第第2 2講導(dǎo)數(shù)的簡(jiǎn)單應(yīng)用講導(dǎo)數(shù)的簡(jiǎn)單應(yīng)用熱點(diǎn)突破熱點(diǎn)突破高考導(dǎo)航高考導(dǎo)航閱卷評(píng)析閱卷評(píng)析高考導(dǎo)航高考導(dǎo)航 演真題演真題明備考明備考高考體驗(yàn)高考體驗(yàn)1.(1.(20142014全國(guó)全國(guó)卷卷, ,文文1111) )若函數(shù)若函數(shù)f(x)=kx-lnf(x)=kx-ln x x在區(qū)間在區(qū)間(1,+)(1,+)上單調(diào)遞增上單調(diào)遞增, ,則則k k的取值范圍是的取值范圍是( ( ) )(A)(-,-2(A)(-,-2(B)(-,-1(B)(-,-1 (C)2,+)(C)2,+) (D)1,+) (D)1,+)D D2.(2.(20132013全國(guó)全國(guó)卷卷, ,文文1111) )已知函數(shù)已知函數(shù)f(xf(x)=x)=x3 3+ax+ax2 2+bx+c,+bx+c,下列結(jié)論中錯(cuò)誤的是下列結(jié)論中錯(cuò)誤的是( ( ) )(A)(A)x x0 0R R,f(x,f(x0 0)=0)=0(B)(B)函數(shù)函數(shù)y=f(xy=f(x) )的圖象是中心對(duì)稱(chēng)圖形的圖象是中心對(duì)稱(chēng)圖形(C)(C)若若x x0 0是是f(xf(x) )的極小值點(diǎn)的極小值點(diǎn), ,則則f(xf(x) )在區(qū)間在區(qū)間(-,x(-,x0 0) )上單調(diào)遞減上單調(diào)遞減(D)(D)若若x x0 0是是f(xf(x) )的極值點(diǎn)的極值點(diǎn), ,則則f(xf(x0 0)=0)=0C C解析解析: :因?yàn)楹瘮?shù)因?yàn)楹瘮?shù)f(xf(x) )的值域?yàn)榈闹涤驗(yàn)镽 R, ,故選項(xiàng)故選項(xiàng)A A正確正確. .假設(shè)函數(shù)假設(shè)函數(shù)y=f(xy=f(x) )的對(duì)的對(duì)稱(chēng)中心為稱(chēng)中心為(m,n(m,n),),按向量按向量a a=(-m,-n=(-m,-n) )將函數(shù)的圖象平移將函數(shù)的圖象平移, ,則所得函數(shù)則所得函數(shù)y=f(x+m)-ny=f(x+m)-n為奇函數(shù)為奇函數(shù), ,因此因此f(x+m)+f(-x+m)-2n=0,f(x+m)+f(-x+m)-2n=0,代入化簡(jiǎn)得代入化簡(jiǎn)得(3m+a)x(3m+a)x2 2+m+m3 3+am+am2 2+bm+c-n=0+bm+c-n=0對(duì)對(duì)xxR R恒成立恒成立. .3.(3.(20152015全國(guó)全國(guó)卷卷, ,文文1414) )已知函數(shù)已知函數(shù)f(xf(x)=ax)=ax3 3+x+1+x+1的圖象在點(diǎn)的圖象在點(diǎn)(1,f(1)(1,f(1)處的處的切線過(guò)點(diǎn)切線過(guò)點(diǎn)(2,7),(2,7),則則a=a= .解析解析: :因?yàn)橐驗(yàn)閒(xf(x)=ax)=ax3 3+x+1,+x+1,所以所以f(xf(x)=3ax)=3ax2 2+1,+1,所以所以f(xf(x) )在點(diǎn)在點(diǎn)(1,f(1)(1,f(1)處的切線斜率為處的切線斜率為k=3a+1,k=3a+1,又又f(1)=a+2,f(1)=a+2,所以切線方程為所以切線方程為y-(a+2)=(3a+1)(x-1),y-(a+2)=(3a+1)(x-1),因?yàn)辄c(diǎn)因?yàn)辄c(diǎn)(2,7)(2,7)在切線上在切線上, ,所以所以7-(a+2)=3a+1,7-(a+2)=3a+1,解得解得a=1.a=1.答案答案: :1 14.(4.