3.1.1函数的概念课件-2024-2025学年高一上学期数学人教A版(2019)必修第一册

2024-09-20
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 函数的概念与性质 3.1 函数的概念及其表示 第一课时 函数的概念 1 2 PPT模板:www.1ppt.com/moban/ PPT素材:www.1ppt.com/sucai/ PPT背景:www.1ppt.com/beijing/ PPT图表:www.1ppt.com/tubiao/ PPT下载:www.1ppt.com/xiazai/ PPT教程: www.1ppt.com/powerpoint/ 资料下载:www.1ppt.com/ziliao/ 范文下载:www.1ppt.com/fanwen/ 试卷下载:www.1ppt.com/shiti/ 教案下载:www.1ppt.com/jiaoan/ PPT论坛:www.1ppt.cn PPT课件:www.1ppt.com/kejian/ 语文课件:www.1ppt.com/kejian/yuwen/ 数学课件:www.1ppt.com/kejian/shuxue/ 英语课件:www.1ppt.com/kejian/yingyu/ 美术课件:www.1ppt.com/kejian/meishu/ 科学课件:www.1ppt.com/kejian/kexue/ 物理课件:www.1ppt.com/kejian/wuli/ 化学课件:www.1ppt.com/kejian/huaxue/ 生物课件:www.1ppt.com/kejian/shengwu/ 地理课件:www.1ppt.com/kejian/dili/ 历史课件:www.1ppt.com/kejian/lishi/ 2 客观世界中有各种各样的运动变化现象. 例如,天宫二号在发射过程中,离发射点的距离随时间的变化而变化;一个装满水的蓄水池在使用过程中,水面高度随时间的变化而不断降低;我国高速铁路营业里程逐年增加,已突破2万公里…… 所有这些都表现为变量间的对应关系,这种关系常常可用函数模型来描述, 并且通过研究函数模型就可以把握相应的运动变化规律. 在初中我们已经接触过函数的概念,知道函数是刻画变量之间对应关系的数学模型和工具. 初中所学的函数的概念: 在一个变化过程中有两个变量x和y,如果对于x的每一个值,y都有唯一的值与它对应.那么就说y是x的函数,其中x叫做自变量. 初中已经学过的函数: 正比例函数:y=kx (k≠0); 反比例函数: y = (k≠0); 一次函数: y=kx+b (k≠0); 二次函数:y=ax2+bx+c (a≠0) 然而,客观世界纷繁复杂,变量之间的函数模型远不止这些,复杂程度也不是初中的函数定义所能涵盖的. 先分析以下问题: 问题1 某“复兴号”高速列车加速到350km/h后保持匀速运行半小时.这段时间内,列车行进的路程S(单位:km)与运行时间t(单位:h)的关系可以表示为: S=350t ① 这里S和t是两个变量,而且对于t的每一个确定的值,S都有唯一确定的值与之对应,所以S是t的函数. t的变化范围:A1={t|0≤t≤0.5}, S的变化范围:B1={S|0≤S≤175}. A1中任一t,按照①,B1中都有唯一确定的S和它对应. 问题2 某电气维修公司要求工人每周工作至少1天,至多不超过6天.如果公司确定的工资标准是每人每天350元,而且每周付一次工资,那么你认为该怎样确定一个工人每周的工资? 显然,工资w 是一周工作天数d 的函数,其对应关系是 w=350d ② d的变化范围:A2={1, 2, 3, 4, 5, 6}, w的变化范围:B2={350, 700, 1050, 1400, 1750, 2100}. 对于A2中任一d,按照②,B2中都有唯一确定的w和它对应. 问题3 图1是北京市2016年11月23日的空气质量指数变化图.如何根据该图确定这一天内任一时刻th的空气质量指数的值I? 图1 t的变化范围:A3={t|0≤t≤24}, I的变化范围:B3={I|0<I<150}. 对于A3中任一t,按照图1,B3中都有唯一确定的I和它对应. 问题4 国际上常用恩格尔系数r(r=食物支出金额/总支出金额)反映一个地区人民生活质量的高低,恩格尔系数越低,生活质量越高.表1是我国某省城镇居民恩格尔系数变化情况,从中可以看出,该省城镇居民的生活质量越来越高. 年份y的范围:A4={2006,2007,…,2015}, 系数r的范围:B4={r|0<r≤1}. 对于A4中任一y,按照表1,B4中都有唯一确定的r和它对应. 上述问题1~问题4中的函数有哪些不同点和共同特征? 