📚 Exercise 2A: Functions – Domain, Range, and Notation | 练习2A：函数——定义域、值域与符号
Exercise 2A in IB Mathematics typically forms the first set of practice problems in Topic 2: Functions. It covers the fundamental definitions of a function, domain and range, function notation, the vertical line test, and basic operations with functions such as composition and finding inverses. Mastering this exercise is essential for success in IB Math Analysis and Approaches or Applications and Interpretation, as these concepts reappear in calculus and graphing questions.
IB数学中的练习2A通常出现在主题2：函数的第一套练习题中。它涵盖了函数的定义、定义域与值域、函数符号、垂直线检验以及函数的复合与反函数等基础运算。掌握这套练习对于IB数学分析与方法或应用与解释的成功至关重要，因为这些概念会反复出现在微积分和函数作图题中。
1. What is a Function? | 什么是函数？
uts is the domain, and the set of all outputs is the range. In IB, a function can be represented by an equation, a graph, a table, or a mapping diagram.
函数是一种规则，将每个输入恰好分配给一个输出。我们通常用x表示输入，y表示输出，写作y = f(x)。所有允许的输入的集合称为定义域，输出的集合称为值域。在IB中，函数可以用方程、图像、表格或映射图来表示。
2. Domain and Range Basics | 定义域与值域基础
The domain of a function is the set of all real numbers (ℝ) for which the function is defined, unless otherwise restricted. For example, f(x) = 1/(x – 2) is undefined when x = 2, so its domain is {x | x ∈ ℝ l possible resulting y-values. Visualizing the graph often helps in identifying the range.
函数的定义域是使得函数有意义的所有实数（ℝ）的集合，除非有额外限制。例如，f(x) = 1/(x – 2) 在x = 2时无定义，因此它的定义域是{x | x ∈ ℝ, x ≠ 2}。值域是所有可能的y值的集合。借助图像往往有助于确定值域。
3. Function Notation | 函数符号
nds on the input. When evaluating f(3), we substitute 3 into the expression. For f(x) = x² – 4x, f(3) = 3² – 4(3) = 9 – 12 = –3. This notation is also used when composing functions or when defining inverses.
函数符号如f(x)强调了输出依赖于输入。计算f(3)时，我们将3代入表达式。例如对于f(x) = x² – 4x，f(3) = 3² – 4(3) = 9 – 12 = –3。在进行函数复合或定义反函数时，这种符号同样会被使用。
4. Vertical Line Test | 垂直线检验
gle-output rule. For instance, a circle fails the test, while a parabola y = x² passes it.
要检验一个图像是否表示函数，可以使用垂直线检验：如果任意一条垂直线与图像相交超过一次，那么该关系就不是函数。这确保了单一输出规则。例如，圆就无法通过检验，而抛物线y = x²则可以通过。
5. Finding Domain from Equation | 从方程求定义域
inators and even roots. Denominators cannot be zero: for f(x) = 2/(x + 1), x + 1 ≠ 0 → x ≠ –1. Even roots require non-negative radicands: g(x) = √x – 4 gives x – 4 ≥ 0 → x ≥ 4. Logarithmic functions require positive arguments: ln(x) has domain x > 0.
从方程求定义域时，要留意分母与偶次根式。分母不能为零：对于f(x) = 2/(x + 1)，x + 1 ≠ 0 即 x ≠ –1。偶次根式的被开方数必须非负：g(x) = √x – 4 给出 x – 4 ≥ 0 即 x ≥ 4。对数函数要求自变量为正：ln(x) 的定义域为 x > 0。
6. Range of a Function | 函数的值域
en requires analyzing the shape of the function. For a quadratic in vertex form f(x) = a(x – ge is all real numbers except y = 0. Using a graph or calculus can help confirm the range.
