第三章函數(shù)-3.3的奇偶性

3.3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)的圖象特征.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(wèn)題.【知識(shí)回顧】1.函數(shù)奇偶性的定義(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.圖象特征(1)奇函數(shù)的圖象關(guān)于坐標(biāo)原點(diǎn)成中心對(duì)稱(chēng)圖形;反之,如一個(gè)函數(shù)的圖象關(guān)于坐標(biāo)原點(diǎn)成中心對(duì)稱(chēng)圖形,那么函數(shù)是奇函數(shù).(2)偶函數(shù)的圖象關(guān)于y軸成軸對(duì)稱(chēng)圖形;反之,一個(gè)函數(shù)的圖象關(guān)于y軸成軸對(duì)稱(chēng)圖形,那么函數(shù)是偶函數(shù).3.判斷奇偶性的步驟(1)寫(xiě)出定義域.(明確奇函數(shù)、偶函數(shù)定義域關(guān)于對(duì)稱(chēng))(2)求f(-x).(3)對(duì)f(-x)與f(x)進(jìn)行比較.【例題精解】【解】(1)函數(shù)f(x)=x3-2x的定義域?yàn)镽,當(dāng)x∈R時(shí),-x∈R∵f(-x)=(-x)3-2(-x)=-x3+2x=-f(x)

∴f(x)=x3-2x為奇函數(shù).(2)函數(shù)f(x)=-3x6-x2的定義域?yàn)镽∵f(-x)=-3(-x)6-(-x)2=-3x6-x2=f(x)

∴f(x)=-3x6-x2為偶函數(shù).(3)f(x)=x2+2x-5的定義域?yàn)镽,當(dāng)x∈R時(shí),-x∈R∵f(-x)=(-x)2+2(-x)-5=x2-2x-5可以看出f(-x)≠f(x)且f(-x)≠-f(x)∴f(x)=x2+2x-5為非奇非偶函數(shù).【點(diǎn)評(píng)】判定函數(shù)奇偶性的步驟:(1)判定函數(shù)的定義域A.(2)判定A是否關(guān)于原點(diǎn)對(duì)稱(chēng).A是否關(guān)于原點(diǎn)對(duì)稱(chēng),是判斷一個(gè)函數(shù)奇偶性的必要條件,若函數(shù)定義域關(guān)于原點(diǎn)不對(duì)稱(chēng),則函數(shù)為非奇非偶函數(shù).①只要有一對(duì)相反數(shù)不同時(shí)屬于定義域A,則A關(guān)于原點(diǎn)不對(duì)稱(chēng).如(4)②若A關(guān)于原點(diǎn)不對(duì)稱(chēng),則函數(shù)一定為非奇非偶函數(shù).如(4)(3)觀察是否有f(-x)=-f(x)或f(-x)=f(x)的成立.(4)若f(-x)=-f(x),則函數(shù)為奇函數(shù);若f(-x)=f(x),則函數(shù)為偶函數(shù).【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).答案為C.【例3】(1)函數(shù)y=-ax2+bx+5(a≠0)為偶函數(shù)充要條件是

.

(2)已知f(x)=(m2+m-6)x2+(m-2)x+(n+7)為奇函數(shù),則m=

,n=

.

【例4】函數(shù)f(x)是奇函數(shù),且在x>0上是增函數(shù)且有最大值6;函數(shù)g(x)是偶函數(shù),且在x>0上是減函數(shù)且有最大值-5.那么當(dāng)x<0時(shí),它們的增減性和最值情況是 (

)A.f(x)是減函數(shù)最小值為6,g(x)是增函數(shù)最小值為-5B.f(x)是增函數(shù)最大值為6,g(x)是增函數(shù)最小值為5C.f(x)是減函數(shù)最大值為-6,g(x)也是減函數(shù)最大值為5D.f(x)是增函數(shù)且最小值為-6,g(x)是增函數(shù)且最大值為-5【答案】D【點(diǎn)評(píng)】(1)若f(x)為奇函數(shù),則f(x)在兩對(duì)稱(chēng)區(qū)間上單調(diào)性相同,若f(x)為偶函數(shù),則f(x)在兩對(duì)稱(chēng)區(qū)間上單調(diào)性相反.(2)若奇函數(shù)在某區(qū)間上有最大值,則在對(duì)稱(chēng)區(qū)間有最小值,且兩值相反;若偶函數(shù)在某區(qū)間上有最大值,則在對(duì)稱(chēng)區(qū)間有最大值,且兩值相等.【例5】已知f(x)=ax5+bx3-x+3,且f(2)=7,求f(-2).【解】令g(x)=ax5+bx3-x,則f(x)=g(x)+3

