2015高中数学一轮复习04函数的定义域和值域
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求函数定义域和值域专题1 知识点拨一、函数的定义域及求法1、分式的分母≠0;偶次方根的被开方数≥0;2、对数函数的真数>0;对数函数的底数>0且≠1;3、正切函数:x ≠kπ+ π/2 ,k∈Z;4、一次函数、二次函数、指数函数的定义域为R;5、复合函数定义域的求法:取交集及分类讨论;6、抽象函数定义域的求法;二、含分式的函数在求含分式的函数的定义域时,要注意两点:(1)分式的分母一定不能为0;(2)绝对不能先化简后求函数定义域。
例1:求函数f(x)=211xx-+的定义域.三、含偶次根式的函数注意(1)求含偶次根式的函数的定义域时,注意偶次根式的被开方数不小于0,通过求不等式来求其定义域;(2)在研究函数时,常常用到区间的概念,它是数学中常用的术语和符号,注意区间的开闭情况.例2 :求函数y =3-ax (a 为不等于0的常数)的定义域.四、复合型函数注意 函数是由一些基本初等函数通过四则运算而得到的,则它的定义域是各基本函数定义域的交集,通过列不等式组来实现.例3:求函数y =23-x +30323-+x x )(的定义域.练习1、求下列函数的定义域。
⑴y=xx -||1 ⑵y=3102++x x (3)y=||11x - (4)y=2121---x x (5)2143)(2-+--=x x x x f五、抽象函数1.已知)(x f 的定义域,求复合函数()][x g f 的定义域由复合函数的定义我们可知,要构成复合函数,则内层函数的值域必须包含于外层函数的定义域之中,因此可得其方法为:若)(x f 的定义域为()b a x ,∈,求出)]([x g f 中b x g a <<)(的解x 的范围,即为)]([x g f 的定义域。
2.已知复合函数()][x g f 的定义域,求)(x f 的定义域方法是:若()][x g f 的定义域为()b a x ,∈,则由b x a <<确定)(x g 的范围即为)(x f 的定义域。
精品基础教育教学资料,仅供参考,需要可下载使用!第二章函数的概念与基本初等函数Ⅰ第一节函数及其表示一、基础知识1.函数与映射的概念2.函数的有关概念(1)函数的定义域、值域:在函数y=f(x),x∈A中,x叫做自变量,x的取值范围A叫做函数的定义域;与x的值相对应的y值叫做函数值,函数值的集合{f(x)|x∈A}叫做函数的值域.求函数定义域的策略(1)确定函数的定义域常从解析式本身有意义,或从实际出发.(2)如果函数y=f(x)是用表格给出,则表格中x的集合即为定义域.(3)如果函数y=f(x)是用图象给出,则图象在x轴上的投影所覆盖的x的集合即为定义域.(2)函数的三要素:定义域、值域和对应关系.(3)相等函数:如果两个函数的定义域和对应关系完全一致,则这两个函数相等,这是判断两函数相等的依据.两函数值域与对应关系相同时,两函数不一定相同.(4)函数的表示法:表示函数的常用方法有:解析法、图象法、列表法.3.分段函数若函数在其定义域内,对于定义域内的不同取值区间,有着不同的对应关系,这样的函数通常叫做分段函数.关于分段函数的3个注意(1)分段函数虽然由几个部分构成,但它表示同一个函数.(2)分段函数的定义域是各段定义域的并集,值域是各段值域的并集.(3)各段函数的定义域不可以相交.考点一函数的定义域[典例] (1)(2019·长春质检)函数y =ln (1-x )x +1+1x 的定义域是( )A .[-1,0)∪(0,1)B .[-1,0)∪(0,1]C .(-1,0)∪(0,1]D .(-1,0)∪(0,1)(2)已知函数f (x )的定义域为(-1,0),则函数f (2x +1)的定义域为( ) A .(-1,1) B.⎝⎛⎭⎫-1,-12 C .(-1,0)D.⎝⎛⎭⎫12,1[解析] (1)由题意得⎩⎪⎨⎪⎧1-x >0,x +1>0,x ≠0,解得-1<x <0或0<x <1.所以原函数的定义域为(-1,0)∪(0,1).(2)令u =2x +1,由f (x )的定义域为(-1,0),可知-1<u <0,即-1<2x +1<0, 得-1<x <-12.[答案] (1)D (2)B [解题技法]1.使函数解析式有意义的一般准则 (1)分式中的分母不为0; (2)偶次根式的被开方数非负; (3)y =x 0要求x ≠0;(4)对数式中的真数大于0,底数大于0且不等于1; (5)正切函数y =tan x ,x ≠k π+π2(k ∈Z);(6)实际问题中除考虑函数解析式有意义外,还应考虑实际问题本身的要求. 2.抽象函数的定义域问题(1)若已知函数f (x )的定义域为[a ,b ],其复合函数f (g (x ))的定义域由不等式a ≤g (x )≤b 求出;(2)若已知函数f (g (x ))的定义域为[a ,b ],则f (x )的定义域为g (x )在x ∈[a ,b ]上的值域.[题组训练]1.函数f (x )=1ln (x +1)+4-x 2的定义域为( )A .[-2,0)∪(0,2]B .(-1,0)∪(0,2]C .[-2,2]D .(-1,2]解析:选B 由⎩⎪⎨⎪⎧x +1>0,ln (x +1)≠0,4-x 2≥0,得-1<x ≤2,且x ≠0.2.若函数y =f (x )的定义域是[1,2 019],则函数g (x )=f (x +1)x -1的定义域是________________.解析:因为y =f (x )的定义域是[1,2 019],所以若g (x )有意义,应满足⎩⎪⎨⎪⎧1≤x +1≤2 019,x -1≠0,所以0≤x ≤2 018,且x ≠1.因此g (x )的定义域是{x |0≤x ≤2 018,且x ≠1}. 答案:{x |0≤x ≤2 018,且x ≠1}考点二 求函数的解析式[典例] (1)已知二次函数f (2x +1)=4x 2-6x +5,求f (x ); (2)已知函数f (x )满足f (-x )+2f (x )=2x ,求f (x ). [解] (1)法一:待定系数法因为f (x )是二次函数,所以设f (x )=ax 2+bx +c (a ≠0),则f (2x +1)=a (2x +1)2+b (2x +1)+c =4ax 2+(4a +2b )x +a +b +c .因为f (2x +1)=4x 2-6x +5, 所以⎩⎪⎨⎪⎧4a =4,4a +2b =-6,a +b +c =5,解得⎩⎪⎨⎪⎧a =1,b =-5,c =9,所以f (x )=x 2-5x +9(x ∈R). 法二:换元法令2x +1=t (t ∈R),则x =t -12,所以f (t )=4⎝⎛⎭⎫t -122-6·t -12+5=t 2-5t +9(t ∈R),所以f (x )=x 2-5x +9(x ∈R). 法三:配凑法因为f (2x +1)=4x 2-6x +5=(2x +1)2-10x +4=(2x +1)2-5(2x +1)+9, 所以f (x )=x 2-5x +9(x ∈R).(2)解方程组法由f (-x )+2f (x )=2x , ① 得f (x )+2f (-x )=2-x ,② ①×2-②,得3f (x )=2x +1-2-x . 即f (x )=2x +1-2-x3.故f (x )的解析式是f (x )=2x +1-2-x3(x ∈R).[解题技法] 求函数解析式的4种方法及适用条件 (1)待定系数法先设出含有待定系数的解析式,再利用恒等式的性质,或将已知条件代入,建立方程(组),通过解方程(组)求出相应的待定系数.(2)换元法对于形如y =f (g (x ))的函数解析式,令t =g (x ),从中求出x =φ(t ),然后代入表达式求出f (t ),再将t 换成x ,得到f (x )的解析式,要注意新元的取值范围.(3)配凑法由已知条件f (g (x ))=F (x ),可将F (x )改写成关于g (x )的表达式,然后以x 替代g (x ),便得f (x )的解析式.(4)解方程组法已知关于f (x )与f ⎝⎛⎭⎫1x 或f (-x )的表达式,可根据已知条件再构造出另外一个等式组成方程组,通过解方程组求出f (x ).[提醒] 由于函数的解析式相同,定义域不同,则为不相同的函数,因此求函数的解析式时,如果定义域不是R ,一定要注明函数的定义域.[题组训练]1.[口诀第2句]已知f (x )是二次函数,且f (0)=0,f (x +1)=f (x )+x +1,则f (x )=________________.解析:设f (x )=ax 2+bx +c (a ≠0), 由f (0)=0,知c =0,f (x )=ax 2+bx . 又由f (x +1)=f (x )+x +1,得a (x +1)2+b (x +1)=ax 2+bx +x +1, 即ax 2+(2a +b )x +a +b =ax 2+(b +1)x +1,所以⎩⎪⎨⎪⎧2a +b =b +1,a +b =1,解得a =b =12.所以f (x )=12x 2+12x (x ∈R).答案:12x 2+12x (x ∈R)2.[口诀第3句]已知f ⎝⎛⎭⎫2x +1=lg x ,则f (x )=________________.解析:令2x +1=t ,得x =2t -1,则f (t )=lg 2t -1,又x >0,所以t >1,故f (x )的解析式是f (x )=lg2x -1(x >1). 答案:lg2x -1(x >1) 3.[口诀第4句]已知f (x )满足2f (x )+f ⎝⎛⎭⎫1x =3x ,则f (x )=________. 解析:∵2f (x )+f ⎝⎛⎭⎫1x =3x ,①把①中的x 换成1x ,得2f ⎝⎛⎭⎫1x +f (x )=3x.② 联立①②可得⎩⎨⎧2f (x )+f ⎝⎛⎭⎫1x =3x ,2f ⎝⎛⎭⎫1x +f (x )=3x,解此方程组可得f (x )=2x -1x(x ≠0).答案:2x -1x (x ≠0)考点三 分段函数考法(一) 求函数值[典例] (2019·石家庄模拟)已知f (x )=⎩⎪⎨⎪⎧log 3x ,x >0,a x +b ,x ≤0(0<a <1),且f (-2)=5,f (-1)=3,则f (f (-3))=( )A .-2B .2C .3D .-3[解析] 由题意得,f (-2)=a -2+b =5,① f (-1)=a -1+b =3,②联立①②,结合0<a <1,得a =12,b =1,所以f (x )=⎩⎪⎨⎪⎧log 3x ,x >0,⎝⎛⎭⎫12x +1,x ≤0,则f (-3)=⎝⎛⎭⎫12-3+1=9,f (f (-3))=f (9)=log 39=2. [答案] B[解题技法] 求分段函数的函数值的策略(1)求分段函数的函数值时,要先确定要求值的自变量属于哪一区间,然后代入该区间对应的解析式求值;(2)当出现f (f (a ))的形式时,应从内到外依次求值;(3)当自变量的值所在区间不确定时,要分类讨论,分类标准应参照分段函数不同段的端点.考法(二) 求参数或自变量的值(或范围)[典例] (2018·全国卷Ⅰ)设函数f (x )=⎩⎪⎨⎪⎧2-x ,x ≤0,1,x >0,则满足f (x +1)<f (2x )的x 的取值范围是( )A .(-∞,-1]B .(0,+∞)C .(-1,0)D .(-∞,0)[解析] 法一:分类讨论法①当⎩⎪⎨⎪⎧x +1≤0,2x ≤0,即x ≤-1时,f (x +1)<f (2x ),即为2-(x +1)<2-2x,即-(x +1)<-2x ,解得x <1. 因此不等式的解集为(-∞,-1].②当⎩⎪⎨⎪⎧x +1≤0,2x >0时,不等式组无解.③当⎩⎪⎨⎪⎧x +1>0,2x ≤0,即-1<x ≤0时,f (x +1)<f (2x ),即为1<2-2x,解得x <0.因此不等式的解集为(-1,0).④当⎩⎪⎨⎪⎧x +1>0,2x >0,即x >0时,f (x +1)=1,f (2x )=1,不合题意.综上,不等式f (x +1)<f (2x )的解集为(-∞,0). 法二:数形结合法∵f (x )=⎩⎪⎨⎪⎧2-x ,x ≤0,1,x >0,∴函数f (x )的图象如图所示. 结合图象知,要使f (x +1)<f (2x ), 则需⎩⎪⎨⎪⎧x +1<0,2x <0,2x <x +1或⎩⎪⎨⎪⎧x +1≥0,2x <0,∴x <0,故选D. [答案] D[解题技法]已知函数值(或范围)求自变量的值(或范围)的方法(1)根据每一段的解析式分别求解,但要注意检验所求自变量的值(或范围)是否符合相应段的自变量的取值范围,最后将各段的结果合起来(求并集)即可;(2)如果分段函数的图象易得,也可以画出函数图象后结合图象求解.[题组训练]1.设f (x )=⎩⎨⎧x ,0<x <1,2(x -1),x ≥1,若f (a )=f (a +1),则f ⎝⎛⎭⎫1a =( ) A .2 B .4 C .6D .8解析:选C 当0<a <1时,a +1>1,f (a )=a ,f (a +1)=2(a +1-1)=2a , ∵f (a )=f (a +1),∴a =2a , 解得a =14或a =0(舍去).∴f ⎝⎛⎭⎫1a =f (4)=2×(4-1)=6.当a ≥1时,a +1≥2,f (a )=2(a -1),f (a +1)=2(a +1-1)=2a , ∵f (a )=f (a +1),∴2(a -1)=2a ,无解. 综上,f ⎝⎛⎭⎫1a =6.2.已知函数f (x )=⎩⎪⎨⎪⎧2x ,x ≤1,f (x -1),x >1,则f (f (3))=________.解析:由题意,得f (3)=f (2)=f (1)=21=2,∴f (f (3))=f (2)=2. 答案:23.(2017·全国卷Ⅲ)设函数f (x )=⎩⎪⎨⎪⎧x +1,x ≤0,2x ,x >0,则满足f (x )+f ⎝⎛⎭⎫x -12>1的x 的取值范围是________.解析:由题意知,可对不等式分x ≤0,0<x ≤12,x >12讨论.①当x ≤0时,原不等式为x +1+x +12>1,解得x >-14,故-14<x ≤0.②当0<x ≤12时,原不等式为2x +x +12>1,显然成立.③当x >12时,原不等式为2x +2x -12>1,显然成立.综上可知,所求x 的取值范围是⎝⎛⎭⎫-14,+∞. 答案:⎝⎛⎭⎫-14,+∞ 4.设函数f (x )=⎩⎪⎨⎪⎧⎝⎛⎭⎫12x -7,x <0,x ,x ≥0,若f (a )<1,则实数a 的取值范围是____________.解析:若a <0,则f (a )<1⇔⎝⎛⎭⎫12a-7<1⇔⎝⎛⎭⎫12a <8,解得a >-3,故-3<a <0; 若a ≥0,则f (a )<1⇔a <1,解得a <1,故0≤a <1. 综上可得-3<a <1. 答案:(-3,1)[课时跟踪检测]1.下列所给图象是函数图象的个数为( )A .1B .2C .3D .4解析:选B ①中当x >0时,每一个x 的值对应两个不同的y 值,因此不是函数图象;②中当x =x 0时,y 的值有两个,因此不是函数图象;③④中每一个x 的值对应唯一的y 值,因此是函数图象.故选B.2.函数f (x )=2x -1+1x -2的定义域为( ) A .[0,2)B .(2,+∞)C .[0,2)∪(2,+∞)D .(-∞,2)∪(2,+∞)解析:选C 由题意得⎩⎪⎨⎪⎧2x -1≥0,x -2≠0,解得x ≥0,且x ≠2.3.已知f ⎝⎛⎭⎫12x -1=2x -5,且f (a )=6,则a 等于( ) A.74 B .-74C.43D .-43解析:选A 令t =12x -1,则x =2t +2,f (t )=2(2t +2)-5=4t -1,则4a -1=6,解得a =74.4.(2019·贵阳检测)下列函数中,同一个函数的定义域与值域相同的是( ) A .y =x -1 B .y =ln x C .y =13x -1D .y =x +1x -1解析:选D 对于A ,定义域为[1,+∞),值域为[0,+∞),不满足题意;对于B ,定义域为(0,+∞),值域为R ,不满足题意;对于C ,定义域为(-∞,0)∪(0,+∞),值域为(-∞,-1)∪(0,+∞),不满足题意;对于D ,y =x +1x -1=1+2x -1,定义域为(-∞,1)∪(1,+∞),值域也是(-∞,1)∪(1,+∞).5.(2018·福建期末)已知函数f (x )=⎩⎪⎨⎪⎧log 2x +a ,x >0,4x -2-1,x ≤0.