4.4.2.1对数函数的图象和性质 （第一课时）课件-2025-2026学年高一上学期数学人教A版（2019）必修第一册

4.4.2.1对数函数的图象和性质 2019人教A版第一册第四章 1 复习回顾 回顾:如何研究一类新函数?研究一个函数的性质主要是研究哪些方面? 函数图象 值域 单调性 奇偶性 定点/关键点 定义域 复习回顾 问题:如何研究对数函数的图象和性质? 研究指数函数性质的过程与方法 研究对数函数性质的过程与方法 类比学习 探究新知 探究1:请同学们完成x,y的对应表4.2-2,并用描点法画出函数y=log2x的图象. x y 0.5 1 2 4 6 8 12 16 x y O 描点法步骤:列表—描点—连线 探究新知 探究新知 探究2:请同学们用描点法画出y=x的图象. x y O x y 0.5 1 2 4 6 8 12 16 探究新知 探究新知 问题1:把y=log2x和y=x的图象放在同一个直角坐标系中,观察它们的图象有什么关系? 由图可观察出y=log2x和y=x的图象关于x轴对称 探究新知 追问1: y=log2x和y=x的图象关于x轴对称,你能对此进行证明吗? 且y=x= - log2x 证明:因为点(x,y)与(x,-y)关于x轴对称, 所以函数图象上的任意一点P(x,y)关于y轴的对称点P1(x,-y)都在函数y=x的图象上.反之亦然. 探究新知 追问2:除了描点法画出y=x的图象,你还能用什么方法作出y=x的图象? 因为 y=log2x和y=x的图象关于x轴对称.可利用函数y=log2x的图象,画出y=x的图象 追问4:当a取其他值时,它们的图象是否都有这样的关系? 追问3: y=log2x和y=x的底数有什么关系? 利用对称性作图 底数互为倒数,图象关于x轴对称 探究新知 探究3:请同学们选取底数a(a>0,且a≠1)的若干个不同的值,在同一直角坐标系内画出相应的对数函数的图象. x y O 探究新知 探究新知 问题2:观察这些图象的位置、公共点和变化趋势,它们有哪些共性?由此你能概括出对数函数y=logax(a>0,且a≠1)的值域和性质吗? 项目 0<a<1 a>1 图象 定义域 值域 性质 R (0,+) (1)过定点(1,0),即x=1时,y=0 (2)减函数 (2)增函数 记忆口诀:大“1”增,小“1”减,图象恒过(1,0)点. 探究新知 追问:观察这些图象,对数函数的图象与底数有什么关系? (1)底数互为倒数的两个对数函数的图象关于x轴对称. 即y=logax和y=x(a>0,且a≠1)的图象关于x轴对称. (2)a>1时,图象:“底大图低”; 0<a<1时,图象:“底大图低”; 探究新知 √ √ 辨析:判断下列说法是否正确,正确的在它后面的括号里画“√”,错误的画“ ”. (1)对数函数的图象一定在y轴的右侧.( ) (2)对数函数y = logax(a>0,且a≠1)在(0,+∞)上是增函数.( ) (3)当a>1时,若0<x<1,则logax<0.( ) (4)函数 y=x与y = logax(a>0,且a≠1)的图象关于y轴对称.( ) 例题讲解 例3 比较下列各题中两个值的大小: (1)log23.4, log28.5 (2) log0.31.8, log0.32.7 (3) loga5.1, loga5.9 (a>0,且a≠1) 分析:对于(1)(2),要比较的两个值可以看作一个对数函数的两个函数值,因此可以直接利用对数函数的单调性进行比较;对于(3) loga5.1, loga5.9 (a>0,且a≠1)需讨论a的范围,再直接利用对数函数的单调性进行比较. 例题讲解 例3 比较下列各题中两个值的大小: 解: (1) log23.4和 log28.5可看作函数y=log2x当x取3.4和8.5时所对应的两个函数值. ∵底数2>1 ∴对数函数y=log2x是增函数 ∵ 3.4<8.5 ∴ log23.4 < log28.5 底数相同,真数不同: 构造对数函数,利用对数函数单调性进行比较. (1)log23.4, log28.5 (2) log0.31.8, log0.32.7 (3)loga5.1, loga5.9 (a>0,且a≠1) 例题讲解 例3 比较下列各题中两个值的大小: 解: (2) log0.31.8和log0.32.7可看作函数y=log0.3x当x取和时所对应的两个函数值. ∵底数0<0.3<1 ∴指数函数y=log0.3x是减函数 ∵1.8<2.7 ∴ log0.31.8>log0.32.7 底数相同,真数不同: 构造对数函数,利用对数函数单调性进行比较. (1)log23.4, log28.5 (2) log0.31.8, log0.32.7 (3)loga5.1, loga5.9 (a>0,且a≠1) 例题讲解 例3 比较下列各题中两个值的大小: 解: (3) loga5.1和loga5.9可看作函数y= logax当x取和时所对应的两个函数值.