不同点: 问题1和问题2是用解析式刻画变量之间的对应关系, 问题3是用图象刻画变量之间的对应关系, 问题4是用表格刻画变量之间的对应关系. 共同特征: (1)都包含两个非空数集,用A,B来表示; (2)都有一个对应关系; (3)尽管对应关系表示方法不同,但它们都有如下特性:对于数集A中的任意一个数x,按照对应关系,在数集B中都有唯一确定的数y和它对应. 11 PPT模板:www.1ppt.com/moban/ PPT素材:www.1ppt.com/sucai/ PPT背景:www.1ppt.com/beijing/ PPT图表:www.1ppt.com/tubiao/ PPT下载:www.1ppt.com/xiazai/ PPT教程: www.1ppt.com/powerpoint/ 资料下载:www.1ppt.com/ziliao/ 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PPT课件:www.1ppt.com/kejian/ 语文课件:www.1ppt.com/kejian/yuwen/ 数学课件:www.1ppt.com/kejian/shuxue/ 英语课件:www.1ppt.com/kejian/yingyu/ 美术课件:www.1ppt.com/kejian/meishu/ 科学课件:www.1ppt.com/kejian/kexue/ 物理课件:www.1ppt.com/kejian/wuli/ 化学课件:www.1ppt.com/kejian/huaxue/ 生物课件:www.1ppt.com/kejian/shengwu/ 地理课件:www.1ppt.com/kejian/dili/ 历史课件:www.1ppt.com/kejian/lishi/ 43 学 习 目 标 核 心 素 养 1.进一步体会函数是描述变量之间的依赖关系的重要数学模型.能用集合与对应的语言刻画出函数,体会对应关系在刻画数学概念中的作用.(重点、难点) 2.了解构成函数的要素,会求一些简单函数的定义域和值域.(重点) 3.能够正确使用区间表示数集.(易混点) 1.通过学习函数的概念,培养数学抽象素养. 2.借助函数定义域的求解,培养数学运算素养. 3.借助f(x)与f(a)的关系,培养逻辑推理素养. 唯一确定 自变量x {f(x)|x∈A} 1.函数的概念 定义 一般地,设A,B是非空的   ,如果对于集合A中的      按照某种确定的对应关系f,在集合B中都有    的数y和它对应,那么就称f:A→B为从集合A到集合B的一个函数 三要素 对应关系 y=f(x),x∈A 定义域     的取值范围 值域 与x的值相对应的y的函数值的集合        实数集 任意一个数x 思考1:(1)有人认为“y=f(x)”表示的是“y等于f与x的乘积”,这种看法对吗? (2)f(x)与f(a)有何区别与联系? 提示:(1)这种看法不对. 符号y=f(x)是“y是x的函数”的数学表示,应理解为x是自变量,它是关系所施加的对象;f是对应关系,它可以是一个或几个解析式,可以是图象、表格,也可以是文字描述;y是自变量的函数,当x允许取某一具体值时,相应的y值为与该自变量值对应的函数值.y=f(x)仅仅是函数符号,不表示“y等于f与x的乘积”.在研究函数时,除用符号f(x)外,还常用g(x),F(x),G(x)等来表示函数. (2)f(x)与f(a)的区别与联系:f(a)表示当x=a时,函数f(x)的值,是一个常量,而f(x)是自变量x的函数,一般情况下,它是一个变量,f(a)是f(x)的一个特殊值,如一次函数f(x)=3x+4,当x=8时,f(8)=3×8+4=28是一个常数. 2.区间及有关概念 (1)一般区间的表示 设a,b∈R,且a<b,规定如下: 定义 名称 符号 数轴表示 {x|a≤x≤b} 闭区间 __________________ {x|a<x<b} 开区间 __________________ {x|a≤x<b} 半开半闭区间 [a,b) {x|a<x≤b} 半开半闭区间 (a,b] [a,b] (a,b) (2)特殊区间的表示 定义 R {x|x≥a} {x|x>a} {x|x≤a} {x|x<a} 符号        [a,+∞) (a,+∞) (-∞,a] (-∞,a) (-∞,+∞) 思考2:(1)区间是数集的另一种表示方法,那么任何数集都能用区间表示吗? (2)“∞”是数吗?如何正确使用“∞”? 提示:(1)不是任何数集都能用区间表示,如集合{0}就不能用区间表示. (2)“∞”读作“无穷大”,是一个符号,不是数.以“-∞”或“+∞”作为区间一端时,这一端必须是小括号. 