若f (a )=3,则f (a -2)=( )A .-1516B .3C .-6364或3D .-1516或3解析:选A 当a >0时,若f (a )=3,则log 2a +a =3,解得a =2(满足a >0);当a ≤0时,若f (a )=3,则4a -2-1=3,解得a =3,不满足a ≤0,所以舍去.于是,可得a =2.故f (a -2)=f (0)=4-2-1=-1516.6.已知函数y =f (2x -1)的定义域是[0,1],则函数f (2x +1)log 2(x +1)的定义域是( )A .[1,2]B .(-1,1] C.⎣⎡⎦⎤-12,0 D .(-1,0)解析:选D 由f (2x -1)的定义域是[0,1], 得0≤x ≤1,故-1≤2x -1≤1, ∴f (x )的定义域是[-1,1], ∴要使函数f (2x +1)log 2(x +1)有意义,需满足⎩⎪⎨⎪⎧-1≤2x +1≤1,x +1>0,x +1≠1,解得-1<x <0.7.下列函数中,不满足f (2 018x )=2 018f (x )的是( ) A .f (x )=|x | B .f (x )=x -|x | C .f (x )=x +2D .f (x )=-2x解析:选C 若f (x )=|x |,则f (2 018x )=|2 018x |=2 018|x |=2 018f (x );若f (x )=x -|x |,则f (2 018x )=2 018x -|2 018x |=2 018(x -|x |)=2 018f (x );若f (x )=x +2,则f (2 018x )=2 018x +2,而2 018f (x )=2 018x +2 018×2,故f (x )=x +2不满足f (2 018x )=2 018f (x );若f (x )=-2x ,则f (2 018x )=-2×2 018x =2 018×(-2x )=2 018f (x ).故选C.8.已知具有性质:f ⎝⎛⎭⎫1x =-f (x )的函数,我们称为满足“倒负”变换的函数,下列函数: ①f (x )=x -1x ;②f (x )=x +1x ;③f (x )=⎩⎪⎨⎪⎧x ,0<x <1,0,x =1,-1x ,x >1.其中满足“倒负”变换的函数是( ) A .①② B .①③ C .②③D .①解析:选B 对于①,f (x )=x -1x ,f ⎝⎛⎭⎫1x =1x-x =-f (x ),满足题意;对于②,f ⎝⎛⎭⎫1x =1x +x =f (x ),不满足题意;对于③,f ⎝⎛⎭⎫1x =⎩⎪⎨⎪⎧1x ,0<1x<1,0,1x =1,-x ,1x >1,即f ⎝⎛⎭⎫1x =⎩⎪⎨⎪⎧1x,x >1,0,x =1,-x ,0<x <1,故f ⎝⎛⎭⎫1x =-f (x ),满足题意.综上可知,满足“倒负”变换的函数是①③.9.(2019·青岛模拟)函数y =ln ⎝⎛⎭⎫1+1x +1-x 2的定义域为________. 解析:由⎩⎪⎨⎪⎧1+1x >0,1-x 2≥0⇒⎩⎪⎨⎪⎧x <-1或x >0,-1≤x ≤1⇒0<x ≤1.所以该函数的定义域为(0,1]. 答案:(0,1]10.(2019·益阳、湘潭调研)若函数f (x )=⎩⎨⎧lg (1-x ),x <0,-2x ,x ≥0,则f (f (-9))=________.解析:∵函数f (x )=⎩⎨⎧lg (1-x ),x <0,-2x ,x ≥0,∴f (-9)=lg 10=1,∴f (f (-9))=f (1)=-2.答案:-211.(2018·张掖一诊)已知函数f (x )=⎩⎪⎨⎪⎧2x ,x >0,x +1,x ≤0,若f (a )+f (1)=0,则实数a 的值等于________.解析:∵f (1)=2,且f (1)+f (a )=0,∴f (a )=-2<0,故a ≤0. 依题知a +1=-2,解得a =-3. 答案:-312.已知f (x )=⎩⎪⎨⎪⎧12x +1,x ≤0,-(x -1)2,x >0,使f (x )≥-1成立的x 的取值范围是________.解析:由题意知⎩⎪⎨⎪⎧x ≤0,12x +1≥-1或⎩⎪⎨⎪⎧x >0,-(x -1)2≥-1, 解得-4≤x ≤0或0<x ≤2, 故所求x 的取值范围是[-4,2]. 答案:[-4,2]13.设函数f (x )=⎩⎪⎨⎪⎧ax +b ,x <0,2x ,x ≥0,且f (-2)=3,f (-1)=f (1).(1)求函数f (x )的解析式;(2)在如图所示的直角坐标系中画出f (x )的图象.解:(1)由f (-2)=3,f (-1)=f (1),得⎩⎪⎨⎪⎧-2a +b =3,-a +b =2,解得⎩⎪⎨⎪⎧ a =-1,b =1,所以f (x )=⎩⎪⎨⎪⎧-x +1,x <0,2x ,x ≥0.(2)函数f (x )的图象如图所示.第二节函数的单调性与最值一、基础知识1.增函数、减函数定义:设函数f(x)的定义域为I:(1)增函数:如果对于定义域I内某个区间D上的任意两个自变量的值x1,x2,当x1<x2时,都有f(x1)<f(x2),那么就说函数f(x)在区间D上是增函数.(2)减函数:如果对于定义域I内某个区间D上的任意两个自变量的值x1,x2,当x1<x2时,都有f(x1)>f(x2),那么就说函数f(x)在区间D上是减函数.增(减)函数定义中的x1,x2的三个特征一是任意性;二是有大小,即x1<x2(x1>x2);三是同属于一个单调区间,三者缺一不可.2.单调性、单调区间若函数y=f(x)在区间D上是增函数或减函数,则称函数y=f(x)在这一区间具有(严格的)单调性,区间D叫做函数y=f(x)的单调区间.有关单调区间的两个防范(1)单调区间只能用区间表示,不能用不等式表示.(2)有多个单调区间应分别写,不能用符号“∪”连接,也不能用“或”连接,只能用“逗号”或“和”连接.3.函数的最值设函数y=f(x)的定义域为I,如果存在实数M满足:(1)对于任意的x∈I,都有f(x)≤M或f(x)≥M.(2)存在x0∈I,使得f(x0)=M.那么,我们称M是函数y=f(x)的最大值或最小值.函数最值存在的两条结论(1)闭区间上的连续函数一定存在最大值和最小值.当函数在闭区间上单调时最值一定在端点取到.(2)开区间上的“单峰”函数一定存在最大(小)值.二、常用结论在公共定义域内:(1)函数f(x)单调递增,g(x)单调递增,则f(x)+g(x)是增函数;(2)函数f (x )单调递减,g (x )单调递减,则f (x )+g (x )是减函数; (3)函数f (x )单调递增,g (x )单调递减,则f (x )-g (x )是增函数; (4)函数f (x )单调递减,g (x )单调递增,则f (x )-g (x )是减函数;(5)若k >0,则kf (x )与f (x )单调性相同;若k <0,则kf (x )与f (x )单调性相反; (6)函数y =f (x )(f (x )>0)在公共定义域内与y =-f (x ),y =1f (x )的单调性相反;(7)复合函数y =f [g (x )]的单调性与y =f (u )和u =g (x )的单调性有关.简记:“同增异减”.考点一 确定函数的单调性(区间))[典例] (1)求函数f (x )=-x 2+2|x |+1的单调区间. (2)试讨论函数f (x )=ax x -1(a ≠0)在(-1,1)上的单调性.[解] (1)易知f (x )=⎩⎪⎨⎪⎧-x 2+2x +1,x ≥0,-x 2-2x +1,x <0=⎩⎪⎨⎪⎧-(x -1)2+2,x ≥0,-(x +1)2+2,x <0. 画出函数图象如图所示,可知单调递增区间为(-∞,-1]和[0,1],单调递减区间为[-1,0]和[1,+∞).(2)法一:定义法 设-1<x 1<x 2<1, f (x )=a ⎝⎛⎭⎪⎫x -1+1x -1=a ⎝⎛⎭⎫1+1x -1,则f (x 1)-f (x 2)=a ⎝⎛⎭⎫1+1x 1-1-a ⎝⎛⎭⎫1+1x 2-1=a (x 2-x 1)(x 1-1)(x 2-1).由于-1<x 1<x 2<1,所以x 2-x 1>0,x 1-1<0,x 2-1<0, 故当a >0时,f (x 1)-f (x 2)>0,即f (x 1)>f (x 2), 函数f (x )在(-1,1)上单调递减;当a <0时,f (x 1)-f (x 2)<0,即f (x 1)<f (x 2), 函数f (x )在(-1,1)上单调递增. 法二:导数法f ′(x )=(ax )′(x -1)-ax (x -1)′(x -1)2=a (x -1)-ax (x -1)2=-a(x -1)2. 当a >0时,f ′(x )<0,函数f (x )在(-1,1)上单调递减; 当a <0时,f ′(x )>0,函数f (x )在(-1,1)上单调递增.[解题技法] 判断函数单调性和求单调区间的方法(1)定义法:一般步骤为设元―→作差―→变形―→判断符号―→得出结论.(2)图象法:如果f (x )是以图象形式给出的,或者f (x )的图象易作出,则可由图象的上升或下降确定单调性.(3)导数法:先求导数,利用导数值的正负确定函数的单调性及区间.(4)性质法:对于由基本初等函数的和、差构成的函数,根据各初等函数的增减性及复合函数单调性性质进行判断;复合函数单调性,可用同增异减来确定.[题组训练]1.下列函数中,满足“∀x 1,x 2∈(0,+∞)且x 1≠x 2,(x 1-x 2)·[f (x 1)-f (x 2)]<0”的是( ) A .f (x )=2x B .f (x )=|x -1| C .f (x )=1x-xD .f (x )=ln(x +1)解析:选C 由(x 1-x 2)·[f (x 1)-f (x 2)]<0可知,f (x )在(0,+∞)上是减函数,A 、D 选项中,f (x )为增函数;B 中,f (x )=|x -1|在(0,+∞)上不单调;对于f (x )=1x -x ,因为y =1x 与y=-x 在(0,+∞)上单调递减,因此f (x )在(0,+∞)上是减函数.2.函数f (x )=log 12(x 2-4)的单调递增区间是( )A .(0,+∞)B .(-∞,0)C .(2,+∞)D .(-∞,-2)解析:选D 令t =x 2-4,则y =log 12t .因为y =log 12t 在定义域上是减函数,所以求原函数的单调递增区间,即求函数t =x 2-4的单调递减区间,结合函数的定义域,可知所求区间为(-∞,-2).3.判断函数f (x )=x +ax (a >0)在(0,+∞)上的单调性.解:设x 1,x 2是任意两个正数,且x 1<x 2,则f (x 1)-f (x 2)=⎝⎛⎭⎫x 1+a x 1-⎝⎛⎭⎫x 2+a x 2=x 1-x 2x 1x 2(x 1x 2-a ). 当0<x 1<x 2≤a 时,0<x 1x 2<a ,x 1-x 2<0,所以f (x 1)-f (x 2)>0,即f (x 1)>f (x 2), 所以函数f (x )在(0,a ]上是减函数; 当a ≤x 1<x 2时,x 1x 2>a ,x 1-x 2<0, 所以f (x 1)-f (x 2)<0,即f (x 1)<f (x 2), 所以函数f (x )在[a ,+∞)上是增函数.综上可知,函数f (x )=x +ax (a >0)在(0,a ]上是减函数,在[a ,+∞)上是增函数.考点二 求函数的值域(最值))[典例] (1)(2019•深圳调研)函数y =|x +1|+|x -2|的值域为________.(2)若函数f (x )=-ax+b (a >0)在⎣⎡⎦⎤12,2上的值域为⎣⎡⎦⎤12,2,则a =________,b =________. (3)函数f (x )=⎩⎪⎨⎪⎧-x 2-4x ,x ≤0,sin x ,x >0的最大值为________.[解析] (1)图象法函数y =⎩⎪⎨⎪⎧-2x +1,x ≤-1,3,-1<x <2,2x -1,x ≥2.作出函数的图象如图所示.根据图象可知,函数y =|x +1|+|x -2|的值域为[3,+∞). (2)单调性法∵f (x )=-ax +b (a >0)在⎣⎡⎦⎤12,2上是增函数, ∴f (x )min =f ⎝⎛⎭⎫12=12,f (x )max =f (2)=2.即⎩⎨⎧-2a +b =12,-a2+b =2,解得a =1,b =52.(3)当x ≤0时,f (x )=-x 2-4x =-(x +2)2+4,而-2∈(-∞,0],此时f (x )在x =-2处取得最大值,且f (-2)=4;当x >0时,f (x )=sin x ,此时f (x )在区间(0,+∞)上的最大值为1.综上所述,函数f (x )的最大值为4.[答案] (1)[3,+∞) (2)1 52 (3)4[提醒] (1)求函数的最值时,应先确定函数的定义域.(2)求分段函数的最值时,应先求出每一段上的最值,再选取其中最大的作为分段函数的最大值,最小的作为分段函数的最小值.[题组训练]1.函数f (x )=x 2+4x 的值域为________.解析:当x >0时,f (x )=x +4x ≥4,当且仅当x =2时取等号; 当x <0时,-x +⎝⎛⎭⎫-4x ≥4, 即f (x )=x +4x ≤-4,当且仅当x =-2取等号,所以函数f (x )的值域为(-∞,-4]∪[4,+∞). 答案:(-∞,-4]∪[4,+∞)2.若x ∈⎣⎡⎦⎤-π6,2π3,则函数y =4sin 2x -12sin x -1的最大值为________,最小值为________.解析:令t =sin x ,因为x ∈⎣⎡⎦⎤-π6,2π3, 所以t ∈⎣⎡⎦⎤-12,1,y =f (t )=4t 2-12t -1, 因为该二次函数的图象开口向上,且对称轴为t =32,所以当t ∈⎣⎡⎦⎤-12,1时,函数f (t )单调递减,所以当t =-12时,y max =6;当t =1时,y min =-9. 答案:6 -93.已知f (x )=x 2+2x +ax ,x ∈[1,+∞),且a ≤1.若对任意x ∈[1,+∞),f (x )>0恒成立,则实数a 的取值范围是________.解析:对任意x ∈[1,+∞),f (x )>0恒成立等价于x 2+2x +a >0在x ∈[1,+∞)上恒成立,即a >-x 2-2x 在x ∈[1,+∞)上恒成立.又函数y =-x 2-2x 在[1,+∞)上单调递减, ∴(-x 2-2x )max =-3,故a >-3,又∵a ≤1,∴-3<a ≤1. 答案:(-3,1]考点三 函数单调性的应用考法(一) 比较函数值的大小[典例] 设偶函数f (x )的定义域为R ,当x ∈[0,+∞)时,f (x )是增函数,则f (-2),f (π),f (-3)的大小关系是( )A .f (π)>f (-3)>f (-2)B .f (π)>f (-2)>f (-3)C .f (π)<f (-3)<f (-2)D .f (π)<f (-2)<f (-3)[解析] 因为f (x )是偶函数,所以f (-3)=f (3),f (-2)=f (2). 又因为函数f (x )在[0,+∞)上是增函数. 所以f (π)>f (3)>f (2),即f (π)>f (-3)>f (-2). [答案] A[解题技法] 比较函数值大小的解题思路比较函数值的大小时,若自变量的值不在同一个单调区间内,要利用其函数性质,转化到同一个单调区间内进行比较,对于选择题、填空题能数形结合的尽量用图象法求解.考法(二) 解函数不等式[典例] 设函数f (x )=⎩⎪⎨⎪⎧2x ,x <2,x 2,x ≥2.若f (a +1)≥f (2a -1),则实数a 的取值范围是( )A .(-∞,1]B .(-∞,2]C .[2,6]D .[2,+∞)[解析] 易知函数f (x )在定义域(-∞,+∞)上是增函数,∵f (a +1)≥f (2a -1), ∴a +1≥2a -1,解得a ≤2.故实数a 的取值范围是(-∞,2]. [答案] B[解题技法] 求解含“f ”的函数不等式的解题思路先利用函数的相关性质将不等式转化为f (g (x ))>f (h (x ))的形式,再根据函数的单调性去掉“f ”,得到一般的不等式g (x )>h (x )(或g (x )<h (x )).考法(三) 利用单调性求参数的范围(或值)[典例] (2019•南京调研)已知函数f (x )=x -a x +a2在(1,+∞)上是增函数,则实数a 的取值范围是________.