2025年高考數(shù)學(xué)一輪復(fù)習(xí)-4.4.2-導(dǎo)數(shù)的函數(shù)零點(diǎn)問(wèn)題【課件】

第2課時(shí)導(dǎo)數(shù)的函數(shù)零點(diǎn)問(wèn)題核心考點(diǎn)·分類突破第四章一元函數(shù)的導(dǎo)數(shù)及其應(yīng)用【命題分析】函數(shù)零點(diǎn)問(wèn)題在高考中占有很重要的地位,主要涉及判斷函數(shù)零點(diǎn)的個(gè)數(shù)或范圍.高考常考查基本初等函數(shù)、三次函數(shù)與復(fù)合函數(shù)的零點(diǎn)問(wèn)題,以及函數(shù)零點(diǎn)與其他知識(shí)的交匯問(wèn)題,一般作為解答題的壓軸題出現(xiàn).核心考點(diǎn)·分類突破

解題技法利用導(dǎo)數(shù)確定函數(shù)零點(diǎn)或方程的根的個(gè)數(shù)的方法(1)構(gòu)造函數(shù):構(gòu)建函數(shù)g(x)(要求g'(x)易求,g'(x)=0可解),轉(zhuǎn)化為確定g(x)的零點(diǎn)個(gè)數(shù)問(wèn)題求解,利用導(dǎo)數(shù)研究該函數(shù)的單調(diào)性、極值,并確定定義區(qū)間端點(diǎn)值的符號(hào)(或變化趨勢(shì))等,畫(huà)出g(x)的圖象草圖,數(shù)形結(jié)合求解函數(shù)零點(diǎn)的個(gè)數(shù).(2)應(yīng)用定理:利用零點(diǎn)存在定理,先用該定理判斷函數(shù)在某區(qū)間上有零點(diǎn),然后利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性、極值(最值)及區(qū)間端點(diǎn)值的符號(hào),進(jìn)而判斷函數(shù)在該區(qū)間上零點(diǎn)的個(gè)數(shù).對(duì)點(diǎn)訓(xùn)練(2023·鄭州質(zhì)檢)已知函數(shù)f(x)=ex-ax+2a,a∈R.(1)討論函數(shù)f(x)的單調(diào)性;【解析】(1)f(x)=ex-ax+2a,定義域?yàn)镽,且f'(x)=ex-a,當(dāng)a≤0時(shí),f'(x)>0,則f(x)在R上單調(diào)遞增;當(dāng)a>0時(shí),令f'(x)=0,則x=lna,當(dāng)x<lna時(shí),f'(x)<0,f(x)單調(diào)遞減;當(dāng)x>lna時(shí),f'(x)>0,f(x)單調(diào)遞增.綜上所述,當(dāng)a≤0時(shí),f(x)在R上單調(diào)遞增;當(dāng)a>0時(shí),f(x)在(-∞,lna)上單調(diào)遞減,在(lna,+∞)上單調(diào)遞增.(2023·鄭州質(zhì)檢)已知函數(shù)f(x)=ex-ax+2a,a∈R.(2)求函數(shù)f(x)的零點(diǎn)個(gè)數(shù).

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已知函數(shù)f(x)=xex+ex.(1)求函數(shù)f(x)的單調(diào)區(qū)間和極值;【解析】(1)函數(shù)f(x)的定義域?yàn)镽,且f'(x)=(x+2)ex,令f'(x)=0得x=-2,則f'(x),f(x)的變化情況如表所示:x(-∞,-2)-2(-2,+∞)f'(x)-0+f(x)單調(diào)遞減單調(diào)遞增

已知函數(shù)f(x)=xex+ex.(2)討論函數(shù)g(x)=f(x)-a(a∈R)的零點(diǎn)的個(gè)數(shù).

考點(diǎn)二利用函數(shù)零點(diǎn)問(wèn)題求參數(shù)范圍[例2]已知函數(shù)f(x)=ex-a(x+2).(1)當(dāng)a=1時(shí),討論f(x)的單調(diào)性;【解析】(1)當(dāng)a=1時(shí),f(x)=ex-x-2,則f'(x)=ex-1.當(dāng)x<0時(shí),f'(x)<0;當(dāng)x>0時(shí),f'(x)>0.所以f(x)在(-∞,0)上單調(diào)遞減,在(0,+∞)上單調(diào)遞增;[例2]已知函數(shù)f(x)=ex-a(x+2).

