函數(shù)與方程復(fù)習(xí)課件

第8課時(shí)函數(shù)與方程Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2004-2011AsposePtyLtd.2014高考導(dǎo)航考綱展示備考指南1.結(jié)合二次函數(shù)的圖象，了解函數(shù)的零點(diǎn)與方程根的聯(lián)系，判斷一元二次方程根的存在性及根的個(gè)數(shù).2.根據(jù)具體函數(shù)的圖象，能夠用二分法求相應(yīng)方程的近似解.1.函數(shù)的零點(diǎn)、方程根的個(gè)數(shù)是歷年高考的重要考點(diǎn).2.利用函數(shù)的圖象及性質(zhì)判斷函數(shù)的零點(diǎn)，及利用它們求參數(shù)的取值范圍問題是重點(diǎn)，也是難點(diǎn).3.題型以選擇題和填空題為主，常與函數(shù)的圖象與性質(zhì)交匯命題.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2004-2011AsposePtyLtd.本節(jié)目錄教材回顧夯實(shí)雙基考點(diǎn)探究講練互動(dòng)名師講壇精彩呈現(xiàn)知能演練輕松闖關(guān)Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2004-2011AsposePtyLtd.教材回顧夯實(shí)雙基基礎(chǔ)梳理1.函數(shù)的零點(diǎn)(1)函數(shù)零點(diǎn)的定義對(duì)于函數(shù)y＝f(x)(x∈D)，把使___________成立的實(shí)數(shù)x叫做函數(shù)y＝f(x)(x∈D)的零點(diǎn)．(2)幾個(gè)等價(jià)關(guān)系方程f(x)＝0有實(shí)數(shù)根?函數(shù)y＝f(x)的圖象與_______有交點(diǎn)?函數(shù)y＝f(x)有______．f(x)＝0x軸零點(diǎn)Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2004-2011AsposePtyLtd.思考探究1．是否任意函數(shù)都有零點(diǎn)？提示：并非任意函數(shù)都有零點(diǎn)，只有f(x)＝0有根的函數(shù)y＝f(x)才有零點(diǎn)．Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2004-2011AsposePtyLtd.(3)函數(shù)零點(diǎn)的判定(零點(diǎn)存在性定理)如果函數(shù)y＝f(x)在區(qū)間[a，b]上的圖象是連續(xù)不斷的一條曲線，并且有____________，則函數(shù)y＝f(x)在區(qū)間(a，b)內(nèi)有零點(diǎn)，即存在c∈(a，b)，使得___________，這個(gè)c也就是f(x)＝0的根．思考探究2．在上面的條件下，(a，b)內(nèi)的零點(diǎn)有幾個(gè)？提示：在上面的條件下，(a，b)內(nèi)的零點(diǎn)至少有一個(gè)c，還可能有其他零點(diǎn)，個(gè)數(shù)不確定．f(a)·f(b)<0f(c)＝0Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2004-2011AsposePtyLtd.2.二次函數(shù)y＝ax2＋bx＋c(a＞0)的圖象與零點(diǎn)的關(guān)系Δ＞0Δ＝0Δ＜0二次函數(shù)y＝ax2＋bx＋c(a＞0)的圖象與x軸的交點(diǎn)______，________(x1，0)或(x2，0)無交點(diǎn)零點(diǎn)個(gè)數(shù)兩個(gè)一個(gè)零個(gè)(x1，0)(x2，0)Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2004-2011AsposePtyLtd.3.二分法的定義對(duì)于在區(qū)間[a，b]上連續(xù)不斷且____________的函數(shù)y＝f(x)，通過不斷地把函數(shù)f(x)的零點(diǎn)所在的區(qū)間一分為二，使區(qū)間的兩個(gè)端點(diǎn)逐步逼近_________，進(jìn)而得到零點(diǎn)近似值的方法叫做二分法．f(a)·f(b)<0零點(diǎn)Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2004-2011AsposePtyLtd.課前熱身1.如圖所示的函數(shù)圖象與x軸均有交點(diǎn)，其中不能用二分法求圖中交點(diǎn)橫坐標(biāo)的是(

)A．①②

B．①③C．①④ D．③④答案：BEvaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2004-2011AsposePtyLtd.4.函數(shù)f(x)＝x3－x的零點(diǎn)是________．答案：－1，0，15.若二次函數(shù)f(x)＝ax2＋bx＋c中，a·c＜0，則其零點(diǎn)個(gè)數(shù)是________．答案：2Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2004-2011AsposePtyLtd.考點(diǎn)探究講練互動(dòng)例1考點(diǎn)突破Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2004-2011AsposePtyLtd.【解析】