[解析] 设1<x 1<x 2,∴x 1x 2>1. ∵函数f (x )在(1,+∞)上是增函数, ∴f (x 1)-f (x 2)=x 1-a x 1+a2-⎝⎛⎭⎫x 2-a x 2+a 2 =(x 1-x 2)⎝⎛⎭⎫1+a x 1x 2<0.∵x 1-x 2<0,∴1+ax 1x 2>0,即a >-x 1x 2.∵1<x 1<x 2,x 1x 2>1,∴-x 1x 2<-1,∴a ≥-1. ∴a 的取值范围是[-1,+∞). [答案] [-1,+∞)[解题技法]利用单调性求参数的范围(或值)的方法(1)视参数为已知数,依据函数的图象或单调性定义,确定函数的单调区间,与已知单调区间比较求参数;(2)需注意若函数在区间[a ,b ]上是单调的,则该函数在此区间的任意子集上也是单调的.[题组训练]1.已知函数f (x )的图象向左平移1个单位后关于y 轴对称,当x 2>x 1>1时,[f (x 2)-f (x 1)]·(x 2-x 1)<0恒成立,设a =f ⎝⎛⎭⎫-12,b =f (2),c =f (3),则a ,b ,c 的大小关系为( ) A .c >a >b B .c >b >a C .a >c >bD .b >a >c解析:选D 由于函数f (x )的图象向左平移1个单位后得到的图象关于y 轴对称,故函数y =f (x )的图象关于直线x =1对称,所以a =f ⎝⎛⎭⎫-12=f ⎝⎛⎭⎫52.当x 2>x 1>1时,[f (x 2)-f (x 1)](x 2-x 1)<0恒成立,等价于函数f (x )在(1,+∞)上单调递减,所以b >a >c .2.已知函数f (x )=⎩⎪⎨⎪⎧ax 2-x -14,x ≤1,log a x -1,x >1是R 上的单调函数,则实数a 的取值范围是( )A.⎣⎡⎭⎫14,12 B.⎣⎡⎦⎤14,12 C.⎝⎛⎦⎤0,12 D.⎣⎡⎭⎫12,1解析:选B 由对数函数的定义可得a >0,且a ≠1.又函数f (x )在R 上单调,而二次函数y =ax 2-x -14的图象开口向上,所以函数f (x )在R 上单调递减,故有⎩⎪⎨⎪⎧0<a <1,12a≥1,a ×12-1-14≥log a1-1,即⎩⎪⎨⎪⎧0<a <1,0<a ≤12,a ≥14.所以a ∈⎣⎡⎦⎤14,12.[课时跟踪检测]A 级1.下列四个函数中,在x ∈(0,+∞)上为增函数的是( ) A .f (x )=3-x B .f (x )=x 2-3x C .f (x )=-1x +1D .f (x )=-|x |解析:选C 当x >0时,f (x )=3-x 为减函数;当x ∈⎝⎛⎭⎫0,32时,f (x )=x 2-3x 为减函数,当x ∈⎝⎛⎭⎫32,+∞时,f (x )=x 2-3x 为增函数;当x ∈(0,+∞)时,f (x )=-1x +1为增函数;当x ∈(0,+∞)时,f (x )=-|x |为减函数.2.若函数f (x )=ax +1在R 上单调递减,则函数g (x )=a (x 2-4x +3)的单调递增区间是( )A .(2,+∞)B .(-∞,2)C .(4,+∞)D .(-∞,4)解析:选B 因为f (x )=ax +1在R 上单调递减,所以a <0. 而g (x )=a (x 2-4x +3)=a (x -2)2-a .因为a <0,所以g (x )在(-∞,2)上单调递增.3.已知函数f (x )是定义在区间[0,+∞)上的函数,且在该区间上单调递增,则满足f (2x -1)<f ⎝⎛⎭⎫13的x 的取值范围是( )A.⎝⎛⎭⎫13,23 B.⎣⎡⎭⎫13,23 C.⎝⎛⎭⎫12,23D.⎣⎡⎭⎫12,23解析:选D 因为函数f (x )是定义在区间[0,+∞)上的增函数,满足f (2x -1)<f ⎝⎛⎭⎫13. 所以0≤2x -1<13,解得12≤x <23.4.(2019·菏泽模拟)定义新运算⊕:当a ≥b 时,a ⊕b =a ;当a <b 时,a ⊕b =b 2,则函数f (x )=(1⊕x )x -(2⊕x ),x ∈[-2,2]的最大值等于( )A .-1B .1C .6D .12解析:选C 由题意知当-2≤x ≤1时,f (x )=x -2,当1<x ≤2时,f (x )=x 3-2,又f (x )=x -2,f (x )=x 3-2在相应的定义域内都为增函数,且f (1)=-1,f (2)=6,∴f (x )的最大值为6.5.已知函数f (x )是R 上的增函数,A (0,-3),B (3,1)是其图象上的两点,那么不等式-3<f (x +1)<1的解集的补集是(全集为R)( )A .(-1,2)B .(1,4)C .(-∞,-1)∪[4,+∞)D .(-∞,-1]∪[2,+∞)解析:选D 由函数f (x )是R 上的增函数,A (0,-3),B (3,1)是其图象上的两点,知不等式-3<f (x +1)<1即为f (0)<f (x +1)<f (3),所以0<x +1<3,所以-1<x <2,故不等式-3<f (x +1)<1的解集的补集是(-∞,-1]∪[2,+∞).6.已知函数f (x )=⎩⎪⎨⎪⎧-x 2-ax -5,x ≤1,a x ,x >1是R 上的增函数,则实数a 的取值范围是( )A .[-3,0)B .(-∞,-2]C .[-3,-2]D .(-∞,0)解析:选C 若f (x )是R 上的增函数,则应满足⎩⎪⎨⎪⎧-a2≥1,a <0,-12-a ×1-5≤a 1,解得-3≤a ≤-2.7.已知函数f (x )=x 2-2x -3,则该函数的单调递增区间为________.解析:设t =x 2-2x -3,由t ≥0,即x 2-2x -3≥0,解得x ≤-1或x ≥3,所以函数f (x )的定义域为(-∞,-1]∪[3,+∞).因为函数t =x 2-2x -3的图象的对称轴为x =1,所以函数t =x 2-2x -3在(-∞,-1]上单调递减,在[3,+∞)上单调递增,所以函数f (x )的单调递增区间为[3,+∞).答案:[3,+∞)8.函数f (x )=⎩⎪⎨⎪⎧1x ,x ≥1,-x 2+2,x <1的最大值为________.解析:当x ≥1时,函数f (x )=1x 为减函数,所以f (x )在x =1处取得最大值,为f (1)=1;当x <1时,易知函数f (x )=-x 2+2在x =0处取得最大值,为f (0)=2.故函数f (x )的最大值为2.答案:29.若函数f (x )=1x 在区间[2,a ]上的最大值与最小值的和为34,则a =________.解析:由f (x )=1x 的图象知,f (x )=1x 在(0,+∞)上是减函数,∵[2,a ]⊆(0,+∞),∴f (x )=1x 在[2,a ]上也是减函数,∴f (x )max =f (2)=12,f (x )min =f (a )=1a ,∴12+1a =34,∴a =4. 答案:410.(2019·甘肃会宁联考)若f (x )=x +a -1x +2在区间(-2,+∞)上是增函数,则实数a 的取值范围是________.解析:f (x )=x +a -1x +2=x +2+a -3x +2=1+a -3x +2,要使函数在区间(-2,+∞)上是增函数,需使a -3<0,解得a <3.答案:(-∞,3)11.已知函数f (x )=1a -1x (a >0,x >0).(1)求证:f (x )在(0,+∞)上是增函数;(2)若f (x )在⎣⎡⎦⎤12,2上的值域是⎣⎡⎦⎤12,2,求a 的值. 解:(1)证明:任取x 1>x 2>0, 则f (x 1)-f (x 2)=1a -1x 1-1a +1x 2=x 1-x 2x 1x 2,∵x 1>x 2>0,∴x 1-x 2>0,x 1x 2>0, ∴f (x 1)-f (x 2)>0, 即f (x 1)>f (x 2),∴f (x )在(0,+∞)上是增函数.(2)由(1)可知,f (x )在⎣⎡⎦⎤12,2上是增函数, ∴f ⎝⎛⎭⎫12=1a -2=12,f (2)=1a -12=2, 解得a =25.12.已知f (x )=xx -a(x ≠a ).(1)若a =-2,试证f (x )在(-∞,-2)内单调递增;(2)若a >0且f (x )在(1,+∞)内单调递减,求a 的取值范围. 解:(1)证明:当a =-2时,f (x )=xx +2.任取x 1,x 2∈(-∞,-2),且x 1<x 2, 则f (x 1)-f (x 2)=x 1x 1+2-x 2x 2+2=2(x 1-x 2)(x 1+2)(x 2+2). 因为(x 1+2)(x 2+2)>0,x 1-x 2<0, 所以f (x 1)-f (x 2)<0,即f (x 1)<f (x 2), 所以f (x )在(-∞,-2)内单调递增. (2)任取x 1,x 2∈(1,+∞),且x 1<x 2, 则f (x 1)-f (x 2)=x 1x 1-a -x 2x 2-a =a (x 2-x 1)(x 1-a )(x 2-a ). 因为a >0,x 2-x 1>0,又由题意知f (x 1)-f (x 2)>0, 所以(x 1-a )(x 2-a )>0恒成立,所以a ≤1. 所以0<a ≤1.所以a 的取值范围为(0,1].B 级1.若f (x )=-x 2+4mx 与g (x )=2mx +1在区间[2,4]上都是减函数,则m 的取值范围是( )A .(-∞,0)∪(0,1]B .(-1,0)∪(0,1]C .(0,+∞)D .(0,1]解析:选D 函数f (x )=-x 2+4mx 的图象开口向下,且以直线x =2m 为对称轴,若在区间[2,4]上是减函数,则2m ≤2,解得m ≤1;g (x )=2m x +1的图象由y =2mx 的图象向左平移一个单位长度得到,若在区间[2,4]上是减函数,则2m >0,解得m >0.综上可得,m 的取值范围是(0,1].2.已知函数f (x )=ln x +x ,若f (a 2-a )>f (a +3),则正数a 的取值范围是________. 解析:因为f (x )=ln x +x 在(0,+∞)上是增函数,所以⎩⎪⎨⎪⎧a 2-a >a +3,a 2-a >0,a +3>0,解得-3<a <-1或a >3.又a >0,所以a >3. 答案:(3,+∞)3.已知定义在R 上的函数f (x )满足:①f (x +y )=f (x )+f (y )+1,②当x >0时,f (x )>-1. (1)求f (0)的值,并证明f (x )在R 上是单调增函数; (2)若f (1)=1,解关于x 的不等式f (x 2+2x )+f (1-x )>4. 解:(1)令x =y =0,得f (0)=-1.在R 上任取x 1>x 2,则x 1-x 2>0,f (x 1-x 2)>-1. 又f (x 1)=f [(x 1-x 2)+x 2]=f (x 1-x 2)+f (x 2)+1>f (x 2), 所以函数f (x )在R 上是单调增函数. (2)由f (1)=1,得f (2)=3,f (3)=5.由f (x 2+2x )+f (1-x )>4得f (x 2+x +1)>f (3), 又函数f (x )在R 上是增函数,故x 2+x +1>3, 解得x <-2或x >1,故原不等式的解集为{x |x <-2或x >1}.第三节 函数的奇偶性与周期性一、基础知1.函数的奇偶性函数的定义域关于原点对称是函数具有奇偶性的前提条件.若f (x )≠0,则奇(偶)函数定义的等价形式如下:(1)f (-x )=f (x )⇔f (-x )-f (x )=0⇔f (-x )f (x )=1⇔f (x )为偶函数;(2)f (-x )=-f (x )⇔f (-x )+f (x )=0⇔f (-x )f (x )=-1⇔f (x )为奇函数.2.函数的周期性 (1)周期函数对于函数f (x ),如果存在一个非零常数T ,使得当x 取定义域内的任何值时,都有f (x +T )=f (x ),那么就称函数f (x )为周期函数,称T 为这个函数的周期.周期函数定义的实质存在一个非零常数T ,使f (x +T )=f (x )为恒等式,即自变量x 每增加一个T 后,函数值就会重复出现一次.(2)最小正周期如果在周期函数f (x )的所有周期中存在一个最小的正数,那么这个最小正数就叫做f (x )的最小正周期.二、常用结论1.函数奇偶性常用结论(1)如果函数f (x )是奇函数且在x =0处有定义,则一定有f (0)=0;如果函数f (x )是偶函数,那么f (x )=f (|x |).(2)奇函数在两个对称的区间上具有相同的单调性;偶函数在两个对称的区间上具有相反的单调性.(3)在公共定义域内有:奇±奇=奇,偶±偶=偶,奇×奇=偶,偶×偶=偶,奇×偶=奇.2.函数周期性常用结论 对f (x )定义域内任一自变量x : (1)若f (x +a )=-f (x ),则T =2a (a >0). (2)若f (x +a )=1f (x ),则T =2a (a >0). (3)若f (x +a )=-1f (x ),则T =2a (a >0).3.函数图象的对称性(1)若函数y =f (x +a )是偶函数,即f (a -x )=f (a +x ),则函数y =f (x )的图象关于直线x =a 对称.(2)若对于R 上的任意x 都有f (2a -x )=f (x )或f (-x )=f (2a +x ),则y =f (x )的图象关于直线x =a 对称.(3)若函数y =f (x +b )是奇函数,即f (-x +b )+f (x +b )=0,则函数y =f (x )关于点(b,0)中心对称.考点一 函数奇偶性的判断[典例] 判断下列函数的奇偶性: (1)f (x )=36-x 2|x +3|-3;(2)f (x )=1-x 2+x 2-1; (3)f (x )=log 2(1-x 2)|x -2|-2;(4)f (x )=⎩⎪⎨⎪⎧x 2+x ,x <0,x 2-x ,x >0.[解] (1)由f (x )=36-x 2|x +3|-3,可知⎩⎪⎨⎪⎧ 36-x 2≥0,|x +3|-3≠0⇒⎩⎪⎨⎪⎧-6≤x ≤6,x ≠0且x ≠-6,故函数f (x )的定义域为(-6,0)∪(0,6],定义域不关于原点对称,故f (x )为非奇非偶函数.(2)由⎩⎪⎨⎪⎧1-x 2≥0,x 2-1≥0⇒x 2=1⇒x =±1,故函数f (x )的定义域为{-1,1},关于原点对称,且f (x )=0,所以f (-x )=f (x )=-f (x ),所以函数f (x )既是奇函数又是偶函数.(3)由⎩⎪⎨⎪⎧1-x 2>0,|x -2|-2≠0⇒-1<x <0或0<x <1,定义域关于原点对称.此时f (x )=log 2(1-x 2)|x -2|-2=log 2(1-x 2)2-x -2=-log 2(1-x 2)x ,故有f (-x )=-log 2[1-(-x )2]-x =log 2(1-x 2)x =-f (x ),所以函数f (x )为奇函数. (4)法一:图象法画出函数f (x )=⎩⎪⎨⎪⎧x 2+x ,x <0,x 2-x ,x >0的图象如图所示,图象关于y 轴对称,故f (x )为偶函数.法二:定义法易知函数f (x )的定义域为(-∞,0)∪(0,+∞),关于原点对称,当x >0时,f (x )=x 2-x ,则当x <0时,-x >0,故f (-x )=x 2+x =f (x );当x <0时,f (x )=x 2+x ,则当x >0时,-x <0,故f (-x )=x 2-x =f (x ),故原函数是偶函数.法三:f (x )还可以写成f (x )=x 2-|x |(x ≠0),故f (x )为偶函数.[题组训练]1.(2018·福建期末)下列函数为偶函数的是( ) A .y =tan ⎝⎛⎭⎫x +π4 B .y =x 2+e |x | C .y =x cos xD .y =ln|x |-sin x解析:选B 对于选项A ,易知y =tan ⎝⎛⎭⎫x +π4为非奇非偶函数;对于选项B ,设f (x )=x 2+e |x |,则f (-x )=(-x )2+e |-x |=x 2+e |x |=f (x ),所以y =x 2+e |x |为偶函数;对于选项C ,设f (x )=x cos x ,则f (-x )=-x cos(-x )=-x cos x =-f (x ),所以y =x cos x 为奇函数;对于选项D ,设f (x )=ln|x |-sin x ,则f (2)=ln 2-sin 2,f (-2)=ln 2-sin(-2)=ln 2+sin 2≠f (2),所以y =ln|x |-sin x 为非奇非偶函数,故选B.2.