)1)為偶函數(shù)為偶函數(shù) 函數(shù)函數(shù)g g( (x x) )f f( (x x1

12、)1)的圖象關(guān)于直線的圖象關(guān)于直線x x0 0對(duì)稱對(duì)稱 又函數(shù)又函數(shù)f f( (x x) )的圖象是由函數(shù)的圖象是由函數(shù)g g( (x x) )f f( (x x1)1)的圖象向右平移一個(gè)單位而得的圖象向右平移一個(gè)單位而得 函數(shù)函數(shù)f f( (x x) )的圖象關(guān)于直線的圖象關(guān)于直線x x1 1對(duì)稱對(duì)稱 探究探究2 2奇偶函數(shù)的性質(zhì)主要體現(xiàn)在奇偶函數(shù)的性質(zhì)主要體現(xiàn)在 若若f f( (x x) )為奇函數(shù)，則為奇函數(shù)，則f f( (x x) )f f( (x x) ) 若若f f( (x x) )為偶函數(shù)，則為偶函數(shù)，則f f( (x x) )f f( (x x) ) 奇偶函數(shù)的對(duì)稱性奇偶函數(shù)的

13、對(duì)稱性 奇偶函數(shù)在關(guān)于原點(diǎn)對(duì)稱的區(qū)間上的單調(diào)性奇偶函數(shù)在關(guān)于原點(diǎn)對(duì)稱的區(qū)間上的單調(diào)性 思考題思考題2(1)若函數(shù)若函數(shù)f(x)是是R上的偶函數(shù)，且在上的偶函數(shù)，且在0，)上是減函數(shù)，滿上是減函數(shù)，滿足足f()f(a)的實(shí)數(shù)的實(shí)數(shù)a的取值范圍是的取值范圍是_ 【解析解析】若若a0，f(x)在在0，)上是減函數(shù)，且上是減函數(shù)，且f()f(a)，得，得a. 若若a0，f()f(), 則由則由f(x)在在0，)上是減函數(shù)，得知上是減函數(shù)，得知f(x)在在(，0上是增函數(shù)上是增函數(shù) 由于由于f() 即即a0. 由上述兩種情況知由上述兩種情況知a(，) 【答案答案】(，) (2)函數(shù)函數(shù)yf(x2)為奇函

14、數(shù)，則函數(shù)為奇函數(shù)，則函數(shù)yf(x)的圖象的對(duì)稱中心為的圖象的對(duì)稱中心為_(kāi) 【解析解析】f f( (x x2)2)為奇函數(shù)為奇函數(shù) f f( (x x2)2)的圖象的對(duì)稱中心為的圖象的對(duì)稱中心為(0,0)(0,0) 又又f f( (x x) )的圖象可由函數(shù)的圖象可由函數(shù)f f( (x x2)2)的圖象向左平移兩個(gè)單位而得的圖象向左平移兩個(gè)單位而得 f f( (x x) )的圖象的對(duì)稱中心為的圖象的對(duì)稱中心為( (2,0)2,0) 題型三題型三 函數(shù)的周期性函數(shù)的周期性 例例3 3(09(09山東山東) )已知定義在已知定義在R R上的奇函數(shù)上的奇函數(shù)f f( (x x) )滿足滿足f f(

15、(x x4)4)f f( (x x) )，且，且在區(qū)間在區(qū)間0,20,2上是增函數(shù)若方程上是增函數(shù)若方程f f( (x x) )m m( (m m0)0)在區(qū)間在區(qū)間 8,88,8上有四上有四個(gè)不同的根個(gè)不同的根x x1 1，x x2 2，x x3 3，x x4 4，則，則x x1 1x x2 2x x3 3x x4 4_ 【解析解析】由由f f( (x x4)4)f f( (x x) )f f(4(4x x) )f f( (x x) )，故函數(shù)圖象關(guān)于直線，故函數(shù)圖象關(guān)于直線x x2 2對(duì)稱，又函數(shù)對(duì)稱，又函數(shù)f f( (x x) )在在0,20,2上是增函數(shù)，且為奇函數(shù)，故上是增函數(shù)，且為

