蘇教版高中數(shù)學(xué)教案-函數(shù)教師版第5課時(shí)

第5函數(shù)的單調(diào)性（二111f(x

xR設(shè)x1x2f(x)f(x)

x1111112x211 (x111

1x21x2112

12x12

11 11x1 12x (xx) ∵x1x2，∴x1x20∵x|x

，∴x

011111111212

01111

x

0f(xf(x0f(xf(x12 121f(x) x12f(x)ax1(a1在(2,x 解f(xaxxax2a1x11x設(shè)2x1x2(x22)(x12)0,x2x1∴f(x2)f12a1x2

(x1x2(x22)(x1 (x1x2) 0(x22)(x12)1時(shí)f1時(shí)f(xf(x)，此時(shí)函數(shù)f(x)ax(a121 時(shí)f(xf(x)，此時(shí)函數(shù)f(x)xax(a212 x2a【隨堂

在(2,)在(2,)f(x是定義域上單調(diào)遞減函數(shù)，且過點(diǎn)(32和(12)，則|f(x|2x的取值范圍是（(3,(C)

(D)(, 已知函數(shù)f(x)是區(qū)間(0，＋∞)上的減函數(shù)，那么小于等

－a＋1)f()4f(x)x

2x在(74

xx7 ∴f(x2)f(x2

2x2)(x1

(x2x1)(2x2 2x222x22

(x2x1 2x222x22x112x22 0例3若函數(shù)f(x)4x2mx5m在[2,)上是增函數(shù)，在(,2]上是減函數(shù)，則實(shí)數(shù)m的值為 x2即m28m例4若函數(shù)f(x)4x2mx5m在[2,)上是增函數(shù)，則實(shí)數(shù)m的取值范圍 解：由題意可以知道m(xù)2即m8例5若函數(shù)f(x)4x2mx5m的單調(diào)遞增區(qū)間為[2,)，則實(shí)數(shù)m的值 x2即m28m，f(x)axg(x)b在(0上都是減函數(shù)，則h(xax2bxcx(,0) （ (A)是增函 (B)是減函 (D)h(x)的單調(diào)性不能確f(x)x22(a1)x2在區(qū)間(4上是減函數(shù)，則實(shí)數(shù)aa f(x)(2a1)xb在Ｒ上是減函數(shù)，則a

a2yf(x在Rf(m2f(m，則實(shí)數(shù)m,10,f(x)4x2mx1在(3上遞減，在[2,上遞增，則實(shí)數(shù)m圍[24, 設(shè)fx是定義在R上的偶函數(shù),且在(－∞,0)上是增函數(shù),f2與fa22a（aR）

f2≥fa22ax2x2

的值域 fxax2a1，當(dāng)1x1y的值有正有負(fù)，則實(shí)數(shù)a1,13 3 f(xx3x在上是增函數(shù)．x1x2∴f(x2)f (x3x)(x3 (x3x3)(xx )(x2xxx2)(xx 1 x)[(x1x)23x2 2 40f(x2ff(xx3x在f(xx3ax在0,1上是增函數(shù)，求實(shí)數(shù)a的取值范圍解析：利用單調(diào)性求出實(shí)數(shù)a的取值范圍3f(xxaa在1上是增函數(shù)，求實(shí)數(shù)a 解析：利用單調(diào)性求出實(shí)數(shù)a的取值范圍1,yf(xR，且對(duì)任意的正數(shù)df(xd

f(