设函数f (x )=e x -e -x2,则下列结论错误的是( )A .|f (x )|是偶函数B .-f (x )是奇函数C .f (x )|f (x )|是奇函数D .f (|x |)f (x )是偶函数解析:选D ∵f (x )=e x -e -x2,则f (-x )=e -x -e x2=-f (x ).∴f (x )是奇函数. ∵f (|-x |)=f (|x |),∴f (|x |)是偶函数,∴f (|x |)f (x )是奇函数.考点二 函数奇偶性的应用[典例] (1)(2019·福建三明模拟)函数y =f (x )是R 上的奇函数,当x <0时,f (x )=2x ,则当x >0时,f (x )=( )A .-2xB .2-x C .-2-xD .2x(2)(2018·贵阳摸底考试)已知函数f (x )=a -2e x +1(a ∈R)是奇函数,则函数f (x )的值域为( )A .(-1,1)B .(-2,2)C .(-3,3)D .(-4,4)[解析] (1)当x >0时,-x <0,∵x <0时,f (x )=2x ,∴当x >0时,f (-x )=2-x .∵f (x )是R 上的奇函数,∴当x >0时,f (x )=-f (-x )=-2-x .(2)法一:由f (x )是奇函数知f (-x )=-f (x ),所以a -2e -x+1=-a +2e x +1,得2a =2e x+1+2e -x +1,所以a =1e x +1+e x e x +1=1,所以f (x )=1-2e x +1.因为e x +1>1,所以0<1e x +1<1,-1<1-2e x +1<1,所以函数f (x )的值域为(-1,1).法二:函数f (x )的定义域为R ,且函数f (x )是奇函数,所以f (0)=a -1=0,即a =1,所以f (x )=1-2e x +1.因为e x +1>1,所以0<1e x +1<1,-1<1-2e x +1<1,所以函数f (x )的值域为(-1,1).[答案] (1)C (2)A[解题技法]应用函数奇偶性可解决的四类问题及解题方法(1)求函数值将待求值利用奇偶性转化为已知区间上的函数值求解.(2)求解析式先将待求区间上的自变量转化到已知区间上,再利用奇偶性求解,或充分利用奇偶性构造关于f (x )的方程(组),从而得到f (x )的解析式.(3)求函数解析式中参数的值利用待定系数法求解,根据f (x )±f (-x )=0得到关于待求参数的恒等式,由系数的对等性得参数的值或方程(组),进而得出参数的值.(4)画函数图象和判断单调性利用奇偶性可画出另一对称区间上的图象及判断另一区间上的单调性.[题组训练]1.(2019·贵阳检测)若函数f (x )是定义在R 上的奇函数,当x ≥0时,f (x )=log 2(x +2)-1,则f (-6)=( )A .2B .4C .-2D .-4解析:选C 根据题意得f (-6)=-f (6)=1-log 2(6+2)=1-3=-2.2.已知函数f (x )为奇函数,当x >0时,f (x )=x 2-x ,则当x <0时,函数f (x )的最大值为________.解析:法一:当x <0时,-x >0,所以f (-x )=x 2+x .又因为函数f (x )为奇函数,所以f (x )=-f (-x )=-x 2-x =-⎝⎛⎭⎫x +122+14,所以当x <0时,函数f (x )的最大值为14. 法二:当x >0时,f (x )=x 2-x =⎝⎛⎭⎫x -122-14,最小值为-14,因为函数f (x )为奇函数,所以当x <0时,函数f (x )的最大值为14.答案:143.(2018·合肥八中模拟)若函数f (x )=x ln(x +a +x 2)为偶函数,则a =________. 解析:∵f (x )=x ln(x +a +x 2)为偶函数,∴f (-x )=f (x ),即-x ln(a +x 2-x )=x ln(x +a +x 2),从而ln[(a +x 2)2-x 2]=0,即ln a =0,故a =1.答案:1考点三 函数的周期性[典例] (1)(2018·开封期末)已知定义在R 上的函数f (x )满足f (x )=-f (x +2),当x ∈(0,2]时,f (x )=2x +log 2x ,则f (2 019)=( )A .5 B.12C .2D .-2(2)(2018·江苏高考)函数f (x )满足f (x +4)=f (x )(x ∈R),且在区间(-2,2]上,f (x )=⎩⎨⎧cos πx2,0<x ≤2,⎪⎪⎪⎪x +12,-2<x ≤0,则f (f (15))的值为________.[解析] (1)由f (x )=-f (x +2),得f (x +4)=f (x ),所以函数f (x )是周期为4的周期函数,所以f (2 019)=f (504×4+3)=f (3)=f (1+2)=-f (1)=-(2+0)=-2.(2)由函数f (x )满足f (x +4)=f (x )(x ∈R), 可知函数f (x )的周期是4, 所以f (15)=f (-1)=⎪⎪⎪⎪-1+12=12, 所以f (f (15))=f ⎝⎛⎭⎫12=cos π4=22. [答案] (1)D (2)22[题组训练]1.(2019·山西八校联考)已知f (x )是定义在R 上的函数,且满足f (x +2)=-1f (x ),当2≤x ≤3时,f (x )=x ,则f ⎝⎛⎭⎫-112=________. 解析:∵f (x +2)=-1f (x ),∴f (x +4)=f (x ), ∴f ⎝⎛⎭⎫-112=f ⎝⎛⎭⎫52,又2≤x ≤3时,f (x )=x , ∴f ⎝⎛⎭⎫52=52,∴f ⎝⎛⎭⎫-112=52. 答案:522.(2019·哈尔滨六中期中)设f (x )是定义在R 上的周期为3的函数,当x ∈[-2,1)时,f (x )=⎩⎪⎨⎪⎧4x 2-2,-2≤x ≤0,x ,0<x <1,则f ⎝⎛⎭⎫f ⎝⎛⎭⎫214=________. 解析:由题意可得f ⎝⎛⎭⎫214=f ⎝⎛⎭⎫6-34=f ⎝⎛⎭⎫-34=4×⎝⎛⎭⎫-342-2=14,f ⎝⎛⎭⎫14=14.答案:14[课时跟踪检测]A 级1.下列函数为奇函数的是( ) A .f (x )=x 3+1 B .f (x )=ln 1-x1+xC .f (x )=e xD .f (x )=x sin x解析:选B 对于A ,f (-x )=-x 3+1≠-f (x ),所以其不是奇函数;对于B ,f (-x )=ln 1+x 1-x=-ln1-x 1+x=-f (x ),所以其是奇函数;对于C ,f (-x )=e -x ≠-f (x ),所以其不是奇函数;对于D ,f (-x )=-x sin(-x )=x sin x =f (x ),所以其不是奇函数.故选B.2.(2019·南昌联考)函数f (x )=9x +13x 的图象( )A .关于x 轴对称B .关于y 轴对称C .关于坐标原点对称D .关于直线y =x 对称解析:选B 因为f (x )=9x +13x =3x +3-x ,易知f (x )为偶函数,所以函数f (x )的图象关于y轴对称.3.设函数f (x )是定义在R 上的奇函数,且f (x )=⎩⎪⎨⎪⎧log 2(x +1),x ≥0,g (x ),x <0,则f (-7)=( )A .3B .-3C .2D .-2解析:选B 因为函数f (x )是定义在R 上的奇函数,且f (x )=⎩⎪⎨⎪⎧log 2(x +1),x ≥0,g (x ),x <0,所以f (-7)=-f (7)=-log 2(7+1)=-3.4.若定义在R 上的偶函数f (x )和奇函数g (x )满足f (x )+g (x )=e x ,则g (x )=( ) A .e x -e -x B.12(e x +e -x )C.12(e -x -e x ) D.12(e x -e -x )解析:选D 因为f (x )+g (x )=e x ,所以f (-x )+g (-x )=f (x )-g (x )=e -x , 所以g (x )=12(e x -e -x ).。
高一数学教案复习函数的基本概念与性质函数是数学中一种重要的概念,它在数理科学的研究和实际应用中都有着广泛的应用。
高一学生正处于数学基础知识的学习和掌握阶段,因此对于函数的基本概念与性质的复习显得尤为重要。
本篇教案将细致地介绍函数的基本概念和常见的性质,以帮助学生加深对该知识点的理解和运用。
一、函数的基本概念函数是指两个集合之间的一种特殊关系,其中每个元素(自变量)在定义域内只对应一个元素(因变量)。
为了确定一个函数,我们需要明确以下几个要素:1.1 定义域和值域函数的定义域是指自变量可能取值的集合,而值域则是函数的所有可能输出值的集合。
需要注意的是,函数的定义域可以是实数集、整数集或自然数集等不同数集。
1.2 关系式或图表函数可以通过关系式或图表的形式来表示。
关系式是指将自变量和因变量之间的关系用式子表示出来,如y = 2x + 3;图表则是将自变量和因变量的对应关系用表格或图像呈现出来。
1.3 函数的特性函数可以通过一些特性来描述和判断,比如奇偶性、单调性、周期性等。
这些特性可以帮助我们更好地理解函数的性质和行为。
二、函数的性质与图像除了基本概念之外,函数还具有一些常见的性质。
下面我们将介绍一些关于函数性质的重要内容,并通过图像来进一步说明。
2.1 奇偶性一个函数可以是奇函数、偶函数或者既不是奇函数也不是偶函数。
奇函数的图像关于原点对称,即f(-x) = -f(x);偶函数的图像关于y轴对称,即f(-x) = f(x)。
2.2 单调性单调函数是指在定义域上具有单调性的函数。
如果函数在某一区间上递增,那么它是递增函数;如果函数在某一区间上递减,那么它是递减函数。
2.3 周期性周期函数是指在一定区间内,函数的值按照一定规律重复出现。
常见的周期函数有正弦函数和余弦函数等。
周期可以通过函数的图像来观察和确定。
三、函数的应用函数的概念和性质在数学和实际应用中都有广泛的应用。
在数学上,函数可以用于解决各种数学问题,如方程的求解、不等式的证明等。
高中数学教学备课教案函数的定义域与值域高中数学教学备课教案函数的定义域与值域介绍:函数是数学中的重要概念,对于高中数学教学来说,理解函数的定义域与值域是非常关键的。
本教案将围绕函数的定义域与值域展开,旨在帮助学生深入理解函数的特性和应用。
一、函数的基本概念1.1 函数的定义函数是两个集合之间的对应关系,其中一个集合称为定义域,另一个集合称为值域。
在数学中,我们常以字母f表示函数,用x表示定义域中的元素。
1.2 定义域的确定定义域是函数中可以取得实际意义的自变量的取值范围。
它由函数的解析式、图像、实际问题和常识共同确定。
1.3 值域的确定值域是函数在定义域上所有可能的取值的集合。
通过函数的解析式、图像以及实际问题,我们可以较为准确地确定函数的值域。
二、定义域的常见类型有理函数是指可以表示为两个多项式的比值的函数。
有理函数的定义域通常由其分母的零点确定。
2.2 幂函数及其定义域幂函数是指以x为底数的指数函数,形如f(x) = x^a。
对于幂函数,定义域为实数集。
2.3 指数函数及其定义域指数函数是以一个正实数为底的指数函数,形如f(x) = a^x。
对于指数函数,定义域为实数集。
2.4 对数函数及其定义域对数函数是指以一个正实数为底的对数函数,形如f(x) = loga(x)。
对于对数函数,定义域为正实数集。
三、值域的常见类型3.1 有界函数及其值域有界函数是指在定义域上,函数的值上下都有限制的函数。
值域是一个有限的区间。
3.2 无界函数及其值域无界函数是指函数在定义域上,函数的值没有上下限的函数。
值域为整个实数集。
单调递增函数是指在定义域上,随着自变量的增大,函数值也随之增大的函数。
值域为一个区间。
3.4 单调递减函数及其值域单调递减函数是指在定义域上,随着自变量的增大,函数值反而减小的函数。
值域为一个区间。
结论:通过本教案,我们对高中数学中函数的定义域和值域有了更深入的理解。
定义域是函数自变量的取值范围,它由函数的解析式、图像、实际问题和常识共同确定。
高中数学函数的定义域及值域1500字函数是数学中常用的概念,它描述了两个集合之间的对应关系。
函数的定义域是指输入的值的集合,而值域是函数输出的值的集合。
在高中数学中,我们经常需要确定函数的定义域和值域,以便了解函数的性质和行为。
为了确定一个函数的定义域,我们需要考虑两个因素:函数的解析式和函数的定义限制。
函数的解析式告诉我们函数如何计算输出值,而定义限制告诉我们输入值可以是哪些数。
首先,让我们考虑一些常见的函数类型及其定义域和值域。
1. 线性函数:线性函数的解析式可以写为y = mx + c,其中m是斜率,c是截距。
线性函数的定义域是所有实数集合,值域也是所有实数集合。
2. 幂函数:幂函数的解析式可以写为y = x^n,其中n是一个实数。
幂函数的定义域是所有实数集合,但值域取决于指数n的值。
例如,如果n是正偶数,那么幂函数的值域是非负实数集合;如果n是负偶数,那么幂函数的值域是正实数集合;如果n是奇数,那么幂函数的值域是所有实数集合。
3. 指数函数:指数函数的解析式可以写为y = a^x,其中a是一个正实数且不等于1。
指数函数的定义域是所有实数集合,值域是正实数集合。
4. 对数函数:对数函数的解析式可以写为y = log_a(x),其中a是一个正实数且不等于1。
对数函数的定义域是正实数集合,值域是所有实数集合。
5. 三角函数:三角函数包括正弦函数、余弦函数和正切函数等。
三角函数的定义域是所有实数集合,值域取决于具体的函数类型。
例如,正弦函数的值域是[-1, 1];余弦函数的值域也是[-1, 1];正切函数的值域是所有实数集合。
除了上述函数类型外,还有其他函数类型的定义域和值域也需要特别注意。
例如,有理函数的定义域由分母的零点确定,值域取决于分子的次数和分母的次数;反比例函数的定义域是除了零的所有实数,值域也是除了零的所有实数。
在确定函数的定义域和值域时,我们还需要注意一些常见的限制,如根式的奇次指数、分母不能为零、对数的底不能为1等。
高一数学定义域和值域
定义域和值域是高中数学课程中很重要的概念。
学习者必须了解它们,以完成一些数学函数的图形求解。
定义域是指函数f(x)的可能输入值,它是一个非空的子集,包含预期的输入
的所有可能的值,这是寻找函数的输出的第一步。
定义域有时也被称为实际域(或集合)。
根据所涉及的函数,定义域可以是所有实数,即定义域为R ,它也可以
是所有自然数或其他子集。
值域是函数f(x)映射至定义域之外的可能值,一般而言,值域被称为函数f(x)的可能输出或可能出现的值,可以是积分或有理数。
值域也可以是所有实数,或者是任何比实数更小的集合,包括有理数,自然数和负数。
定义域和值域的概念对于学习者来说很重要,因为需要使用它们来解决一些数学函数的图形求解问题。
学生需要对这两个概念有足够的了解,才能够联系起来,从而解决求解图标和函数问题。
通过定义域和值域,可以更好地理解数学函数,甚至可以绘制函数的图形,这给学生的学习带来帮助。
总的来说,定义域和值域是高中数学学习的重要概念,它能帮助学生有更良好的理解和掌握数学函数的图形求解的原理和知识,从而有助于提高学生的学习和解决图形问题的能力。
大方向教育个性化辅导教案教师:徐琨学生:学科:数学时间:课题(课型)函数概念与基本初等函数教学方法:知识梳理、例题讲解、归纳总结、巩固训练(一)函数1.了解构成函数的要素,了解映射的概念,会求一些简单函数的定义域和值域2.理解函数的三种表示法:解析法、图象法和列表法,能根据不同的要求选择恰当的方法表示简单的函数。
3.了解分段函数,能用分段函数来解决一些简单的数学问题。
4.理解函数的单调性,会讨论和证明一些简单的函数的单调性;理解函数奇偶性的含义,会判断简单的函数奇偶性。
5.理解函数的最大(小)值及其几何意义,并能求出一些简单的函数的最大(小)值6.会运用函数图像理解和研究函数的性质(二)指数函数1.了解指数函数模型的实际背景。
2.理解有理指数幂的含义,了解实数指数幂的意义,掌握幂的运算。
3.理解指数函数的概念,会求与指数函数性质有关的问题。
4.知道指数函数是一类重要的函数模型。
(三)对数函数1.理解对数的概念及其运算性质,知道用换底公式能将一般对数转化成自然对数或常用对数;了解对数在简化运算中的作用。