∴f(2)=g(2)+3∵f(2)=7

f(2)=g(2)+3

∴g(2)=f(2)-3=7-3=4∵g(x)為奇函數(shù)∴g(-2)=-4又∵f(x)=g(x)+3

∴f(-2)=g(-2)+3=-4+3=-1【點(diǎn)評(píng)】一般地,當(dāng)題目已知f(a)的函數(shù)值求f(-a)的函數(shù)值,我們?？紤]利用函數(shù)的奇偶性解題;但這時(shí)如果題中給出的是一個(gè)非奇非偶函數(shù),就需要我們根據(jù)函數(shù)表達(dá)式去構(gòu)造一個(gè)奇函數(shù)或偶函數(shù),如例5中f(x)就是一個(gè)非奇非偶函數(shù),這時(shí)我們根據(jù)函數(shù)表達(dá)式去構(gòu)造一個(gè)奇函數(shù)g(x)=x5+2x3-x,從而順利解題.【同步訓(xùn)練】【答案】A【答案】B【答案】C【答案】A4.已知函數(shù)f(x)在[-7,7]上是奇函數(shù),且f(2)<f(1),則(

) A.f(-1)<f(-2) B.f(-2)>f(1) C.f(-1)>f(-2) D.f(2)<f(-1)【答案】D5.函數(shù)f(x)是奇函數(shù),且在(0,+∞)上是增函數(shù);函數(shù)g(x)是偶函數(shù),且在(0,+∞)上是減函數(shù),那么在(-∞,0)上,它們的增減性是 (

) A.f(x)是減函數(shù),g(x)是增函數(shù) B.f(x)是增函數(shù),g(x)是減函數(shù) C.f(x)是減函數(shù),g(x)是減函數(shù) D.f(x)是增函數(shù),g(x)是增函數(shù)【答案】B6.如果奇函數(shù)f(x)在區(qū)間[3,7]上是增函數(shù),且最小值為5,那么f(x)在區(qū)間[-7,-3]上是 (

) A.增函數(shù)且最小值為-5 B.增函數(shù)且最大值為-5 C.減函數(shù)且最小值為-5 D.減函數(shù)且最大值為-5【答案】A7.設(shè)f(x)為定義在(-∞,+∞)上的偶函數(shù),且f(x)在[0,+∞)上為增函數(shù),則f(-2),f(-π),f(3)的大小順序是 (

) A.f(-π)>f(3)>f(-2) B.f(-π)>f(-2)>f(3) C.f(-π)<f(3)<f(-2) D.f(-π)<f(-2)<f(3)【答案】C【答案】D9.已知函數(shù)f(x)、g(x)都是奇函數(shù),且F(x)=f(x)+g(x)+2在(0,+∞)上有最大值8,則在(-∞,0)上,F(x)有 (

) A.最小值-8 B.最大值-8 C.最小值-6 D.最小值-42或1-3非奇非偶非奇非偶函數(shù)二、填空題10.已知一次函數(shù)f(x)=x+(k2-3k+2)是奇函數(shù),則k的值是

.

11.已知二次函數(shù)f(x)=(a-1)x2+(a2+2a-3)x+5為偶函數(shù),則a的值為

.

12.已知函數(shù)f(x)=x2,x∈[-2,4],則函數(shù)的奇偶性

.

13.已知f(x)、g(x)是定義域相同的兩個(gè)函數(shù),且f(x)是奇函數(shù),g(x)是偶函數(shù),則G(x)=f(x)+g(x)為

.

三、解答題【證明】(1)函數(shù)的定義域?yàn)镽f(-x)=4(-x)3-(-x)=-4x3+x=-f(x)∴f(x)=4x3-x是奇函數(shù).15.證明下列函數(shù)是偶函數(shù):(1)f(x)=-x2+1 (2)f(x)=lgx2【證明】(1)函數(shù)的定義域?yàn)镽f(-x)=-(-x)2+1=-x2+1=f(x)∴f(x)是偶函數(shù).(2)函數(shù)的定義域?yàn)锳={x|x≠0},當(dāng)x∈A時(shí),-x∈A.f(-x)=lg(-x)2=lgx2∴f(x)是偶函數(shù).16.已知函數(shù)f(x)=-ax3+bx-1,且f(3)=-6,求f(-3).【解】設(shè)g(x)=-ax3+bx則f(x)=g(x)-1∵f(3)=g(3)-1=-6

∴g(3)=-5又∵g(x)是奇函數(shù)∴g(-3)=5f(-3)=g(-3)-1=417.已知函數(shù)f(x)=ax5-bx+1,且f(3)=-7,試求f(-3).【解】設(shè)g(x)=ax5-bx,則f(x)=g(x)+1∵f(3)=g(3)+1=-7∴g(3)=-8又∵g(x)是奇函數(shù)∴g(-3)=8f(-3)=g(-3)+1=9.18.已知函數(shù)f(x)為奇函數(shù),g(x)是偶函數(shù),且它們的定義域相同,試證明:(1)f(x)·g(x)是奇函數(shù);(2)f(x)+g