2015专题五:函数与导数
在解题中常用的有关结论(需要熟记):
(1)曲线()y f x =在0x x =处的切线的斜率等于0()f x ',切线方程为000()()()y f x x x f x '=-+ (2)若可导函数()y f x =在 0x x = 处取得极值,则0()0f x '=。反之,不成立。
(3)对于可导函数()f x ,不等式()f x '0>0<()的解集决定函数()f x 的递增(减)区间。
(4)函数()f x 在区间I 上递增(减)的充要条件是:x I ?∈()f x '0≥(0)≤恒成立 (5)函数()f x 在区间I 上不单调等价于()f x 在区间I 上有极值,则可等价转化为方程
()0f x '=在区间I 上有实根且为非二重根。(若()f x '为二次函数且I=R ,则有0?>)。
(6) ()f x 在区间I 上无极值等价于()f x 在区间在上是单调函数,进而得到()f x '0≥或
()f x '0≤在I 上恒成立
(7)若x I ?∈,()f x 0>恒成立,则min ()f x 0>; 若x I ?∈,()f x 0<恒成立,则max ()f x 0< (8)若0x I ?∈,使得0()f x 0>,则max ()f x 0>;若0x I ?∈,使得0()f x 0<,则min ()f x 0<. (9)设()f x 与()g x 的定义域的交集为D 若x ?∈D ()()f x g x >恒成立则有[]min
()()0f x g x ->
(10)若对11x I ?∈、22x I ∈ ,12()()f x g x >恒成立,则min max ()()f x g x >.
若对11x I ?∈,22x I ?∈,使得12()()f x g x >,则min min ()()f x g x >. 若对11x I ?∈,22x I ?∈,使得12()()f x g x <,则max max ()()f x g x <. (11)已知()f x 在区间1I 上的值域为A,,()g x 在区间2I 上值域为B ,
若对11x I ?∈,22x I ?∈,使得1()f x =2()g x 成立,则A B ?。
(12)若三次函数f(x)有三个零点,则方程()0f x '=有两个不等实根12x x 、,且极大值大
于0,极小值小于0. (13)证题中常用的不等式:
① ln 1(0)x x x ≤-> ② ln +1(1)x x x ≤>-() ③ 1x e x ≥+ ④ 1x
e
x -≥- ⑤
ln 1
(1)12
x x x x -<>+ ⑥
22
ln 11
(0)22x x x x
<->
考点一:导数几何意义:
角度一 求切线方程
1.(2014·洛阳统考)已知函数f (x )=3x +cos 2x +sin 2x ,a =f ′????π4,f ′(x )是f (x )的导函数,则过曲线y =x 3
上一点P (a ,b )的切线方程为( )
A .3x -y -2=0
B .4x -3y +1=0
C .3x -y -2=0或3x -4y +1=0
D .3x -y -2=0或4x -3y +1=0
解析:选A 由f (x )=3x +cos 2x +sin 2x 得f ′(x )=3-2sin 2x +2cos 2x ,则a =f ′????π4=3-2sin π2+2cos π2=1.由y =x 3得y ′=3x 2,过曲线y =x 3上一点P (a ,b )的切线的斜率k =3a 2=3×12=3.又b =a 3,则b =1,所以切点P 的坐标为(1,1),故过曲线y =x 3上的点P 的切线方程为y -1=3(x -1),即3x -y -2=0.
角度二 求切点坐标
2.(2013·辽宁五校第二次联考)曲线y =3ln x +x +2在点P 0处的切线方程为4x -y -1=0,则点P 0的坐标是( )
A .(0,1)
B .(1,-1)
C .(1,3)
D .(1,0)
解析:选C 由题意知y ′=3
x +1=4,解得x =1,此时4×1-y -1=0,解得y =3,∴点P 0的坐标是(1,3).
角度三 求参数的值
3.已知f (x )=ln x ,g (x )=12x 2+mx +7
2(m <0),直线l 与函数f (x ),g (x )的图像都相切,且与f (x )图像的切点为
(1,f (1)),则m 等于( )
A .-1
B .-3
C .-4
D .-2
解析:选D ∵f ′(x )=1
x ,
∴直线l 的斜率为k =f ′(1)=1, 又f (1)=0,
∴切线l 的方程为y =x -1.
g ′(x )=x +m ,设直线l 与g (x )的图像的切点为(x 0,y 0), 则有x 0+m =1,y 0=x 0-1,y 0=12x 20+mx 0+7
2,m <0,
于是解得m =-2,故选D.