2.理解对数函数的概念;会求与对数函数性质有关的问题3.知道对数函数是一类重要的函数模型4.了解指数函数与对数函数互为反函数()。
(四)幂函数1.了解幂函数的概念。
2.结合函数的图像,了解它们的变化情况。
(五)函数与方程1.了解函数零点的概念,结合二次函数的图像,了解函数的零点与方程根的联系。
2.理解并掌握连续函数在某个区间上存在零点的判定方法。
能利用函数的图象和性质判别函数零点的个数函数的定义域和值一、定义域:1.函数的定义域就是使函数式的集合.2.常见的三种题型确定定义域:①已知函数的解析式,就是 .②复合函数f [g(x)]的有关定义域,就要保证内函数g(x)的域是外函数f (x)的域.③实际应用问题的定义域,就是要使得有意义的自变量的取值集合.二、值域:onfere nces, learne d superi ors policie s; reports, you ca n learn t o deal wit h problems, art, just wante d to lear n, to lear n at any time. To continuously expa nd the scope of the study. T he complexity and comprehensive ness of our w ork, deci ded to not o nly do we want to be "professionals", or if "generalists", to dabble i n different areas of knowle dge. Mastery of knowle dge not only t o have knowle dge to rational analysis. T his on requirements w e, not only to le arning national of route, and approach, a nd policy, also to lear ning national of legal, and regulati ons; not only to wit h political t heory armed mind, al so to wit h econom ic the ory gui de work; not only to rea d hist ory, al so to rea d hist ory; not only to a bsorpti on SM of mirror, also t o gets is of roa d; not only t o from local lear n, also to to outsi de learn; not only to t o books lear n, also to field st udy lear n. Therefore, I would e ncourage you, and more to the out side w orld w hen I have time to take a look at, take a stroll, outside devel opme nts, learn other people's development a pproach, and the n come back to g uide our w ork. Second, we must think. "Learni ng without thought to over sha dow, thi nking wit hout learni ng is peri lous." We wa nt to foster the character of advancing wit h the times, forge ahead, often reflect themse lves in a dapti ng to the new situati on on w hether t here are delays, have a nythi ng to fear in the reform and innovation, whether i n terms of accelerati ng the development of a relaxation, keep pr omising State of mind. In order t o do i nvestigations, investigati ons i s the roa d to heaven, God dispose s of the base. o wa nts to real he ard truth, and justi ce to trut h, and get truth, re ceived effective ness, on m ust in-de pth a ctual, a nd outreach, a nd in-depth ma sses, w orship masses for Divisi on, a nd masses heart, with masses with discussion everyone care of problem, listeni ng they of calls, understand they of mood, feel they of suffering, summary they of experience, conce ntrated they of wisdom, e spe cially on masse s most hope, and most urgent, a nd most w orries, a nd most com plaine d of hot, and difficultie s and focus problem, more to a ctive resear ch, caug ht not put. Dare t o innovate, in our w ork, often remain "the old way doe sn't work, t he ne w approach w oul d not" situation, the rea son, the key is that many people k now that copying ot her pe opl e's idea s, mechani cally, usi ng some one else's old way s. To change thi s predicament, re quire s us t o break the sha ckles of traditional conce pts to over come conservative, conformist, g ood at finding new problems a nd to a cce pt ne w thing s, summarize new experience, new idea s, deve loping ne w programmes a nd Maveri ck, a ne w approa ch to solve pr oblem s, speed up development. Espe cially in impl ement superi or policy spirit of process in the, bot h implementati on, a nd cannot m echani call y, to to extraordi nary of courage a nd develop enterpri sing of spirit, put flexibility a nd pri nci ple combi ned up, put superi or of spirit a nd l ocal reality combi ned up, ong conspiracy to breakthr ough of poli cy, a nd exercise innovati on of lift, devel ope d out devel opme nt works of, and grass-root s welcomes ofonfere nces, learne d superi ors policie s; reports, you ca n learn t o deal wit h problems, art, just wante d to lear n, to lear n at any time. To continuously expa nd the scope of the study. T he complexity and comprehensive ness of our w ork, deci ded to not only do we want to be "professionals", or if "generalists", to dabble i n different areas of knowle dge. Mastery of knowle dge not only t o have knowle dge to rational analysis. T his on requirements w e, not only to le arning national of route , and approach, a nd policy, also to lear ning national of legal, and regulati ons; not only to wit h political t heory armed mind, al so to wit h economic the ory gui de work ; not only to rea d hist ory, al so to rea d hist ory; not only to a bsorpti on SM of mirror, also t o gets is of roa d; not only t o from local lear n, also to to outsi de learn; not only to t o books lear n, also to field st udy lear n. Therefore, I would e ncourage you, and more to the out side w orld w hen I have time to take a look at, take a stroll, outside devel opme nts, learn other people's development a pproach, and the n come back to g uide our w ork. Second, we must think. "Learni ng without thought to over sha dow, thi nking wit hout learni ng is peri lous." We wa nt to foster the character of advancing wit h the times, forge ahead, often reflect themse lves in a dapti ng to the new situati on on w hether t here are delays, have a nythi ng to fear in the reform and innovation, whether i n terms of accelerati ng the development of a relaxat ion, keep pr omising State of mind. In order t o do i nvestigations, investigati ons i s the roa d to heaven, God dispose s of the base. o wa nts to real he ard truth, and justi ce to trut h, and get truth, re ceived effective ness, on m ust in -de pth a ctual, a nd outreach, a nd in -depth ma sses, w orship masses for Divisi on, a nd masses heart, with masses with discussion everyone care of problem, listeni ng they of calls, understand they of mood, feel they of suffering, summary they of experience, conce ntrated they of wisdom , e spe cially on masse s most hope , and most urgent, a nd most w orries, a nd most com plaine d of hot, and difficultie s and focus problem, more to a ctive resear ch, caug ht not put. Dare t o innovate, in our w ork, often remain "the old way doe sn't work, t he ne wapproach w oul d not" situation, the rea son, the key is that many people k now that copying ot her pe opl e's idea s, mechani cally, usi ng some one else's old way s. To change thi s predicament, re quire s us t o break the sha ckles of traditional conce pts to over come conservative, conformist, g ood at finding new problems a nd to a cce pt ne w thing s, summarize new experience, new idea s, deve loping ne w programmes a nd Maveri ck, a ne w approa ch to solve pr oblem s, speed up development. Espe cially in impl ement superi or policy spirit of process in the , bot h implementati on, a nd cannot mechani call y, to to extraordi nary of courage a nd develop enterpri sing of spirit, put flexibility a nd pri nci ple combi ned up, put superi or of spirit a nd l ocal reality combi ned up, ong conspiracy to breakthr ough of poli cy, a nd exercise innovati on of lift, devel ope d out devel opme nt works of, and grass -root s welcomes of- 2 -1.函数y =f (x )中,与自变量x 的值 的集合.2.