高考數(shù)學(xué)北師大（理）一輪復(fù)習(xí)ppt課件28函數(shù)與方程

2.8

函數(shù)與方程2.8函數(shù)與方程-2-知識(shí)梳理考點(diǎn)自診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)與函數(shù)零點(diǎn)有關(guān)的等價(jià)關(guān)系方程f(x)=0有實(shí)數(shù)根?函數(shù)y=f(x)的圖像與

有交點(diǎn)?函數(shù)y=f(x)有

.

(3)函數(shù)零點(diǎn)的判定(零點(diǎn)存在性定理)f(x)=0x軸

零點(diǎn)

連續(xù)不斷的

f(a)·f(b)<0f(x0)=0-2-知識(shí)梳理考點(diǎn)自診1.函數(shù)的零點(diǎn)f(x)=0x軸零點(diǎn)-3-知識(shí)梳理考點(diǎn)自診2.二次函數(shù)y=ax2+bx+c(a>0)的圖像與零點(diǎn)的關(guān)系(x1,0),(x2,0)(x1,0)2103.二分法函數(shù)y=f(x)的圖像在區(qū)間[a,b]上連續(xù)不斷,且

,通過不斷地把函數(shù)f(x)的零點(diǎn)所在的區(qū)間

,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)<0一分為二

零點(diǎn)

-3-知識(shí)梳理考點(diǎn)自診2.二次函數(shù)y=ax2+bx+-4-知識(shí)梳理考點(diǎn)自診1.若y=f(x)在閉區(qū)間[a,b]上的圖像連續(xù)不斷,且有f(a)·f(b)<0,則函數(shù)y=f(x)一定有零點(diǎn).2.f(a)·f(b)<0是y=f(x)在閉區(qū)間[a,b]上有零點(diǎn)的充分不必要條件.3.若函數(shù)f(x)在[a,b]上是單調(diào)函數(shù),且f(x)的圖像連續(xù)不斷,則f(a)·f(b)<0?函數(shù)f(x)在區(qū)間[a,b]上只有一個(gè)零點(diǎn).-4-知識(shí)梳理考點(diǎn)自診1.若y=f(x)在閉區(qū)間[a,b]上-5-知識(shí)梳理考點(diǎn)自診1.判斷下列結(jié)論是否正確,正確的畫“√”,錯(cuò)誤的畫“×”.(1)函數(shù)f(x)=x2-1的零點(diǎn)是(-1,0)和(1,0).(

)(2)二次函數(shù)y=ax2+bx+c(a≠0)在b2-4ac<0時(shí)沒有零點(diǎn).(

)(3)只要函數(shù)有零點(diǎn),我們就可以用二分法求出零點(diǎn)的近似值.(

)(4)已知函數(shù)f(x)在(a,b)內(nèi)圖像連續(xù)且單調(diào),若f(a)·f(b)<0,則函數(shù)f(x)在[a,b]上有且只有一個(gè)零點(diǎn).(

)(5)函數(shù)y=2sinx-1的零點(diǎn)有無數(shù)多個(gè).(

)×√×√√-5-知識(shí)梳理考點(diǎn)自診1.判斷下列結(jié)論是否正確,正確的畫“√-6-知識(shí)梳理考點(diǎn)自診2.已知函數(shù)y=x2-2x+m無零點(diǎn),則m的取值范圍為(

)A.m<1 B.m<-1C.m>1 D.m>-1C解析:由Δ=(-2)2-4m<0,得m>1,故選C.3.函數(shù)f(x)=lnx+2x-6的零點(diǎn)所在的區(qū)間是(

)A.(0,1) B.(1,2)C.(2,3) D.(3,4)C解析:∵y=ln

x與y=2x-6在(0,+∞)內(nèi)都是增函數(shù),∴f(x)=ln

x+2x-6在(0,+∞)內(nèi)是增函數(shù).又f(1)=-4,f(2)=ln

2-2<ln

e-2<0,f(3)=ln

3>0,∴零點(diǎn)在區(qū)間(2,3)內(nèi),故選C.-6-知識(shí)梳理考點(diǎn)自診2.已知函數(shù)y=x2-2x+m無零點(diǎn),-7-知識(shí)梳理考點(diǎn)自診4.方程2x+3x=k的解都在[1,2)內(nèi),則k的取值范圍為(