【例1】 (1)下列各组函数是同一函数的是(  ) ①f(x)=eq \r(-2x3)与g(x)=xeq \r(-2x); ②f(x)=x与g(x)=eq \r(x2); ③f(x)=x0与g(x)=eq \f(1,x0); ④f(x)=x2-2x-1与g(t)=t2-2t-1. A.①②    B.①③ C.③④ D.①④ 函数的概念 (1)C [①f(x)=eq \r(-2x3)=|x|eq \r(-2x)与g(x)=xeq \r(-2x)的对应法则和值域不同,故不是同一函数. ②g(x)=eq \r(x2)=|x|与f(x)=x的对应法则和值域不同,故不是同一函数. ③f(x)=x0与g(x)=eq \f(1,x0)都可化为y=1且定义域是{x|x≠0},故是同一函数. ④f(x)=x2-2x-1与g(t)=t2-2t-1的定义域都是R,对应法则也相同,而与用什么字母表示无关,故是同一函数. 由上可知是同一函数的是③④. 故选C.] (2)判断下列对应是不是从集合A到集合B的函数. ①A=N,B=N*,对应法则f:对集合A中的元素取绝对值与B中元素对应; ②A={-1,1,2,-2},B={1,4},对应法则f:x→y=x2,x∈A,y∈B; ③A={-1,1,2,-2},B={1,2,4},对应法则f:x→y=x2,x∈A,y∈B; ④A={三角形},B={x|x>0},对应法则f:对A中元素求面积与B中元素对应. [解] ①对于A中的元素0,在f的作用下得0,但0不属于B,即A中的元素0在B中没有元素与之对应,所以不是函数. ②对于A中的元素±1,在f的作用下与B中的1对应,A中的元素±2,在f的作用下与B中的4对应,所以满足A中的任一元素与B中唯一元素对应,是“多对一”的对应,故是函数. ③对于A中的任一元素,在对应关系f的作用下,B中都有唯一的元素与之对应,如±1对应1,±2对应4,所以是函数. ④集合A不是数集,故不是函数.] 1.判断对应关系是否为函数的2个条件 (1)A,B必须是非空实数集. (2)A中任意一元素在B中有且只有一个元素与之对应. 对应关系是“一对一”或“多对一”的是函数关系,“一对多”的不是函数关系. 2.判断函数相等的方法 (1)先看定义域,若定义域不同,则不相等; (2)若定义域相同,再化简函数的解析式,看对应关系是否相同. 1.下列四个图象中,不是函数图象的是(  ) A   B   C   D B [根据函数的定义知:y是x的函数中,x确定一个值,y就随之确定一个值,体现在图象上,图象与平行于y轴的直线最多只能有一个交点,对照选项,可知只有B不符合此条件.故选B.] 2.下列各组函数中是相等函数的是(  ) A.y=x+1与y=eq \f(x2-1,x-1) B.y=x2+1与s=t2+1 C.y=2x与y=2x(x≥0) D.y=(x+1)2与y=x2 B [A,C选项中两函数的定义域不同,D选项中两函数的对应关系不同,故A,C,D错误,选B.] 【例2】 设f(x)=2x2+2,g(x)=eq \f(1,x+2), (1)求f(2),f(a+3),g(a)+g(0)(a≠-2),g(f(2)). (2)求g(f(x)). [思路点拨] (1)直接把变量的取值代入相应函数解析式,求值即可; (2)把f(x)直接代入g(x)中便可得到g(f(x)). 求函数值 [解] (1)因为f(x)=2x2+2, 所以f(2)=2×22+2=10, f(a+3)=2(a+3)2+2=2a2+12a+20.因为g(x)=eq \f(1,x+2), 所以g(a)+g(0)=eq \f(1,a+2)+eq \f(1,0+2)=eq \f(1,a+2)+eq \f(1,2)(a≠-2). g(f(2))=g(10)=eq \f(1,10+2)=eq \f(1,12). (2)g(f(x))=eq \f(1,fx+2)=eq \f(1,2x2+2+2)=eq \f(1,2x2+4). 函数求值的方法 1已知fx的表达式时,只需用a替换表达式中的x即得fa的值. 2求fga的值应遵循由里往外的原则. 3.已知f(x)=x3+2x+3,求f(1),f(t),f(2a-1)和f(f(-1))的值. [解] f(1)=13+2×1+3=6; f(t)=t3+2t+3; f(2a-1)=(2a-1)3+2(2a-1)+3=8a3-12a2+10a; f(f(-1))=f((-1)3+2×(-1)+3)=f(0)=3. [探究问题] 1.已知函数的解析式,求其定义域时,能否可以对其先化简再求定义域? 提示:不可以.如f(x)=eq \f(x+1,x2-1).倘若先化简,则f(x)=eq \f(1,x-1),从而定义域与原函数不等价. 求函数的定义域 2.