考点二:判断函数单调性,求函数的单调区间。
[典例1]已知函数f (x )=x 2-e x 试判断f (x )的单调性并给予证明. 解:f (x )=x 2-e x ,f (x )在R 上单调递减,
f ′(x )=2x -e x ,只要证明f ′(x )≤0恒成立即可. 设
g (x )=f ′(x )=2x -e x ,则g ′(x )=2-e x , 当x =ln 2时,g ′(x )=0, 当x ∈(-∞,ln 2)时,g ′(x )>0, 当x ∈(ln 2,+∞)时,g ′(x )<0.
∴f ′(x )max =g (x )max =g (ln 2)=2ln 2-2<0, ∴f ′(x )<0恒成立, ∴f (x )在R 上单调递减.
[典例2] (2012·北京高考改编)已知函数f (x )=ax 2+1(a >0),g (x )=x 3+bx .
(1)若曲线y =f (x )与曲线y =g (x )在它们的交点(1,c )处具有公共切线,求a ,b 的值; (2)当a 2=4b 时,求函数f (x )+g (x )的单调区间. [解] (1)f ′(x )=2ax ,g ′(x )=3x 2+b , 由已知可得????
?
f (1)=a +1=c ,
g (1)=1+b =c ,
2a =3+b ,
解得a =b =3.
(2)令F (x )=f (x )+g (x )=x 3
+ax 2
+a 24x +1,F ′(x )=3x 2
+2ax +a 24,令F ′(x )=0,得x 1=-a 2,x 2=-a 6,
∵a >0,∴x 1 由F ′(x )>0得,x <-a 2或x >-a 6; 由F ′(x )<0得,-a 2 6 . ∴单调递增区间是????-∞,-a 2,????-a 6,+∞;单调递减区间为????-a 2,-a 6. [针对训练] (2013·重庆高考)设f (x ) =a (x -5)2+6ln x ,其中a ∈R ,曲线y =f (x )在点(1,f (1))处的切线与y 轴相交于点(0,6). (1)确定a 的值; (2)求函数f (x )的单调区间与极值. 解:(1)因为f (x )=a (x -5)2+6ln x ,故f ′(x )=2a (x -5)+6 x . 令x =1,得f (1)=16a ,f ′(1)=6-8a ,所以曲线y =f (x )在点(1,f (1))处的切线方程为y -16a =(6-8a )·(x -1),由点(0,6)在切线上可得6-16a =8a -6, 故a =12 . (2)由(1)知,f (x )=1 2(x -5)2+6ln x (x >0), f ′(x )=x -5+6x =(x -2)(x -3) x . 令f ′(x )=0,解得x 1=2,x 2=3. 当0 上为减函数. 由此可知f (x )在x =2处取得极大值f (2)=9 2+6ln 2,在x =3处取得极小值f (3)=2+6ln 3. 考点三:已知函数的单调性求参数的范围 [典例] (2014·山西诊断)已知函数f (x )=ln x -a 2x 2+ax (a ∈R). (1)当a =1时,求函数f (x )的单调区间; (2)若函数f (x )在区间(1,+∞)上是减函数,求实数a 的取值范围. [解] (1)当a =1时,f (x )=ln x -x 2+x ,其定义域是(0,+∞), f ′(x )=1 x -2x +1=-2x 2-x -1x , 令f ′(x )=0,即-2x 2-x -1x =0,解得x =-1 2或x =1. ∵x >0,∴x =1. 当0 ∴函数f (x )在区间(0,1)上单调递增,在区间(1,+∞)上单调递减. (2)显然函数f (x )=ln x -a 2x 2+ax 的定义域为(0,+∞), ∴f ′(x )=1x -2a 2 x +a =-2a 2x 2+ax +1x =-(2ax +1)(ax -1)x . ①当a =0时,f ′(x )=1 x >0, ∴f (x )在区间(1,+∞)上为增函数,不合题意. ②当a >0时,f ′(x )≤0(x >0)等价于(2ax +1)〃(ax -1)≥0(x >0),即x ≥1 a , 此时f (x )的单调递减区间为????1a ,+∞. 由????? 1a ≤1, a >0, 得a ≥1. ③当a <0时,f ′(x )≤0(x >0)等价于(2ax +1)〃(ax -1)≥0(x >0),即x ≥- 1 2a ,此时f (x )的单调递减区间为??? ?