确定值域通常需要分析函数的形状。对于顶点式f(x) = a(x – h)² + k的二次函数，若a > 0，最小y值为k，则值域为y ≥ k。对于有理函数如h(x) = 1/x，值域为除y=0外的所有实数。借助图像或微积分可以帮助确认值域。
7. Composite Functions | 复合函数
A composite function (f ◦ ormed by applying g first and then f. For example, if f(x) = 2x + 1 and g(x) = x², then (f ◦ f the composite consists of all x in the domain of g such that g(x) is in the domain of f.
复合函数 (f ◦ g)(x) = f(g(x)) 是先作用g再作用f。例如，若f(x) = 2x + 1，g(x) = x²，则 (f ◦ g)(x) = 2(x²) + 1 = 2x² + 1。复合函数的定义域由所有使得g(x)在f的定义域内的、属于g的定义域的x组成。
8. Inverse Functions | 反函数
ction must be one-to-one (pass the horizontal line test) to have an inverse. For f(x) = 3x – 4, solving y = 3x – 4 → x = 3y – 4 → y = (x + 4)/3 gives f⁻¹(x) = (x + 4)/3.
反函数 f⁻¹ 逆转了 f 的作用：若 f(a) = b，则 f⁻¹(b) = a。求反函数时，在方程 y = f(x) 中交换 x 和 y，然后解出 y。原函数必须是一对一的（通过水平线检验）才能有反函数。对于 f(x) = 3x – 4，解 y = 3x – 4 → x = 3y – 4 → y = (x + 4)/3，得 f⁻¹(x) = (x + 4)/3。
9. Domain and Range of Inverse | 反函数的定义域与值域
nge of f⁻¹ is the domain of f. This interchange is a powerful tool. For example, if f(x) = √x – 2 with domain x ≥ 2 and range y ≥ 0, then f⁻¹(x) = x² + 2 has domain x ≥ 0 and range y ≥ 2.
反函数 f⁻¹ 的定义域正是原函数 f 的值域，反函数的值域则是原函数的定义域。这种互换是一个强有力的工具。例如，如果 f(x) = √x – 2，定义域为 x ≥ 2，值域为 y ≥ 0，那么 f⁻¹(x) = x² + 2 的定义域为 x ≥ 0，值域为 y ≥ 2。
10. Practice with Exercise 2A | 练习2A实战
A typical problem from Exercise 2A might be: Given f(x) = √2x – 6 d its domain, range, and inverse function. Solution: The radicand must be non-negative: 2x – 6 ≥ 0 → x ≥ 3. Domain: {x | }. Since the square root gives outputs ≥ 0, the range is y ≥ 0. For the inverse, write y = √2x – 6, swap: x = √2y – 6, square both sides: x² = 2y – )/2, with the restriction that x ≥ 0 (the range of f). Thus f⁻¹(x) = (x² + 6)/2 for x ≥ 0.
练习2A中的一道典型题目可能是：已知 f(x) = √2x – 6，求它的定义域、值域和反函数。解答：被开方数必须非负：2x – 6 ≥ 0 → x ≥ 3。定义域：{x | x ≥ 3}。由于平方根输出 ≥ 0，值域为 y ≥ 0。求反函数：写出 y = √2x – 6，交换得 x = √2y – 6，两边平方：x² = 2y – 6 → 2y = x² + 6 → y = (x² + 6)/2，并附加条件 x ≥ 0（原函数的值域）。因此 f⁻¹(x) = (x² + 6)/2（x ≥ 0）。
11. Common Mistakes | 常见错误
Mistake 1: Forgetting to restrict the domain of the inverse. After swapping and solving, you must limit the inverse's domain to the range of the original function. Without that, f⁻¹ would not be a function.
错误1：忘记限制反函数的定义域。交换并求解后，必须将反函数的定义域限定为原函数的值域。否则 f⁻¹ 就不是一个函数。
ng domains. Always set the denominator not equal to zero before simplifying. For f(x) = (x – 1)/(x – , the domain is all real numbers except x = 1, even though the expression simplifies to 1.
错误2：在求定义域时错误约分有理表达式。在化简之前始终要令分母不等于零。对于 f(x) = (x – 1)/(x – 1)，定义域是除 x = 1
Published by TutorHao | IB Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图