(20162016全國(guó)全國(guó)卷卷, ,文文1616) )已知已知f(xf(x) )為偶函數(shù)為偶函數(shù), ,當(dāng)當(dāng)x0 x0時(shí)時(shí),f(x,f(x)=e)=e-x-1-x-1-x,-x,則曲則曲線線y=f(xy=f(x) )在點(diǎn)在點(diǎn)(1,2)(1,2)處的切線方程是處的切線方程是 .解析解析: :令令x0,x0,則則-x0,f(-x)=e-x0,f(-x)=ex-1x-1+x,+x,又又f(xf(x) )為偶函數(shù)為偶函數(shù), ,所以所以x0 x0時(shí)時(shí),f(x,f(x)=e)=ex-1x-1+x,+x,所以所以f(1)=2,f(1)=2,f(xf(x)=e)=ex-1x-1+1,+1,f(1)=2,f(1)=2,所求切線方程為所求切線方程為y-2=2(x-1),y-2=2(x-1),即即y=2x.y=2x.答案答案: :y=2xy=2x5.(5.(20132013全國(guó)全國(guó)卷卷, ,文文2020) )已知函數(shù)已知函數(shù)f(xf(x)=e)=ex x(ax+b)-x(ax+b)-x2 2-4x,-4x,曲線曲線y=f(xy=f(x) )在點(diǎn)在點(diǎn)(0,(0,f(0)f(0)處的切線方程為處的切線方程為y=4x+4.y=4x+4.(1)(1)求求a,ba,b的值的值; ;(2)(2)討論討論f(xf(x) )的單調(diào)性的單調(diào)性, ,并求并求f(xf(x) )的極大值的極大值. .解解: :(1)f(x)=e(1)f(x)=ex x(ax+a+b)-2x-4.(ax+a+b)-2x-4.由已知得由已知得f(0)=4,f(0)=4.f(0)=4,f(0)=4.故故b=4,a+b=8,b=4,a+b=8,從而從而a=4,b=4.a=4,b=4.(2)(2)由由(1)(1)知知,f(x)=4e,f(x)=4ex x(x+1)-x(x+1)-x2 2-4x.-4x.f(x)=4ef(x)=4ex x(x+2)-2x-4=4(x+2)(x+2)-2x-4=4(x+2)( (e ex x- - ) ). .令令f(x)=0f(x)=0得得,x=-ln 2,x=-ln 2或或x=-2.x=-2.從而當(dāng)從而當(dāng)x(-,-2)(-ln 2,+)x(-,-2)(-ln 2,+)時(shí)時(shí),f(x)0;,f(x)0;當(dāng)當(dāng)x(-2,-ln 2)x(-2,-ln 2)時(shí)時(shí),f(x,f(x)0.)0.故故f(xf(x) )在在(-,-2),(-ln 2,+)(-,-2),(-ln 2,+)上單調(diào)遞增上單調(diào)遞增, ,在在(-2,-ln 2)(-2,-ln 2)上單調(diào)遞減上單調(diào)遞減. .當(dāng)當(dāng)x=-2x=-2時(shí)時(shí), ,函數(shù)函數(shù)f(xf(x) )取得極大值取得極大值, ,極大值為極大值為f(-2)=4(1-ef(-2)=4(1-e-2-2).).12高考感悟高考感悟1.1.考查角度考查角度(1)(1)導(dǎo)數(shù)的幾何意義導(dǎo)數(shù)的幾何意義: :求切線方程或求參數(shù)求切線方程或求參數(shù). .(2)(2)利用導(dǎo)數(shù)求函數(shù)的單調(diào)區(qū)間或由單調(diào)性求參數(shù)范圍利用導(dǎo)數(shù)求函數(shù)的單調(diào)區(qū)間或由單調(diào)性求參數(shù)范圍. .(3)(3)利用導(dǎo)數(shù)求函數(shù)的極值、最值或由極值、最值求參數(shù)范圍利用導(dǎo)數(shù)求函數(shù)的極值、最值或由極值、最值求參數(shù)范圍. .2.2.題型及難易度題型及難易度選擇題、填空題、解答題、難度中檔偏上選擇題、填空題、解答題、難度中檔偏上. .