)A.5≤k<10 B.5<k≤10C.5≤k≤10 D.5<k<10A解析:令函數(shù)f(x)=2x+3x-k,則f(x)在R上是增函數(shù).當(dāng)方程2x+3x=k的解在(1,2)內(nèi)時(shí),f(1)·f(2)<0,即(5-k)(10-k)<0,解得5<k<10.當(dāng)f(1)=0時(shí),k=5,故選A.2解析:當(dāng)x≤0時(shí),由f(x)=x2+2x-3=0,得x1=1(舍去),x2=-3;當(dāng)x>0時(shí),由f(x)=-2+ln

x=0,得x=e2,所以函數(shù)f(x)的零點(diǎn)個(gè)數(shù)為2.-7-知識(shí)梳理考點(diǎn)自診4.方程2x+3x=k的解都在[1,2-8-考點(diǎn)1考點(diǎn)2考點(diǎn)3判斷函數(shù)零點(diǎn)所在的區(qū)間例1(1)如圖是二次函數(shù)f(x)=x2-bx+a的部分圖像,則函數(shù)g(x)=ex+f'(x)的零點(diǎn)所在的大致區(qū)間是(

)A.(-1,0) B.(0,1)C.(1,2) D.(2,3)(2)已知函數(shù)f(x)=logax+x-b(a>0,且a≠1).當(dāng)2<a<3<b<4時(shí),函數(shù)f(x)的零點(diǎn)x0∈(n,n+1),n∈N+,則n=

.

B2-8-考點(diǎn)1考點(diǎn)2考點(diǎn)3判斷函數(shù)零點(diǎn)所在的區(qū)間B2-9-考點(diǎn)1考點(diǎn)2考點(diǎn)3所以g(0)g(1)<0.故選B.(2)∵2<a<3<b<4,∴f(1)=loga1+1-b=1-b<0,f(2)=loga2+2-b<0,f(3)=loga3+3-b,又∵loga3>1,-1<3-b<0,∴f(3)>0,即f(2)f(3)<0,故x0∈(2,3),即n=2.-9-考點(diǎn)1考點(diǎn)2考點(diǎn)3所以g(0)g(1)<0.故選B.-10-考點(diǎn)1考點(diǎn)2考點(diǎn)3思考判斷函數(shù)y=f(x)在某個(gè)區(qū)間上是否存在零點(diǎn)的常用方法有哪些?解題心得判斷函數(shù)y=f(x)在某個(gè)區(qū)間上是否存在零點(diǎn),常用以下方法:(1)解方程:當(dāng)對(duì)應(yīng)方程易解時(shí),可通過解方程,觀察方程是否有根落在給定區(qū)間上.(2)利用函數(shù)零點(diǎn)的存在性定理進(jìn)行判斷:首先看函數(shù)y=f(x)在區(qū)間[a,b]上的圖像是否連續(xù),然后看是否有f(a)·f(b)<0.若有,則函數(shù)y=f(x)在區(qū)間(a,b)內(nèi)必有零點(diǎn);若沒有,則不一定有零點(diǎn).(3)通過畫函數(shù)圖像,觀察圖像與x軸在給定區(qū)間上是否有交點(diǎn)來判斷.-10-考點(diǎn)1考點(diǎn)2考點(diǎn)3思考判斷函數(shù)y=f(x)在某個(gè)區(qū)間-11-考點(diǎn)1考點(diǎn)2考點(diǎn)3對(duì)點(diǎn)訓(xùn)練1(1)函數(shù)f(x)=πx+log2x的零點(diǎn)所在的區(qū)間為(

)(2)(2017浙江嘉興模擬)已知函數(shù)y=x3與

的圖像的交點(diǎn)為(x0,y0).若x0∈(n,n+1),n∈N,則x0所在的區(qū)間是

.