-12a ,+∞. 由????? -12a ≤1,a <0, 得a ≤-1 2 . 综上,实数a 的取值范围是????-∞,-1 2∪[1,+∞). [针对训练] (2014·荆州质检)设函数f (x )=13x 3-a 2x 2+bx +c ,曲线y =f (x )在点(0,f (0))处的切线方程为y =1. (1)求b ,c 的值; (2)若a >0,求函数f (x )的单调区间; (3)设函数g (x )=f (x )+2x ,且g (x )在区间(-2,-1)内存在单调递减区间,求实数a 的取值范围. 解:(1)f ′(x )=x 2-ax +b , 由题意得????? f (0)=1,f ′(0)=0,即????? c =1,b =0. (2)由(1)得,f ′(x )=x 2-ax =x (x -a )(a >0), 当x ∈(-∞,0)时,f ′(x )>0, 当x ∈(0,a )时,f ′(x )<0, 当x ∈(a ,+∞)时,f ′(x )>0. 所以函数f (x )的单调递增区间为(-∞,0),(a ,+∞),单调递减区间为(0,a ). (3)g ′(x )=x 2-ax +2, 依题意,存在x ∈(-2,-1),使不等式g ′(x )=x 2-ax +2<0成立, 即x ∈(-2,-1)时,a ???x +2 x max =-22, 当且仅当“x =2 x ”即x =-2时等号成立, 所以满足要求的a 的取值范围是(-∞,-22). 考点四:用导数解决函数的极值问题 [典例] (2013·福建高考节选)已知函数f (x )=x -1+a e x (a ∈R ,e 为自然对数的底数). (1)若曲线y =f (x )在点(1,f (1))处的切线平行于x 轴,求a 的值; (2)求函数f (x )的极值. [解] (1)由f (x )=x -1+a e x ,得f ′(x )=1-a e x . 又曲线y =f (x )在点(1,f (1))处的切线平行于x 轴, 得f ′(1)=0,即1-a e =0,解得a =e. (2)f ′(x )=1-a e x , ①当a ≤0时,f ′(x )>0,f (x )为(-∞,+∞)上的增函数,所以函数f (x )无极值. ②当a >0时,令f ′(x )=0,得e x =a ,即x =ln a . x ∈(-∞,ln a ),f ′(x )<0;x ∈(ln a ,+∞),f ′(x )>0, 所以f (x )在(-∞,ln a )上单调递减,在(ln a ,+∞)上单调递增, 故f (x )在x =ln a 处取得极小值, 且极小值为f (ln a )=ln a ,无极大值. 综上,当a ≤0时,函数f (x )无极值; 当a >0时,f (x )在x =ln a 处取得极小值ln a ,无极大值. [针对训练] 设f (x )=2x 3+ax 2+bx +1的导数为f ′(x ),若函数y =f ′(x )的图像关于直线x =-1 2对称,且f ′(1)=0. (1)求实数a ,b 的值; (2)求函数f (x )的极值. 解:(1)因为f (x )=2x 3+ax 2+bx +1, 故f ′(x )=6x 2+2ax +b , 从而f ′(x )=6????x +a 62+b -a 2 6, 即y =f ′(x )关于直线x =-a 6对称. 从而由题设条件知-a 6=-1 2,即a =3. 又由于f ′(1)=0,即6+2a +b =0, 得b =-12. (2)由(1)知f (x )=2x 3+3x 2-12x +1, 所以f ′(x )=6x 2+6x -12=6(x -1)(x +2), 令f ′(x )=0, 即6(x -1)(x +2)=0, 解得x =-2或x =1, 当x ∈(-∞,-2)时,f ′(x )>0, 即f (x )在(-∞,-2)上单调递增; 当x ∈(-2,1)时,f ′(x )<0, 即f (x )在(-2,1)上单调递减; 当x ∈(1,+∞)时,f ′(x )>0, 即f (x )在(1,+∞)上单调递增. 从而函数f (x )在x =-2处取得极大值f (-2)=21, 在x =1处取得极小值f (1)=-6. 