高一数学必修一二第章小结.ppt

Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 本章内容 2.1 指数函数 2.2 对数函数 2.3 幂函数 第二章 小结 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 知识要点 自我检测题 复习参考题 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 1. 指数幂的运算 负指数: 分数指数: 同底数幂相乘除: 幂的乘方: 积的乘方: aman=am+n. (am)n=amn. (ab)n=anbn. 返回目录 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 2. 指数函数 解析式: 图象特点: y = ax (a0, 且a1). x y O 1 y=ax (01) Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 3. 指数函数的性质 定义域: 值域: 01, 负指数幂小于1, 正指数幂大于1. 01, (-, +)上是增函数. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 4. 对数运算 对数式与指数式的互化: 常用对数: aN = b N = logab. 自然对数: 1 的对数等 0, 底的对数等于 1. 以10为底, log10a = lga. 以 e=2.71828为底, logea=lna. 两个特殊对数值: Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 5. 对数的运算性质 换底公式: (1) loga(MN) = logaM + logaN. (3) logaMn = nlogaM (nR). (2) loga = logaM - logaN. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 6. 对数函数 解析式: 图象特点: y = logax (a0, 且a1). x y o 1 y=logax a1 x y o 1 y=logax 01, (0, +)上是增函数. 底数、真数一个大于1, 一个小于1, 对数值为负. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 8. 幂函数 解析式: 几种幂函数的图象特点: y = xa (a为常数). x y o y=x 1 1 x y o y=x2 1 1 x y o y=x3 1 1 x y o 1 1 x y o y=x-1 1 1 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 9. 幂函数的性质 定义域 值域 单调性 过定点 y=xa 奇偶性 (1, 1)(1, 1) (-, 0) (0, +) (-, +)0, +) 0, +) (-, 0)减 (-, +)增 y=x-1y=x2y=x3y=x (1, 1)(1, 1)(1, 1) (-, +) (-, +) (-, 0) (0, +) 0, +) (-, +) (-, +) (0, +)减 0, +)增 (-, 0减 0, +)增 (-, +)增 奇非奇偶奇偶奇 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 10. 反函数 由于习惯用 x 表示自变量, 所以将变换后 函数中的字母 x, y 相交换得 将一个函数 y=f(x) 中的 y 表示成 x 的函数 x=g(y), 我们把 x=g(y) 叫做 y=f(x) 的反函数. y=g(x). 指数函数与对数函数互为反函数. 如果两函数互为反函数, 则它们的图象关 于直线 y=x 即称. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 返回目录 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. A 组 1. 求下列各式的值: (1) (2) (3) (4) 解: (1)=11. (2) (3) (4) Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 2. 化简下列各式: (1) (2) (a2-2+a-2)(a2-a-2). 解: (1) 原式 = (2) 原式 = Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 3. (1) 已知 lg2=a, lg3=b, 试用 a、b 表示 log125; (2) 已知 log23=a, log37=b, 试用 a、b 表示 log1456. 解: (1) Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 3. (1) 已知 lg2=a, lg3=b, 试用 a、b 表示 log125; (2) 已知 log23=a, log37=b, 试用 a、b 表示 log1456. 解: (2)由 log23=a, Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 4. 求下列函数的定义域: (1) (2) 解: (1) 要使函数有定义, 只需 2x-10, 即 函数的定义域为 (2) 要使函数有定义, 需 x0. 函数的定义域为 x|x0. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 5. 求下列函数的定义域: (1) (2) y=loga(2-x) (a0, 且a1); (3) y=loga(1-x)2 (a0, 且a1). 解: (1) 要使函数有定义, 需 原函数的定义域为 (2) 要使函数有意义, 需 2-x0, 得 x0, 且a1); (3) y=loga(1-x)2 (a0, 且a1). 解: (3) 要使函数有定义, 需 (1-x)20, 即 1-x0, 原函数的定义域为 xR|x1. 得 x1, Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 6. 比较下列各组中两个值的大小: (1) log67, log76; (2) log3p, log20.8. 解: (1) log67log66 =1, log76log76. (2) 31, p 1, log3p0, 又 21, 0.8log20.8. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 7. 