A(1,2)-11-考點(diǎn)1考點(diǎn)2考點(diǎn)3對(duì)點(diǎn)訓(xùn)練1(1)函數(shù)f(x)=πx-12-考點(diǎn)1考點(diǎn)2考點(diǎn)3-12-考點(diǎn)1考點(diǎn)2考點(diǎn)3-13-考點(diǎn)1考點(diǎn)2考點(diǎn)3判斷函數(shù)零點(diǎn)的個(gè)數(shù)例2(1)函數(shù)f(x)=2x|log0.5x|-1的零點(diǎn)個(gè)數(shù)為

(

)A.1 B.2 C.3 D.4(2)(2018湖南長(zhǎng)郡中學(xué)一模,11)已知函數(shù)y=f(x)是定義域?yàn)镽的周期為3的奇函數(shù),且當(dāng)

時(shí),f(x)=ln(x2-x+1),則方程f(x)=0在區(qū)間[0,6]上解的個(gè)數(shù)是(

)A.5 B.6C.7 D.9BD-13-考點(diǎn)1考點(diǎn)2考點(diǎn)3判斷函數(shù)零點(diǎn)的個(gè)數(shù)BD-14-考點(diǎn)1考點(diǎn)2考點(diǎn)3-14-考點(diǎn)1考點(diǎn)2考點(diǎn)3-15-考點(diǎn)1考點(diǎn)2考點(diǎn)3(2)∵當(dāng)x∈(0,1.5)時(shí),f(x)=ln(x2-x+1),令f(x)=0,則x2-x+1=1,解得x=1.∵函數(shù)f(x)是定義域?yàn)镽的奇函數(shù),∴在[-1.5,1.5]內(nèi),f(-1)=f(1)=0,f(0)=0,f(1.5)=f(-1.5+3)=f(-1.5)=-f(1.5),∴f(1.5)=0.∴f(-1)=f(1)=f(0)=f(1.5)=0,∵函數(shù)f(x)是周期為3的周期函數(shù),則方程f(x)=0在區(qū)間[0,6]上的解有0,1,1.5,2,3,4,4.5,5,6共9個(gè),故選D.-15-考點(diǎn)1考點(diǎn)2考點(diǎn)3(2)∵當(dāng)x∈(0,1.5)時(shí),f-16-考點(diǎn)1考點(diǎn)2考點(diǎn)3思考判斷函數(shù)零點(diǎn)個(gè)數(shù)的常用方法有哪些?解題心得判斷函數(shù)零點(diǎn)個(gè)數(shù)的方法:(1)解方程法:若對(duì)應(yīng)方程f(x)=0可解時(shí),通過解方程,有幾個(gè)解就有幾個(gè)零點(diǎn).(2)零點(diǎn)存在性定理法:利用定理不僅要判斷函數(shù)的圖像在區(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)的圖像與性質(zhì)(如單調(diào)性、奇偶性、周期性、對(duì)稱性)才能確定函數(shù)有多少個(gè)零點(diǎn).(3)數(shù)形結(jié)合法:轉(zhuǎn)化為兩個(gè)函數(shù)的圖像的交點(diǎn)個(gè)數(shù)問題.先畫出兩個(gè)函數(shù)的圖像,再看其交點(diǎn)的個(gè)數(shù),其中交點(diǎn)的個(gè)數(shù)就是函數(shù)零點(diǎn)的個(gè)數(shù).-16-考點(diǎn)1考點(diǎn)2考點(diǎn)3思考判斷函數(shù)零點(diǎn)個(gè)數(shù)的常用方法有哪-17-考點(diǎn)1考點(diǎn)2考點(diǎn)3對(duì)點(diǎn)訓(xùn)練2(1)函數(shù)f(x)=sin(πcosx)在區(qū)間[0,2π]上的零點(diǎn)個(gè)數(shù)是(