考点五 运用导数解决函数的最值问题 [典例] 已知函数f (x )=ln x -ax (a ∈R). (1)求函数f (x )的单调区间; (2)当a >0时,求函数f (x )在[1,2]上的最小值. [解] (1)f ′(x )=1 x -a (x >0), ①当a ≤0时,f ′(x )=1 x -a >0, 即函数f (x )的单调增区间为(0,+∞). ②当a >0时,令f ′(x )=1x -a =0,可得x =1 a , 当0 a 时,f ′(x )=1-ax x >0; 当x >1 a 时,f ′(x )=1-ax x <0, 故函数f (x )的单调递增区间为??? ?0,1 a , 单调递减区间为??? ?1 a ,+∞. (2)①当1 a ≤1,即a ≥1时,函数f (x )在区间[1,2]上是减函数,∴f (x )的最小值是f (2)=ln 2-2a . ②当1a ≥2,即0 2 时,函数f (x )在区间[1,2]上是增函数,∴f (x )的最小值是f (1)=-a . ③当1<1a <2,即1 2 2 当0 设函数f (x )=a ln x -bx 2(x >0),若函数f (x )在x =1处与直线y =-1 2相切, (1)求实数a ,b 的值; (2)求函数f (x )在???? 1e ,e 上的最大值. 解:(1)f ′(x )=a x -2bx , ∵函数f (x )在x =1处与直线y =-1 2相切, ∴????? f ′(1)=a -2b =0,f (1)=-b =-1 2,解得????? a =1, b =12. (2)f (x )=ln x -12x 2,f ′(x )=1 x -x =1-x 2x , ∵当1e ≤x ≤e 时,令f ′(x )>0得1 e ≤x <1; 令f ′(x )<0,得1 2. 考点六:用导数解决函数极值、最值问题 [典例] (2013·北京丰台高三期末)已知函数f (x )=ax 2+bx +c e x (a >0)的导函数y = f ′(x )的两个零点为-3和0. (1)求f (x )的单调区间; (2)若f (x )的极小值为-e 3,求f (x )在区间[-5,+∞)上的最大值. [解] (1)f ′(x )=(2ax +b )e x -(ax 2+bx +c )e x (e x )2 =-ax 2+(2a -b )x +b -c e x , 令g (x )=-ax 2+(2a -b )x +b -c , 因为e x >0,所以y =f ′(x )的零点就是g (x )=-ax 2+(2a -b )x +b -c 的零点,且f ′(x )与g (x )符号相同. 又因为a >0,所以-3 当x <-3或x >0时,g (x )<0,即f ′(x )<0,所以f (x )的单调增区间是(-3,0),单调减区间是(-∞,-3),(0,+∞). (2)由(1)知,x =-3是f (x )的极小值点,所以有 ????? 9a -3b +c e -3 =-e 3 ,g (0)=b -c =0,g (-3)=-9a -3(2a -b )+b -c =0, 解得a =1,b =5,c =5, 所以f (x )=x 2+5x +5e x . 因为f (x )的单调增区间是(-3,0),单调减区间是(-∞,-3),(0,+∞), 所以f (0)=5为函数f (x )的极大值, 故f (x )在区间[-5,+∞)上的最大值取f (-5)和f (0)中的最大者. 而f (-5)= 5e -5=5e 5>5=f (0),所以函数f (x )在区间[-5,+∞)上的最大值是5e 5. [针对训练] 已知函数f (x )=x 3+ax 2+bx +c ,曲线y =f (x )在点x =1处的切线为l :3x -y +1=0,若x =2 3时,y =f (x )有极 值. (1)求a ,b ,c 的值; (2)求y =f (x )在[-3,1]上的最大值和最小值. 解:(1)由f (x )=x 3+ax 2+bx +c ,得f ′(x )=3x 2+2ax +b .当x =1时,切线l 的斜率为3,可得2a +b =0,① 当x =2 3时,y =f (x )有极值,则f ′????23=0,可得4a +3b +4=0,② 由①②,解得a =2,b =-4.由于切点的横坐标为1, 所以f (1)=4. 所以1+a +b +c =4.所以c =5. (2)由(1),可得f (x )=x 3+2x 2-4x +5,f ′(x )=3x 2+4x -4.令f ′(x )=0,解之,得x 1=-2,x 2=23. 