∵f(x)＝ex＋x－4，∴f′(x)＝ex＋1＞0，∴函數(shù)f(x)在R上單調(diào)遞增，對(duì)于A項(xiàng)，f(－1)＝e－1＋(－1)－4＝－5＋e－1＜0，f(0)＝－3＜0，f(－1)f(0)＞0，A不正確，同理可驗(yàn)證B、D不正確．對(duì)于C項(xiàng)，∵f(1)＝e＋1－4＝e－3＜0，f(2)＝e2＋2－4＝e2－2＞0，f(1)f(2)＜0，故選C.【答案】CEvaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2004-2011AsposePtyLtd.【規(guī)律小結(jié)】判定函數(shù)零點(diǎn)個(gè)數(shù)的幾種方法：(1)直接求零點(diǎn)：令f(x)＝0，如果能求出解，則有幾個(gè)解就有幾個(gè)零點(diǎn)；(2)零點(diǎn)存在性定理：利用定理不僅要求函數(shù)在區(qū)間[a，b]上是連續(xù)不斷的曲線，且f(a)·f(b)<0，還必須結(jié)合函數(shù)的圖象與性質(zhì)(如單調(diào)性、奇偶性)才能確定函數(shù)有多少個(gè)零點(diǎn)；(3)利用圖象交點(diǎn)的個(gè)數(shù)：畫出兩個(gè)函數(shù)的圖象，看其交點(diǎn)的個(gè)數(shù)，其中交點(diǎn)的橫坐標(biāo)有幾個(gè)不同的值，就有幾個(gè)不同的零點(diǎn)．Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2004-2011AsposePtyLtd.跟蹤訓(xùn)練Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2004-2011AsposePtyLtd.例2f(1)＝－2f(1.5)＝0.625f(1.25)＝－0.984f(1.375)＝－0.260f(1.4375)

＝0.162f(1.40625)＝－0.054Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2004-2011AsposePtyLtd.【解析】

由參考數(shù)據(jù)知f(1.4375)＞0，f(1.40625)＜0，f(1.4375)·f(1.40625)＜0，且精確到0.1時(shí)，1.4375≈1.4，1.40625≈1.4，所以函數(shù)f(x)的一個(gè)零點(diǎn)的近似值是1.4，也就是方程x3＋x2－2x－2＝0的一個(gè)近似根．【答案】1.4Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2004-2011AsposePtyLtd.【規(guī)律小結(jié)】利用二分法求近似解需注意的問題：(1)第一步中：①區(qū)間長(zhǎng)度盡量??；②f(a)、f(b)的值比較容易計(jì)算且f(a)·f(b)＜0；(2)根據(jù)函數(shù)的零點(diǎn)與相應(yīng)方程根的關(guān)系，求函數(shù)的零點(diǎn)與相應(yīng)方程的根是等價(jià)的．提醒：對(duì)于方程f(x)＝g(x)的根，可以構(gòu)造函數(shù)F(x)＝f(x)－g(x)，函數(shù)F(x)的零點(diǎn)即為方程f(x)＝g(x)的根．Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2004-2011AsposePtyLtd.跟蹤訓(xùn)練2.(2013·武漢模擬)若函數(shù)f(x)在(1，2)內(nèi)有一個(gè)零點(diǎn)，要使零點(diǎn)的近似值滿足精確度為0.01，則對(duì)區(qū)間(1，2)至少二等分(

)A．5次 B．6次C．7次 D．8次Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2004-2011AsposePtyLtd.例3Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2004-2011AsposePtyLtd.【答案】DEvaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2004-2011AsposePtyLtd.【規(guī)律小結(jié)】已知函數(shù)有零點(diǎn)(方程有根)求參數(shù)值常用的方法和思路：(1)直接法：直接求解方程得到方程的根，再通過解不等式確定參數(shù)范圍；(2)分離參數(shù)法：先將參數(shù)分離，轉(zhuǎn)化成求函數(shù)值域問題加以解決；(3)數(shù)形結(jié)合：先對(duì)解析式變形，在同一平面直角坐標(biāo)系中，畫出函數(shù)的圖象，然后觀察求解．Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2004-2011AsposePtyLtd.跟蹤訓(xùn)練3.是否存在這樣的實(shí)數(shù)a，使函數(shù)f(x)＝x2＋(3a－2)x＋a－1在區(qū)間[－1，3]上恒有一個(gè)零點(diǎn)，且只有一個(gè)零點(diǎn)？若存在，求出a的范圍；若不存在，說明理由．Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2004-2011AsposePtyLtd.方法感悟1.函數(shù)y＝f(x)的零點(diǎn)即方程f(x)＝0的實(shí)根，是數(shù)不是點(diǎn)．2.若函數(shù)y＝f(x)在閉區(qū)間[a，b]上的圖象是連續(xù)不間斷的，并且在區(qū)間端點(diǎn)的函數(shù)值符號(hào)相反，即f(a)·f(b)＜0，滿足這些條件一定有零點(diǎn)，不滿足這些條件也不能說就沒有零點(diǎn)．如圖，f(a)·f(b)＞0，f(x)在區(qū)間(a，b)上照樣存在零點(diǎn)，而且有兩個(gè)．所以說零點(diǎn)存在性定理的條件是充分條件,但并不必要．Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2004-2011AsposePtyLtd.名師講壇精彩呈現(xiàn)例數(shù)學(xué)思想Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2004-2011AsposePtyLtd.【答案】BEvaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2004-2011AsposePtyLtd.【感悟提高】解答本題利用了轉(zhuǎn)化與化歸、數(shù)形結(jié)合的思想．所謂轉(zhuǎn)化與化歸思想方法，就是在研究和解決有關(guān)數(shù)學(xué)問題時(shí)采用某種手段將問題通過變換使