《高中數(shù)學 第三章 導數(shù)應用 3.1 函數(shù)的單調(diào)性與極值 3.1.1.1 導數(shù)與函數(shù)的單調(diào)性課件 北師大版選修22》由會員分享，可在線閱讀，更多相關《高中數(shù)學 第三章 導數(shù)應用 3.1 函數(shù)的單調(diào)性與極值 3.1.1.1 導數(shù)與函數(shù)的單調(diào)性課件 北師大版選修22（22頁珍藏版）》請在裝配圖網(wǎng)上搜索。

1、第三章導數(shù)應用1 1函數(shù)的單調(diào)性與極值1.1導數(shù)與函數(shù)的單調(diào)性第1 1課時導數(shù)與函數(shù)的單調(diào)性MUBIAODAOHANG目標導航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLI

2、AN隨堂演練ZHISHISHULI知識梳理1.結合實例,借助幾何直觀探索并了解函數(shù)的單調(diào)性與導數(shù)的關系.2.能利用導數(shù)研究函數(shù)的單調(diào)性,會求一些多項式函數(shù)的單調(diào)區(qū)間.MUBIAODAOHANG目標導航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHU

3、LI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理導函數(shù)符號與函數(shù)的單調(diào)性之間的關系如果在某個區(qū)間內(nèi),函數(shù)y=f(x)的導數(shù)f(x)0,那么在這個區(qū)間上,函數(shù)y=f(x)是增加的.如果在某個區(qū)間內(nèi),函數(shù)y=f(x)的導數(shù)f(x)0和f(x)0,(x-2)20.由f(x)0,得x3,所以函數(shù)f(x)的遞增區(qū)間為(3,+);由f(x)0,得x3,又定義域為(-,2)(2,+),所以函數(shù)f(x)的遞減區(qū)間為(-,2),(2,3).MUBIAODAOHANG目標導航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZH

4、ISHI SHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理題型一題型二題型三分析以前我們用定義法證明函數(shù)的單調(diào)性,即任取定義域內(nèi)的x1,x2,若x1x2,f(x1)0,且f(a)0,則在區(qū)間(a,b)內(nèi)有f(x)(

5、)A.大于0 B.小于0C.等于0 D.不能確定取值符號解析:若f(x)0在區(qū)間(a,b)內(nèi)成立,即在(a,b)內(nèi),f(x)是增加的.對任意x(a,b),若xa,則f(x)f(a)0.故f(x)0.答案:AMUBIAODAOHANG目標導航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANG

6、YANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理1 2 3 4 52函數(shù)f(x)=(x-3)ex的遞增區(qū)間為()A.(-,2) B.(0,3)C.(1,4)D.(2,+)解析:f(x)=(x-3)ex+(x-3)(ex)=(x-2)ex,令f(x)=(x-2)ex0,解得x2,即函數(shù)f(x)的遞增區(qū)間為(2,+).故選D.答案:DMUBIAODAOHANG目標導航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標導航DIANLITOU

7、XI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理1 2 3 4 53函數(shù)f(x)=-x3+x在區(qū)間(1,+)上是()A.減少的B.增加的C.常數(shù)函數(shù)D.不單調(diào)的解析:當x(1,+)時,f(x)=-3x2+10,即2x(x-2)0,解得x2,所以函數(shù)的遞增區(qū)間

8、為(-,0)和(2,+).令y0,即2x(x-2)0,解得0 x2,所以函數(shù)的遞減區(qū)間為(0,2).(2)函數(shù)的定義域為(-,0)(0,+), 令y0,得x0,得x1.故函數(shù)的遞減區(qū)間為(-,0)和(0,1),遞增區(qū)間為(1,+).MUBIAODAOHANG目標導航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理1 2 3 4 5