熱點(diǎn)突破熱點(diǎn)突破 剖典例剖典例促遷移促遷移導(dǎo)數(shù)的幾何意義導(dǎo)數(shù)的幾何意義熱點(diǎn)一熱點(diǎn)一【例【例1 1】 (1)(1)曲線曲線y=x(3ln x+1)y=x(3ln x+1)在點(diǎn)在點(diǎn)(1,1)(1,1)處的切線方程為處的切線方程為;解析解析: :(1)(1)由由y=x(3ln x+1)y=x(3ln x+1)得得y=3ln x+4,y=3ln x+4,則所求切線斜率為則所求切線斜率為4,4,則所求切線方程為則所求切線方程為y=4x-3.y=4x-3.答案答案: :(1)y=4x-3(1)y=4x-3(2)(2)(20162016福建福建“四地六校四地六校”聯(lián)考聯(lián)考) )已知曲線已知曲線f(xf(x)= x)= x3 3-x-x2 2+ax-1+ax-1存在兩條斜存在兩條斜率為率為3 3的切線的切線, ,且切點(diǎn)的橫坐標(biāo)都大于零且切點(diǎn)的橫坐標(biāo)都大于零, ,則實(shí)數(shù)則實(shí)數(shù)a a的取值范圍為的取值范圍為.23【方法技巧【方法技巧】 求曲線求曲線y=f(xy=f(x) )的切線方程的三種類(lèi)型及方法的切線方程的三種類(lèi)型及方法. .(1)(1)已知切點(diǎn)已知切點(diǎn)P(xP(x0 0,y,y0 0),),求求y=f(xy=f(x) )過(guò)點(diǎn)過(guò)點(diǎn)P P的切線方程的切線方程. .求出切線的斜率求出切線的斜率f(xf(x0 0),),由點(diǎn)斜式寫(xiě)出方程由點(diǎn)斜式寫(xiě)出方程. .(2)(2)已知切線的斜率為已知切線的斜率為k,k,求求y=f(xy=f(x) )的切線方程的切線方程. .設(shè)切點(diǎn)設(shè)切點(diǎn)P(xP(x0 0,y,y0 0),),通過(guò)方程通過(guò)方程k=f(xk=f(x0 0) )解得解得x x0 0, ,再由點(diǎn)斜式寫(xiě)出方程再由點(diǎn)斜式寫(xiě)出方程. .(3)(3)已知切線上一點(diǎn)已知切線上一點(diǎn)( (非切點(diǎn)非切點(diǎn)),),求求y=f(xy=f(x) )的切線方程的切線方程. .設(shè)切點(diǎn)設(shè)切點(diǎn)P(xP(x0 0,y,y0 0),),利用利用導(dǎo)數(shù)求得切線斜率導(dǎo)數(shù)求得切線斜率f(xf(x0 0),),然后由斜率公式求得切線斜率然后由斜率公式求得切線斜率, ,列方程列方程( (組組) )解解得得x x0 0, ,再由點(diǎn)斜式或兩點(diǎn)式寫(xiě)出方程再由點(diǎn)斜式或兩點(diǎn)式寫(xiě)出方程. .(2)(2)(20162016貴陽(yáng)二模貴陽(yáng)二模) )過(guò)點(diǎn)過(guò)點(diǎn)(-1,0)(-1,0)作拋物線作拋物線y=xy=x2 2+x+1+x+1的切線的切線, ,則其中一條切線則其中一條切線為為( () )(A)2x+y+2=0(A)2x+y+2=0(B)3x-y+3=0(B)3x-y+3=0(C)x+y+1=0(C)x+y+1=0(D)x-y+1=0(D)x-y+1=0解析解析: : (2) (2)因?yàn)橐驗(yàn)閥=2x+1,y=2x+1,設(shè)切點(diǎn)坐標(biāo)為設(shè)切點(diǎn)坐標(biāo)為(x(x0 0,y,y0 0),),則切線斜率為則切線斜率為2x2x0 0+1,+1,且且y y0 0= +x= +x0 0+1,+1,所以切線方程為所以切線方程為y- -xy- -x0 0-1=(2x-1=(2x0 0+1)(x-x+1)(x-x0 0),),又點(diǎn)又點(diǎn)(-1,0)(-1,0)在切線上在切線上, ,代入上式可解得代入上式可解得x x0 0=0=0或或x x0 0=-2,=-2,當(dāng)當(dāng)x x0 0=0=0時(shí)時(shí),y,y0 0=1,=1,當(dāng)當(dāng)x x0 0=-2=-2時(shí)時(shí),y,y0 0=3.=3.驗(yàn)證知驗(yàn)證知D D正確正確. .