)A.3 B.4 C.5 D.6(2)(2018河北衡水中學(xué)十模,10)設(shè)函數(shù)f(x)為定義域?yàn)镽的奇函數(shù),且f(x)=f(2-x),當(dāng)x∈[0,1]時(shí),f(x)=sinx,則函數(shù)g(x)=|cos(πx)|-f(x)在區(qū)間

上的所有零點(diǎn)的和為(

)A.6 B.7C.13 D.14CA-17-考點(diǎn)1考點(diǎn)2考點(diǎn)3對(duì)點(diǎn)訓(xùn)練2(1)函數(shù)f(x)=si-18-考點(diǎn)1考點(diǎn)2考點(diǎn)3解析:(1)令f(x)=0,得πcos

x=kπ(k∈Z),即cos

x=k(k∈Z),故k=0,1,-1.則x=π,故零點(diǎn)個(gè)數(shù)為5.(2)由題意,函數(shù)f(-x)=-f(x),f(x)=f(2-x),則-f(-x)=f(2-x),可得f(x+4)=f(x),即函數(shù)的周期為4,且y=f(x)的圖像關(guān)于直線x=1對(duì)稱.即方程|cos(πx)|=f(x)的零點(diǎn),分別畫出y=|cos(πx)|與y=f(x)的大致圖像,∵兩個(gè)函數(shù)的圖像都關(guān)于直線x=1對(duì)稱,∴方程|cos(πx)|=f(x)的零點(diǎn)關(guān)于直線x=1對(duì)稱,由圖像可知交點(diǎn)個(gè)數(shù)為6個(gè),可得所有零點(diǎn)的和為6,故選A.-18-考點(diǎn)1考點(diǎn)2考點(diǎn)3解析:(1)令f(x)=0,得π-19-考點(diǎn)1考點(diǎn)2考點(diǎn)3函數(shù)零點(diǎn)的應(yīng)用(多考向)考向1

已知函數(shù)零點(diǎn)所在區(qū)間求參數(shù)例3若函數(shù)f(x)=log2x+x-k(k∈Z)在區(qū)間(2,3)內(nèi)有零點(diǎn),則k=

.4解析:由題意可得f(2)f(3)<0,即(log22+2-k)·(log23+3-k)<0,整理得(3-k)(log23+3-k)<0,解得3<k<3+log23,而4<3+log23<5.因?yàn)閗∈Z,所以k=4.思考已知函數(shù)零點(diǎn)所在的區(qū)間,怎樣求參數(shù)的取值范圍?-19-考點(diǎn)1考點(diǎn)2考點(diǎn)3函數(shù)零點(diǎn)的應(yīng)用(多考向)4解析:由-20-考點(diǎn)1考點(diǎn)2考點(diǎn)3考向2

已知函數(shù)零點(diǎn)個(gè)數(shù)求參數(shù)問題

例4(2018全國(guó)1,理9)已知函數(shù)

,若g(x)存在2個(gè)零點(diǎn),則a的取值范圍是(

)A.[-1,0) B.[0,+∞)

C.[-1,+∞) D.[1,+∞)C解析:要使得方程g(x)=f(x)+x+a有兩個(gè)零點(diǎn),等價(jià)于方程f(x)=-x-a有兩個(gè)實(shí)根,即函數(shù)y=f(x)的圖像與直線y=-x-a的圖像有兩個(gè)交點(diǎn),從圖像可知,必須使得直線y=-x-a位于直線y=-x+1的下方,所以-a≤1,即a≥-1.故選C.-20-考點(diǎn)1考點(diǎn)2考點(diǎn)3考向2已知函數(shù)零點(diǎn)個(gè)數(shù)求參數(shù)問題-21-考點(diǎn)1考點(diǎn)2考點(diǎn)3思考已知函數(shù)有零點(diǎn)(方程有根),求參數(shù)的取值范圍常用的方法有哪些?解題心得已知函數(shù)有零點(diǎn)(方程有根),求參數(shù)的取值范圍常用的方法:(1)直接法:直接根據(jù)題設(shè)條件構(gòu)建關(guān)于參數(shù)的不等式,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,再轉(zhuǎn)化成求函數(shù)值域問題加以解決.(3)數(shù)形結(jié)合法: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)的圖像,再數(shù)形結(jié)合求解.-21-考點(diǎn)1考點(diǎn)2考點(diǎn)3思考已知函數(shù)有零點(diǎn)(方程有根),求-22-考點(diǎn)1考點(diǎn)2考點(diǎn)3對(duì)點(diǎn)訓(xùn)練3(1)已知函數(shù)f(x)=2ax-a+3,若存在x0∈(-1,1),f(x0)=0,則實(shí)數(shù)a的取值范圍是(

