📚 Exercise 3A: Functions and Their Notations | 练习3A：函数及其表示法
Welcome to Exercise 3A, a focused drill on the foundational concept of functions in IB Mathematics. In this set of exercises, you will deepen your understanding of what makes a relation a function, how to use function notation fluently, and how to determine domains and ranges from algebraic expressions and graphs. Mastering these skills is essential not only for success in the Functions topic but also for later work in calculus, vectors, and probability, where functional relationships underpin most models.
欢迎来到练习3A，它是IB数学中针对函数基础概念的专项训练。通过这组练习，你将深入理解什么关系才能构成函数、如何熟练使用函数符号，以及如何从代数表达式和图像中确定定义域与值域。掌握这些技能不仅对函数主题至关重要，对后续的微积分、向量和概率也必不可少，因为函数关系是大多数模型的基石。
1. Introduction to Functions | 函数导论
A function is a special type of relation between two sets, typically called the domain (input set) and the codomain (possible output set). Every element in the domain must be assigned to exactly one element in the codomain. This 'one and only one' rule is the heart of the function concept. For example, pairing each student with their height in centimeters defines a function if each student has a single height recorded, but pairing a student with their favourite subjects would not be a function if a student can list multiple subjects.
函数是两个集合（通常称为定义域（输入集）和陪域（可能输出集））之间的一种特殊关系。定义域中的每个元素必须恰好分配给陪域中的一个元素。这种“唯一且仅有一个”的规则是函数概念的核心。例如，将每位学生与他们的身高（厘米）进行配对就定义了一个函数，因为每位学生只记录一个身高；但如果把学生与他们最喜欢的科目配对，由于学生可能列出多门科目，这就不是一个函数。
2. Function Notation f(x) | 函数符号 f(x)
s uses f(x) to name a function and indicate its input variable. The symbol f(x) is read as 'f of x' to talk about the function as an object and to perform operations like composition and inversion.
现代数学使用 f(x) 来命名函数并指明其输入变量。符号 f(x) 读作“关于 x 的 f”。可以把函数想象成一台机器：你输入一个 x 值，机器根据某个规则处理，然后输出 f(x)。例如，f(x) = 2x + 1 告诉我们，输入 3 时输出为 f(3) = 2(3) + 1 = 7。这种紧凑的表示法让我们能将函数作为对象讨论，并对其进行像复合、求逆这样的运算。
3. Domain of a Function | 函数的定义域
mber. In IB Mathematics, you will often write the domain using interval notation, set-builder notation, or simply stating the condition. For example, the domain of f(x) = √x – 2 is x ≥ 2, since the radicand must be non-negative.
函数的定义域是函数有定义的所有可能输入值（x 值）的完整集合。当函数以代数公式给出时，我们必须排除导致分母为零、偶次根号下出现负数或对数为非正数的所有 x 值。在IB数学中，你经常会用区间记号、集合构造记号或直接陈述条件来写出定义域。例如，f(x) = √x – 2 的定义域为 x ≥ 2，因为被开方数必须非负。
In set-builder form we write: { x ∈ ℝ | close attention to whether endpoints are included or excluded. In the function g(x) = 1/(x – 3), the domain is x ≠ 3, written as (-∞, 3) ∪ (3, ∞).
用集合构造记号我们写作：{ x ∈ ℝ | x ≥ 2 }。用区间记号则是 [2, ∞)。务必注意端点是包含还是不包含。对于函数 g(x) = 1/(x – 3)，定义域为 x ≠ 3，写作 (-∞, 3) ∪ (3, ∞)。
4. Range of a Function | 函数的值域
e squaring never produces a negative number. For rational functions, you might need to consider horizontal asymptotes or solve y = f(x) for x to find restrictions on y.