16、奇函數(shù)，故f f(0)(0)0 0，故，故函數(shù)函數(shù)f f( (x x) )在在(0,2)(0,2)上大于上大于0 0，根據(jù)對(duì)稱性知函數(shù)，根據(jù)對(duì)稱性知函數(shù)f f( (x x) )在在2,42,4上大于上大于0 0，同，同理推知函數(shù)理推知函數(shù)f f( (x x) )在在4,84,8上小于上小于0 0，故在區(qū)間，故在區(qū)間0,80,8上方程上方程f f( (x x) )m m( (m m0)0)的兩根關(guān)于直線的兩根關(guān)于直線x x2 2對(duì)稱，故此兩根之和等于對(duì)稱，故此兩根之和等于4 4，根據(jù)，根據(jù)f f( (x x4)4)f f( (x x) )f f( (x x8)8)f f( (x x4)4)f f

17、( (x x) )，函數(shù)，函數(shù)f f( (x x) )以以8 8為周期，故在區(qū)間為周期，故在區(qū)間( (8,0)8,0)上方程上方程f f( (x x) )m m( (m m0)0)的兩根關(guān)于直線的兩根關(guān)于直線x x6 6對(duì)稱，此兩根之和等于對(duì)稱，此兩根之和等于1212，綜上四個(gè)根之和等于，綜上四個(gè)根之和等于8.8. 【答案答案】8 8 探究探究3 3證明函數(shù)是周期函數(shù)應(yīng)緊扣周期函數(shù)的定義。證明函數(shù)是周期函數(shù)應(yīng)緊扣周期函數(shù)的定義。 若函數(shù)若函數(shù)f f( (x x) )對(duì)任意對(duì)任意x x滿足滿足f f( (x xa a) )f f( (x xb b) )，則，則f f( (x x) )為周期函數(shù)，

18、若函為周期函數(shù)，若函數(shù)數(shù)f f( (x x) )對(duì)任意對(duì)任意x x滿足滿足f f( (x xa a) )f f( (b bx x) )，則函數(shù)圖象為軸對(duì)稱圖形，則函數(shù)圖象為軸對(duì)稱圖形 思考題思考題3 3f f( (x x) )的定義域?yàn)榈亩x域?yàn)镽 R的奇函數(shù)，且圖象關(guān)于直線的奇函數(shù)，且圖象關(guān)于直線x x1 1對(duì)稱，試對(duì)稱，試判斷判斷f f( (x x) )的周期性的周期性 【答案答案】T T4 4【答案答案】T T2 2 求求x x5,75,7時(shí)，時(shí)，f f( (x x) )的解析式的解析式 【解析解析】解析一解析一f f( (1 1x x) )f f(1(1x x) ) f f( (x x

19、) )f f(2(2x x) )，f f( (x x) )為周期函數(shù)，為周期函數(shù)，T T2 2 f f( (x x) )為偶函數(shù)為偶函數(shù) x x1,01,0時(shí)，時(shí)，x x0,10,1f f( (x x) )f f( (x x) )x x1 1 x x5,65,6時(shí)，時(shí)，x x661,01,0f f( (x x) )f f( (x x6)6)( (x x6)6)1 1x x5 5x x6,76,7時(shí)，時(shí)，x x60,160,1f f( (x x) )f f( (x x6)6)( (x x6)6)1 1x x7 7 思考題思考題4 4已知定義在已知定義在R R上的函數(shù)上的函數(shù)f f( (x x) )是偶函數(shù)，對(duì)是偶函數(shù)，對(duì)x xRR，f f(2(2x x) )f f(2(2x x) )，當(dāng)，當(dāng)f f( (1)1)2 2時(shí)，時(shí)，f f(2011)(2011)的值為的值為_(kāi) 【解析解析】因?yàn)槎x在因?yàn)槎x在R R上的函數(shù)上的函數(shù)f f( (x x) )是偶函數(shù)，所以是偶函數(shù)，所以f f(2(2x x) )f f(2(2x x) )f f( (x x2)2)，故函數(shù)，故函數(shù)f f( (x x) )是以是以4 4為周期的函數(shù)，所以為周期的函數(shù)，所以f f(2011)(2011)f f( (1 14 4503)503)f f( (1)1)2.2. 【答案答案】2 2本課總結(jié)本課總結(jié)