(2)若f(x)有兩個(gè)零點(diǎn),求a的取值范圍.【解析】(2)f'(x)=ex-a.當(dāng)a≤0時(shí),f'(x)>0,所以f(x)在(-∞,+∞)上單調(diào)遞增,故f(x)至多存在1個(gè)零點(diǎn),不合題意;當(dāng)a>0時(shí),由f'(x)=0可得x=lna.當(dāng)x∈(-∞,lna)時(shí),f'(x)<0;當(dāng)x∈(lna,+∞)時(shí),f'(x)>0.所以f(x)在(-∞,lna)上單調(diào)遞減,在(lna,+∞)上單調(diào)遞增,故當(dāng)x=lna時(shí),f(x)取得最小值,最小值為f(lna)=-a(1+lna).

解題技法由函數(shù)零點(diǎn)求參數(shù)范圍的策略(1)涉及函數(shù)的零點(diǎn)(方程的根)問(wèn)題,主要利用導(dǎo)數(shù)確定函數(shù)的單調(diào)區(qū)間和極值點(diǎn),根據(jù)函數(shù)零點(diǎn)的個(gè)數(shù)尋找函數(shù)在給定區(qū)間的極值以及區(qū)間端點(diǎn)的函數(shù)值與0的關(guān)系,從而求得參數(shù)的取值范圍.(2)解決此類問(wèn)題的關(guān)鍵是將函數(shù)零點(diǎn)、方程的根、曲線交點(diǎn)相互轉(zhuǎn)化,突出導(dǎo)數(shù)的工具作用,體現(xiàn)轉(zhuǎn)化與化歸的思想方法.(3)含參數(shù)的函數(shù)零點(diǎn)個(gè)數(shù),可轉(zhuǎn)化為方程解的個(gè)數(shù),若能分離參數(shù),可將參數(shù)分離出來(lái)后,用x表示參數(shù)的函數(shù),作出該函數(shù)圖象,根據(jù)圖象特征求參數(shù)的范圍.對(duì)點(diǎn)訓(xùn)練

已知函數(shù)f(x)=xlnx,g(x)=(-x2+ax-3)ex(a∈R).(1)當(dāng)a=4時(shí),求曲線y=g(x)在x=0處的切線方程;【解析】(1)當(dāng)a=4時(shí),g(x)=(-x2+4x-3)ex,g(0)=-3,g'(x)=(-x2+2x+1)ex,g'(0)=1,所以所求的切線方程為y+3=x-0,即y=x-3.

x1(1,e]h'(x)-0+h(x)單調(diào)遞減極小值(最小值)單調(diào)遞增【加練備選】

(一題多法)(2020·全國(guó)Ⅰ卷)已知函數(shù)f(x)=ex-a(x+2).(1)當(dāng)a=1時(shí),討論f(x)的單調(diào)性;【解析】(1)當(dāng)a=1時(shí),f(x)=ex-(x+2),f'(x)=ex-1,令f'(x)<0,解得x<0,令f'(x)>0,解得x>0,所以f(x)在(-∞,0)上單調(diào)遞減,在(0,+∞)上單調(diào)遞增.

(一題多法)(2020·全國(guó)Ⅰ卷)已知函數(shù)f(x)=ex-a(x+2).

(2)若f(x)有兩個(gè)零點(diǎn),求a的取值范圍.

考點(diǎn)三與函數(shù)零點(diǎn)相關(guān)的綜合問(wèn)題[例3](2024·錦州模擬)設(shè)函數(shù)f(x)=e2x-alnx.(1)討論f(x)的導(dǎo)函數(shù)f'(x)零點(diǎn)的個(gè)數(shù);

解題技法1.證明與零點(diǎn)有關(guān)的不等式,函數(shù)的零點(diǎn)本身就是一個(gè)條件,即零點(diǎn)對(duì)應(yīng)的函數(shù)值為0;2.證明的思路一般對(duì)條件等價(jià)轉(zhuǎn)化,構(gòu)