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Calculus Core Concepts Review

MATH1131 Lab Test Preparation

Domain & Range, Limits & Continuity, Core Theorems

目录

第一章： 函数的定义域与值域

第二章： 极限与连续性

第三章： 微积分核心定理

1. 函数的定义域与值域

这是微积分的基础，核心是理解不同函数类型带来的限制。

知识点 1：根号函数 √g(x)

  • 定义域限制： 根号下的表达式必须大于或等于零 (g(x) ≥ 0)。
  • 值域限制： 主平方根的输出值永远大于或等于零 (√g(x) ≥ 0)。

知识点 2：分式函数 f(x)/g(x)

  • 定义域限制： 分母不能为零 (g(x) ≠ 0)。

知识点 3：极限中的"抓大头"原则

在计算 x → ∞ 的有理函数极限时，极限值由分子和分母的最高次项决定，所有低次项都可以忽略不计。

解题方法与例题分析

类型一：根号在分母里，如 f(x) = 1 / √(-x² + 3x)

  1. 建立不等式： 这是"根号"和"分母"的双重限制，所以根号内的部分必须严格大于零。即 -x² + 3x > 0。
  2. 解不等式： 解二次不等式 x(3-x) > 0，得到定义域为 (0, 3)。
  3. 求值域： 问题转化为求 f(x) 在 (0, 3) 上的值域。我们知道 f(x) 的最小值会在内层函数 g(x) = -x² + 3x 取最大值时出现。
  4. 找顶点： g(x) 是开口向下的抛物线，在顶点 x = 1.5 处取最大值 g(1.5) = 2.25。
  5. 计算结果： 所以 f(x) 的最小值为 1 / √2.25 = 1 / 1.5 = 2/3。当x趋近边界0或3时，f(x) 趋近无穷。所以值域是 [2/3, ∞)。

类型二：根号不在分母里，如 f(x) = √(-x² + 3x)

  1. 建立不等式： 限制条件变宽松，根号内的部分只需大于或等于零。即 -x² + 3x ≥ 0，解得定义域为 [0, 3]。
  2. 求值域： 值域的最小值显然是0（当x=0或x=3时）。最大值同样在内层函数 g(x) 取最大值时出现。
  3. 计算结果： g(x) 的最大值是2.25，所以 f(x) 的最大值是 √2.25 = 1.5。值域为 [0, 1.5]。

类型三：x趋于无穷的极限，如 f(x) = (-2x⁷ + 3x² ...) / (-3x² + 4x ...)

  1. 抓大头： 分子的"大头"是 -2x⁷，分母的"大头"是 -3x²。
  2. 化简： f(x) ≈ (-2x⁷) / (-3x²) = (2/3)x⁵。
  3. 看趋势： 当 x → ∞ 时，(2/3)x⁵ 也趋向于 ∞。所以极限是 ∞。

💡 醍醐灌顶

  • 定义域： 就像给函数找一个"合法的居住地"。根号要求"必须在零度以上地区"，分母要求"不能住在门牌号为0的房子里"。当根号住进分母家时，就必须住在"严格高于零度"的地区。
  • 值域： 对于复合函数，采用"先内后外"的策略。先分析内层函数的取值范围（找到它的最高点和最低点），再看外层函数会如何"加工"这些值，从而得到最终的输出范围。
  • 无穷极限： 在无穷的世界里，低次项就像是"蚂蚁"，高次项是"大象"。当大象跑起来时，你只需要关心大象的方向，蚂蚁可以完全忽略。

2. 极限与连续性

考察函数在某一点附近的行为，特别是"洞"和"断崖"的区别。

知识点

  • 连续的定义： 函数在 c 点连续，当且仅当 左极限 = 右极限 = 函数在该点的值。
  • 可去间断点： 左右极限相等，但不等于函数值。这是一个可以被"填补"的"小坑"。
  • 跳跃间断点： 左右极限都存在，但不相等。这是一个无法修复的"断崖"。
  • 绝对值处理： |u| 在 u > 0 时等于 u，在 u < 0 时等于 -u。

解题方法与例题分析

解题步骤

  1. 尝试代入： 永远先将目标点 c 代入函数。
  2. 分析结果：
    • 如果得到一个确定的数值，那么极限就是这个值。
    • 如果得到 0/0（不定式），这是一个强烈的信号：极限可能存在，需要化简！
    • 如果得到 (非零数)/0，这是一个强烈的信号：极限是无穷大，存在一个垂直渐近线。
  3. 化简 (针对0/0)： 对分子和分母进行因式分解，然后约掉导致0/0的公因式。
  4. 再次代入： 将 c 代入化简后的表达式，得到极限值。
  5. 判断连续性。