函数的值域是当 x 遍历定义域时实际出现的所有输出值 f(x) 的集合。确定值域通常需要将代数洞察与图像理解结合起来。对于像 f(x) = x² 这样简单的二次函数，定义域为全体实数，但值域为 y ≥ 0，因为平方运算不会产生负数。对于有理函数，你可能需要考虑水平渐近线，或者通过解出 x 关于 y 的表达式来找出 y 的限制条件。
ssible y-values that yield real x-solutions. For instance, to find the range of f(x) = (2x – 1)/(x + 3), set y = (2x – or x. You will obtain a condition that y ≠ 2, giving a range of all real numbers except 2.
一个强大的技巧是将 y = f(x) 改写为关于 x 的方程，然后求出能得到实数 x 解的 y 值。例如，要找到 f(x) = (2x – 1)/(x + 3) 的值域，设 y = (2x – 1)/(x + 3)，两边乘以 (x + 3) 并解出 x，你将得到 y ≠ 2 的条件，从而值域为除 2 外的全体实数。
5. The Vertical Line Test | 垂直线检验
Graphically, a relation drawn in the xy-plane represents a function if and only if any vertical line cuts the graph at most once. This visual tool, called the vertical line test, instantly flags a violation of the 'unique output' rule: if a vertical line intersects the graph in two or more points, then one x-value corresponds to multiple y-values, so the relation is not a function. Circles, sideways parabolas, and many closed curves fail this test.
从图形上看，如果一个关系在 xy 平面上画出的图像满足任意一条垂直线最多与该图像相交一次，那么它就表示一个函数。这种称为垂直线检验的直观工具可以立即标记出对“唯一输出”规则的违反：如果某条垂直线与图像相交于两点或更多，那么一个 x 值就对应了多个 y 值，该关系就不是函数。圆、横向抛物线以及许多封闭曲线都无法通过这一检验。
6. Evaluating Functions | 计算函数值
the formula and simplifying. When you encounter composite expressions like f(a + h) or f(– eat the entire quantity inside the parentheses as the new input. For example, if f(x) = x² – 3x, then f(t + 2) = (t + 2)² – 3(t + 2) = t² + 4t + 4 – 3t – 6 = t² + t – hanical substitution is vital preparation for difference quotients and limits in calculus.
在特定数 a 处对函数 f(x) 求值，就是把 x = a 代入表达式并化简。当你遇见诸如 f(a + h) 或 f(–x) 这样的复合表达式时，请将括号内的整个量视为新的输入。例如，若 f(x) = x² – 3x，则 f(t + 2) = (t + 2)² – 3(t + 2) = t² + 4t + 4 – 3t – 6 = t² + t – 2。这种机械化的代入是为微积分中的差商和极限做的重要准备。
Always maintain correct use of brackets to avoid sign errors. Evaluating f(x) = 1/(x – 4) at x = 2h + 1 gives f(2h + 1) = 1/[(2h + 1) – 4] = 1/(2h – g the parentheses around (2h + 1) ensures that the subtraction applies to the whole input.
务必正确使用括号以避免符号错误。对 f(x) = 1/(x – 4) 在 x = 2h + 1 处求值，得到 f(2h + 1) = 1/[(2h + 1) – 4] = 1/(2h – 3)。注意保留 (2h + 1) 周围的括号可以确保减法作用于整个输入量。
7. Composite Functions | 复合函数
n combines two functions by applying one function to the output of another. The notation f ∘ g (read 'f composed with g') means (f ∘ g)(x) = f(g(x)). The domain of f ∘ in the domain of g for which g(x) is in the domain of f. This ordering matters greatly: f ∘ g and g ∘ f are generally different. For instance, with f(x) = √x and g(x) = 5 – x, (f ∘ g)(x) = √5 – x requires 5 – x ≥ 0, i.e., x ≤ 5, whereas (g ∘ f)(x) = 5 – √x requires x ≥ 0.