对数函数的单调性取决于底数a是大于1还是小于1,因此需要对底数a进行讨论. ①当a>1时,y= logax是增函数,且5.1<5.9,所以loga5.1<loga5.9 (1)log23.4, log28.5 (2) log0.31.8, log0.32.7 (3)loga5.1, loga5.9 (a>0,且a≠1) ②当0<a<1时,y= logax是减函数,且5.1<5.9,所以loga5.1>loga5.9 如果底数为字母,那么要分类讨论,进行分类讨论时,要做到不重不漏. 例题讲解 例3 比较下列各题中两个值的大小: 变式1:比较log23.4, log33.4的大小 方法1解: log23.4可看作对数函数y=log2x当x取3.4时对应的函数值, log33.4可看作对数函数y=log3x当x取3.4时对应的函数值. 在同一直角坐标系作出y=log2x和y=log3x的图象(底大图低) 由图象可知log23.4 > log33.4 底数不同,真数相同: 构造两个对数函数,利用“底大图低”进行比较 例题讲解 例3 比较下列各题中两个值的大小: 变式1:比较log23.4, log33.4的大小 方法2 解: 利用换底公式,将两个对数值换成同底.log23.4=, log33.4= 因为>0, 3>2>0.两个对数值分子相同. 所以log23.4 > log33.4 底数不同,真数相同: 利用换底公式,换成同底对数再比较. 例题讲解 例3 比较下列各题中两个值的大小: 变式2:比较log 3, log3 的大小 解: 因为函数y=log3x是增函数,且 >3,所以log3 >log33=1.同理,1=log >log 3,所以log3 >log 3. 底数不同,真数不同: 引入中间量进行比较,常见中间量有0,1等 方法总结: 例题讲解 底数相同真数不同 构造对数函数,利用对数函数单调性进行比较. 底数不同真数不同 引入中间量进行比较.常用的中间量有0,1等. 底数不同真数相同 利用换底公式换成同底再比较 构造两个对数函数,利用“底大图低”进行比较 学以致用 (书本135页练习2)[练习1].比较下列各题中两个值的大小: 解: 学以致用 (书本140页复习巩固2)[练习2].比较满足下列条件的两个正数m,n的大小: 例题讲解 例4 溶液酸碱度的测量. 溶液酸碱度是通过pH计量的.pH的计算公式为pH=-lg[H+],其中[H+]表示溶液中氢离子的浓度,单位是摩尔/升. (1)根据对数函数性质及上述pH的计算公式,说明溶液酸碱度与溶液中氢离子的浓度之间的变化关系; (2)已知纯净水中氢离子的浓度[H+]=10-7摩尔/升,计算纯净水的pH. 例题讲解 例4 溶液酸碱度的测量. 溶液酸碱度是通过pH计量的.pH的计算公式为pH=lg[H+],其中[H+]表示溶液中氢离子的浓度,单位是摩尔/升. (1)根据对数函数性质及上述pH的计算公式,说明溶液酸碱度与溶液中氢离子的浓度之间的变化关系; 解:根据对数的运算性质,有pH=lg[H+]=lg[H+]-1=lg 在(0,+)上,随着[H+]的增大, 减小,相应地, lg也减小,即pH减小.所以,随着[H+]的增大, pH减小,即溶液中氢离子的浓度越大,溶液的酸性就越强. 例题讲解 例4 溶液酸碱度的测量. 溶液酸碱度是通过pH计量的.pH的计算公式为pH=-lg[H+],其中[H+]表示溶液中氢离子的浓度,单位是摩尔/升. (2)已知纯净水中氢离子的浓度[H+]=10-7摩尔/升,计算纯净水的pH. 解:(2)当[H+]=10-7时, pH=lg10-7=7.所以,纯净水的pH是7. 学以致用 (书本135页复习巩固2)[练习3]. 某地去年的GDP(国内生产总值)为3000亿元人民币,预计未来五年的平均增长率为6.8%. (1)设经过x年达到的年GDP为y亿元,试写出未来5年内,y关于x的函数解析式; (2)经过几年该地GDP能达到3900亿元人民币. 所以经过4年该地GDP能达到3 900亿元人民币. 解: (1)由题意得y=3000(1+ 6.8%) x, 1≤ x≤5,且x∈N* (2)由3000(1+ 6.8%) x=3900,解得x≈3.99, 课堂小结 知识点 对数函数的图象和性质 数学思想 ①由特殊到一般 ②分类讨论 ③数形结合 核心素养 ①直观想象 ②逻辑推理 ③数学运算 布置作业 课后探究:书本135页探究与发现:探究互为反函数的两个函数图象间的关系. 基础作业:书本199页综合运用第5题 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