常见函数的值域求法,就是优先考虑 ,取决于 ,常用的方法有:①观察法;②配方法;③反函数法;④不等式法;⑤单调性法;⑥数形法;⑦判别式法;⑧有界性法;⑨换元法(又分为 法和 法) 例如:① 形如y =221x +,可采用 法;② y =)32(2312-≠++x x x ,可采用 法或 法;③ y =a [f (x )]2+bf (x )+c ,可采用 法;④ y =x -x -1,可采用 法;⑤ y =x -21x -,可采用 法;⑥ y =xx cos 2sin -可采用 法等.函数的单调性一、单调性1.定义:如果函数y =f (x )对于属于定义域I 内某个区间上的任意两个自变量的值x 1、、x 2,当x 1、<x 2时,①都有 ,则称f (x )在这个区间上是增函数,而这个区间称函数的一个 ;②都有 ,则称f (x )在这个区间上是减函数,而这个区间称函数的一个 . 若函数f (x )在整个定义域l 内只有唯一的一个单调区间,则f (x )称为 . 2.判断单调性的方法:(1) 定义法,其步骤为:① ;② ;③ .(2) 导数法,若函数y =f (x )在定义域内的某个区间上可导,①若 ,则f (x )在这个区间上是增函数;②若 ,则f (x )在这个区间上是减函数. 二、单调性的有关结论1.若f (x ), g (x )均为增(减)函数,则f (x )+g (x ) 函数; 2.若f (x )为增(减)函数,则-f (x )为 ; 3.互为反函数的两个函数有 的单调性; 4.复合函数y =f [g(x )]是定义在M 上的函数,若f (x )与g(x )的单调相同,则f [g(x )]为 ,若f (x ), g(x )的单调性相反,则f [g(x )]为 .5.奇函数在其对称区间上的单调性 ,偶函数在其对称区间上的单调性 .函数的奇偶性1.奇偶性:① 定义:如果对于函数 f (x )定义域内的任意x 都有 ,则称 f (x )为奇函数;若 ,则称f (x )为偶函数. 如果函数f (x )不具有上述性质,则f (x )不具有 . 如果函数同时具有上述两条性质,则f (x ) . ② 简单性质:1) 图象的对称性质:一个函数是奇函数的充要条件是它的图象关于 对称;一个函数是偶函数的充要条件是它的图象关于 对称.2) 函数f (x )具有奇偶性的必要条件是其定义域关于 对称. 2.与函数周期有关的结论:①已知条件中如果出现)()(x f a x f -=+、或m x f a x f =+)()((a 、m 均为非零常数,0>a ),都可以得出)(x f 的周期为 ;②)(x f y =的图象关于点)0,(),0,(b a 中心对称或)(x f y =的图象关于直线b x a x ==,轴对称,关up, from masses most care, nd most directly, nd most reality of intere sts pr oblem grabbed, real do ove for people by De partment, nd right for pe opl e by with, nd Le e for people by conspira cy to. Thr ee with the fame of mind. "Non -indifferent not insi st your dream, wit hout serenity not go far." Asce nsi on to fame is morality, is the sublimati on of the soul. Tow nshi p and village, be sure to maintai n a normal state of mind, establ ish a correct view poi nt on pow er, position and value and Outl ook, as fame is light like water, depe ndi ng ... 50 km, also red and like hundreds of thousands of the KMT military combat but he w ent on to col lect a variety of new spa pers and magazines, and the n race against time to pore over. long as the study of mind, it is not nece ssary to come to the libr ary, workshops, int o the village s, pe ople lear ned farming techni ques, pr oblem -solving methods, partici pationin大方向教育——值得您信赖的专业化个性化辅导学校onfere nces, learne d superi ors policie s; reports, you ca n learn t o deal wit h problems, art, just wante d to lear n, to lear n at any time. To continuously expa nd the scope of the study. T he complexity and comprehensive ness of our w ork, deci ded to not only do we want to be "professionals", or if "generalists", to dabble i n different areas of knowle dge. Mastery of knowle dge not only t o have knowle dge to rational analysis. T his on requirements w e, not only to le arning national of route , and approach, a nd policy, also to lear ning national of legal, and regulati ons; not only to wit h political t heory armed mind, al so to wit h economic the ory gui de work ; not only to rea d hist ory, al so to rea d hist ory; not only to a bsorpti on SM of mirror, also t o gets is of roa d; not only t o from local lear n, also to to outsi de learn; not only to t o books lear n, also to field st udy lear n. Therefore, I would e ncourage you, and more to the out side w orld w hen I have time to take a look at, take a stroll, outside devel opme nts, learn other people's development a pproach, and the n come back to g uide our w ork. Sec ond, we must think. "Learni ng without thought to over sha dow, thi nking wit hout learni ng is peri lous." We wa nt to foster the character of advancing wit h the times, forge ahead, often reflect themse lves in a dapti ng to the new situati on on w hether t here are delays, have a nythi ng to fear in the reform and innovation, whether i n terms of accelerati ng the development of a relaxation, k eep pr omising State of mind. In order t o do i nvestigations, investigati ons i s the roa d to heaven, God dispose s of the base. o wa nts to real he ard truth, and justi ce to trut h, and get truth, re ceived effective ness, on m ust in -de pth a ctual, a nd outreach, a nd in -depth ma sses, w orship masses for Divisi on, a nd masses heart, with masses with discussion everyone care of problem, listeni ng they of calls, understand they of mood, feel they of suffering, summary they of experience, conce ntrated they of wisdom, e spe cially on masse s most hope , and most urgent, a nd most w orries, a nd most com plaine d of hot, and difficultie s and focus problem, more to a ctive resear ch, caug ht not put. Dare t o innovate, in our w ork, often remain "the old way doe sn't work, t he ne wapproach w oul d not" situation, the rea son, the key is that many people k now that copying ot her pe opl e's idea s, mechani cally, usi ng some one else's old way s. To change thi s predicament, re quire s us t o break the sha ckles of traditional conce pts to over come conservative, conformist, g ood at finding new problems a nd to a cce pt ne w thing s, summarize new experience, new idea s, deve loping ne w programmes a nd Maveri ck, a ne w approa ch to solve pr oblem s, speed up development. Espe cially in impl ement superi or policy spirit of process in the , bot h implementati on, a nd cannot mechani call y, to to extraordi nary of courage a nd develop enterpri sing of spirit, put flexibility a nd pri nci ple combi ned up, put superi or of spirit a nd l ocal reality combi ned up, ong conspiracy to breakthr ough of poli cy, a nd exercise innovati on of lift, devel ope d out devel opme nt works of, and grass -root s welcomes of- 3 -均可以得到)(x f 周期指数函数1.根式:(1) 定义:若a x n=,则x 称为a 的n 次方根① 当n 为奇数时,n a 的次方根记作__________;② 当n 为偶数时,负数a 没有n 次方根,而正数a 有两个n 次方根且互为相反数,记作________(a >0). (2) 性质:① a a n n=)(;② 当n 为奇数时,a a n n =;③ 当n 为偶数时,=n n a _______= ⎩⎨⎧<-≥)0()0(a a a a 2.指数: (1) 规定:① a 0= (a ≠0); ② a -p = ;③ (0,mn m na a a m => .(2) 运算性质:① a a a a sr s r ,0(>=⋅+ (a>0, r 、∈s Q ) ② a a a sr s r ,0()(>=⋅ (a>0, r 、∈s Q ) ③ >>⋅=⋅r b a b a b a rr r ,0,0()((a>0, r 、∈s Q ) 注:上述性质对r 、∈s R 均适用.3.指数函数: ① 定义:函数 称为指数函数,1) 函数的定义域为 ;2) 函数的值域为 ;3) 当________时函数为减函数,当_______时为增函数. ② 函数图像:1) 过点 ,图象在 ;2) 指数函数以 为渐近线(当10<<a 时,图象向 无限接近x 轴,当1>a 时,图象向 无限接近x 轴);3)函数xx a y a y -==与的图象关于 对称.③ 函数值的变化特征:10<<a1>a① 时0>x ② 时0=x ③ 时0<x ① 时0>x ② 时0=x ③ 时0<xfrom masses most care, a nd m ost dire ctly, a nd most reality of interests pr oblem grabbed, real do l ove for pe ople by Departme nt, and right for pe ople by with, a nd Lee for pe opl e by conspiracy t o. Thre e with the fame of mind. "Non-i ndifferent not insi st your dream, without sere nity not go far." Ascension t o fame is morality, is the sublimation of the soul. A s T ownship a nd vil lage, be sure to mai ntain a normal state of mind, e stablish a correct view poi nt on pow er, positi on a nd val ues and Outlook, as fame is light lik e water, de pending ... 