例题1 (小坑)：f(x) = (x+3) / (x² + 2x - 3) 在 x=-3 处

  • 化简后为 1/(x-1)，极限值为 -1/4。
  • 而已知 f(-3) = 0。
  • 因为极限值 -1/4 ≠ 函数值 0，所以不连续。但因为左右极限相等，所以可以通过重新定义 f(-3) = -1/4 来修复它。

例题2 (断崖)：g(x) = |x² - x - 20| / (x - 5) 在 x=5 处

  • 因为绝对值的存在，必须讨论左右极限。
  • 在 x→5⁺ 时，x² - x - 20 > 0，函数化简为 x+4，右极限为 9。
  • 在 x→5⁻ 时，x² - x - 20 < 0，函数化简为 -(x+4)，左极限为 -9。
  • 因为左右极限不相等，所以这是一个无法修复的"断崖"，不能通过重新定义 g(5) 使其连续。

💡 醍醐灌顶

极限只关心"路"的终点在哪里，而不在乎终点那个地方的"标记"画的是什么。如果左右两条路能汇合到同一个点（左右极限相等），但那个点的标记却画歪了（函数值不等于极限值），这就是一个可以修复的"小坑"。如果左右两条路通往了不同的地方（左右极限不相等），这就是一个无法修复的"断崖"。

3. 微积分核心定理

理解这些定理的"承诺"和生效的"前提条件"。

知识点 1：介值定理 (Intermediate Value Theorem, IVT)

  • 承诺： 如果一条连续的路径起点在河的一边，终点在另一边，那么途中必然至少有一次穿过了河流。
  • 前提条件： 函数在闭区间 [a, b] 上连续。
  • 应用： 找一个 a 和一个 b，使得 f(a) 和 f(b) 的符号相反（一个正，一个负），就能证明在 (a,b) 之间至少存在一个根。

知识点 2：最大-最小值定理 (Maximum-Minimum Theorem / Extreme Value Theorem)

  • 承诺： 如果你被关在一个四壁齐全的封闭房间里，沿着一条连续的路径行走，那么你一定能找到房间里的最高点和最低点。
  • 前提条件：
    1. 函数在定义域上连续。
    2. 定义域是闭区间 [a, b]（必须是方括号）。
  • 应用： 逐一检查选项，只有同时满足"连续"和"闭区间"两个条件的，才能保证最大最小值的存在。

💡 醍醐灌顶

定理就像一份"保险合同"，你必须仔细阅读它的"生效条款"（前提条件）。介值定理的条款是"连续"，它保证了"根"的存在性。最大-最小值定理的条款是"连续"并且"闭区间"，它保证了"最值"的存在性。做题时，你的任务就是严格地检查这些条款是否被满足。

Calculus Core Concepts Review

MATH1131 Calculus Lab Test Review

Domain & Range, Limits & Continuity, Core Theorems

Table of Contents

Chapter 1: Domain and Range of Functions

Chapter 2: Limits and Continuity

Chapter 3: Core Calculus Theorems

1. Domain and Range of Functions

This is the foundation of calculus, with the core being understanding the restrictions imposed by different function types.

Key Concept 1: Square Root Functions √g(x)

  • Domain Restriction: The expression under the square root must be greater than or equal to zero (g(x) ≥ 0).
  • Range Restriction: The principal square root output is always greater than or equal to zero (√g(x) ≥ 0).

Key Concept 2: Rational Functions f(x)/g(x)

  • Domain Restriction: The denominator cannot be zero (g(x) ≠ 0).

Key Concept 3: "Dominant Term" Principle for Limits

When calculating limits of rational functions as x → ∞, the limit is determined by the highest degree terms in the numerator and denominator, with all lower-order terms becoming negligible.

Solution Methods and Example Analysis

Type 1: Square Root in Denominator, e.g., f(x) = 1 / √(-x² + 3x)

  1. Set up inequality: This has both "square root" and "denominator" restrictions, so the expression under the root must be strictly greater than zero: -x² + 3x > 0.
  2. Solve inequality: Solve x(3-x) > 0, giving domain (0, 3).
  3. Find range: Transform to finding range of f(x) on (0, 3). The minimum of f(x) occurs when inner function g(x) = -x² + 3x reaches its maximum.
  4. Find vertex: g(x) is a downward parabola with maximum at x = 1.5, where g(1.5) = 2.25.
  5. Calculate result: So minimum of f(x) is 1 / √2.25 = 1 / 1.5 = 2/3. As x approaches boundaries 0 or 3, f(x) approaches infinity. Range is [2/3, ∞).

Type 2: Square Root not in Denominator, e.g., f(x) = √(-x² + 3x)

  1. Set up inequality: Restrictions are relaxed, expression under root only needs to be greater than or equal to zero: -x² + 3x ≥ 0, giving domain [0, 3].
  2. Find range: Minimum is clearly 0 (when x=0 or x=3). Maximum occurs when inner function g(x) reaches its maximum.
  3. Calculate result: Maximum of g(x) is 2.25, so maximum of f(x) is √2.25 = 1.5. Range is [0, 1.5].