選選D.D.20 x20 x利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性熱點(diǎn)二熱點(diǎn)二考向考向1 1確定函數(shù)的單調(diào)性確定函數(shù)的單調(diào)性( (區(qū)間區(qū)間) )【例【例2 2】 ( (20162016甘肅河西部分高中聯(lián)考甘肅河西部分高中聯(lián)考) )已知函數(shù)已知函數(shù)f(xf(x)= -1.)= -1.(1)(1)試判斷函數(shù)試判斷函數(shù)f(xf(x) )的單調(diào)性的單調(diào)性; ;ln xx解解: :(1)(1)函數(shù)函數(shù)f(xf(x) )的定義域是的定義域是(0,+),(0,+),由已知得由已知得f(xf(x)= ,)= ,令令f(xf(x)=0)=0得得x=e,x=e,當(dāng)當(dāng)0 xe0 x0;)0;當(dāng)當(dāng)xexe時(shí)時(shí),f(x,f(x)0;)0,m0,求求f(x)f(x)在在m,2mm,2m上的最大值上的最大值. .考向考向2 2由函數(shù)的單調(diào)性求參數(shù)的范圍由函數(shù)的單調(diào)性求參數(shù)的范圍【方法技巧【方法技巧】 (1)(1)求函數(shù)的單調(diào)區(qū)間的方法求函數(shù)的單調(diào)區(qū)間的方法. .確定函數(shù)確定函數(shù)y=f(xy=f(x) )的定義域的定義域; ;求導(dǎo)數(shù)求導(dǎo)數(shù)y=f(xy=f(x););解不等式解不等式f(xf(x)0)0或或f(x)0,f(x)0 x0時(shí)時(shí),xf(x)-f(x,xf(x)-f(x)0,)0)0成立的成立的x x的取值范圍是的取值范圍是( () )(A)(-,-1)(0,1)(A)(-,-1)(0,1)(B)(-1,0)(1,+)(B)(-1,0)(1,+)(C)(-,-1)(-1,0)(C)(-,-1)(-1,0)(D)(0,1)(1,+)(D)(0,1)(1,+)利用導(dǎo)數(shù)研究函數(shù)的極值、最值利用導(dǎo)數(shù)研究函數(shù)的極值、最值熱點(diǎn)三熱點(diǎn)三【例【例4 4】 ( (20162016閩粵部分名校聯(lián)考閩粵部分名校聯(lián)考) )已知函數(shù)已知函數(shù)f(xf(x)= x)= x3 3- x- x2 2+cx+d+cx+d有極值有極值. .(1)(1)求求c c的取值范圍的取值范圍; ;1312(2)(2)若若f(xf(x) )在在x=2x=2處取得極值處取得極值, ,且當(dāng)且當(dāng)x0 x0時(shí)時(shí),f(x,f(x) d)0,f(x)0,f(x)單調(diào)遞增單調(diào)遞增;x(1,+);x(1,+)時(shí)時(shí),f(x,f(x)0,f(x)0,)0,求求a a的取值范圍的取值范圍. .當(dāng)當(dāng)a2a2時(shí)時(shí),f(1)=2-a0,f(1)=2-a0,故存在故存在x x0 0(1,+),f(x(1,+),f(x0 0)=0,)=0,此時(shí)函數(shù)此時(shí)函數(shù)f(xf(x) )在在(1,x(1,x0 0) )上單調(diào)遞減上單調(diào)遞減, ,在在(x(x0 0,+),+)上單調(diào)遞增上單調(diào)遞增, ,又又f(1)=0,f(1)=0,可得存在可得存在x x0 0(1,+),f(x(1,+),f(x0 0)0,)00求求a a的范圍的范圍. .而發(fā)現(xiàn)而發(fā)現(xiàn)f(1)=0f(1)=0是解決問(wèn)題的關(guān)鍵是解決問(wèn)題的關(guān)鍵. .3.3.在求解此類(lèi)問(wèn)題時(shí)注意數(shù)形結(jié)合思想的應(yīng)用在求解此類(lèi)問(wèn)題時(shí)注意數(shù)形結(jié)合思想的應(yīng)用. .