已知 f(x)=3x, 求证: (1) f(x)f(y)=f(x+y); (2) f(x)f(y)=f(x-y). 证明: (1) f(x)=3x, f(x)f(y)=3x3y=3x+y, f(x+y)=3x+y, 则 f(x)f(y)=f(x+y)成立. (2) f(x)f(y)=3x3y =3x-y, f(x-y)=3x-y, f(x)f(y)=f(x-y)成立. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 8. 已知 f(x)= a, b(-1, 1), 求证: 证明: 即 成立. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 9. 牛奶保鲜时间因储藏时温度的不同而不同, 假 定保鲜时间与储藏温度是一种指数关系, 若牛奶放在 0的冰箱中, 保鲜时间约是192 h, 而在22的厨房 中则约是42 h. (1) 写出保鲜时间 y 关于储藏温度 x 的函数解析式 ; (2) 利用 (1) 中结论, 指出温度在30和16的保 鲜时间 (精确到 1 h); (3) 运用上面的数据, 作此函数的图象. 解: (1) 设保鲜时间与温度的指数关系为 y=kax, 当 x=0 时, y=192; 当 x=22 时, y=42. 则 k=192, a0.93. 于是得保鲜时间与温度的函数式为 y=1920.93x. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 9. 牛奶保鲜时间因储藏时温度的不同而不同, 假 定保鲜时间与储藏温度是一种指数关系, 若牛奶放在 0的冰箱中, 保鲜时间约是192 h, 而在22的厨房 中则约是42 h. (1) 写出保鲜时间 y 关于储藏温度 x 的函数解析式 ; (2) 利用 (1) 中结论, 指出温度在30和16的保 鲜时间 (精确到 1 h); (3) 运用上面的数据, 作此函数的图象. 解: (2) 由(1)得函数式为 y=1920.93x. 当 x=30 时, y=1920.933022 (h); 当 x=16 时, y=1920.931660 (h). 答: 在30温度下, 可保鲜22小时, 在16温 度下, 可保鲜60小时. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 9. 牛奶保鲜时间因储藏时温度的不同而不同, 假 定保鲜时间与储藏温度是一种指数关系, 若牛奶放在 0的冰箱中, 保鲜时间约是192 h, 而在22的厨房 中则约是42 h. (1) 写出保鲜时间 y 关于储藏温度 x 的函数解析式 ; (2) 利用 (1) 中结论, 指出温度在30和16的保 鲜时间 (精确到 1 h); (3) 运用上面的数据, 作此函数的图象. 解: (3) 图象经过点 (30, 22).(22, 42), (16, 60),(0, 192), x y o 16 2230 22 42 60 192 y=1920.93x Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 10. 已知幂函数 y=f(x) 的图象过点(2, ), 试求 此函数的解析式, 并作出图象, 判断奇偶性、单调性. 解: 幂函数 y=xa 经过点 则有 得 即函数解析式为 定义域为(0, +), 图象过点 (1, 1), x y o 1 1 24 函数非奇非偶, 在(0, +)上是减函数. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. B 组 1. 已知集合A=y | y=log2x, x1, B=y | y= x1, 则AB = ( ) (A) (B) y|01 时, log2x0, A=y|y0, 则 A Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 2. 若 2a=5b=10, 则 解: 2a=10, a=log210, 5b=10, b=log510, 则 = lg(25) =1. 1 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 3. 对于函数 f(x)=a- (aR): (1) 探索函数 f(x) 的单调性; (2) 是否存在实数 a 使函数 f(x) 为奇函数? 解: (1) 2x是 (-, +)上的增函数, 当 x1x2 时, 则 所以得 f(x1)f(x2), 函数在(-, +)上是增函数. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 3. 对于函数 f(x)=a- (aR): (1) 探索函数 f(x) 的单调性; (2) 是否存在实数 a 使函数 f(x) 为奇函数? 解: (2) 要使 f(x)为奇函数, 需 f(-x) = -f(x) 即 整理得 解得 即当 a=1 时, f(x)为奇函数. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 4. 设 求证: (1) g(x)2-f(x)2=1; (2) f(2x)=2f(x)g(x); (3) g(2x)=g(x)2+f(x)2. 证明: 原等式成立. (1) =1, Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 4. 设 求证: (1) g(x)2-f(x)2=1; (2) f(2x)=2f(x)g(x); (3) g(2x)=g(x)2+f(x)2. 证明: 原等式成立. (2) 又 2f(x)g(x) Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 4. 设 求证: (1) g(x)2-f(x)2=1; (2) f(2x)=2f(x)g(x); (3) g(2x)=g(x)2+f(x)2. 证明: (3) g(2x)=g(x)2+f(x)2 成立. 又 g(x)2+f(x)2 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 5. 把物体放在冷空气中冷却, 如果物体原来的温度是 q1, 空气的温度是q0. t min 后物体的温度q可由公式 q = q0+(q1-q0)e-kt 求得, 这里 k 是一个随着物体与空气的接触状况而定的正的常 量, 现有62的物体, 放在15的空气中冷却, 1 min以后物体 的温度是52, 求上式中 k 的值 (精确到0.01), 然后计算开始 冷却后多长时间物体的温度是42, 32. 物体会不会冷却到 12? 