50 km, also red a nd like hundreds of thousa nds of the KMT military combat, but he went on to colle ct a variety of newspapers and magazine s, and the n race agai nst time to pore over. A s long a s the st udy of mind, it is not ne cessary t o come t o the li brary, w orkshops, into the villages, people lear ned farming techni que s, pr oblem -solvi ng methods, partici pation i nonfere nces, learne d superi ors policie s; reports, you ca n learn t o deal wit h problems, art, just wante d to lear n, to lear n at any time. To continuously expa nd the scope of the study. T he complexity and comprehensive ness of our w ork, deci ded to not only do we want to be "professionals", or if "generalists", to dabble i n different areas of knowle dge. Mastery of knowle dge not only t o have knowle dge to rational analysis. T his on requirements w e, not only to le arning national of route , and approach, a nd policy, also to lear ning national of legal, and regulati ons; not only to wit h political t heory armed mind, al so to wit h economic the ory gui de work ; not only to rea d hist ory, al so to rea d hist ory; not only to a bsorpti on SM of mirror, also t o gets is of roa d; not only t o from local lear n, also to to outsi de learn; not only to t o books lear n, also to field st udy lear n. Therefore, I would e ncourage you, and more to the out side w orld w hen I have time to take a look at, take a stroll, outside devel opme nts, learn other people's development a pproach, and the n come back to g uide our w ork. Second, we must think. "Learni ng without thought to over sha dow, thi nking wit hout learni ng is peri lous." We wa nt to foster the character of advancing wit h the times, forge ahead, often reflect themse lves in a dapti ng to the new situati on on w hether t here are delays, have a nythi ng to fear in the reform and innovation, whether i n terms of accelerati ng the development of a relaxat ion, keep pr omising State of mind. In order t o do i nvestigations, investigati ons i s the roa d to heaven, God dispose s of the base. o wa nts to real he ard truth, and justi ce to trut h, and get truth, re ceived effective ness, on m ust in -de pth a ctual, a nd outreach, a nd in -depth ma sses, w orship masses for Divisi on, a nd masses heart, with masses with discussion everyone care of problem, listeni ng they of calls, understand they of mood, feel they of suffering, summary they of experience, conce ntrated they of wisdom , e spe cially on masse s most hope , and most urgent, a nd most w orries, a nd most com plaine d of hot, and difficultie s and focus problem, more to a ctive resear ch, caug ht not put. Dare t o innovate, in our w ork, often remain "the old way doe sn't work, t he ne w approach w oul d not" situation, the rea son, the key is that many people k now that copying ot her pe opl e's idea s, mechani cally, usi ng some one else's old way s. To change thi s predicament, re quire s us t o break the sha ckles of traditional conce pts to over come conservative, conformist, g ood at finding new problems a nd to a cce pt ne w thing s, summarize new experience, new idea s, deve loping ne w programmes a nd Maveri ck, a ne w approa ch to solve pr oblem s, speed up development. Espe cially in impl ement superi or policy spirit of process in the , bot h implementati on, a nd cannot mechani call y, to to extraordi nary of courage a nd develop enterpri sing of spirit, put flexibility a nd pri nci ple combi ned up, put superi or of spirit a nd l ocal reality combi ned up, ong conspiracy to breakthr ough of poli cy, a nd exercise innovati on of lift, devel ope d out devel opme nt works of, and grass -root s welcomes of- 4 -对数函数1.对数:(1) 定义:如果N a b =)1,0(≠>a a 且,那么称 为 ,记作 ,其中a 称为对数的底,N 称为真数.① 以10为底的对数称为常用对数,N 10log 记作___________.② 以无理数)71828.2( =e e 为底的对数称为自然对数,N e log 记作_________. (2) 基本性质:① 真数N 为 (负数和零无对数);② 01log =a ;③ 1log =a a ; ④ 对数恒等式:N a N a =log . (3) 运算性质:① log a (MN)=___________________________; ② log a NM =____________________________;③ log a M n= (n ∈R).④ 换底公式:log a N = (a >0,a ≠1,m >0,m ≠1,N>0)⑤ log mna a nb b m = .2.对数函数:① 定义:函数 称为对数函数,1) 函数的定义域为( ;2) 函数的值域为 ;3) 当______时,函数为减函数,当______时为增函数;4) 函数x y a log =与函数)1,0(≠>=a a a y x且互为反函数. ② 1) 图象经过点( ),图象在 ;2) 对数函数以 为渐近线(当10<<a 时,图象向上无限接近y 轴;当1>a 时,图象向下无限接近y 轴); 4) 函数y =log a x 与 的图象关于x 轴对称. ③ 函数值的变化特征:10<<a1>a① 时1>x ② 时1=x ③ 时10<<x ① 时1>x ② 时1=x ③ 时10<<x典型例题:例1.求下列函数的值域: (1)y=521+-x x (2)y=|x|21x-例2:已知函数f(x)=x 2-(1)求函数的值域为[0,+∞)时的a 的值;(2)若函数的值均为非负值,求函数f(a)=2-a|a+3|的值域.up, from masses most care, nd most directly, nd most reality of intere sts pr oblem grabbed, real do ove for people by De partment, nd right for pe opl e by with, nd Le e for people by conspira cy to. Thr ee with the fame of mind. "Non -indifferent not insi st your dream, wit hout serenity not go far." Asce nsi on to fame is morality, is the sublimati on of the soul. Tow nshi p and village, be sure to maintai n a normal state of mind, establ ish a correct view poi nt on pow er, position and value and Outl ook, as fame is light like water, depe ndi ng ... 50 km, also red and like hundreds of thousands of the KMT military combat but he w ent on to col lect a variety of new spa pers and magazines, and the n race against time to pore over. long as the study of mind, it is not nece ssary to come to the libr ary, workshops, int o the village s, pe ople lear ned farming techni ques, pr oblem -solving methods, partici pationin大方向教育——值得您信赖的专业化个性化辅导学校onfere nces, learne d superi ors policie s; reports, you ca n learn t o deal wit h problems, art, just wante d to lear n, to lear n at any time. To continuously expa nd the scope of the study. T he complexity and comprehensive ness of our w ork, deci ded to not only do we want to be "professionals", or if "generalists", to dabble i n different areas of knowle dge. Mastery of knowle dge not only t o have knowle dge to rational analysis. T his on requirements w e, not only to le arning national of route , and approach, a nd policy, also to lear ning national of legal, and regulati ons; not only to wit h political t heory armed mind, al so to wit h economic the ory gui de work ; not only to rea d hist ory, al so to rea d hist ory; not only to a bsorpti on SM of mirror, also t o gets is of roa d; not only t o from local lear n, also to to outsi de learn; not only to t o books lear n, also to field st udy lear n. Therefore, I would e ncourage you, and more to the out side w orld w hen I have time to take a look at, take a stroll, outside devel opme nts, learn other people's development a pproach, and the n come back to g uide our w ork. Sec ond, we must think. "Learni ng without thought to over sha dow, thi nking wit hout learni ng is peri lous." We wa nt to foster the character of advancing wit h the times, forge ahead, often reflect themse lves in a dapti ng to the new situati on on w hether t here are delays, have a nythi ng to fear in the reform and innovation, whether i n terms of accelerati ng the development of a relaxation, k eep pr omising State of mind. In order t o do i nvestigations, investigati ons i s the roa d to heaven, God dispose s of the base. o wa nts to real he ard truth, and justi ce to trut h, and get truth, re ceived effective ness, on m ust in -de pth a ctual, a nd outreach, a nd in -depth ma sses, w orship masses for Divisi on, a nd masses heart, with masses with discussion everyone care of problem, listeni ng they of calls, understand they of mood, feel they of suffering, summary