必修1函數(shù)深化拓展講義第7的奇偶性

7講1yf(xxDxD,都有xDf(xf(xf(x為偶函數(shù).2yf(xxDxD,都有xDf(xf(xf(x為奇函數(shù)（1）f(x)c(c0)（2）f(x0f(xff(x是奇函數(shù)函數(shù)圖像關(guān)于原點對稱；函數(shù)f(x)是偶函數(shù)函數(shù)圖像關(guān)于y軸對稱.

ff

1等例2．若奇函數(shù)f(x)在[3,7]上的最小值是5，那么f(x)在[7,3]上 A.最小值是 B.最小值是 C.最大值是 D.最大值是例4．設(shè)f(x)是(－∞,+∞)上的奇函數(shù)，f(x2)f(x),當0x1時，f(x)x,則f(7.5)等于 A. B.－0. C. D.5f(xR上為奇函數(shù)，在[1,0)f(x2)f(x), f(),f(),f(1的大小關(guān)系 1f(),f(8.5),f3例6．若f(x)為奇函數(shù)，且在0,內(nèi)是增函數(shù)，又f(3)0，則xf(x)0的解集 如右圖，則不等式f(x)<0的解

f(x8yfxxyRfxyfxfyf(xf9.(1)f(xRx0fxx2xf(x的解析式.并畫出函數(shù)圖(2)f(xRx0f(xx24x3f(x的解析式10f(x是定義在區(qū)間[2,2]x[0,2]f(xf(1mf(m成立，求實數(shù)ｍ的取值范圍例11f(x)定義域R為對任意的x1x2R都有f(x1x2f(x1f(x2x0思考:已知函數(shù)f(x)R為對任意的x1x2R都有f(x1x20f(x1,f(0)x0f(x的取值范圍

f(x1f(x2x0f(xRx0f(xx22x，

f(x)在R上的表達式是 A.f(x)x2C.f(x)|x|(x

B.f(x)x(|x|D.f(x)x(|x|函數(shù)f(x)是定義在[6,6]上的偶函數(shù)，且在[6,0]上是減函數(shù)，則 f(3)f(4)C.f(2)f(5)

f(3)f(2)D.f(4)f(1)設(shè)f(x)是定義在R上的任意一個增函數(shù)，令F(x)f(x)f(x)，則F(x)必是 B.增函數(shù)且是偶函C.減函數(shù)且是奇函 D.減函數(shù)且是偶函設(shè)f(x),g(x)都是奇函數(shù)，F(xiàn)(x)f(x)g(x)3，若F(5)9,則F(5) A. B. D)）A.最小值B.最大值C.最小值D.最小值f(x)

1|x31

是 A.奇函 B.偶函 C.既奇又偶函 D.沒有確定的奇偶若f(x)ax2bxc(a0)是偶函數(shù)，則g(x)ax3bx2cx是 A.奇函 B.偶函 C.既奇又偶函 D.沒有確定的奇偶軸對稱（4）f(x0，其中真命題個數(shù)是（B. C. D. B.增函數(shù)且最大值為C.減函數(shù)且最小值為 D.減函數(shù)且最大值為（1）f(x)=1是偶函數(shù) （2）g(x)x3,x(1,1]是奇函數(shù)f(xg(xH(xf(xg(x函數(shù)yf(x)的圖象關(guān)于y軸對稱。 B. C. D.已知f(x)x4ax3bx8，且f(2)10，則f(2) x1(x（1）f(x) (xx1(x（2）f(x0a、bRf(abf(af已知f(x)是定義在R上的奇函數(shù),當x0時,f(x) 1,則f(x) f(x)f(xyf(xy)2f(x(

數(shù)yf(x為奇函數(shù)，在（0，+∞）f(x0F(x

f

1f(3)f(π)2x[7,3]，則x[3,7]f(x5f(xf(xf(x，則f(x5f(x5，－5f(x在[7,3]3f f

x1；[2]xfxf1f1

1x f(7.5)f(5.52)f(5.5)f(3.52)f(3.5)f(1.5f(1.5)f(0.52)f(0.5)f(0.5)5f(xR

1f(

1f(

，f(1)ff(

f( ))f(x在(1,0)1>2>－1f())

f(

f(1)，∴1f(2)ff( f(x

xf(x)0xf(x)或x

或xf(x) x

或x

f(x)f f(x)f x 不等式的解集是(2,0)Uyy025x例xy0f(0)f(00f(0)f(0)2f(0f(0)0yxf(0)fxx)fxfx0f(x為奇函數(shù)，且不是偶函數(shù)9x0，則x0f(xx)2xx2xf(xf(xf(xf(xx2xf(x所以f(x

x2xx0.圖像如前頁圖，值域為[0,f(xx0yx2x軸對稱的性質(zhì)可知，f(x)x0時的圖像是二次函數(shù) x1，頂點坐標為11y f(x)

x2xx0.值域為[0,因為是奇函數(shù)，且0Df(0)0x0x0f(x)x)24(x3x24xf(xf(xx24x3f(xx24x

f(x)

xxx10、此題關(guān)鍵是如何去掉函數(shù)符號“f”，與分類討論思想的應(yīng)用f(x在[0,2]上是減函數(shù)，所以在[2,0](1)當21m01m ，得m,故此情況不存在

2m01m

2m1m當0m

0m

，得0mf(x在[0,2]f(1mf(m可轉(zhuǎn)化為1mmm2∴0m2(3)當01m21m1，得1m2m

2mf(1mf(m1f(1mf(mf(m1)f∵f(x)在[2,0]上是增函數(shù)，∴m1 ∴1m

21m

1m當0m 0m2，得1m ∴0m11，∵f(1m)f(m1)f(1mf(mf(m1f綜上所述，滿足條件的實數(shù)ｍ的取值范圍為1m2f(1mf(m轉(zhuǎn)化到同一個單調(diào)區(qū)間內(nèi)的兩個函數(shù)值是解題的關(guān)鍵.此題若利用偶函數(shù)的性質(zhì)“f(x)f(|x|)”可使問題變得簡單。11x1x20,f(0)f(0)f(0)f(0)0x1xx2xf(0)f(xx))f(xf(x)0f(x為奇函數(shù)。x1x2R,x1x2x2x10，所以f(x2)f(x1)f(x2)f(x1)f(x2x1)0f(x2f(x1f(x在R上為減函數(shù)f(1)2f(3f(12f(1f(23f(16,f(3)6f(x在3,3上為減函數(shù),x3時f(x)maxf(3)6,x3時,f(x)minf(3)6 解 由圖可知選 6

Gx奇函數(shù)QF5G53 G51f1 解：定義域x|1x1且x

fxx f210得8a2b13（1）偶函數(shù) x

f2168a2b8

x x

Qfx為奇函 fxfx xfx

x x fxf即m21x2m1xn2m21x2n1xnn2nmn2n

mn

xy0f0f02fQf0