高職高考數(shù)學(xué)復(fù)習(xí)§3.4函數(shù)的奇偶性【復(fù)習(xí)目標(biāo)】1.理解和掌握函數(shù)奇偶性的概念.2.掌握奇函數(shù)、偶函數(shù)的圖像特征.3.掌握判斷和證明函數(shù)奇偶性的方法.4.能利用函數(shù)的奇偶性解決簡(jiǎn)單問題.【知識(shí)回顧】1.函數(shù)奇偶性的概念及圖像特征(1)概念①奇函數(shù):如果對(duì)于函數(shù)y=f(x)在定義域內(nèi)的任意一個(gè)x,都有f(-x)=-f(x),那么這個(gè)函數(shù)叫奇函數(shù).②偶函數(shù):如果對(duì)于函數(shù)y=f(x)在定義域內(nèi)的任意一個(gè)x,都有f(-x)=f(x),那么這個(gè)函數(shù)叫偶函數(shù).2.圖像特征奇函數(shù)的圖像關(guān)于坐標(biāo)原點(diǎn)成中心對(duì)稱圖形;反之,一個(gè)函數(shù)的圖像關(guān)于坐標(biāo)原點(diǎn)成中心對(duì)稱圖形,那么這個(gè)函數(shù)是奇函數(shù)偶函數(shù)的圖像關(guān)于y軸成軸對(duì)稱圖形;反之,一個(gè)函數(shù)的圖像關(guān)于y軸成軸對(duì)稱圖形,那么這個(gè)函數(shù)是偶函數(shù)2.判斷函數(shù)奇偶性的方法(1)定義法①判斷定義域是否關(guān)于原點(diǎn)對(duì)稱,若定義域關(guān)于原點(diǎn)不對(duì)稱,則函數(shù)為非奇非偶函數(shù).②若定義域關(guān)于原點(diǎn)對(duì)稱,則判斷f(-x)與f(x)的關(guān)系.若f(-x)=f(x),則函數(shù)為偶函數(shù);若f(-x)=-f(x),則函數(shù)為奇函數(shù).(2)圖像判斷法若函數(shù)的圖像關(guān)于坐標(biāo)原點(diǎn)成中心對(duì)稱圖形,那么該函數(shù)是奇函數(shù);若函數(shù)的圖像關(guān)于y軸成軸對(duì)稱圖形,那么該函數(shù)是偶函數(shù).3.奇、偶函數(shù)的判斷法則

(1)奇函數(shù)±奇函數(shù)=奇函數(shù); (2)偶函數(shù)±偶函數(shù)=偶函數(shù);(3)奇函數(shù)×偶函數(shù)=奇函數(shù); (4)奇函數(shù)×奇函數(shù)=偶函數(shù);(5)偶函數(shù)×偶函數(shù)=偶函數(shù); (6)奇函數(shù)±偶函數(shù)=非奇非偶函數(shù).【例題精解】【例1】(1)已知函數(shù)f(x)是定義在R上的偶函數(shù),且f(2)=5,則f(-2)=

.

(2)已知函數(shù)f(x)是定義在R上的奇函數(shù),且f(1)=-8,則f(-1)=

.

【解】(1)由函數(shù)f(x)是偶函數(shù),可得f(-2)=f(2),即f(-2)=5.故答案為5.(2)由函數(shù)f(x)是奇函數(shù),可得f(-1)=-f(1),即f(-2)=8.故答案為8.【點(diǎn)評(píng)】由函數(shù)的奇偶性定義可得,如果函數(shù)y=f(x)為奇函數(shù),則在定義域內(nèi)的任意一個(gè)x,都有f(-x)=-f(x);如果函數(shù)y=f(x)為偶函數(shù),則在定義域內(nèi)的任意一個(gè)x,都有f(-x)=f(x).【對(duì)點(diǎn)練習(xí)1】(1)已知函數(shù)f(x)是偶函數(shù),且f(3)=-1,則f(-3)=

.

(2)已知函數(shù)f(x)是奇函數(shù),且f(-1)=3,則f(1)=

.

【答案】(1)-1

(2)-3【例2】已知函數(shù)f(x)為奇函數(shù),當(dāng)x<0時(shí),f(x)=x(x-1),則f(2)=(

) A.-6 B.6 C.-2 D.2【解】∵x<0,f(x)=x(x-1),∴f(-2)=-2×(-2-1)=6.又∵函數(shù)f(x)是定義在R上的奇函數(shù),∴f(2)=-f(-2),即f(2)=-6.故選A.【點(diǎn)評(píng)】此題設(shè)置了一個(gè)小陷阱,給出了當(dāng)x<0時(shí)函數(shù)的表達(dá)式,卻要求x=2>0時(shí)的函數(shù)值.我們應(yīng)利用函數(shù)的奇偶性,由f(x)是定義在R上的奇函數(shù)可得f(2)=-f(-2),先求出f(-2)的值,再根據(jù)f(2)=-f(-2)得出答案.