they of experience, conce ntrated they of wisdom, e spe cially on masse s most hope , and most urgent, a nd most w orries, a nd most com plaine d of hot, and difficultie s and focus problem, more to a ctive resear ch, caug ht not put. Dare t o innovate, in our w ork, often remain "the old way doe sn't work, t he ne wapproach w oul d not" situation, the rea son, the key is that many people k now that copying ot her pe opl e's idea s, mechani cally, usi ng some one else's old way s. To change thi s predicament, re quire s us t o break the sha ckles of traditional conce pts to over come conservative, conformist, g ood at finding new problems a nd to a cce pt ne w thing s, summarize new experience, new idea s, deve loping ne w programmes a nd Maveri ck, a ne w approa ch to solve pr oblem s, speed up development. Espe cially in impl ement superi or policy spirit of process in the , bot h implementati on, a nd cannot mechani call y, to to extraordi nary of courage a nd develop enterpri sing of spirit, put flexibility a nd pri nci ple combi ned up, put superi or of spirit a nd l ocal reality combi ned up, ong conspiracy to breakthr ough of poli cy, a nd exercise innovati on of lift, devel ope d out devel opme nt works of, and grass -root s welcomes of- 5 -例3.(1)y=4-223x x -+; (2)y=x+x4;(3)y=4)2(122+-++x x .例4.已知f(x)是以2为周期的偶函数,且当x ∈(0,1)时,f(x)=2x -1,则f(log 212)的值为例5.已知函数y=31++-x x 的最大值为M ,最小值为m ,则Mm的值为例6.f(x)是定义在R 上的以3为周期的偶函数,且f(2)=0,则方程f(x)=0在区间(0,6)内解的个数的最小值是例7.定义在R 上的函数f(x )满足f(x)= ⎩⎨⎧>---≤-0),2()1(0),1(log 2x x f x f x x ,则f (2009)的值为from masses most care, a nd m ost dire ctly, a nd most reality of interests pr oblem grabbed, real do l ove for pe ople by Departme nt, and right for pe ople by with, a nd Lee for pe opl e by conspiracy t o. Thre e with the fame of mind. "Non-i ndifferent not insi st your dream, without sere nity not go far." Ascension t o fame is morality, is the sublimation of the soul. A s T ownship a nd vil lage, be sure to mai ntain a normal state of mind, e stablish a correct view poi nt on pow er, positi on a nd val ues and Outlook, as fame is light lik e water, de pending ... 50 km, also red a nd like hundreds of thousa nds of the KMT military combat, but he went on to colle ct a variety of newspapers and magazine s, and the n race agai nst time to pore over. A s long a s the st udy of mind, it is not ne cessary t o come t o the li brary, w orkshops, into the villages, people lear ned farming techni que s, pr oblem -solvi ng methods, partici pation i nonfere nces, learne d superi ors policie s; reports, you ca n learn t o deal wit h problems, art, just wante d to lear n, to lear n at any time. To continuously expa nd the scope of the study. T he complexity and comprehensive ness of our w ork, deci ded to not only do we want to be "professionals", or if "generalists", to dabble i n different areas of knowle dge. Mastery of knowle dge not only t o have knowle dge to rational analysis. T his on requirements w e, not only to le arning national of route , and approach, a nd policy, also to lear ning national of legal, and regulati ons; not only to wit h political t heory armed mind, al so to wit h economic the ory gui de work ; not only to rea d hist ory, al so to rea d hist ory; not only to a bsorpti on SM of mirror, also t o gets is of roa d; not only t o from local lear n, also to to outsi de learn; not only to t o books lear n, also to field st udy lear n. Therefore, I would e ncourage you, and more to the out side w orld w hen I have time to take a look at, take a stroll, outside devel opme nts, learn other people's development a pproach, and the n come back to g uide our w ork. Second, we must think. "Learni ng without thought to over sha dow, thi nking wit hout learni ng is peri lous." We wa nt to foster the character of advancing wit h the times, forge ahead, often reflect themse lves in a dapti ng to the new situati on on w hether t here are delays, have a nythi ng to fear in the reform and innovation, whether i n terms of accelerati ng the development of a relaxat ion, keep pr omising State of mind. In order t o do i nvestigations, investigati ons i s the roa d to heaven, God dispose s of the base. o wa nts to real he ard truth, and justi ce to trut h, and get truth, re ceived effective ness, on m ust in -de pth a ctual, a nd outreach, a nd in -depth ma sses, w orship masses for Divisi on, a nd masses heart, with masses with discussion everyone care of problem, listeni ng they of calls, understand they of mood, feel they of suffering, summary they of experience, conce ntrated they of wisdom , e spe cially on masse s most hope , and most urgent, a nd most w orries, a nd most com plaine d of hot, and difficultie s and focus problem, more to a ctive resear ch, caug ht not put. Dare t o innovate, in our w ork, often remain "the old way doe sn't work, t he ne wapproach w oul d not" situation, the rea son, the key is that many people k now that copying ot her pe opl e's idea s, mechani cally, usi ng some one else's old way s. To change thi s predicament, re quire s us t o break the sha ckles of traditional conce pts to over come conservative, conformist, g ood at finding new problems a nd to a cce pt ne w thing s, summarize new experience, new idea s, deve loping ne w programmes a nd Maveri ck, a ne w approa ch to solve pr oblem s, speed up development. Espe cially in impl ement superi or policy spirit of process in the , bot h implementati on, a nd cannot mechani call y, to to extraordi nary of courage a nd develop enterpri sing of spirit, put flexibility a nd pri nci ple combi ned up, put superi or of spirit a nd l ocal reality combi ned up, ong conspiracy to breakthr ough of poli cy, a nd exercise innovati on of lift, devel ope d out devel opme nt works of, and grass -root s welcomes of- 6 -例8.定义在R 上的函数f(x )满足f(x)= ⎩⎨⎧>---≤-0),2()1(0),4(log 2x x f x f x x ,则f (3)的值为例9.已知函数()f x 是(,)-∞+∞上的偶函数,若对于0x ≥,都有(2()f x f x +=),且当[0,2)x ∈时,2()log (1f x x =+),则(2008)(2009)f f -+的值例10.设函数⎩⎨⎧<+≥+-=0,60,64)(2x x x x x x f 则不等式)1()(f x f >的解集是例11.已知偶函数()f x 在区间[0,)+∞单调增加,则满足(21)f x -<1()3f 的x 取值范围是。
(二)函数的定义域(1)解决函数问题,优先考虑定义域.若没有标明定义域,则认为定义域是使得函数解析式有意义的x 的取值范围.实际问题中还要考虑自变量的实际意义.(2)分式中分母0≠;偶次根式中被开方数应为非负数;)0(10≠==x x y ;)10(≠>=a a a y x 且;,log x y a =真数,0>x 底数10≠>a a 且;x y sin =定义域为,R x y cos =定义域为,R x y tan =定义域为x {|},2Z k k x ∈+≠ππ.(3)复合函数的定义域方法:①定义域是输入值x 的集合;②同一对应法则下的括号内整体范围一样.例:已知)1(+=x f y 的定义域为],3,2[-则)12(-=x f y 的定义域为.答案:]25,0[小结:①若已知)(x f 的定义域为],,[b a 则复合函数))((x g f 的定义域可由b x g a ≤≤)(解出;②若已知))((x g f 的定义域为],,[b a 则)(x f 的定义域即为],[b a x ∈时)(x g 的值域.(三)函数的值域(数形结合)常用方法法一:图象法(形)1.)10(22≤<+-=x x x y 2..30,113<≤+-=x x x y 3..14,4-≤≤-+=x xx y 法二:换元法+图象法(形)4.3212++=x x y 5.x x y 21-+= 6.1212+-=x x y 7.)0(422>+=x x x y 8.).1(1542>-+-=x x x x y 9.)10(210212≤≤++=x x xy 法三:单调性(导数和单调性的性质)(数)10.x x y 21--=11.2,0[,sin π∈+=x x x y 12.]3,3[,8123-∈+-=x x x y 法四:几何意义(形)13.2cos 1sin --=x x y 答案:1.]81,1[-;2.)2,1[-;3.]4,5[--;4.]21,0(;5.]1,(-∞;6.)1,1(-;7.]21,0(;8.),222[+∞-;9.]10103,22[;10.21,(-∞;11.]12,0[+π;12.]24,8[-;13.34,0[。