当x 变化时,f ′(x ),f (x )的取值及变化情况如下表所示: x -3 (-3,-2) -2 ? ???-2,23 2 3 ??? ?23,1 1 f ′(x ) + + 0 - 0 + + f (x ) 8 13 9527 4 所以y =f (x )在[-3,1]上的最大值为13,最小值为95 27. 考点七:利用导数研究恒成立问题及参数求解 [典例] (2013·全国卷Ⅰ)设函数f (x )=x 2+ax +b ,g (x )=e x (cx +d ).若曲线y =f (x )和曲线y =g (x )都过点P (0,2),且在点P 处有相同的切线y =4x +2. (1)求a ,b ,c ,d 的值; (2)若x ≥-2时,f (x )≤kg (x ),求k 的取值范围. [解] (1)由已知得f (0)=2,g (0)=2, f ′(0)=4,g ′(0)=4. 而f ′(x )=2x +a ,g ′(x )=e x (cx +d +c ),故b =2,d =2,a =4,d +c =4. 从而a =4,b =2,c =2,d =2. (2)由(1)知,f (x )=x 2+4x +2,g (x )=2e x (x +1). 设函数F (x )=kg (x )-f (x )=2k e x (x +1)-x 2-4x -2, 则F ′(x )=2k e x (x +2)-2x -4=2(x +2)(k e x -1). 由题设可得F (0)≥0,即k ≥1. 令F ′(x )=0得x 1=-ln k ,x 2=-2. (ⅰ)若1≤k <e 2,则-2<x 1≤0.从而当x ∈(-2,x 1)时,F ′(x )<0;当x ∈(x 1,+∞)时,F ′(x )>0,即F (x )在(-2,x 1)上单调递减,在(x 1,+∞)上单调递增,故F (x )在[-2,+∞)上的最小值为F (x 1).而F (x 1)=2x 1+2 -x 21-4x 1-2=-x 1(x 1+2)≥0. 故当x ≥-2时,F (x )≥0,即f (x )≤kg (x )恒成立. (ⅱ)若k =e 2,则F ′(x )=2e 2(x +2)(e x -e - 2).从而当x >-2时,F ′(x )>0,即F (x )在(-2,+∞)上单调递 增, 而F (-2)=0,故当x ≥-2时,F (x )≥0,即f (x )≤kg (x )恒成立. (ⅲ)若k >e 2,则F (-2)=-2k e - 2+2=-2e - 2〃(k -e 2)<0.从而当x ≥-2时,f (x )≤kg (x )不可能恒成立. 综上,k 的取值范围是[1,e 2]. [针对训练] 设函数f (x )=1 2x 2+e x -x e x . (1)求f (x )的单调区间; (2)若当x ∈[-2,2]时,不等式f (x )>m 恒成立,求实数m 的取值范围. 解:(1)函数f (x )的定义域为(-∞,+∞), ∵f ′(x )=x +e x -(e x +x e x )=x (1-e x ), 若x =0,则f ′(x )=0; 若x <0,则1-e x >0,所以f ′(x )<0; 若x >0,则1-e x <0,所以f ′(x )<0. ∴f (x )在(-∞,+∞)上为减函数, 即f (x )的单调减区间为(-∞,+∞). (2)由(1)知,f (x )在[-2,2]上单调递减. 故[f (x )]min =f (2)=2-e 2, ∴m <2-e 2时,不等式f (x )>m 恒成立. 故m 的取值范围为(-∞,2-e 2). 考点八、利用导数证明不等式问题 [典例] (2013·河南省三市调研)已知函数f (x )=ax -e x (a >0). (1)若a =1 2,求函数f (x )的单调区间; (2)当1≤a ≤1+e 时,求证:f (x )≤x . [解] (1)当a =12时,f (x )=1 2x -e x . f ′(x )=1 2-e x ,令f ′(x )=0,得x =-ln 2. 当x <-ln 2时,f ′(x )>0; 当x >-ln 2时,f ′(x )<0, ∴函数f (x )的单调递增区间为(-∞,-ln 2),单调递减区间为(-ln 2,+∞). (2)证明:法一:令F (x )=x -f (x )=e x -(a -1)x , (ⅰ)当a =1时,F (x )=e x >0, ∴f (x )≤x 成立.