解: 当 q1= 62, q0= 15, t =1时, q = 52, 则得 52=15+(62-15)e-k,解得 0.24. 得此物体的冷却公式为q =15+47e-0.24t. 当 q =42 时, 解得 t 2.3(min); 当 q =32 时, 解得 t 4.2(min). 当 q =12 时, 得 t = lg0.79(-0.06), 对数无意义. (答略) 事实上, 物 体不可能冷 却到比空气 的温度还低. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 6. 某工厂产生的废气经过过滤后排放, 过滤过程中废气的 污染物数量 P mg/L 与时间 t h 间的关系为 p = p0e-kt. 如果在前5小时消除了10%的污染物, 试回答: (1) 10小时后还剩百分之几的污染物? (2) 污染物减少50%需要花多少时间 (精确到 1 h)? (3) 画出污染物数量关于时间变化的函数图象, 并在图象 上表示计算结果. 解: 当 t=5 时, P=90%P0, 则 90%P0=P0e-5k, 解得 k0.02, 得 P 与 t 的关系式为 P=P0e-0.02t. (1) 当 t =10 时, P=P0e-0.20.82P0, 答: 10小时后大约还剩百分之八十二的污染物. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 6. 某工厂产生的废气经过过滤后排放, 过滤过程中废气的 污染物数量 P mg/L 与时间 t h 间的关系为 p = p0e-kt. 如果在前5小时消除了10%的污染物, 试回答: (1) 10小时后还剩百分之几的污染物? (2) 污染物减少50%需要花多少时间 (精确到 1 h)? (3) 画出污染物数量关于时间变化的函数图象, 并在图象 上表示计算结果. 解: 当 t=5 时, P=90%P0, 则 90%P0=P0e-5k, 解得 k0.02, 得 P 与 t 的关系式为 P=P0e-0.02t. (2) 当 P =0.5P0 时, 答: 污染物减少50%, 大约需要花35小时. 得 0.5P0=P0e-0.02t. 解得 t 35, Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 6. 某工厂产生的废气经过过滤后排放, 过滤过程中废气的 污染物数量 P mg/L 与时间 t h 间的关系为 p = p0e-kt. 如果在前5小时消除了10%的污染物, 试回答: (1) 10小时后还剩百分之几的污染物? (2) 污染物减少50%需要花多少时间 (精确到 1 h)? (3) 画出污染物数量关于时间变化的函数图象, 并在图象 上表示计算结果. 解:图象过点 (5, 0.9),(3) (10, 0.82),(35, 0.5). x y o 5 1035 0.5 0.82 0.9 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 自我检测题 返回目录 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 检测题 一、选择题(每小题只有一个正确选项) 1. 已知集合A=y|y=log2x, x1, B=y|y= , x1, 则AB=( ) (A) (B) y|0b, 则 ( ) (A) a2b2 (B) (C) lg(a-b)0 (D) 3. 如果a1, bf(1), 则x的取值范围是( ) (A) (B) (C) (D) (0,1) (10,+) 二、填空题 6. 1992年底世界人口达到54.8亿, 若人口的年平均增长率为1%, 经过x年后世界人口数为y(亿), 则y与x的函数 解析式为 . 7. 函数y=logx-1(3-x)的定义域是 . 8. 设0x2,则函数 的最大值是 , 最小值是 . 三、解答题 9. 已知函数f(x)=loga(ax-1) (a0, 且a1). (1) 求f(x)的定义域; (2) 讨论函数f(x)的增减性. 10. 某电器公司生产A型电脑, 1993年这种电脑每台平均生产成本为5000元, 并以纯利润20%确定出厂价, 从 1994年开始, 公司通过更新设备和加强管理,使生产成本逐年降低, 到1997年, 尽管A型电脑出厂价是1993 年出厂价的80%, 但却实现了50%纯利润的高效益. (1) 求1997年每台A型电脑的生产成本; (2) 以1993年的生产成本为基数, 求19931997年生产成本平均每年降低的百分数 (精确到0.01, 以下数据 可供参考: ). Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 检测题 一、选择题(每小题只有一个正确选项) 1. 已知集合A=y| y=log2x, x1, B=y| y= , x1, 则AB= ( ) (A) (B) y|00, A Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 2. 若 a, b 是任意实数, 且 ab, 则 ( ) (A) a2b2 (B) (C) lg(a-b)0 (D) 分析: 用函数的思想判断 A、D 选项, A 选项看作二次函数, 在任意实数范围内不是一 个单调区间, 不能确定大小. D 选项看作指数函数, 底数小于 1, 在 (-, +) 上是减函数, ab, D 也可用具体实数检验: 1-2 12(-2)2, -1-2 a-b=0.1 lg(a-b)0, A不对;B不对; C不对. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 3. 如果 a1, bf(1), 则 x 的取值范围是 ( ) (A) (B) (C) (D) (0, 1) (10, +) 分析: 由 f(x) 在 0, +) 上是减函数, 且是偶函数, 则大概图象如图: x y O1 -1 f(1)=f(-1), 要使 f(lgx)f(1), 需 -10, x-11, 3-x0, 解得 10, 且a1). (1) 求 f(x) 的定义域; (2) 讨论函数 f(x) 的增减性. 解: (1) 要使对数有意义, 需 ax-10, 即 ax1, 当 01 时, x0, 此时函数的定义域为 (0, +). Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 三、解答题 9. 已知函数 f(x)=loga(ax-1) (a0, 且a1). (1) 求 f(x) 的定义域; (2) 讨论函数 f(x) 的增减性. 解: (2) 当 00, 且a1). (1) 求 f(x) 的定义域; (2) 讨论函数 f(x) 的增减性. 解: (2)