复合函数通过将一个函数作用于另一个函数的输出来组合两个函数。记号 f ∘ g（读作“f 复合 g”）表示 (f ∘ g)(x) = f(g(x))。f ∘ g 的定义域由那些 g(x) 落在 f 的定义域内的 x 组成，且 x 本身在 g 的定义域中。复合的顺序非常重要：f ∘ g 与 g ∘ f 通常不同。例如，对于 f(x) = √x 和 g(x) = 5 – x，(f ∘ g)(x) = √5 – x 要求 5 – x ≥ 0，即 x ≤ 5；而 (g ∘ f)(x) = 5 – √x 要求 x ≥ 0。
To form compositions correctly, always write the inner function first and then substitute it into the outer function’s formula. Practice with linear, quadratic, and radical functions is common in Exercise 3A to build fluency.
要正确构建复合函数，务必先写出内层函数，再将其代入外层函数的公式中。练习3A通常会练习线性、二次和根式函数的复合，以培养熟练度。
8. Inverse Functions | 反函数
al graph across the line y = x. To find the algebraic form of an inverse, swap x and y in the equation y = f(x) and then solve for y.
函数 f 的反函数记作 f⁻¹，它会逆转 f 的作用，使得对于 f 定义域中的所有 x 都有 f⁻¹(f(x)) = x，并且对于 f 值域中的所有 y 都有 f(f⁻¹(y)) = y。一个函数必须是一一对应（单射）才能拥有仍为函数的反函数。从图像上看，反函数是原图像关于直线 y = x 的反射。要求出反函数的代数形式，可在 y = f(x) 的方程中交换 x 和 y，然后解出 y。
For example, to invert f(x) = (3x – 5)/2, write y = (3x – 5)/2, swap to x = (3y – 5)/2, multiply by 2: 2x = 3y – lifies to x. The domain of f⁻¹ is the range of f, and the range of f⁻¹ is the domain of f.
例如，为求 f(x) = (3x – 5)/2 的反函数，先写出 y = (3x – 5)/2，交换得 x = (3y – 5)/2，乘以 2 得 2x = 3y – 5，于是 3y = 2x + 5，得到 f⁻¹(x) = (2x + 5)/3。务必验证复合 f(f⁻¹(x)) 能否化简为 x。f⁻¹ 的定义域是 f 的值域，而 f⁻¹ 的值域是 f 的定义域。
9. Piecewise-Defined Functions | 分段函数
r input x, then applying the corresponding sub-formula. The absolute value function f(x) = |x| is a classic example: f(x) = { x, if x ≥ 0; – frequently appear in modeling real-world situations where rules change after a threshold.
分段函数在定义域的不同部分使用不同的公式。求值时，先判断输入 x 落在哪个区间，然后应用对应的子公式。绝对值函数 f(x) = |x| 就是一个经典例子：f(x) = { x，若 x ≥ 0；–x，若 x < 0 }。分段函数经常出现在对现实情境的建模中，此时规则在某个阈值后会发生变化。
ndicate whether the endpoint is included. For a function defined as f(x) = { x², x < 2 ; 4 – at (2,4)) and a line starting at (2,2) with a closed circle, showing a jump discontinuity.
绘制分段函数图像时，注意在边界点处使用空心圆和实心圆，以表明端点是否包含在内。对于定义为 f(x) = { x²，x < 2 ；4 – x，x ≥ 2 } 的函数，图像在 x = 2 之前是一条抛物线（在 (2,4) 处为空心圆），之后是一条从 (2,2) 处开始并带实心圆的直线，显示出跳跃间断。
10. Applications and Exercise 3A Examples | 应用与练习3A例题
Let us consolidate these ideas with a few worked examples typical of Exercise 3A. These problems integrate notation, domain and range reasoning, composition, and inversion—exactly the blend you will face in IB examinations.
让我们通过一些练习3A中常见的典型例题来巩固这些概念。这些问题综合了符号运用、定义域和值域的推理、复合与反函数——正是你在IB考试中会遇到的那种混合类型。
Example 1: Let f(x) = √4 – x² and g(x) = 2x + 1. Find the domain of f, the domain of g, and the composite function (f ∘ g)(x) along with its domain.