Type 3: Limit as x approaches infinity, e.g., f(x) = (-2x⁷ + 3x² ...) / (-3x² + 4x ...)

  1. Identify dominant terms: Numerator's "dominant term" is -2x⁷, denominator's is -3x².
  2. Simplify: f(x) ≈ (-2x⁷) / (-3x²) = (2/3)x⁵.
  3. Analyze behavior: As x → ∞, (2/3)x⁵ also approaches ∞. So the limit is ∞.

💡 Key Insights

  • Domain: Like finding a "legal residence" for the function. Square roots require "above-zero temperature zones," denominators require "non-zero address houses." When square roots move into denominator's house, they must live in "strictly above-zero temperature" areas.
  • Range: For composite functions, use "inside-out" strategy. First analyze the range of the inner function (find its highest and lowest points), then see how the outer function will "process" these values to get final output range.
  • Infinite limits: In the infinite world, lower-order terms are like "ants" and higher-order terms are "elephants." When elephants run, you only need to care about the elephant's direction; ants can be completely ignored.

2. Limits and Continuity

Examining function behavior near a point, particularly distinguishing between "holes" and "cliffs."

Key Concepts

  • Definition of Continuity: A function is continuous at point c if and only if left limit = right limit = function value at that point.
  • Removable Discontinuity: Left and right limits are equal but not equal to the function value. This is a fillable "small hole".
  • Jump Discontinuity: Left and right limits both exist but are not equal. This is an unfixable "cliff".
  • Absolute Value Handling: |u| equals u when u > 0, equals -u when u < 0.

Solution Methods and Example Analysis

Solution Steps

  1. Try direct substitution: Always first substitute target point c into the function.
  2. Analyze result:
    • If you get a definite number, that's the limit.
    • If you get 0/0 (indeterminate), this is a strong signal: limit may exist, need to simplify!
    • If you get (non-zero)/0, this is a strong signal: limit is infinite, there's a vertical asymptote.
  3. Simplify (for 0/0): Factor numerator and denominator, then cancel the common factor causing 0/0.
  4. Substitute again: Substitute c into the simplified expression to get the limit.
  5. Determine continuity.

Example 1 (Small Hole): f(x) = (x+3) / (x² + 2x - 3) at x=-3

  • After simplification: 1/(x-1), limit value is -1/4.
  • Given f(-3) = 0.
  • Since limit value -1/4 ≠ function value 0, it's discontinuous. But since left and right limits are equal, it can be fixed by redefining f(-3) = -1/4.

Example 2 (Cliff): g(x) = |x² - x - 20| / (x - 5) at x=5

  • Due to absolute value, must consider left and right limits separately.
  • As x→5⁺, x² - x - 20 > 0, function simplifies to x+4, right limit is 9.
  • As x→5⁻, x² - x - 20 < 0, function simplifies to -(x+4), left limit is -9.
  • Since left and right limits are unequal, this is an unfixable "cliff" that cannot be made continuous by redefining g(5).

💡 Key Insights

Limits only care about where the "path" leads, not what "sign" is posted at the destination. If left and right paths converge to the same point (equal left/right limits) but the sign at that point is wrong (function value ≠ limit), that's a fixable "small hole." If left and right paths lead to different places (unequal left/right limits), that's an unfixable "cliff."

3. Core Calculus Theorems

Understanding these theorems' "promises" and their "prerequisite conditions."

Key Concept 1: Intermediate Value Theorem (IVT)

  • Promise: If a continuous path starts on one side of a river and ends on the other side, it must cross the river at least once.
  • Prerequisite: Function is continuous on closed interval [a, b].
  • Application: Find an a and a b such that f(a) and f(b) have opposite signs (one positive, one negative), proving at least one root exists in (a,b).

Key Concept 2: Extreme Value Theorem (Maximum-Minimum Theorem)

  • Promise: If you're locked in a completely enclosed room following a continuous path, you can definitely find the highest and lowest points in the room.
  • Prerequisites:
    1. Function is continuous on its domain.
    2. Domain is a closed interval [a, b] (must use square brackets).
  • Application: Check each option individually; only those satisfying both "continuous" and "closed interval" conditions guarantee existence of maximum and minimum values.

💡 Key Insights

Theorems are like "insurance contracts" - you must carefully read their "terms and conditions" (prerequisites). The Intermediate Value Theorem's condition is "continuity," guaranteeing existence of roots. The Extreme Value Theorem's conditions are "continuity" and "closed interval," guaranteeing existence of extreme values. When solving problems, your task is to rigorously check whether these conditions are satisfied.