若函数y=f(x+1)的定义域是[1,2],这里的“[1,2]”是指谁的取值范围?函数y=f(x)的定义域是什么? 提示:[1,2]是自变量x的取值范围. 函数y=f(x)的定义域是x+1的范围[2,3]. 【例3】 求下列函数的定义域: (1)f(x)=2+eq \f(3,x-2); (2)f(x)=(x-1)0+eq \r(\f(2,x+1)); (3)f(x)=eq \r(3-x)·eq \r(x-1); (4)f(x)=eq \f(x+12,x+1)-eq \r(1-x). [思路点拨] 要求函数的定义域,只需分母不为0,偶次方根中被开方数大于等于0即可. [解] (1)当且仅当x-2≠0,即x≠2时, 函数f(x)=2+eq \f(3,x-2)有意义, 所以这个函数的定义域为{x|x≠2}. (2)函数有意义,当且仅当eq \b\lc\{\rc\ (\a\vs4\al\co1(x-1≠0,,\f(2,x+1)≥0,,x+1≠0,)) 解得x>-1且x≠1, 所以这个函数的定义域为{x|x>-1且x≠1}. (3)函数有意义,当且仅当eq \b\lc\{\rc\ (\a\vs4\al\co1(3-x≥0,,x-1≥0,))解得1≤x≤3, 所以这个函数的定义域为{x|1≤x≤3}. (4)要使函数有意义,自变量x的取值必须满足eq \b\lc\{\rc\ (\a\vs4\al\co1(x+1≠0,,1-x≥0,))解得x≤1且x≠-1, 即函数定义域为{x|x≤1且x≠-1}. (变结论)在本例(3)条件不变的前提下,求函数y=f(x+1)的定义域. [解] 由1≤x+1≤3得0≤x≤2. 所以函数y=f(x+1)的定义域为[0,2]. 求函数定义域的常用方法 1若fx是分式,则应考虑使分母不为零. 2若fx是偶次根式,则被开方数大于或等于零. 3若fx是指数幂,则函数的定义域是使幂运算有意义的实数集合. 4若fx是由几个式子构成的,则函数的定义域是几个部分定义域的交集. 5若fx是实际问题的解析式,则应符合实际问题,使实际问题有意义. 1.对于用关系式表示的函数.如果没有给出定义域,那么就认为函数的定义域是指使函数表达式有意义的自变量取值的集合.这也是求某函数定义域的依据. 2.函数的定义主要包括定义域和定义域到值域的对应法则,因此,判定两个函数是否相同时,就看定义域和对应法则是否完全一致,完全一致的两个函数才算相同. 3.函数符号y=f(x)是学习的难点,它是抽象符号之一.首先明确符号“y=f(x)”为y是x的函数,它仅仅是函数符号,不是表示“y等于f与x的乘积”. 1.思考辨析 (1)区间表示数集,数集一定能用区间表示.(  ) (2)数集{x|x≥2}可用区间表示为[2,+∞].(  ) (3)函数的定义域和对应关系确定后,函数的值域也就确定了.(  ) (4)函数值域中每一个数在定义域中一定只有一个数与之对应.(  ) (5)函数的定义域和值域一定是无限集合.(  ) [答案] (1)× (2)× (3)√ (4)× (5)× 2.下列函数中,与函数y=x相等的是(  ) A.y=(eq \r(x))2   B.y=eq \r(x2) C.y=|x| D.y=eq \r(3,x3) D [函数y=x的定义域为R;y=(eq \r(x))2的定义域为[0,+∞);y=eq \r(x2)=|x|,对应关系不同;y=|x|对应关系不同;y=eq \r(3,x3)=x,且定义域为R.故选D.] 3.将函数y=eq \f(3,1-\r(1-x))的定义域用区间表示为________. (-∞,0)∪(0,1] [由eq \b\lc\{\rc\ (\a\vs4\al\co1(1-x≥0,,1-\r(1-x)≠0,)) 解得x≤1且x≠0, 用区间表示为(-∞,0)∪(0,1].] 4.已知函数f(x)=x+eq \f(1,x), (1)求f(x)的定义域; (2)求f(-1),f(2)的值; (3)当a≠-1时,求f(a+1)的值. [解] (1)要使函数f(x)有意义,必须使x≠0, ∴f(x)的定义域是(-∞,0)∪(0,+∞). (2)f(-1)=-1+eq \f(1,-1)=-2,f(2)=2+eq \f(1,2)=eq \f(5,2). (3)当a≠-1时,a+1≠0, ∴f(a+1)=a+1+eq \f(1,a+1). $$

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3.1.1函数的概念课件-2024-2025学年高一上学期数学人教A版(2019)必修第一册
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