《高中數(shù)學(xué) 第三章 導(dǎo)數(shù)應(yīng)用 3.1 函數(shù)的單調(diào)性與極值 3.1.1.1 導(dǎo)數(shù)與函數(shù)的單調(diào)性課件 北師大版選修22》由會(huì)員分享，可在線閱讀，更多相關(guān)《高中數(shù)學(xué) 第三章 導(dǎo)數(shù)應(yīng)用 3.1 函數(shù)的單調(diào)性與極值 3.1.1.1 導(dǎo)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(qǐng)?jiān)谘b配圖網(wǎng)上搜索。
1、第三章導(dǎo)數(shù)應(yīng)用1 1函數(shù)的單調(diào)性與極值1.1導(dǎo)數(shù)與函數(shù)的單調(diào)性第1 1課時(shí)導(dǎo)數(shù)與函數(shù)的單調(diào)性MUBIAODAOHANG目標(biāo)導(dǎo)航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLI
2、AN隨堂演練ZHISHISHULI知識(shí)梳理1.結(jié)合實(shí)例,借助幾何直觀探索并了解函數(shù)的單調(diào)性與導(dǎo)數(shù)的關(guān)系.2.能利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性,會(huì)求一些多項(xiàng)式函數(shù)的單調(diào)區(qū)間.MUBIAODAOHANG目標(biāo)導(dǎo)航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHU
3、LI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理導(dǎo)函數(shù)符號(hào)與函數(shù)的單調(diào)性之間的關(guān)系如果在某個(gè)區(qū)間內(nèi),函數(shù)y=f(x)的導(dǎo)數(shù)f(x)0,那么在這個(gè)區(qū)間上,函數(shù)y=f(x)是增加的.如果在某個(gè)區(qū)間內(nèi),函數(shù)y=f(x)的導(dǎo)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?yàn)?-,2)(2,+),所以函數(shù)f(x)的遞減區(qū)間為(-,2),(2,3).MUBIAODAOHANG目標(biāo)導(dǎo)航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZH
4、ISHI SHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)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(hào)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(duì)任意x(a,b),若xa,則f(x)f(a)0.故f(x)0.答案:AMUBIAODAOHANG目標(biāo)導(dǎo)航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANG
6、YANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)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(biāo)導(dǎo)航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOU
7、XI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)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(dāng)x(1,+)時(shí),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?yàn)?-,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(biāo)導(dǎo)航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理1 2 3 4 5
高中數(shù)學(xué) 第三章 導(dǎo)數(shù)應(yīng)用 3.1 函數(shù)的單調(diào)性與極值 3.1.1.1 導(dǎo)數(shù)與函數(shù)的單調(diào)性課件 北師大版選修22