)A.(-∞,-3)∪(1,+∞) B.(-∞,-3)C.(-3,1) D.(1,+∞)(2)(2018山東師大附中一模,12)函數(shù)f(x)是定義在R上的偶函數(shù),且滿足f(x+2)=f(x),當(dāng)x∈[0,1]時(shí),f(x)=2x,若在區(qū)間[-2,3]上方程ax+2a-f(x)=0恰有四個(gè)不相等的實(shí)數(shù)根,則實(shí)數(shù)a的取值范圍是(

)AD-22-考點(diǎn)1考點(diǎn)2考點(diǎn)3對(duì)點(diǎn)訓(xùn)練3(1)已知函數(shù)f(x)=-23-考點(diǎn)1考點(diǎn)2考點(diǎn)3解析:(1)函數(shù)f(x)=2ax-a+3,若存在x0∈(-1,1),f(x0)=0,可得(-3a+3)(a+3)<0,解得a∈(-∞,-3)∪(1,+∞).(2)若在區(qū)間[-2,3]上方程ax+2a-f(x)=0恰有四個(gè)不相等的實(shí)數(shù)根,等價(jià)于f(x)=a(x+2)有四個(gè)不相等的實(shí)數(shù)根,即函數(shù)y=f(x)和g(x)=a(x+2)有四個(gè)不同的交點(diǎn),∵f(x+2)=f(x),∴函數(shù)f(x)的周期為2,當(dāng)-1≤x≤0時(shí),0≤-x≤1,此時(shí)f(-x)=-2x.∵f(x)是定義在R上的偶函數(shù),∴f(-x)=-2x=f(x),即f(x)=-2x,-1≤x≤0.作出函數(shù)f(x)和g(x)的圖像,-23-考點(diǎn)1考點(diǎn)2考點(diǎn)3解析:(1)函數(shù)f(x)=2ax-24-考點(diǎn)1考點(diǎn)2考點(diǎn)3-24-考點(diǎn)1考點(diǎn)2考點(diǎn)3-25-考點(diǎn)1考點(diǎn)2考點(diǎn)31.函數(shù)零點(diǎn)的常用判定方法:(1)零點(diǎn)存在性定理;(2)數(shù)形結(jié)合;(3)解方程f(x)=0.2.研究方程f(x)=g(x)的解,實(shí)質(zhì)就是研究G(x)=f(x)-g(x)的零點(diǎn).3.轉(zhuǎn)化思想:方程解的個(gè)數(shù)問題可轉(zhuǎn)化為兩個(gè)函數(shù)圖像交點(diǎn)的個(gè)數(shù)問題;已知方程有解求參數(shù)范圍問題可轉(zhuǎn)化為函數(shù)值域問題.1.函數(shù)f(x)的零點(diǎn)是一個(gè)實(shí)數(shù),是方程f(x)=0的根,也是函數(shù)y=f(x)的圖像與x軸交點(diǎn)的橫坐標(biāo).2.函數(shù)零點(diǎn)存在性定理是零點(diǎn)存在的一個(gè)充分條件,而不是必要條件;判斷零點(diǎn)個(gè)數(shù)還要根據(jù)函數(shù)的單調(diào)性、對(duì)稱性或結(jié)合函數(shù)圖像等綜合考慮.-25-考點(diǎn)1考點(diǎn)2考點(diǎn)31.函數(shù)零點(diǎn)的常用判定方法:1.函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