例题1：设 f(x) = √4 – x²，g(x) = 2x + 1。求 f 的定义域、g 的定义域，以及复合函数 (f ∘ g)(x) 及其定义域。
Solution: The domain of f requires 4 – x² ≥ 0 → x² ≤ 4 → –2 ≤ x ≤ 2, so domain of f is [–2, 2]. g is a linear polynomial, so its domain is all real numbers, ℝ. For (f ∘ g)(x) = f(g(x)) = √(4 – (2x + 1)²), we need (2x + 1)² ≤ 4. Solving gives –2 ≤ 2x + 1 ≤ 2. Subtract 1: –3 ≤ 2x ≤ 1 → –1.5 ≤ x ≤ 0.5. Thus the domain of f ∘ g is [–1.5, 0.5].
解答：f 的定义域要求 4 – x² ≥ 0 → x² ≤ 4 → –2 ≤ x ≤ 2，因此 f 的定义域为 [–2, 2]。g 是线性多项式，其定义域为全体实数 ℝ。对于 (f ∘ g)(x) = f(g(x)) = √(4 – (2x + 1)²)，我们需要 (2x + 1)² ≤ 4。解此不等式得 –2 ≤ 2x + 1 ≤ 2。减去 1：–3 ≤ 2x ≤ 1 → –1.5 ≤ x ≤ 0.5。故 f ∘ g 的定义域为 [–1.5, 0.5]。
Example 2: Find the inverse of the function h(x) = 4 – 3x, and verify your result.
例题2：求函数 h(x) = 4 – 3x 的反函数，并验证结果。
Solution: Write y = 4 – 3x. Swap x and y: x = 4 – 3y. Solve for y: 3y = 4 – x → y = (4 – x)/3. Thus h⁻¹(x) = (4 – x)/3. Check: h(h⁻¹(x)) = 4 – 3[(4 – x)/3] = 4 – (4 – x) = x. Also h⁻¹(h(x)) = (4 – (4 – 3x))/3 = (3x)/3 = x. Verified.
解答：写出 y = 4 – 3x。交换 x 和 y：x = 4 – 3y。解出 y：3y = 4 – x → y = (4 – x)/3。因此 h⁻¹(x) = (4 – x)/3。检验：h(h⁻¹(x)) = 4 – 3[(4 – x)/3] = 4 – (4 – x) = x。同样 h⁻¹(h(x)) = (4 – (4 – 3x))/3 = (3x)/3 = x。验证通过。
Example 3: Determine the range of f(x) = x² + 2x – 3, where the domain is restricted to x ≤ –1.
例题3：确定函数 f(x) = x² + 2x – 3 在定义域限制为 x ≤ –1 时的值域。
Solution: Complete the square: f(x) = (x² + 2x + 1) – 4 = (x + 1)² – 4. The vertex of this upward-opening parabola is at (–1, –4). Since the domain is x ≤ –1, we only consider the left branch from the vertex downwards. As x decreases towards –∞, (x + 1)² grows without bound, so f(x) → ∞. The minimum value occurs at the vertex, f(–1) = –4. Hence the range is [–4, ∞).
解答：配方：f(x) = (x² + 2x + 1) – 4 = (x + 1)² – 4。这个开口向上的抛物线顶点为 (–1, –4)。由于定义域为 x ≤ –1，我们只考虑从顶点向左的分支。当 x 向 –∞ 递减时，(x + 1)² 无限增大，故 f(x) → ∞。最小值出现在顶点处，f(–1) = –4。因此值域为 [–4, ∞)。
As you work through Exercise 3A, remember to draw graphs where helpful, annotate domain restrictions, and practise swapping between algebraic and graphical reasoning. These skills form the backbone of your IB analysis and approaches course.
当你在做练习3A时，记得在有用的地方画出图像，标注定义域的限制，并练习在代数推理和图像推理之间切换。这些技能构成了你的IB分析与方法课程的主干。
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