【解】A為偶函數(shù),B為非奇非偶函數(shù),C為奇函數(shù),D為非奇非偶函數(shù).故選C.

【對(duì)點(diǎn)練習(xí)3】下列函數(shù)是偶函數(shù)的是 (

)

A.f(x)=x3

B.f(x)=x2

C.f(x)=x3+1 D.f(x)=x-1【答案】B【例4】若函數(shù)f(x)=x3+(m-2)x2+x為奇函數(shù),則實(shí)數(shù)m的值為(

) A.-1 B.1 C.0 D.2【解】由題可得m-2=0,解得m=2.故選D.

【對(duì)點(diǎn)練習(xí)4】若二次函數(shù)f(x)=2x2+(m-3)x+5為偶函數(shù),則實(shí)數(shù)m=

.

【答案】3

【解析】

由m-3=0,解得m=3.【例5】已知函數(shù)f(x)是奇函數(shù),且對(duì)于任意實(shí)數(shù)x,都有f(x+4)=f(x),若f(1)=1,則f(7)= (

) A.-1 B.1 C.2 D.0【解】∵函數(shù)f(x)是定義在R上的奇函數(shù),且f(1)=1,∴f(-1)=-1.又∵f(x+4)=f(x),∴f(-1+4)=f(-1),即f(3)=f(-1).同理f(3+4)=f(3),即f(7)=f(3).∴f(7)=f(3)=f(-1)=-1.故選A.【點(diǎn)評(píng)】本題綜合考查了函數(shù)的奇偶性與函數(shù)的周期性.【對(duì)點(diǎn)練習(xí)5】若函數(shù)f(x)為奇函數(shù),且對(duì)于任意實(shí)數(shù)x,都有f(x+4)=f(x),若f(-1)=3,則f(-3)= (

) A.-3 B.3 C.4 D.-4【答案】A

【解析】∵f(x+4)=f(x),∴f(-1+4)=f(-1),即f(3)=f(-1)=3.又∵函數(shù)f(x)是定義在R上的奇函數(shù),且f(3)=3,∴f(-3)=-f(3)=-3.故選A.【仿真訓(xùn)練】一、選擇題1.下列函數(shù)中是奇函數(shù)的是 (

)

A.f(x)=x2-2x

B.f(x)=2x+1

C.f(x)=x3+x

D.f(x)=x2+1【答案】C2.下列函數(shù)中是偶函數(shù)的是 (

)

A.f(x)=2x2-3 B.f(x)=x

C.f(x)=x3+2 D.f(x)=x2,x∈[0,1]【答案】A3.若函數(shù)f(x)為偶函數(shù),且f(-1)=2,則f(1)= (

) A.-1 B.-2 C.1 D.2【答案】D4.若函數(shù)f(x)=ax3+bx,且f(2)=5,則f(-2)= (

) A.5 B.-5 C.2 D.-2【答案】B5.已知函數(shù)f(x)是奇函數(shù),且f(3)=2,則[f(-3)]3= (

) A.-8 B.-1 C.1 D.8【答案】A6.已知函數(shù)f(x)為奇函數(shù),當(dāng)x>0時(shí),f(x)=2x2-x,則f(-1)= (

) A.-3 B.-1 C.1 D.3

7.已知函數(shù)f(x)為偶函數(shù),當(dāng)x<0時(shí),f(x)=x(x-1),則f(2)= (

) A.-6 B.6 C.-2 D.2

8.若函數(shù)f(x)=x3+(m-3)x2+x為奇函數(shù),則實(shí)數(shù)m的值為 (

) A.-1 B.1 C.0 D.3

9.若函數(shù)f(x)=x2+(1-a)x-a為偶函數(shù),則實(shí)數(shù)a的值為 (

) A.1 B.-1 C.2 D.-2

10.若奇函數(shù)f(x)對(duì)于任意實(shí)數(shù)x,都有f(x+4)=f(x),若f(3)=3,則f(1)=(

) A.-4 B.4 C.-3 D.3

二、填空題11.已知函數(shù)f(x)是奇函數(shù),且f(5)=16,則f(-5)=

.【答案】-1612.已知函數(shù)f(x)是偶函數(shù),且f(1)=3,則f(-1)=

.

13.已知函數(shù)f(x)是定義在R上的奇函數(shù),則f(0)=

.

【答案】014.若函數(shù)f(x)=x3+(m2-1)x2+x為奇函數(shù),則m=

.

【答案】 ±115.已知函數(shù)f(x)是奇函數(shù),當(dāng)x<0時(shí),f(x)=x2-6,則f(4)=

.

【答案】-10

17.若函數(shù)f(x)是定義在R上的奇函數(shù),當(dāng)x>0時(shí),f(x)=x2+1,求f(-2)+f(0)的值.【解】∵當(dāng)x>0時(shí),f(x)=x2+1,

∴f(2)=22+1=5.又∵函數(shù)f(x)是定義在R上的奇函數(shù),

∴f(-2)=-f(2)=-5,f(0)=0.則f(-2)+f(0)=-5+